Nonlinear Differential Equations on Bounded and Unbounded Domains

CENTRO INTERNACIONAL DE ESTUDOS DE DOUTORAMENTO E AVANZADOS DA USC (CIEDUS) TESE DE DOUTORAMENTO NONLINEAR DIFFERENTIAL EQUATIONS ON BOUNDED AND UNBOUNDED DOMAINS Luc´ ıa L´ opez Somoza ESCOLA DE DOUTORAMENTO INTERNACIONAL PROGRAMA DE DOUTORAMENTO EN MATEM ´ ATICAS SANTIAGO DE COMPOSTELA ANO 2018 . DECLARACI ´ ON DO AUTOR DA TESE Nonlinear differential equations on bounded and unbounded domains Dna. Luc´ ıa L´ opez Somoza Presento a mi˜ na tese, seguindo o procedemento adecuado ao Regulamento, e declaro que: 1) A tese abarca os resultados da elaboraci´ on do meu traballo. 2) No seu caso, na tese faise referencia ´ as colaboraci´ ons que tivo este traballo. 3) A tese ´ e a versi´ on definitiva presentada para a s´ ua defensa e coincide coa versi´ on enviada en formato electr´ onico. 4) Confirmo que a tese non incorre en ning´ un tipo de plaxio doutros autores nin de traballos presentados por min para a obtenci´ on doutros t´ ıtulos. En Santiago de Compostela, 22 de novembro de 2018 Asdo. Luc´ ıa L´ opez Somoza . DECLARACI ´ ON DO DIRECTOR DA TESE Nonlinear differential equations on bounded and unbounded domains D. Alberto Cabada Fern´ andez INFORMA: Que a presente tese se corresponde co traballo realizado por Dna. Luc´ ıa L´ opez Somoza, baixo a mi˜ na direcci´ on, e autorizo a s´ ua presentaci´ on, considerando que re´ une os requisitos esixidos no Regulamento de Estudos de Doutoramento da USC, e que como director desta non incorro nas causas de abstenci´ on establecidas na lei 40/2015. En Santiago de Compostela, 22 de novembro de 2018 Asdo. Alberto Cabada Fern´ andez . To Marcos . Acknowledgements As a final highlight of the work conducing to this Thesis, I would like to thank all those people who, in one way of another, have made it possible. First of all, as it could not be any in other way, I would like to thank my supervisor, Alberto Cabada, for the inestimable support that he has provided me, both professionally and personally, during the more than five years that we have been working together. I would also like to show my gratitude to Professors Rach˚ unkov` a and Feliz Minh´ os for their warm welcome during my stays in Olomouc and ´ Evora. Working with them has been a really enriching experience for me. I acknowledge all my colleagues of Mathematical Analysis at University of Santiago de Compostela, for their good personal treatment during these years. I would like to especially mention Adri´ an, for being always willing to lend a hand and helping me to solve many of my crises (and also for all the coffee liquor shots). I can not forget my fellows, Ignacio, Jorge and Lorena, for having made the work something enjoyable. It is likely that without our gossiping sessions this Thesis would have been finished before, but I’m sure it would not have been as great as it was. Thanks to my family, especially to my parents, Deme and Aurora. You have always supported me, both in my professional and personal decisions, and I know for sure that I could have never reached this without you. Thank you for being always there. I also want to thank my brother Pablo, for being the person who better knows me. I really hope that you never stop trying to make me smile with your bad jokes. Thanks to Marcos, for accompanying me throughout this adventure and all the rest. Thank you simply for being this way, “intrepid”. I would also like to thank my lifetime friends, for staying by my side even if they think that I am crazy: to Iago, for understanding me sometimes better than anyone else and for being the best travel guide; to Luc´ ıa, for being that constant in my life and for taking all the necessary photos to put it on record; to Yaiza, for begin a breath of fresh air and for stopping thinking that I am cold and calculating; to Mar´ ıa, for always having the right words to say and for having chased a famous’ double once with me; to Pita and ´ Alvaro, because even if they are far away, they made every reunion worthwhile. I could not finish without mentioning my flatmates, who have been able to tolerate me during these last months and have made me feel at home: Suso, who always has the perfect song for every moment, and Javi, with whom I have shared many great colacaos. Finally, thank you to all the rest of my doctorate mates from the Faculty, especially to Franco, Andrea and Ar´ ıs, for having been part of this beautiful stage that now finishes. Glossary N: Set of natural numbers, that is, {1,2, . . . }. Z: Set of integer numbers. R: Set of real numbers. R+: Set of positive real numbers. R: Extended real line, that is R= [−∞,∞]. C(I)≡ C(I, R): Space of continuous real functions defined on an interval I. Cn(I)≡ Cn(I, R), n ∈N: Space of n-times differentiable real functions defined on an interval Isuch that the j-th derivative is continuous for j= 0, . . . , n. C∞(I)≡ C∞(I, R): Space of infinitely differentiable real functions defined on an interval I. Cn(R,R): Space of n-times differentiable real functions with real limits at ±∞, that is, Cn(R,R)=f:R→R:f|R∈Cn(R,R),∃lim t→±∞f(j)(t)∈R, j = 0, . . . , n. e Cn ϕ: Space of continuously n-differentiable ϕ-extensions to infinity: e Cn ϕ≡e Cn ϕ(R,R) = nf∈ Cn(R,R): ∃e f∈ Cn(R,R), f =ϕe f|Ro. Lα(I),1≤α < ∞: Space of the measurable functions fon the interval I such that the Lebesgue integral of |f|αis finite. kfkα,1≤α < ∞: Norm of fin the space Lα(I), that is, kfkα=ZI|f(t)|αdt1 α . V Contents L∞(I): Space of the measurable functions fon the interval Isuch that are essentially bounded. kfk∞: Norm of fin the space L∞(I), that is, kfk∞= sup {|f(t)|, t ∈I}. AC(I): Space of absolutely continuous functions, that is, AC(I)=u∈ C(I): ∃f∈L1(I), u(t) = u(t0) + Zt t0 f(s) d s, t, t0∈I. Wk,p(I), k, p ∈N: Sobolev space k−pon the set I, that is, Wk,p(I) = nu∈ Ck−1(I): u(k−1) ∈ AC(I), u(k)∈Lp(I)o. a+: Positive part of a:X→R, that is, a+(t) = max{a(t),0}. a−: Negative part of a:X→R, that is, a−(t) = max{−a(t),0}. α∗: Conjugate of the real number α≥1, that is, the real number such that 1 α+1 α∗= 1. If α= 1 then α∗=∞and vice-versa. a0:a∈Lα(I)such that a(t)≥0for a. e. t∈Iand a6≡ 0on I. Ω: Closure of the set Ω. ∂Ω: Boundary of the set Ω. iK(T,Ω): Index of operator Trelative to Ωin the cone K. r(T): Spectral radius of operator T. µ(T): Principal characteristic value of operator T. VI Preface Differential equations represent one of the strongest connections between Mathematics and real life. This is due to the fact that almost all the physical phenomena, as well as many other in economy, biology or chemistry, are modelled by differential equations. This seems thus a good reason to dedicate our efforts to trying to solve this kind of problems. In particular, we will analyze the qualitative properties of the solutions of nonlinear equations, focusing on the study of constant sign solutions on the whole domain of definition or, at least, on some subset of it. The interest of this property is due to the fact that many of the physical magnitudes which appear in differential problems can not take negative values (typical examples would be pressure, power or temperature in Kelvin degrees). Moreover, in many problems in engineering, models study the deviation of certain structures from their equilibrium point. In this context, if we want to maintain the structure stable, the deformation must occur always in the same direction, which, mathematically speaking, means that the solution must have constant sign. The most common techniques to ensure the existence of solution for these problems are based on the construction of an abstract formulation included into functional analysis, in which the solutions of the differential equations coincide with either the fixed points or the critical ones of certain operators. In this Thesis, we will work with the first method, constructing integral operators which will be determined by some kernel related to the linear part of the equation. This kernel is the so-called Green’s function. In many of the cases, operators will be defined in subsets called cones, which will let us transfer the properties of the Green’s function to the solutions of the considered problem. This fact lets us intuit now the importance that the study of the properties of linear problems (and, specially of Green’s functions) has on the research concerning nonlinear ones. The present Thesis is divided into two parts, which deal with differential problems on bounded and unbounded domains, respectively. It contains most of the work developed by the author in the last years. All these discoveries appear in several VII Preface publications which the reader may consult, namely [22,23, 27,31–34,102,103,131]. VIII Summary The present Thesis, compiled under the title “Nonlinear differential equations on bounded and unbounded domains”, contains almost the whole work developed by the author during the last years. It is divided into two parts: the first of them, which comprises six chapters, covers the study of boundary value problems defined on bounded intervals, as well as the more general case of Hammerstein integral equations. The second part, which comprises three chapters, is focused on the study of both differential and integral problems defined on unbounded domains. It should be noted that, although the title only mentions nonlinear differential equations, the first chapters of the Thesis will be devoted to the study of linear boundary value problems. This is due to the fact that the properties of these linear problems, and particularly those of the related Green’s function, will determine the best way to approach the search for solutions of nonlinear problems. We include now a brief summary of the main results given in each chapter. Chapter 1: Preliminary Results For the purpose of writing a self-contained work, this chapter compiles some preliminary results which will be used throughout the remaining of this Thesis. First, in Section 1.1, we introduce the definition and main properties of the Green’s function. As we will see, this function is a very powerful tool to study both linear and nonlinear differential problems. This is due to the fact that every differential problem can be transformed into an equivalent integral one of which the kernel is, precisely, the aforementioned Green’s function. This way, the problem of finding solutions of differential problems will naturally lead to the more general framework of finding fixed points of integral operators. It is in this context where the results ensuring the existence of fixed points of arbitrary compact operators defined in Banach spaces acquire great importance. Some of these results are collected in Section 1.2, namely the very well-known Schauder’s Fixed Point Theorem and the classical fixed point index theory (which, following the line IX Summary of [64], is introduced for arbitrary open sets, which might be unbounded). Finally, another important tool that we will use in this Thesis to study properties of linear operators is spectral theory. In particular, this theory combined with the fixed point index results, makes it possible to prove the existence of solutions of certain integral problems. Basic results regarding spectral theory are compiled in Section 1.3. Chapter 2: Green’s Functions and Spectral Theory for Even Order Linear BVPs This chapter contains a fully detailed study of even order linear boundary value problems. In particular, we study problems related to the following operator coupled with various boundary conditions: L u(t)≡u(2n)(t) + a2n−1(t)u(2n−1)(t) + ···+a1(t)u0(t) +a0(t)u(t), t ∈I≡[0, T], where ak:I→R, ak∈Lα(I), α ≥1,k= 0,...,2n−1. From this operator we will define two more, concretely e L u(t)≡u(2n)(t) + ˆa2n−1(t)u(2n−1)(t) + ˜a2n−2(t)u(2n−2)(t) +···+ ˆa1(t)u0(t) + ˜a0(t)u(t), t ∈J≡[0,2T], where ˜a2kis the even extension of a2kto the interval Jand ˆa2k+1 is the odd extension of a2k+1 to J, for k= 0, . . . , n −1, and ee L u(t)≡u(2n)(t) + ˆ ˆa2n−1(t)u(2n−1)(t) + ˜ ˜a2n−2(t)u(2n−2)(t) +···+ˆ ˆa1(t)u0(t) + ˜ ˜a0(t)u(t), t ∈[0,4T], where ˜ ˜a2kand ˆ ˆa2k+1 are the even and odd extensions to the interval [0,4T]of ˜a2k and ˆa2k+1, respectively, for k= 0, . . . , n −1. The main idea of this chapter consists of expressing the Green’s function of each Neumann, Dirichlet and mixed problems related to operator Las a sum of Green’s functions of periodic and antiperiodic problems related to e L. This way, the following equalities are proved: GN[T](t, s) = GP[2 T](t, s) + GP[2 T](2 T−t, s),∀(t, s)∈I×I, GD[T](t, s) = GP[2 T](t, s)−GP[2 T](2 T−t, s),∀(t, s)∈I×I, GM1[T](t, s) = GA[2 T](t, s)−GA[2 T](2 T−t, s),∀(t, s)∈I×I, GM2[T](t, s) = GA[2 T](t, s) + GA[2 T](2 T−t, s),∀(t, s)∈I×I, X Summary where GN[T],GD[T],GM1[T]and GM2[T]denote, respectively, the Green’s functions of Neumann, Dirichlet and mixed problems related to operator L. Analogously, GP[2 T]and GA[2 T]denote the Green’s functions of periodic and antiperiodic problems related to e L. Something similar can be done to decompose all the previous Green’s functions as a linear combination of the one related to the periodic problem associated to ee L evaluated in different points. Since the Green’s function is a fundamental tool for studying both linear and nonlinear problems, being able to relate different Green’s functions will let us relate also the spectra and the solutions of the different problems. First, the previous expressions provide a direct relation between the spectra of the considered problems. In particular, we deduce various decompositions of some spectra as the union of others. Moreover, we also obtain a certain order relation between the first eigenvalues of each problem. On the other hand, we are also able to deduce that the constant sign of one Green’s function implies the same constant sign of another one. This can be seen in the following result. Corollary 1 (Corollary 2.4.1).The following properties hold for any coefficients a0, . . . , a2n−1∈L1(I): 1. If GP[2T]≤0on J×J, then GN[T]≤0on I×I. 2. If GP[2T]≥0on J×J, then GN[T]≥0on I×I. 3. If GN[2T]≤0on J×J, then GN[T]≤0on I×I. 4. If GN[2T]≥0on J×J, then GN[T]≥0on I×I. 5. If GD[2T]≤0on J×J, then GM2[T]≤0on I×I. 6. If GD[2T]≥0on J×J, then GM2[T]≥0on I×I. With respect to previous corollary, it must be pointed out that it can be improved for order n= 1, something that will be done in Chapter 3. On the other hand, it is proved in this chapter that the converse of Assertions 1and 2holds when all of the coefficients a0, . . . , a2n−1are constants, whereas the converse of the other assertions does not hold, not even in the constant case, for n > 1. Moreover, a counterexample is given to show that the converse of Assertion 2is not true in general for n > 1. Finally, it remains as an open problem to see if Assertion 1is an equivalence or not when n > 1. Finally, in Section 2.5, under the assumption of constant sign of some Green’s function, we obtain some point by point inequalities between two different Green’s XI Summary functions. This lets us deduce that the solution of the problem under certain boundary conditions is smaller at every point than the solution of another problem with the same operator but different boundary conditions. The results in this chapter are compiled in [31]. Chapter 3: Second Order Equation This chapter considers the problem studied in Chapter 2 in the particular case of the second order equation (that is, we will take n= 1). The reason why this case is studied independently from the general one is the fact that, when working with second order differential equations, it is possible to use Sturm-Liouville’s theory. This theory, which does not hold for differential equations of higher order, provides some properties of oscillation of the solutions of the equations. This will let us obtain stronger results than in previous chapter. In this chapter, two different problems are considered. First, in Section 3.2, we study the problem related to Hill’s operator L u(t)≡u00(t) + a(t)u(t), t ∈I. This will be a particular case of operator Lgiven in Chapter 3 for n= 1 and a1≡0. We note that the fact of considering a1≡0is not an important loss of generality in the results as every second order differential equation written in the form u00(t) + a1(t)u0(t) + a0(t)u(t) = 0, can be transformed into a Hill’s equation through a suitable change of variable, as long as the coefficients a0and a1have enough regularity. The results obtained in this section are more powerful than the corresponding ones given in Chapter 3. An example which illustrates this is the following theorem in which we relate the constant sign of different Green’s functions. Theorem 2 (Theorem 3.2.22).For any a∈L1(I)the following properties hold: 1. GP[2 T]<0on J×Jif and only if GN[T]<0on I×I. This is equivalent to GN[2 T]<0on J×J. 2. GP[2 T]>0on (0,2T)×(0,2T)if and only if GN[T]>0on (0, T)×(0, T). 3. If GN[2 T]>0on (0,2T)×(0,2T)then GN[T]>0on (0, T)×(0, T). 4. If GP[2 T]<0on J×Jthen GD[2 T]<0on (0,2T)×(0,2T). 5. If GP[2 T]>0on (0,2T)×(0,2T)then GD[2 T]<0on (0,2T)×(0,2T). XII Summary 6. If GN[T](or, equivalently, GP[2 T]) has constant sign on I×I, then GD[T]<0 on (0, T)×(0, T),GM1[T]<0on [0, T)×[0, T)and GM2[T]<0on (0, T]×(0, T]. 7. GD[2 T]<0on (0,2T)×(0,2T)if and only if GM2[T]<0on (0, T]×(0, T]. 8. If either GM2[T]<0on (0, T]×(0, T]or GM1[T]<0on [0, T)×[0, T), then GD[T]<0on (0, T)×(0, T). In the same way, the point by point inequalities between various Green’s functions are also more precise, which implies that we obtain more precision when it comes to compare the solutions of different problems. Thus, while in the previous chapter we were only able to ensure that the solution of a problem was smaller at every point than the solution of another one, now we will ensure also that both solutions have constant sign. Moreover, whereas in the previous chapter we could only establish a certain order relation between the first eigenvalues of each problem, we prove in this chapter an alternation between all the eigenvalues of all the considered problems. Finally, we consider some explicit criteria to ensure the constant sign of the Green’s function of the periodic problem and, using the relations between different Green’s function, we will adapt them to the rest of the considered boundary value problems. On the other hand, Section 3.3 deals with the general second order equation given in self-adjoint form, namely (p u0)0(t) + ¯a(t)u(t) = ¯σ(t),a. e. t∈I, with p > 0a. e. t∈I,1 p∈L1(I)and ¯aand ¯σsuch that ¯a pα−1 α,¯σ pα−1 α∈Lα(I), for some α∈[1,∞]. We prove in this section that the Green’s function of any boundary value problem related to the previous equation can be expressed in terms of the Green’s function related to the Hill’s operator coupled with the same boundary conditions. As a consequence, all the results obtained in the previous section can be adapted to this more general framework. This chapter collects results from [22] and [23]. Chapter 4: Solutions for Even Order Nonlinear BVPs with Constant Sign Green’s Functions This chapters considers, for the first time in this Thesis, nonlinear boundary value problems. XIII Summary Then, we will look for fixed points of the integral operator Tu(t) := p(t) + Z∞ −∞ k(t, s)η(s)f(s, u(s)) d s in the Banach space e Cn ϕ, for a given function ϕwhich will precisely represent the asymptotic behavior of the solutions. In other words, the fact that the fixed points of the operator belong to the space e Cn ϕwill imply that such functions will asymptotically behave in a similar way to ϕ. Regarding the method employed to guarantee the existence of fixed points, we consider two different approaches in this chapter: the first of them, developed in Section 8.4, is based on the fixed point index in cones and presents quite restrictive hypotheses for the non linearity f. On the other hand, the second approach, considered in Section 8.5, is based on the definition and spectral properties of several auxiliary linear operators. In particular, if the spectral radius of these operators together with some limits involving the nonlinearity fsatisfy some suitable properties, it will be possible to ensure the existence of fixed points. In this case, the restrictions on the function fare much weaker than the ones imposed with the previous method, but at the expense of requiring the kernel kto satisfy some more restrictive conditions. As it is shown in the chapter with some examples, the two methods are not comparable. All these results are collected in [33] and [34]. Chapter 9: On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian In this last chapter we study a singular initial value problem with φ-Laplacian, with special attention to the existence of unbounded solutions. In this case, since we are dealing with a singular problem, it is not possible to construct an equivalent integral problem, as it was made in previous chapters. Consequently, the techniques used in this chapter will totally differ from the ones considered up to this moment. In particular, we will consider the following nonlinear problem: ((p(t)φ(u0(t)))0+p(t)f(φ(u(t))) = 0, t > 0, u(0) = u0, u0(0) = 0, u0∈[L0, L]. We begin the chapter with the definition of three types of solutions that we may find. This way, denoting usup = sup{u(t): t∈[0,∞)}, XX Summary we will say that A solution uof the problem is damped if usup < L. A solution uof the problem is homoclinic if usup =L. A solution uof the problem will be an escape solution if usup > L. Since both damped and homoclinic solutions are bounded, unbounded solutions will constitute a subset of the escape ones. This motivates the division of the chapter in two parts: 1. Search of conditions to ensure the existence of escape solutions. 2. Search of necessary and sufficient conditions to guarantee that an escape solution is unbounded. Moreover, for the investigation of conditions which assure the existence of escape solutions, we will distinguish two differentiated cases: the first of them, in which both fand φ−1are Lispchitz continuous, is quite simple since, under these conditions, we can guarantee the uniqueness of solution of the problem. On the contrary, the second case (with fand φ−1not necessarily Lispchitz continuous), presents several complications derived from the non uniqueness of solution. To solve these problems, we consider the method of lower and upper solutions. These two cases have another important difference in relation with the results obtained: in the first one, we guarantee the existence of a sequence of escape solutions with different initial values, whereas in the second one, it may occur that all the solutions have the same initial value L0. Finally, the last section of the chapter compiles all of the obtained results. We give there the explicit formulation of some sufficient conditions to assure the existence of unbounded solutions of the problem. Several examples show that all these results are not comparable. All the results in this chapter are given in [131]. XXI Part I Bounded Domains Bounded Domains The first part of this Thesis focuses on the study of linear and nonlinear boundary value problems defined on bounded domains. Our main goal will be the study of nonlinear differential equations. However, when working with this kind of problems, the first step consists of studying their related linear ones. To do this, the main tool will be the so-called Green’s function, whose properties will clearly lead to the best way of dealing with nonlinear problems. In particular, some of the main techniques applied in the recent literature to prove the existence of solutions of nonlinear boundary value problems are, among others, monotone iterative techniques (see [74,96,143]), the lower and upper solutions method (see [16, 46]) or fixed points theorems (see [74, 142]). In all these cases, the constant sign of the associated Green’s functions is usually fundamental to prove such results. All this clearly justifies the necessity of starting this Thesis by studying linear boundary value problems, focusing our attention on the properties which characterize the constant sign of Green’s functions. This part is structured in six chapters as follows: Chapter 1 is dedicated to show some preliminary results and concepts for the sake of constructing a self-contained thesis. First, following [18], we define the Green’s function related to a boundary value problem. Then, we summarize some results which will be used throughout this Thesis to ensure the existence of fixed points of various operators defined on Banach spaces. Finally, we include some definitions and results of spectral theory. Chapter 2 includes a fully-detailed study of even order linear boundary value problems, focusing on finding relations between various Green’s functions. All the results in this chapter are collected in [31]. Chapter 3 particularizes the study developed in Chapter 2 to the second order equation. Since Sturm-Liouville and Oscillation theory is applicable to second order problems, the results in this chapter are more powerful than those in the previous one. This chapter compiles results included in [22] and [23]. Chapter 4 deals, for the first time, with nonlinear boundary value problems. In particular, the results obtained in Chapters 2 and 3 will be shown to be of great importance to ensure the existence of solutions of the nonlinear problems considered in this chapter. A basic assumption will be the constant sign of the considered Green’s functions. Moreover, the tool used to prove the existence of solution will be the lower and upper solutions method. This chapter is based on the last section of [31]. Chapter 5 is completely devoted to find solutions of nonlinear problems in the case when, contrary to Chapter 4, the Green’s functions change sign. In particular, we will deal with second order problems and prove the existence of solutions by means of the fixed point index theory. This results are collected in [27]. 3 Bounded Domains Finally, Chapter 6 considers an integral problem instead of a differential one. As we will see, this is in fact a generalization since differential problems can be transformed into integral ones whose kernel function is, precisely, the Green’s function. In this chapter, we will define a new type of cones which makes it necessary to apply the fixed point index theory in unbounded sets in order to find fixed points of the considered integral operators. Sections 6.1 to 6.6 of this chapter are included in [102], while the particular case given in Section 6.7 is collected in [32]. 4 Chapter 1 Preliminaries In order to construct a self-contained work, we will dedicate this chapter to introduce some definitions and previous results which will be used throughout the different chapters. First of all, since the main tool to study linear problems is the so-called Green’s function, Section 1.1 introduces this concept and establishes how the solutions of linear problems can be explicitly calculated by means of the Green’s function. Next, we will show in Section 1.2 how to transform a nonlinear differential problem into an equivalent integral one, in the sense that the solutions of the aforementioned differential problem correspond to fixed points of a certain integral operator. Moreover, we will include in this section several theorems to prove the existence of fixed points of compact operators defined on Banach spaces. Finally, Section 1.3 compiles some basic results of spectral theory of linear operators. 1.1. Green’s Functions In this section, following [18], we will summarize the definition and main properties of Green’s functions. Consider the general two-point n-th order differential problem (Lnu(t) = σ(t), t ∈[0, T], Ui(u)=0, i = 1, . . . , n, (1.1.1) where Lnu(t)≡u(n)(t) + an−1(t)u(n−1)(t) + ···+a1(t)u0(t) + a0(t)u(t) and Ui(u)≡ n−1 X j=0 αi ju(j)(0) + βi ju(j)(T), i = 1, . . . , n, where αi jand βi jare real constants for all i= 1, . . . , n,j= 0, . . . , n −1and σ, ak∈L1([0, T]) for all k= 1, . . . , n. 5 Preliminaries We will denote I≡[0, T ]. We can characterize the Green’s function for problem (1.1.1) as follows. Definition 1.1.1. [18, Definition 1.4.1] We say that Gis a Green’s function for problem (1.1.1) if it satisfies the following properties: (G1) Gis defined on the square I×I(except at the points with t=sif n= 1). (G2) For k= 0, . . . , n −2, the partial derivatives ∂kG ∂ tkexist and are continuous on I×I. (G3) Both ∂n−1G ∂ tn−1and ∂nG ∂ tnexist and are continuous on the triangles 0≤s<t≤T and 0≤t<s≤T. (G4) For each s∈(0, T), the function G(·, s)is a solution of the differential equation Lny= 0 a. e. on [0, s)∪(s, T ], that is, ∂nG ∂ tn(t, s)+an−1(t)∂n−1G ∂ tn−1(t, s)+···+a1(t)∂ G ∂ t (t, s)+a0(t)G(t, s) = 0, for all t∈I\{s}. (G5) For each t∈(0, T)there exist the lateral limits ∂n−1G ∂ tn−1(t−, t) = ∂n−1G ∂ tn−1(t, t+)and ∂n−1G ∂ tn−1(t, t−) = ∂n−1G ∂ tn−1(t+, t) and, moreover, ∂n−1G ∂ tn−1(t+, t)−∂n−1G ∂ tn−1(t−, t) = ∂n−1G ∂ tn−1(t, t−)−∂n−1G ∂ tn−1(t, t+) = 1. (G6) For each s∈(0, T), the function G(·, s)satisfies the boundary conditions Ui(G(·, s)) = 0,i= 1, . . . , n, that is, n−1 X j=0 αi j ∂jG ∂ tj(0, s) + βi j ∂jG ∂ tj(T, s)= 0, i = 1, . . . , n. Remark 1.1.2. Note that the Green’s function depends on the homogeneous part of problem (1.1.1), but not on the considered function σ. Due to this fact, we will frequently talk about the Green’s function related to the homogeneous problem, namely (Lnu(t)=0, t ∈[0, T], Ui(u) = 0, i = 1, . . . , n. 6 1.1 Green’s Functions We will consider the space Wn,1(I) = nu∈ Cn−1(I) : u(n−1) ∈ AC(I)o, where AC(I)denotes the set of absolutely continuous functions on I. In particular, we will consider a subset X⊂Wn,1(I)defined in the following way X=u∈Wn,1(I) : Ui(u) = 0, i = 1, . . . , n.(1.1.2) It is easy to check that Xis a Banach space with the usual norm kukX= max nu(i):i= 0, . . . , n −1o. Now, we will introduce the following definition. Definition 1.1.3. Given a Banach space X, operator Lnis said to be nonresonant on Xif and only if the homogeneous equation Lnu(t)=0 a. e. t∈I, u ∈X, has only the trivial solution. The following result relates the uniqueness of solution of problem (1.1.1) with the uniqueness of the Green’s function. This can be seen in [18, Corollary 1.2.4 and Theorem 1.2.17]. Theorem 1.1.4. The following assertions are equivalent: 1. Operator Lnis nonresonant on Xgiven in (1.1.2). 2. There exists a unique Green’s function related to problem (1.1.1). 3. Problem (1.1.1) has a unique solution u∈Wn,1(I). In such a case, the unique solution is given by the following expression u(t) = ZT 0 G(t, s)σ(s) d s, ∀t∈I. (1.1.3) Furthermore, it is very well known (see [18, 40, 106]) that operator Lnis selfadjoint on Xif and only if the related Green’s function exists and is symmetrical with respect to the diagonal of its square of definition, that is, G(t, s) = G(s, t),∀(t, s)∈I×I. We will also introduce the following important definitions. In them, we will use the notation h0to denote a function h∈Lα(I)such that h(t)≥0for a. e. t∈I and h6≡ 0on I. 7 Preliminaries Let (N1,k·k1)and (N2,k·k2)be two normed spaces. Let T:N1→N2be a bounded linear operator, that is, such that its norm kTk = sup kuk26=0 kTuk1 kuk2 is finite. We recall the following definitions. Definition 1.3.1. We say that λis an eigenvalue of a linear operator between normed spaces T: (N1,k·k1)→(N2,k·k2), with corresponding eigenfunction φ, if φ6≡ 0 and λ φ =Tφ. The reciprocals of nonzero eigenvalues are called characteristic values of T. Definition 1.3.2. We will define the spectral radius of a bounded linear operator T as r(T) := lim n→∞kT nk1 n, and its principal characteristic value as µ(T) := 1 r(T)if r(T)6= 0. For more properties of this generalized spectral value we refer the reader to [11, 164]. Now we will formulate the very well-known Krein-Rutman Theorem. Theorem 1.3.3 (Krein-Rutman, [52, Theorem 1.1]).Let K⊂Xbe a total cone and T:X→Xa compact linear operator that maps Kto Kwith positive spectral radius r(T). Then r(T)is an eigenvalue with an eigenvector φ∈K\{0}. We will give now the sharper version of this theorem for strongly positive linear operators. Definition 1.3.4. Let K⊂Xbe a cone with nonempty interior and let T:X→X be a compact linear operator. We will say that Tis strongly positive if and only if Tx∈int(K),∀x∈K\{0}, where int(K)denotes the interior of the cone. Theorem 1.3.5 ( [4, Theorem 3.2]).Let K⊂Xbe a cone with nonempty interior and T:X→Xa strongly positive and compact linear operator that maps Kto K. Then, the following assertions hold: The spectral radius r(T)is positive. r(T)is a simple eigenvalue of Twith a positive related eigenfunction and there is no other eigenvalue with a positive eigenfunction. 14 1.3 Spectral Theory Finally, we recall some known results which will let us find some lower and upper bounds for the spectral radius. Theorem 1.3.6 ([148, Theorem 2.7]).Let Tbe a bounded linear operator in a Banach space Xand let Kbe a cone in Xsuch that T(K)⊂K. If there exists λ0>0 and v∈K\{0}such that Tvλ0v, then r(T)≥λ0. Theorem 1.3.7 ( [157, Theorem 1]).Let Tbe a linear and compact operator and let Kbe a cone in X. Assume that Khas non empty interior and that T(K)⊂K. If there exists v, an interior element of the cone, for which the following inequality holds Tvλ0v, then r(T)≤λ0. 15 Chapter 2 Green’s Functions and Spectral Theory for Even Order Linear Boundary Value Problems In this chapter we will develop a fully-detailed study of even order linear boundary value problems. We have seen in the previous chapter that the solutions of a given boundary value problem coincide with the fixed points of related integral operators which have as kernel the associated Green’s function in each case. Thus, the Green’s function plays a very important role in the study of boundary value problems. Traditionally, the most studied boundary value problems have been the periodic and the two-point ones. In this chapter we will take advantage of such studies by finding some connections between the Green’s functions of various separated two point boundary conditions and the Green’s functions of periodic problem. The key idea is that the expression of the Green’s function related to each two points case can be obtained as a linear combination of the Green’s function of periodic problems. From these expressions relating the different Green’s functions, we will be able to compare their constant sign. These results will allow us to obtain some comparison principles which guarantee that, under certain hypotheses, the solution of a boundary value problem under some suitable conditions is bigger in every point than the solution of the same equation under another type of boundary conditions. We will also obtain a decomposition of the spectrum of some problems as a combination of the other ones and some relations of order between the first eigenvalues of the considered problems. The chapter is organized as follows: Section 2.1 includes some preliminary results and proves a symmetry property which will be satisfied by some Green’s functions. In Section 2.2, we detail the aforementioned decomposition of Green’s functions. In Section 2.3, we relate both the spectra and the first eigenvalues of the considered problems. In Section 2.4, we prove some results relating the constant sign of various Green’s functions. Finally, in Section 2.5, we show some point-by-point 17 Green’s Functions and Spectral Theory for Even Order Linear BVPs relations between different Green’s functions and also between solutions of the same operator under several boundary conditions. It must be pointed out that the study developed in Sections 2.3 to 2.5 will be particularized in Chapter 3 for the second order equation. As we will see, in such a case, many results will be stronger than for the general even order problem. The reason is that, for second order equations, Sturm-Liouville’s Theory is applicable, which makes it possible to obtain more information regarding oscillation of the solutions, Green’s functions and spectral theory. All the results in this chapter are collected in [31]. 2.1. Preliminary Results In this section we will introduce three different operators. The first of them, which will be called operator L, will be defined with arbitrary coefficients. On the other hand, the coefficients of the other two operators (denoted by e Land ee L) will be defined as either even or odd extensions of the coefficients of the aforementioned operator L. This way, while the original operator will be defined on the interval [0, T], the two auxiliary operators e Land ee Lwill be defined on [0,2T]and [0,4T], respectively. Furthermore, the symmetries in the coefficients of the operator e Lwill induce also some symmetries on the Green’s functions related to this operator. This property will also be proved in this section. Consider then the 2n-th order general linear operator L u(t)≡u(2n)(t) + a2n−1(t)u(2n−1)(t) + ···+a1(t)u0(t) + a0(t)u(t),(2.1.1) with t∈Iand ak:I→R, ak∈Lα(I), α ≥1,k= 0,...,2n−1. We will introduce now the first auxiliary linear operator, whose coefficients will be defined from those of operator Las follows: e L u(t)≡u(2n)(t) + ˆa2n−1(t)u(2n−1)(t) + ˜a2n−2(t)u(2n−2)(t) +···+ ˆa1(t)u0(t) + ˜a0(t)u(t), t ∈J≡[0,2T], where ˜a2k,k= 0, . . . , n −1, is the even extension of a2kto J, that is, ˜a2k(t) = (a2k(t), t ∈[0, T], a2k(2T−t), t ∈[T, 2T], and ˆa2k+1,k= 0, . . . , n −1, is the odd extension of a2k+1 to J, that is, ˆa2k+1(t) = (a2k+1(t), t ∈[0, T], −a2k+1(2T−t), t ∈(T, 2T]. 18 2.1 Preliminary Results Notation 2.1.1. As we have mentioned before, throughout this chapter we will work with problems defined on different intervals. Because of this reason, we will use the notation G[T],G[2 T]and G[4 T]to indicate that we are working on the interval [0, T],[0,2T]or [0,4T], respectively. This way, we will stress the dependence of the Green’s function on the considered interval. We obtain the following symmetric property for Green’s functions related to operator e L. Lemma 2.1.2. Let X⊂W2n,1(J)be a Banach space such that operator e Lis nonresonant on X. Moreover, suppose that if v∈Xand w∈W2n,1(J)is such that w(t) := v(2 T−t)for all t∈J, then w∈X. Then the following equality holds: G[2 T](t, s) = G[2 T](2 T−t, 2T−s),∀(t, s)∈J×J. (2.1.2) Proof. Let ¯σ∈L1(J)be arbitrarily chosen and consider the problem e L v(t) = ¯σ(t),a. e. t∈J, v ∈X. Since operator e Lis nonresonant on X, this problem has a unique solution vwhich is given by v(t) = Z2T 0 G[2 T](t, s) ¯σ(s) d s. On the other hand, taking into account the fact that ˜a2k(t) = ˜a2k(2 T−t)and ˆa2k+1(t) = −ˆa2k+1(2 T−t), it is easy to verify that w(t) = v(2 T−t)is the unique solution of the problem e L w(t) = ¯σ(2 T−t),a. e. t∈J, w ∈X. Therefore, w(t) = Z2T 0 G[2 T](t, s) ¯σ(2 T−s) d s and, making a suitable change of variable, w(t) = Z2T 0 G[2 T](t, 2T−s) ¯σ(s) d s. Now, since w(t) = v(2 T−t) = Z2T 0 G[2 T](2 T−t, s) ¯σ(s) d s, 19 Green’s Functions and Spectral Theory for Even Order Linear BVPs and ¯σ∈L1(J)is arbitrary, we arrive at the following equality G[2 T](2 T−t, s) = G[2 T](t, 2T−s),∀(t, s)∈J×J, or, which is the same, G[2 T](t, s) = G[2 T](2 T−t, 2T−s),∀(t, s)∈J×J. In addition, we will consider another auxiliary operator ee Lwhich will be constructed from e Lin the same way than e Lhas been constructed from L, that is: ee L u(t)≡u(2n)(t) + ˆ ˆa2n−1(t)u(2n−1)(t) + ˜ ˜a2n−2(t)u(2n−2)(t) +···+ˆ ˆa1(t)u0(t) + ˜ ˜a0(t)u(t), t ∈[0,4T], where ˜ ˜a2kand ˆ ˆa2k+1,k= 0, . . . , n −1, are the even and odd extensions to the interval [0,4T]of ˜a2kand ˆa2k+1, respectively. 2.2. Decomposing Green’s Functions In this section we will obtain the expression of the Green’s function of different two point boundary value problems (Neumann, Dirichlet and Mixed problems) as a sum of Green’s functions of other related problems. This decomposition has been detailed in [22] for the particular case of n= 1 and a1≡0and generalized in [31] for the general case with arbitrary n. In particular, we will work with some problems related to operator L(and, consequently, defined on the interval [0, T]), some others related to operator e L(and, consequently, defined on [0,2T]) and the periodic problem related to ee L(defined on [0,4T]). In the sequel, we describe the different problems and boundary conditions we are dealing with: Neumann problem on the interval I: (L u(t) = σ(t),a. e. t∈I, u(2k+1)(0) = u(2k+1)(T) = 0, k = 0, . . . , n −1.(N, T) Dirichlet problem on the interval I: (L u(t) = σ(t),a. e. t∈I, u(2k)(0) = u(2k)(T) = 0, k = 0, . . . , n −1.(D, T ) 20 2.2 Decomposing Green’s Functions Mixed problem 1 on the interval I: (L u(t) = σ(t),a. e. t∈I, u(2k+1)(0) = u(2k)(T) = 0, k = 0, . . . , n −1.(M1, T) Mixed problem 2 on the interval I: (L u(t) = σ(t),a. e. t∈I, u(2k)(0) = u(2k+1)(T) = 0, k = 0, . . . , n −1.(M2, T) Periodic problem on the interval J: (e L u(t) = ¯σ(t),a. e. t∈J, u(k)(0) = u(k)(2 T), k = 0,...,2n−1.(P, 2T) Antiperiodic problem on the interval J: (e L u(t) = ¯σ(t),a. e. t∈J, u(k)(0) = −u(k)(2 T), k = 0,...,2n−1.(A, 2T) Neumann problem on the interval J: (e L u(t) = ¯σ(t),a. e. t∈J, u(2 k+1)(0) = u(2 k+1)(2 T)=0, k = 0, . . . , n −1.(N, 2T) Dirichlet problem on the interval J: (e L u(t) = ¯σ(t),a. e. t∈J, u(2 k)(0) = u(2 k)(2 T)=0, k = 0, . . . , n −1.(D, 2T) Periodic problem on the interval [0,4T]:   ee L u(t) = ¯ ¯σ(t),a. e. t∈[0,4T], u(k)(0) = u(k)(4 T), k = 0, . . . , 2n−1. (P, 4T) Now, we will show how to relate the expressions of different Green’s functions. We will assume that all the considered operators are nonresonant on the corresponding Banach space with suitable boundary conditions. Later, we will see in Section 2.3 that the aforementioned nonresonant character of all the operators is, in some sense, equivalent. 21 Green’s Functions and Spectral Theory for Even Order Linear BVPs 2.2.1. Neumann Problem To begin with, we will decompose the Green’s function related to problem (N, T) as sum of the Green’s function related to (P, 2T)evaluated in the same point and of the same function evaluated in another point which satisfies a symmetric relation. First, suppose that operator Lis nonresonant on the space XN,T =nu∈W2n,1(I) : u(2k+1)(0) = u(2k+1)(T) = 0, k = 0, . . . , n −1o, that is, problem (N, T)has a unique solution in W2n,1(I)for all σ∈L1(I). Moreover, assume that e Lis nonresonant on XP,2T=nu∈W2n,1(J) : u(k)(0) = u(k)(2 T), k = 0, . . . , 2n−1o, that is, problem (P, 2T)has a unique solution in W2n,1(J)for all ¯σ∈L1(J). Let ube the unique solution of problem (N, T). Then, defining vas the even extension of u, it can be proved that v∈W2n,1(J)satisfies the equation e L v(t) = ¯σ(t) for the particular case of taking ¯σas the even extension of σ. Indeed, for t∈[0, T ], it holds that e L v(t) = L u(t) = σ(t) = ¯σ(t) and, for t∈[T, 2T], e L v(t) = v(2n)(t) + ˆa2n−1(t)v(2n−1)(t) + ˜a2n−2(t)v(2n−2)(t) +···+ ˆa1(t)v0(t) + ˜a0(t)v(t) =u(2n)(2 T−t)−ˆa2n−1(t)u(2n−1)(2 T−t) + ˜a2n−2(t)u(2n−2)(2 T−t) +···−ˆa1(t)u0(2 T−t) + ˜a0(t)u(2 T−t) =u(2n)(2 T−t) + ˆa2n−1(2 T−t)u(2n−1)(2 T−t) + ˜a2n−2(2 T−t)u(2n−2)(2 T−t) +···+ ˆa1(2 T−t)u0(2 T−t) + ˜a0(2 T−t)u(2 T−t) = ¯σ(2 T−t) = ¯σ(t). Moreover, it is clear that v∈XP,2Tand thus vis a solution of problem (P, 2T). Therefore, if we denote by GN[T]and GP[2 T]the Green’s functions related to problems (N, T)and (P, 2T), respectively, we obtain the following equalities for 22 2.2 Decomposing Green’s Functions t∈I: ZT 0 GN[T](t, s)σ(s) d s=u(t) = v(t) = Z2T 0 GP[2 T](t, s) ¯σ(s) d s =ZT 0 GP[2 T](t, s)σ(s) d s +Z2T T GP[2 T](t, s)σ(2 T−s) d s =ZT 0 (GP[2 T](t, s) + GP[2 T](t, 2T−s)) σ(s) d s. Now, since previous equality holds for every σ∈L1(I), we can deduce that GN[T](t, s) = GP[2 T](t, s) + GP[2 T](t, 2T−s),∀(t, s)∈I×I, or, which is the same, using Lemma 2.1.2, GN[T](t, s) = GP[2 T](t, s) + GP[2 T](2 T−t, s),∀(t, s)∈I×I. (2.2.1) The previous expression lets us obtain the exact value at every point of the Green’s function of the Neumann problem by means of the values of the periodic one, as long as both Green’s functions exist. Analogously, assuming e Lis nonresonant on XN,2T=nu∈W2n,1(J) : u(2k+1)(0) = u(2k+1)(2T) = 0, k = 0, . . . , n −1o, it can be also seen that v∈XN,2T, that is, vis a solution of problem (N, 2T). Thus, denoting by GN[2 T]the Green’s function related to (N, 2T)and arguing as in the previous case, it can be deduced that GN[T](t, s) = GN[2 T](t, s) + GN[2 T](2 T−t, s),∀(t, s)∈I×I, (2.2.2) or, using (2.2.1), GN[T](t, s) = GP[4 T](t, s) + GP[4 T](4 T−t, s) +GP[4 T](2 T−t, s) + GP[4 T](2 T+t, s),(2.2.3) for all (t, s)∈I×I. 23 Green’s Functions and Spectral Theory for Even Order Linear BVPs Theorem 2.3.2. Assume that all the previously considered spectra are not empty, the first eigenvalue of each problem (except for (A, 2T)) is simple and its related eigenfunction has constant sign. Then, the following equalities are fulfilled for any a0, . . . , a2n−1∈L1(I): 1. λN 0[T] = λP 0[2 T]< λD 0[T]. 2. λN 0[T] = λN 0[2 T]< λM1 0[T]. 3. λN 0[T] = λP 0[4 T]. 4. λM2 0[T] = λD 0[2 T]< λD 0[T]. 5. λN 0[T]< λM2 0[T]. 6. λA 0[2 T] = min nλM1 0[T], λM2 0[T]o. Proof. Assertion 1is proved in the following way: as we have seen above, the spectrum of (P, 2T)is decomposed as ΛP[2 T]=ΛN[T]∪ΛD[T], which implies that λP 0[2 T] = min λN 0[T], λD 0[T]. Consider now the even extension to Jof the eigenfunction associated to λN 0[T]. This extension has constant sign on Jand, moreover, it satisfies periodic boundary conditions, so it is a constant sign eigenfunction of (P, 2T). On the contrary, the odd extension to Jof the eigenfunction associated to λD 0[T]is a sign changing eigenfunction of (P, 2T). Therefore, since we have assumed that the eigenfunction related to the first eigenvalue of each problem has constant sign, we deduce that λN 0[T] = λP 0[2 T]< λD 0[T]. An analogous argument is valid to prove Assertion 2, by taking into account that ΛN[2 T] = ΛN[T]∪ΛM1[T]. Assertion 3is deduced from the two previous one. Indeed, Assertion 1implies that λN 0[2 T] = λP 0[4 T]and, from Assertion 2, we deduce the equality. Assertion 4is proved analogously to Assertions 1and 2, taking into account the decomposition ΛD[2 T] = ΛD[T]∪ΛM2[T]. Now Assertion 5can be deduced from 1,2and 4. Indeed, Assertion 1implies that λN 0[2 T]< λD 0[2 T]and, using Assertions 2and 4, λN 0[T] = λN 0[2 T]< λD 0[2 T] = λM2 0[T]. Finally, Assertion 6is an immediate consequence of ΛA[2 T] = ΛM1[T]∪ΛM2[T]. 30 2.4 Constant Sign of Green’s Functions Remark 2.3.3. With respect to the hypothesis that all the considered spectra are not empty note that, as a consequence of the relations proved at the beginning of this section, if one of those spectra is not empty, we could ensure that some others are not empty too. On the other hand, there are several results which ensure that, under some suitable conditions, the first eigenvalue of a boundary value problem is simple and its related eigenfunction has constant sign, for instance, Krein-Rutman Theorem. Sufficient conditions to ensure that all the hypotheses required in previous theorem are fulfilled can be found in [87]. First, we can deduce from Theorem 1 in such reference that if there exists some λfor which the Green’s function G[λ, T ]has constant sign and the spectrum of such problem is not empty, then the eigenfunction related to the first eigenvalue has constant sign. Moreover, from Theorem 2 in [87] it is deduced that if there exists some λfor which the Green’s function G[λ, T ]has strict constant sign on [0, T]×(0, T)then the spectrum of such problem is not empty, the first eigenvalue is simple and its related eigenfunction has strict constant sign on (0, T). Finally, from Theorem 2’ in [87] we can ensure that if there exists some λfor which G[λ, T]has strict constant sign on (0, T )×(0, T)and there exists a continuous function φ, positive on (0, T), such that G[λ, T](t, s) φ(t) is continuous on [0, T]×[0, T]and positive on [0, T]×(0, T), then the spectrum of such problem is not empty, the first eigenvalue is simple and its related eigenfunction has strict constant sign on (0, T). Analogously, if conditions given in Lemmas 1.1.8 or 1.1.9 hold for some λ, then we are also able to deduce that the spectrum of such problem is not empty, the first eigenvalue is simple and its related eigenfunction has constant sign. Details of this can be seen in [18], where it is proved that Lemmas 1.1.8 or 1.1.9 imply that KreinRutman Theorem holds. Finally, we must note that, since the eigenfunctions of the considered problems are related, the constant sign of the eigenfunction associated with the first eigenvalue of a problem implies (in some cases) the constant sign of the eigenfunction of other problems. 2.4. Constant Sign of Green’s Functions From all the connecting expressions between different Green’s functions given in Section 2.2, it is possible to deduce that the constant sign of one of them implies the 31 Green’s Functions and Spectral Theory for Even Order Linear BVPs constant sign of another one. In particular, from (2.2.1), (2.2.2) and (2.2.11) we deduce the relations below. Corollary 2.4.1. The following properties hold for any a0, . . . , a2n−1∈L1(I): 1. If GP[2T]≤0on J×J, then GN[T]≤0on I×I. 2. If GP[2T]≥0on J×J, then GN[T]≥0on I×I. 3. If GN[2T]≤0on J×J, then GN[T]≤0on I×I. 4. If GN[2T]≥0on J×J, then GN[T]≥0on I×I. 5. If GD[2T]≤0on J×J, then GM2[T]≤0on I×I. 6. If GD[2T]≥0on J×J, then GM2[T]≥0on I×I. Remark 2.4.2. In the particular case of considering disconjugated operators, the values of λfor which some of the previously considered Green’s functions, related to operator L[λ], have constant sign have been characterized in [38,39]. More specifically, the general boundary conditions considered in that reference include what we have called Dirichlet and Mixed conditions, but do not cover neither Neumann nor periodic and antiperiodic conditions. The reciprocal of Assertions 1 and 2 in the previous corollary holds for constant coefficients. This occurs as a consequence of the following property. Lemma 2.4.3. [18, Section 1.4] Let Lnu(t)≡u(n)(t) + an−1(t)un−1(t) + ···+a1(t)u0(t) + a0(t)u(t), t ∈I, be a n-th order linear operator and let GP[T]denote the Green’s function related to the periodic problem (Lnu(t)=0, t ∈I, u(k)(0) = u(k)(T), k = 0, . . . , n −1. If the coefficients ak,k= 0, . . . , n −1, involved in the definition of operator Lnare constant on I, then the Green’s function is constant over the straight lines of slope one, that is, it satisfies the following property GP[T](t, s) = (GP[T](t−s, 0),0≤s≤t≤T, GP[T](T+t−s, 0),otherwise. As a consequence, we arrive at the following result. 32 2.4 Constant Sign of Green’s Functions Theorem 2.4.4. If all the coefficients a0, . . . , a2n−1are constant, then the following properties hold: 1. GP[2 T]≤0on J×Jif and only if GN[T]≤0on I×I. 2. GP[2 T]≥0on J×Jif and only if GN[T]≥0on I×I. Proof. From Corollary 2.4.1, the assertion is equivalent to prove that if GP[2 T] changes sign, then GN[T]will also change sign. Indeed, assume that there exist two pairs of values (t1, s1)and (t2, s2)such that GP[2 T](t1, s1)<0and GP[2 T](t2, s2)>0. As it is satisfied that GP[2 T](t, s) = GP[2 T](s, t)for all (t, s)∈J×J, we may assume, without loss of generality, that s1≤t1and s2≤t2. Since all the coefficients a0, . . . , a2n−1are constant then, from Lemma 2.4.3, it holds that GP[2 T](t, s) = (GP[2 T](t−s, 0),0≤s≤t≤2T, GP[2 T](2 T+t−s, 0),otherwise. Therefore, it is fulfilled that GP[2 T](t1, s1) = GP[2 T](t1−s1,0) and GP[2 T](t2, s2) = GP[2 T](t2−s2,0). On the other hand, from equality (2.1.2) and the fact that the Green’s function satisfies the periodic boundary conditions (see Definition 1.1.1), it holds that GP[2 T](t1−s1,0) = GP[2 T](2 T−t1+s1,2T) = GP[2 T](2 T−t1+s1,0) and GP[2 T](t2−s2,0) = GP[2 T](2 T−t2+s2,2T) = GP[2 T](2 T−t2+s2,0). Now, we will distinguish two possibilities: If t1−s1≤T, then GN[T](t1−s1,0) = GP[2 T](t1−s1,0) + GP[2 T](2 T−t1+s1,0) = 2 GP[2 T](t1−s1,0) <0. 33 Green’s Functions and Spectral Theory for Even Order Linear BVPs When t1−s1> T, we have GN[T](2 T−t1+s1,0) = GP[2 T](2 T−t1+s1,0) + GP[2 T](t1−s1,0) = 2 GP[2 T](t1−s1,0) <0. Analogously, if t2−s2≤T, then GN[T](t2−s2,0) = 2 GP[2 T](t2−s2,0) >0 and, if t2−s2> T, then GN[T](2 T−t2+s2,0) = 2 GP[2 T](t2−s2,0) >0. It is clear that, in any of the cases, GN[T]changes its sign and the result holds. The following counterexample shows that the converse of Assertion 2 in Corollary 2.4.1 is not true in general for nonconstant coefficients. Example 2.4.5. Consider the Neumann problem on [0, T] = [0,2] related to operator L u(t) = u(4)(t) + ((t−2)4+λ)u(t), t ∈[0,2] ,(2.4.1) and the periodic problem on [0,2T] = [0,4] related to e L u(t)≡u(4)(t) + ((t−2)4+λ)u(t), t ∈[0,4] .(2.4.2) By numerical approach, we find that GN[T]is nonpositive for λ∈λ1, λN 0[T], where λ1≈ −2.26 and λN 0[T] = λP 0[2 T]≈ −1.746. Moreover, it is nonnegative for λ∈λN 0[T], λ2, with λ2≈4.11. However, GP[2 T]is nonpositive for λ∈λ1, λP 0[2 T]and nonnegative for λ∈λP 0[2 T], λ3, with λ3≈5.95. Despite this, we remark that the interval of values of λfor which GN[T]and GP[2 T]are nonpositive is exactly the same. Remark 2.4.6. It must be pointed out that the converse of Assertion 2 in Corollary 2.4.1 also holds for several examples with non constant coefficients. However we have not been able to prove the existence of any general condition under which this assertion holds. Furthermore, up to this moment, we have not been able to find a counterexample for the converse of Assertion 1. So, it remains as an open problem to know if Assertion 1 is or not an equivalence for n≥2. 34 2.4 Constant Sign of Green’s Functions The following counterexample shows that the converse of Assertions 3 and 4 in Corollary 2.4.1 does not hold in general, not even in the constant case. Example 2.4.7. Consider the following Neumann problem with constant coefficients on [0, T] = 0,3 2related to the following operator L u(t)≡u(4)(t) + λ u(t), t ∈0,3 2, and the Neumann problem on [0,2T] = [0,3] related to e L u(t)≡u(4)(t) + λ u(t), t ∈[0,3] , By numerical approach, it can be seen that in this case GN[T]is nonpositive for λ∈λ4, λN 0[T], with λ4≈ −6.1798 and λN 0[T] = 0, and nonnegative for λ∈λN 0[T], λ5, with λ5≈24.7192. However, GN[2 T]is nonpositive for λ∈λ6, λN 0[2 T], with λ6≈ −0.3862 and λN 0[2 T] = 0, and nonnegative for λ∈λN 0[2 T], λ7, with λ7≈1.5449. So, the converse of Assertions 3 and 4 does not hold for these operators. The following counterexample shows that the converse of Assertions 5 and 6 in Corollary 2.4.1 is not true in general, not even in the constant case. Example 2.4.8. Consider the Mixed problem 2 with constant coefficients on the interval [0, T] = [0,1] related to operator L u(t)≡u(4)(t) + λ u(t), t ∈[0,1] , and the Dirichlet problem on [0,2T] = [0,2] related to e L u(t)≡u(4)(t) + λ u(t), t ∈[0,2] . In this case, it can be seen that GM2[T]is nonpositive for λ∈(λ8, λM2 0[T]), with λ8≈ −31.2852 and λM2 0[T] = λD 0[2 T] = −π4 16 ≈ −6.088. Moreover, it is nonnegative for λ∈(λM2 0[T], λ9), with λ9≈389.6365. However, GD[2 T]is nonpositive for λ∈λ10, λD 0[2 T], with λ10 ≈ −14.8576, and nonnegative for λ∈λD 0[2 T], λ11, with λ11 ≈59.4303. Finally, from the relations given in Theorem 2.3.2, together with the general characterization given in Lemmas 1.1.8 and 1.1.9, we can deduce the following corollary. To establish the suitable conditions under which next result is valid, we need to introduce some notation. This way, analogously to what we have done in Section 1.1, 35 Green’s Functions and Spectral Theory for Even Order Linear BVPs consider the parametrized operators defined from Lor e L. In particular, we will denote by L[λ]u(t)≡L u(t) + λ u(t). In this case, to stress also its dependence on λ, we will denote by G[λ, T]the Green’s function related to L[λ], which will also have the corresponding subscript when we refer to one particular problem. Analogous notation can we used for e L[λ]and ee L[λ], whose related Green’s functions will be denoted by G[λ, 2T]and G[λ, 4T], respectively. Corollary 2.4.9. Assume that we are in conditions to apply Lemmas 1.1.8 and 1.1.9, that is, all the considered Green’s functions G[λ, T](or G[λ, 2T],G[λ, 4T], with the suitable subscript for each case) are: nonpositive on I×Iif and only if λ∈(−∞, λ1)or λ∈[−¯µ, λ1), with λ1>0 the first eigenvalue of operator Lncoupled with the corresponding boundary conditions and ¯µ≥0such that Ln[−¯µ]is nonresonant on Xand the related nonpositive Green’s function G[−¯µ]vanishes at some point of the square I×I. nonnegative on I×Iif and only if λ∈(λ1,∞)or λ∈(λ1,¯µ], with λ1<0 the first eigenvalue of operator Lncoupled with the corresponding boundary conditions and ¯µ≥0such that Ln[¯µ]is nonresonant on Xand the related nonnegative Green’s function G[¯µ]vanishes at some point of the square I×I. Then the following relations between the constant sign of Green’s functions are valid for any a0, . . . , a2n−1∈L1(I): If GN[T]is nonpositive on I×I, then GD[T],GM1[T]and GM2[T]either change sign or are nonpositive on I×I. If GN[2 T]is nonpositive on J×J, then GN[T],GD[T],GM1[T]and GM2[T] either change sign or are nonpositive on I×I. If GP[2 T]is nonpositive on J×J, then GN[T],GD[T],GM1[T]and GM2[T] either change sign or are nonpositive on I×I. If GP[4 T]is nonpositive on [0,4T]×[0,4T], then GN[T],GD[T],GM1[T] and GM2[T]either change sign or are nonpositive on I×I. If GM2[T]is nonpositive on I×I, then GD[T]either changes sign or is nonpositive on I×I. If GD[2 T]is nonpositive on J×J, then GD[T]and GM2[T]either change sign or are nonpositive on I×I. 36 2.5 Comparison Principles 2.5. Comparison Principles In this section we will use the connecting expressions for Green’s functions obtained in Section 2.2 to compare the values that several Green’s functions take point by point. First, from (2.2.17), under the hypothesis of the constant sign of GP[2 T], we obtain the following comparison between Green’s functions of problems (N, T)and (D, T). Corollary 2.5.1. If GP[2 T]≥0on J×J, then GN[T](t, s)≥ |GD[T](t, s)|,∀(t, s)∈I×I. If GP[2 T]≤0on J×J, then GN[T](t, s)≤ −|GD[T](t, s)|,∀(t, s)∈I×I. As a consequence, we can compare the solutions of (N, T)and (D, T), as follows. Theorem 2.5.2. Let uNbe the unique solution of problem (N, T )for σ=σ1and uDthe unique solution of problem (D, T)for σ=σ2. Then 1. If GP[2 T]≥0on J×Jand |σ2(t)| ≤ σ1(t)a. e. t∈I, then |uD(t)| ≤ uN(t) for all t∈I. 2. If GP[2 T]≤0on J×Jand 0≤σ2(t)≤σ1(t)a. e. t∈I, then uN(t)≤0 and uN(t)≤uD(t)for all t∈I. 3. If GP[2 T]≤0on J×Jand σ1(t)≤σ2(t)≤0a. e. t∈I, then uN(t)≥0 and uD(t)≤uN(t)for all t∈I. Proof. 1. Since GP[2 T]≥0on J×Jthen, from Corollary 2.5.1, it holds that |uD(t)|=ZT 0 GD[T](t, s)σ2(s) d s≤ZT 0|GD[T](t, s)||σ2(s)|ds ≤ZT 0 GN[T](t, s)σ1(s) d s=uN(t). 2. Since GP[2 T]≤0on J×Jthen, from Corollary 2.5.1, since σ1(s)≥0a. e. s∈I, we have that GN[T](t, s)σ1(s)≤ −|GD[T](t, s)|σ1(s),∀(t, s)∈I×I. 37 Green’s Functions and Spectral Theory for Even Order Linear BVPs Moreover, from σ2(s)≤σ1(s)a. e. s∈I, we deduce that −|GD[T](t, s)|σ1(s)≤ −|GD[T](t, s)|σ2(s),∀(t, s)∈I×I. Finally, since σ2(s)≥0a. e. s∈I, −|GD[T](t, s)|σ2(s)≤GD[T](t, s)σ2(s),∀(t, s)∈I×I. Therefore, for all t∈I, we have uN(t) = ZT 0 GN[T](t, s)σ1(s) d s≤ZT 0−|GD[T](t, s)|σ1(s) d s ≤ZT 0−|GD[T](t, s)|σ2(s) d s≤ZT 0 GD[T](t, s)σ2(s) d s=uD(t). Finally, the fact that uN≤0on Iis a direct consequence from GN[T]≤0and σ1≥0. 3. Since GP[2 T]≤0on J×Jthen, from Corollary 2.5.1, it can be deduced that GN[T](t, s)≤GD[T](t, s)and GN[T](t, s)≤0,∀(t, s)∈I×I and so, since σ2(s)≤0a. e. s∈I, GD[T](t, s)σ2(s)≤GN[T](t, s)σ2(s),∀(t, s)∈I×I and, from σ1(s)≤σ2(s)a. e. s∈I, we deduce that GN[T](t, s)σ2(s)≤GN[T](t, s)σ1(s),∀(t, s)∈I×I. Therefore, uD(t) = ZT 0 GD[T](t, s)σ2(s) d s≤ZT 0 GN[T](t, s)σ2(s) d s ≤ZT 0 GN[T](t, s)σ1(s) d s=uN(t). Finally, the fact that uN≥0on Iis a direct consequence from GN[T]≤0and σ1≤0. 38 2.5 Comparison Principles The situation described in previous theorem is represented in Figures 2.5.1, 2.5.2 and 2.5.3. uD uN −uN Figure 2.5.1: Solutions of (N, T)and (D, T)in Case 1 in Theorem 2.5.2. uD uN Figure 2.5.2: Solutions of (N, T)and (D, T)in Case 2 in Theorem 2.5.2. uD uN Figure 2.5.3: Solutions of (N, T)and (D, T)in Case 3 in Theorem 2.5.2. Analogously, from (2.2.19) and (2.2.20), the constant sign of either GN[2 T]or GD[2 T]lets us deduce some point-by-point relation between various Green’s functions. Corollary 2.5.3. 1. If GN[2 T]≥0on J×J, then GN[T](t, s)≥ |GM1[T](t, s)|,∀(t, s)∈I×I. 39 Second Order Equation 2. If p > 0on (a, b)and the coupled boundary conditions (3.1.5) are fulfilled, then the eigenvalues are bounded from below and can be ordered to satisfy −∞ < λ0≤λ1≤λ2≤ ··· ;λk→ ∞,as k→ ∞ (3.1.8) Each eigenvalue may be simple or double but there cannot be two consecutive equalities in (3.1.8) since, for any value of λ, equation (3.1.4) has exactly two linearly independent solutions. Note that λkis well defined for each k≥0but there is some arbitrariness in the indexing of the eigenfunctions corresponding to a double eigenvalue since every nontrivial solution of the equation for such an eigenvalue is an eigenfunction. Given such an indexing scheme, let ukbe a real-valued eigenfunction of λkfor the coupled conditions (3.1.5), k≥0, then the number of zeros of ukin (a, b)is 0 or 1, if k= 0, and k−1or kor k+ 1 if k≥1. 3. If p > 0and the boundary conditions are the separated ones (3.1.7) then strict inequality holds everywhere in (3.1.8). Furthermore, if ukis an eigenfunction of λk, then ukis unique up to constant multiples and has exactly kzeros in the open interval (a, b). It is important to point out that the coupled conditions (3.1.5) cover the periodic boundary conditions (k11 =k22 = 1,k21 =k12 = 0). In this case, if a,b∈R, KreinRutman Theorem ensures that the least eigenvalue is simple with its corresponding eigenfunction strictly positive on (a, b)and that the rest of the eigenfunctions change its sign on (a, b). Note also that coupled conditions (3.1.5) cover also the antiperiodic boundary conditions (k11 =k22 =−1,k21 =k12 = 0). In this case, Krein-Rutman Theorem is not applicable (because the corresponding Green’s function always changes its sign). On the other hand, the separated conditions (3.1.7) cover Neumann, Dirichlet and mixed conditions. 3.2. Hill’s Equation As we have said before, in this section we will particularize all the results obtained in Chapter 2 to the particular case of considering Hill’s operator defined in (3.0.1). This will be done in Subsections 3.2.3 and 3.2.4. Furthermore, in Subsection 3.2.5 we will complete the study of Hill’s operator by proving that the eigenvalues related to problems (N, T),(D, T),(M1, T)and (M2, T)satisfy a certain order relation. 46 3.2 Hill’s Equation Finally, in Subsection 3.2.6 we will use all the relations between different Green’s functions to deduce some explicit criteria to ensure the constant sign of some Green’s functions, as well as some upper bounds for the first eigenvalues. All the results dealing with this particular case of considering Hill’s equation are included in [22] and [23]. 3.2.1. Historical Background and Applications Hill’s equation (which is named after the pioneering work of the mathematical astronomer George William Hill (1838–1914), see [76]) has numerous applications in engineering and physics. Among them we can find some problems in mechanics, astronomy, circuits, electric conductivity of metals and cyclotrons. As a first example of the Hill’s equation we could consider a mass-spring system, that is, a spring with a mass mhanging from it. It is very well-known that, denoting by x(t)the position of the mass at the instant tand assuming absence of friction, the previous model can be expressed as x00(t) + k mx(t)=0, with k > 0the elastic constant of the string. However, in a real physical system, there exists a friction force which opposes the movement and is proportional to the object’s speed. In this case the situation can be modelled by the equation x00(t) + µ x0(t) + k mx(t) = 0, with µthe so-called friction coefficient. The value of such coefficient is characteristic of the environment where the object oscillates, and depends, among other variables, on the density, temperature and pressure of the environment. However, it could be considered a situation in which the spring moves between two different environments, each one with its particular friction coefficient. Also, the environment could have strong variations of density or temperature that could cause changes in the friction coefficient depending on time. This could be modelled by substituting the friction coefficient µfor a not necessarily constant function µ(t) x00(t) + µ(t)x0(t) + k mx(t)=0. Another possible situation would be that one in which there exists another external force acting periodically on the mass in such a way that it tends to move the 47 Second Order Equation mass back into its position of equilibrium, acting in proportion to the distance to that position. Including this new variable in the previous model we have x00(t) + µ(t)x0(t) + k m+F(t)x(t)=0. In any of the two cases, we obtain an equation in the form (3.0.2) in which, if µ(t)has enough regularity, we could do the following change of variable a(t) = k m+F(t)−1 4µ(t) m2 −1 2 µ0(t) m and transform the equation into one in the form (3.0.1). A second example studied in [42,98] is the inverted pendulum. A mathematical pendulum consists of a particle of mass mconnected to a base through a string (which is supposed to be rigid and of despicable weight) in such a way that the mass moves in a fixed vertical plane. If the particle moves by the force of gravity, then the movement of the pendulum is given by the equation θ00(t)−g lsin (θ(t)) = 0, where gdenotes the gravity, lthe length of the string and θrepresents the angle between the string and the perpendicular line to the base. In the surroundings of the equilibrium point θ= 0, we can approximate sin θ≈θ, so the equation of movement could be rewritten as θ00(t)−g lθ(t)=0. Consider now the case in which the suspension point of the string vibrates vertically with an acceleration a(t). Then, as it is proved in [42], the equation of movement would change into θ00(t)−1 l(g+a(t)) θ(t) = 0, which is of the form (3.0.1). Other equations which fit into the framework of the Hill’s equation are the following ones: Airy’s equation: (see [133]) u00(t) + t u(t)=0. This equation appears in the study of the diffraction of light, the diffraction of radio waves around the Earth’s surface, in aerodynamics and in the swing of an uniform vertical column which bounds under its own weight. 48 3.2 Hill’s Equation Mathieu’s equation: (see [19,142,162]) u00(t)+(c+bcos t)u(t)=0. It is the result of the analysis of the phenomenon of parametric resonance associated with an oscillator whose parameters change with time. It appears in problems related to periodic movements, as the trajectory of an electron in a periodic arrange of atoms. 3.2.2. Preliminary Results Hill’s operator properties have been described in several papers, where existence and multiplicity results, comparison principles, Green’s functions and spectral analysis were studied. Some of these results can be found in [20–22, 142,161]. In particular, the periodic problem related to Hill’s equation, namely (L u(t)=0 a. e. t∈I, u(0) = u(T), u0(0) = u0(T),(P, T) has been widely studied (see [19,21,142,161,162] and references therein). Next we compile some properties which are satisfied by the Green’s function related to problem (P, T),GP[T], and which will be basic tools to prove some of our results. Notation 3.2.1. Note that, as in Chapter 2, we will use the notation G[T]to refer to the Green’s function related to operator L. Moreover, analogously to what we have done in Section 1.1, we will consider the parametrized operators defined from Lor e L. In particular, we will denote by L[λ]u(t)≡L u(t) + λ u(t). In this case, to stress also its dependence on λ, we will denote by G[λ, T]the Green’s function related to L[λ]. Analogous notation will we used for e L[λ]u(t)≡e L u(t) + λ u(t). Lemma 3.2.2. [21, Lemma 2.2] Suppose that the Green’s function GP[T]does not change sign on I×Iand vanishes at some point (t0, s0)∈I×I, then t0=s0, (t0, s0) = (0, T)or (t0, s0) = (T, 0). Lemma 3.2.3. [21, Lemma 2.4] If GP[T]≤0on I×Ithen GP[T]<0on I×I. Lemma 3.2.4. [161, Theorem 1.1] Suppose that a∈L1(I), then: 1. GP[T]<0on I×Iif and only if λP 0[T]>0. 2. GP[T]≥0on I×Iif and only if λP 0[T]<0≤λA 0[T]. 49 Second Order Equation By introducing the parametrized potentials a+λ, with λ∈R, the previous result could be rewritten as follows. Lemma 3.2.5. [161, Theorem 1.2] Suppose that a∈L1(I), then: 1. GP[λ, T]<0on I×Iif and only if λ < λP 0[T]. 2. GP[λ, T]≥0on I×Iif and only if λP 0[T]< λ ≤λA 0[T]. To finish with these preliminary results, we include the following property which is satisfied by Green’s functions related to Hill’s operator coupled with any of the boundary conditions considered in this chapter. Lemma 3.2.6. [21, Lemma 2.8] Let λ1,λ2be such that the Green’s functions of the corresponding problem, G[λ1, T]and G[λ2, T], have the same constant sign on I×I. If λ1> λ2then G[λ1, T ](t, s)< G[λ2, T](t, s)for all (t, s)∈I×I. 3.2.3. Constant Sign of Green’s Functions In this subsection we will study the constant sign of the Green’s functions of various of the boundary value problems which have been previously considered (namely, Neumann, Dirichlet, mixed and periodic). The results in this subsection complement those in Section 2.4 for the particular case of Hill’s equation. First, we will prove a necessary condition that must be satisfied by the Green’s function of a self-adjoint operator. This result generalizes the one obtained for the periodic case in Lemma 3.2.2 and it is valid for periodic, Neumann and Dirichlet problems. Proposition 3.2.7. [22, Proposition 3.1] Assume that operator Lis nonresonant and self-adjoint on a Banach space X. If the Green’s function G[T]does not change sign on I×Iand G[T]vanishes at some point (t0, s0)∈I×I, then either (t0, s0)belongs to the diagonal of the square I×Ior (t0, s0)is in the boundary of I×I, that is, at least one of the three following properties hold: 1. t0=s0∈I. 2. t0= 0 or t0=T. 3. s0= 0 or s0=T. Proof. Suppose, on the contrary, that G[T](t0, s0) = 0 with (t0, s0)∈(0, T)×(0, T) such that t06=s0. Since G[T](t0, s0) = G[T](s0, t0), we may assume that t0> s0. 50 3.2 Hill’s Equation By definition of the Green’s function, we know that x(t)≡G[T](t, s0), t ∈I, solves the equation (x00(t) + a(t)x(t) = 0,a. e. t∈(s0, T], x(t0) = x0(t0) = 0. Then, G[T](t, s0) = 0 for all t∈(s0, T ]and, in consequence, from the symmetric property, G[T](s0, s) = 0 for all s∈(s0, T]. Now, fix s∈(s0, T]. Since G[T]is nonnegative on I×I, we have that function y(t)≡G[T](t, s), t ∈I, is a solution of (y00(t) + a(t)y(t) = 0,a. e. t∈[0, s), y(s0) = y0(s0) = 0. Once again, G[T](t, s) = 0 for all s∈(s0, T]and all t∈[0, s). From symmetry, we deduce G[T](t, s) = 0 for all t∈(s0, T ]and s∈[0, t).This contradicts property (G3) in the definition of the Green’s function (Definition 1.1.1) and so we deduce the result. Remark 3.2.8. Note that in the proof of previous proposition we use the uniqueness of solution of the initial boundary value problem to conclude that a nontrivial solution of a differential equation of order 2 can never have a zero of multiplicity two. Obviously, this does not remain true for differential equations of order higher than two and this is the reason why previous result is not applicable to the general case of the 2n-th order operator. Remark 3.2.9. If we consider the periodic case with a(t) = π T2, using [24] we obtain the following expression for the Green’s function GP[T](t, s) = T 2π   sin π(t−s) T,0≤s≤t≤T, sin π(t−s+T) T,0≤t<s≤T, which is strictly positive on I×Iexcept for the diagonal and the points (0, T)and (T, 0). 51 Second Order Equation On the other hand, when a(t) = k2<π T2and the Dirichlet boundary conditions are studied, we have that the Green’s function is given by the following expression GD[T](t, s) = 1 ksin (k T)(sin (k s) sin (k(t−T)),0≤s≤t≤T, sin (k t) sin (k(s−T)),0≤t<s≤T. We observe that GD[T]is strictly negative on (0, T)×(0, T)and vanishes on the boundary of its square of definition. In consequence, the previous result cannot be improved for general self-adjoint Hill’s operators. In particular, if Neumann boundary conditions are considered, we obtain a more precise localization of the zeros of the related Green’s function. Lemma 3.2.10. [22, Lemma 4.1] Suppose that the Green’s function GN[T]is nonnegative on I×Iand there is some (t0, s0)∈I×Ifor which GN[T](t0, s0)=0, then either (t0, s0) = (0,0) or (t0, s0) = (T, T ). Proof. Suppose that GN[T](t0, s0) = 0 for some (t0, s0)∈I×I. Since GN[T]≥0 on I×I, as operator Lis self-adjoint, Proposition 3.2.7 lets us conclude that (t0, s0) belongs either to the boundary of the square of definition or to its diagonal. In the first case, suppose that t0∈(0, T)and s0= 0. Then we have that x0(t)≡GN[T](t, 0) satisfies the equation (x00 0(t) + a(t)x0(t)=0, t ∈(0, T], x0(t0) = x0 0(t0) = 0, which means that GN[T](t, 0) ≡0on (0, T]. From the symmetry of GN[T], we have that GN[T](0, s)≡0for all s∈(0, T]. As a consequence, xs(t)≡GN[T](t, s)satisfies the equation (x00 s(t) + a(t)xs(t)=0, t ∈[0, s), xs(0) = x0 s(0) = 0, which implies that GN[T](t, s)≡0for all t < s. Using again the symmetry of GN[T]we have that it is identically zero on I×Iand we reach a contradiction. Previous argument is valid for all (t0, s0)in the boundary of I×Iexcept for (0,0) and (T, T ). Assume now that GN[T](t0, t0) = 0 for some t0∈(0, T). In this case, defining xt0(t)as the even extension to Jof GN[T](t, t0), we have that it satisfies the equation (x00 t0(t) + ˜a(t)xt0(t)=0, t ∈(t0,2T−t0), xt0(t0) = xt0(2 T−t0) = 0, 52 3.2 Hill’s Equation where, as usual, ˜adenotes the even extension of ato the interval J. From Sturm’s comparison Theorem (Theorem 3.1.3), we have that for any λ≥0 every nontrivial solution of the equation y00(t) + (˜a(t) + λ)y(t) = 0, t ∈[0,2T],(3.2.1) has as least one zero on [t0,2T−t0]. Now, note that the even extension to Jof the positive eigenfunction on (0, T] associated with λM2 0[T]solves (3.2.1) but does not have any zero on [t0,2T−t0]. Therefore we deduce that λM2 0[T]<0. Furthermore, note that the aforementioned extension is positive on (0,2T)and cancels both at 0and at 2T. As a consequence, for any λ∈(λM2 0[T],0] we have that y0, the even extension to Jof GN[λ, T](t, 0), has at least one zero on (0,2T). Moreover, all the zeros of y0are simple because otherwise GN[λ, T](t, 0) ≡0on (0, T], which cannot happen. Then necessarily y0changes its sign on (0,2T)and, as it is an even function, GN[λ, T](t, 0) changes its sign on (0, T). This contradicts the hypothesis that GN[T] is nonnegative on I×I. This way, we conclude that GN[T]can only vanish at (0,0) or (T, T). Remark 3.2.11. Note that if GN[T](0,0) = 0 we have that x(t)≡GN[T](t, 0) is a solution of (x00(t) + a(t)x(t)=0, t ∈I, x(0) = x0(T) = 0.(3.2.2) Moreover, when GN[T](T, T) = 0,y(t) = GN[T](t, T)is a solution of (y00(t) + a(t)y(t)=0, t ∈I, y0(0) = y(T) = 0.(3.2.3) As a consequence of previous result and equality (2.2.1), we deduce the following corollary. Corollary 3.2.12. If GP[2 T]has constant sign on J×J, then it holds that GN[T] has the same sign as GP[2 T]on I×I. In such a case, GN[T](t, s)is different from zero for all (t, s)∈(I×I)\{(0,0) ∪(T, T )}. Moreover, GN[T](0,0) = 0 if and only if equation (3.2.2) has a non zero and constant sign solution on [0, T), which means that λM2 0[T] = 0. GN[T](T, T ) = 0 if and only if equation (3.2.3) has a non zero and constant sign solution on (0, T], which means that λM1 0[T] = 0. Reasoning as in Lemma 3.2.10, it is deduced the following. 53 Second Order Equation Lemma 3.2.13. Suppose that a∈L1(I)and the Green’s function GD[T]has constant sign on I×Iand there exists some (t0, s0)∈I×Isuch that GD[T](t0, s0) = 0. Then (t0, s0)belongs to the boundary of the square of definition of GD[T]. Remark 3.2.14. From the Dirichlet boundary conditions and property (G5) of the Green’s function (see Definition 1.1.1), it is clear that GD[T]must cancel on the whole boundary of I×I. Previous lemma ensures that, when GD[T]has constant sign, it can not vanish at any other point. In the sequel we will prove that GD[T]can never be nonnegative when working with Hill’s equation and it is negative on (0, T)×(0, T)for λsmaller than the first eigenvalue. Lemma 3.2.15. [23, Lemma 36] Suppose that a∈L1(I), then: GD[λ, T]<0on (0, T)×(0, T)if and only if λ < λD 0[T]. Moreover, if λ > λD 0[T]is such that GD[λ, T]exists, then GD[λ, T]changes sign on I×I. Proof. Choose λ < λD 0[T]. From Theorem 3.1.3 it is clear that any solution of equation u00(t)+(a(t) + λ)u(t)=0, t ∈I, (3.2.4) has at most one zero on I. From Definition 1.1.1, it holds that for each s0∈(0, T),us0(·)≡GD[λ, T](·, s0) satisfies (3.2.4) on [0, s0)∪(s0, T]. Then, if uis the unique solution of (3.2.4) under the initial conditions u(0) = 0, u0(0) = 1, it is clear that there exists a constant k1such that us0(t) = k1u(t)for all t < s0. Obviously, this constant k1depends on the value s0considered, so we could say that there exists a function (which, for the sake of simplicity, will be denoted also by k1) such that GD[λ, T](t, s) = k1(s)u(t)for all t < s. Moreover, since u(0) = 0, we have that u(t)6= 0 for all t∈(0, T]. Analogously, if vis the unique solution of (3.2.4) satisfying the final conditions u(T) = 0, u0(T) = −1, then there exists a constant k2such that us0(t) = k2v(t)for all t > s0. Consequently, GD[λ, T](t, s) = k2(s)v(t)for all s < t. 54 3.2 Hill’s Equation In this case, v(T) = 0 implies that v(t)6= 0 for all t∈[0, T). Now, since GD[λ, T]is a symmetric function, necessarily k1(s) = c v(s)and k2(s) = c u(s)for some non zero constant c, that is, GD[λ, T](t, s) = (c v(s)u(t),0≤t < s ≤T, c u(s)v(t),0≤s < t ≤T, and, since GD[λ, T]is continuous on I×I, it is clear that GD[λ, T](s, s) = c u(s)v(s). Therefore GD[λ, T]has strict constant sign on (0, T)×(0, T)for all λ < λD 0[T]. We will see now that this sign has to be necessarily negative. On the contrary, assume that there exists some value ¯ λ<λD 0[T]for which GD[¯ λ, T]>0on (0, T)×(0, T). From this property, since ∂ G[¯ λ, T] ∂ t (0, s)6= 0 and ∂ G[¯ λ, T] ∂ t (T, s)6= 0,for all s∈(0, T ), it is immediate to verify that, choosing φ(t) = t(T−t), for all s∈(0, T )we have that κ1(s) = min t∈I GD[¯ λ, T](t, s) φ(t)∈(0,∞) and κ2(s) = max t∈I GD[¯ λ, T](t, s) φ(t)∈(0,∞) and are continuous functions on I. Then property (Pg)in Lemma 1.1.9 is fulfilled. Thus, a necessary condition for GD[¯ λ+µ, T]to be nonnegative on I×Iis that µ > λD 0[¯ λ, T], being λD 0[¯ λ, T]the smallest eigenvalue of operator L[¯ λ]coupled with Dirichlet conditions. Now, taking into account that λD 0[¯ λ, T] = λD 0[T]−¯ λ, we have that a necessary condition for GD[¯ λ+µ, T]to be nonnegative on I×Iis that ¯ λ+µ>λD 0[T]or, which is the same, if GD[λ, T ]≥0on I×Ithen λ > λD 0[T]. This facts contradicts the existence of such ¯ λ. As a consequence, GD[λ, T]<0on (0, T)×(0, T)for all λ<λD 0[T]and condition (Ng)is fulfilled. Thus, from Lemma 1.1.8, we can ensure that GD[λ, T]<0on (0, T)×(0, T)if and only if λ < λD 0[T]. Now we will see that for λ>λD 0[T]such that the Green’s function GD[λ, T] exists, it holds that GD[λ, T]changes sign. 55 Second Order Equation uD uN Figure 3.2.4: Solutions of (N, T)and (D, T)in Case 3 in Theorem 3.2.30. Remark 3.2.31. Note that Theorem 3.2.30 is stronger than its corresponding one for the general even order equation (namely, Theorem 2.5.2) as for Hill’s equation we are able to ensure the constant sign of both Neumann and Dirichlet solutions, which did not happen in Theorem 2.5.2. Analogously, Corollary 2.5.3 can be improved in the following way. Corollary 3.2.32. If GN[2 T]≥0on J×J, then GN[T](t, s)≥ −GM1[T](t, s)≥0,∀(t, s)∈I×I. If GN[2 T]<0on J×J, then GN[T](t, s)< GM1[T](t, s)≤0,∀(t, s)∈I×I. If GD[2 T]≤0on J×J, then GM2[T](t, s)< GD[T](t, s)≤0,∀(t, s)∈I×I. Remark 3.2.33. Note that Assertion 4 in Corollary 2.5.3 has not been included in previous result since, as we have seen in Lemma 3.2.15, for Hill’s equation, the Green’s function related to (D, 2T)can never be nonnegative. Now, we can adapt Theorems 2.5.4 and 2.5.5 for the case of Hill’s equation. Theorem 3.2.34. Let uNbe the unique solution of problem (N, T)for σ=σ1and uM1the unique solution of (M1, T)for σ=σ2. Then 1. If GN[2 T]≥0on J×Jand |σ2(t)| ≤ σ1(t)a. e. t∈I, then |uM1(t)| ≤ uN(t)for all t∈I. 1.1 If, moreover, σ2(t)≥0a. e. t∈I, then −uN(t)≤uM1(t)≤0for all t∈I. 62 3.2 Hill’s Equation 1.2 If, moreover, σ2(t)≤0a. e. t∈I, then 0≤uM1(t)≤uN(t)for all t∈I. 2. If GN[2 T]≤0on J×Jand 0≤σ2(t)≤σ1(t)a. e. t∈I, then uN(t)≤uM1(t)≤0for all t∈I. 3. If GN[2 T]≤0on J×Jand σ1(t)≤σ2(t)≤0a. e. t∈I, then 0≤uM1(t)≤uN(t)for all t∈I. Theorem 3.2.35. Suppose that GD[2 T]≤0on J×J. Let uM2be the unique solution of problem (M2, T)for σ=σ1and uDthe unique solution of problem (D, T)for σ=σ2. 1. If 0≤σ2(t)≤σ1(t)a. e. t∈I, then uM2(t)≤uD(t)≤0for all t∈I. 2. If σ1(t)≤σ2(t)≤0a. e. t∈I, then 0≤uD(t)≤uM2(t)for all t∈I. Remark 3.2.36. We note that, since Assertion 4 in Corollary 2.5.3 can never happen for the case studied in this section, it implies that hypotheses in Assertion 1 in Theorem 2.5.5 are never fulfilled in such a case. Therefore, we have not included the corresponding Assertion in Theorem 3.2.35. Moreover, using the characterization given in Subsection 3.2.3, it is possible to rewrite Corollaries 3.2.29 and 3.2.32 in terms of eigenvalues, as follows. Corollary 3.2.37. If (λN[T] =) λP[2 T]<0≤λA[2 T], then GN[T](t, s)≥ −GD[T](t, s)≥0,∀(t, s)∈I×I. If (λN[T] = λN[2 T] = λP[4 T] =) λP[2 T]>0, then GN[T](t, s)< GD[T](t, s)≤0,∀(t, s)∈I×I and GN[T](t, s)< GM1[T](t, s)≤0,∀(t, s)∈I×I. If (λN[T] = λN[2 T] =) λP[4 T]<0≤λA[4 T], then GN[T](t, s)≥ −GM1[T](t, s)≥0,∀(t, s)∈I×I. If (λD[2 T] =) λM2[T]>0, then GM2[T](t, s)< GD[T](t, s)≤0,∀(t, s)∈I×I. 63 Second Order Equation Finally, we are also able to deduce the following result which is not true, in general, for higher order equations. Corollary 3.2.38. If GN[T]≥0on I×I, then GN[T](t, s)≤2GP[2 T](2 T−t, s)on I×I. 0≥GD[T](t, s)≥ −2GP[2 T](2 T−t, s)on I×I. GN[T](t, s)≤2GN[2 T](2 T−t, s)on I×I. 0≥GM1[T](t, s)≥ −2GN[2 T](2 T−t, s)on I×I. In particular, GP[2 T](2 T−t, s)≥0and GN[2 T](2 T−t, s)≥0on I×I. Proof. The inequalities are deduced from expressions (2.2.17) and (2.2.19) by taking into account that if GN[T]≥0on I×Ithen GD[T]≤0and GM1[T]≤0on I×I. 3.2.5. Global Order of Eigenvalues of Hill’s Equation It can also be proved that, when dealing with Hill’s equation, there exists a certain order relation between the eigenvalues related to problems (N, T),(D, T),(M1, T) and (M2, T). Indeed, consider the following facts: (i) Let λN k[T], λN k+1[T]∈ΛN[T]be two consecutive eigenvalues of Neumann problem (N, T)and let uN,T kand uN,T k+1 be their associated eigenfunctions. As we have seen in Section 3.1, the aforementioned eigenfunctions have kand k+ 1 zeros on the interval [0, T ], respectively. If we consider the even extensions of uN,T kand uN,T k+1 to the interval [0,2T], it is clear that they have 2kand 2k+ 2 zeros on [0,2T], respectively, so there must exist an eigenvalue λ∈ΛN[2 T],λN k[T]< λ < λN k+1[T], such that its associated eigenfunction has exactly 2k+1 zeros on the interval [0,2T]. From the decomposition of the Neumann spectrum showed in Section 2.3, we have that, necessarily, λ∈ΛM1[T]. As we know that λN 0[2 T] = λN 0[T]we conclude that λN 0[T]< λM1 0[T]< . . . < λN k[T]< λM1 k[T]< λN k+1[T]< λM1 k+1[T]< . . . (ii) Analogously, we can easily see that ΛM2[T]corresponds with eigenvalues of ΛD[2 T]whose eigenfunctions have an even number of zeros on the interval 64 3.2 Hill’s Equation (0,2T)and ΛD[T]corresponds with eigenvalues of ΛD[2 T]whose eigenfunctions have an odd number of zeros on (0,2T). Taking into account the fact that λD 0[2 T] = λM2 0[T]we conclude that λM2 0[T]< λD 0[T]< . . . < λM2 k[T]< λD k[T]< λM2 k+1[T]< λD k+1[T]< . . . (iii) Oscillation Theorem (Theorem 3.1.6) guarantees that the eigenvalues of periodic and antiperiodic problems related to the same interval always appear in the following order λP 0[T]< λA 0[T]≤λA 1[T]< λP 1[T]≤λP 2[T]< λA 2[T]≤λA 3[T]< . . . Consequently, if we consider Item (iii) for problems (P, 2T)and (A, 2T)and we take into account the inequalities obtained in Items (i) and (ii) we can affirm that In each pair {λP 2k−1[2 T], λP 2k[2 T]}of two consecutive eigenvalues of problem (P, 2T), one of them belongs to ΛN[T]and the other one belongs to ΛD[T]. In particular, if λP 2k−1[2 T] = λP 2k[2 T]is a double eigenvalue, then it belongs to both ΛN[T]and ΛD[T]. In each pair {λA 2k[2 T], λA 2k+1[2 T]}of two consecutive eigenvalues of problem (A, 2T), one of them belongs to ΛM1[T]and the other one belongs to ΛM2[T]. As in the previous case, if λA 2k[2 T] = λA 2k+1[2 T]is a double eigenvalue, then it belongs to both ΛM1[T]and ΛM2[T]. The previous reasoning lets us conclude that the eigenvalues of problem (P, 4T) always appear in the following order: λN 0[T]<{λM1 0[T], λM2 0[T]}<{λD 0[T], λN 1[T]} <{λM1 1[T], λM2 1[T]}<{λD 1[T], λN 2[T]}< . . . As an immediate consequence we can also deduce an alternating relation between eigenvalues of (N, T)and (M2, T)and also between those of (M1, T)and (D, T). Corollary 3.2.39. The following properties hold for any a∈L1(I). 1. λN k[T]< λM2 k[T]< λN k+1[T]< λM2 k+1[T], k = 0,1, . . . 2. λM1 k[T]< λD k[T]< λM1 k+1[T]< λD k+1[T], k = 0,1, . . . 65 Second Order Equation Remark 3.2.40. In [106, Chapter 1] the following equalities are proved in the case of an even potential on [0,2T]: u1(2 T, λ) = 2 u1(T, λ)u0 2(T, λ)−1 = 1 + 2 u0 1(T, λ)u2(T, λ),(3.2.8) u0 1(2 T, λ) = 2 u1(T, λ)u0 1(T, λ),(3.2.9) u2(2 T, λ) = 2 u2(T, λ)u0 2(T, λ),(3.2.10) u0 2(2 T, λ) = u1(2 T, λ),(3.2.11) with u1and u2the fundamental solutions of Hill’s equation defined in Theorem 3.1.6. Moreover, it is easy to verify (see [23, Chapter 2] for the details) that λ∈ΛN[T]if and only if u0 1(T, λ) = 0. λ∈ΛD[T]if and only if u2(T, λ) = 0. λ∈ΛM1[T]if and only if u1(T, λ) = 0. λ∈ΛM2[T]if and only if u0 2(T, λ) = 0. Therefore we deduce that, as ˜ais an even function, the decomposition of Neumann and Dirichlet spectra in 2T, ΛN[2 T] = ΛN[T]∪ΛM1[T]and ΛD[2 T] = ΛD[T]∪ΛM2[T], could also be deduced from the equalities (3.2.9) and (3.2.10). This deduction, despite being more direct than the one presented in this work, does not give any information about the order of eigenvalues. We will see now some examples of the different situations that we could find. To calculate the eigenvalues we will use the characterization of the spectra given in Remark 3.2.40. Example 3.2.41. If we consider the constant case a(t) = 0, it is known that (see [18]) λP 0[2 T] = λN 0[T] = 0 and λA 0[2 T] = λD 0[2 T] = π 2T2. Moreover, denoting λ=m2>0and using [24] we obtain the explicit expressions of the corresponding Green’s functions: GP[m2,2T](t, s) =        cos (m(s−t+T)) 2msin m T ,0≤s≤t≤2T, cos (m(s−t−T)) 2msin m T ,0≤t≤s≤2T, 66 3.2 Hill’s Equation and GN[m2, T](t, s) =        cos (m s) cos (m(T−t)) msin m T ,0≤s≤t≤T, cos (m t) cos (m(T−s)) msin m T ,0≤t≤s≤T. It is obvious that GN[m2, T](0,0) = 2 GP[m2,2T](0,0) = 1 mtan m T . As a consequence, from Theorem 3.2.18, we know that λM1 0[T] = π 2T2. Moreover, from Theorem 3.2.18 and the fact that GN[m2, T](T, T) = 2 GP[m2,2T](T, T) = 1 mtan m T , we deduce that λM2 0[T] = π 2T2. This is also deduced from Corollary 2.3.1. We can use [24] to calculate the Green’s functions for the different boundary conditions GD[m2, T](t, s) =        sin (m s) sin (m(t−T)) msin m T ,0≤s≤t≤T, sin (m t) sin (m(s−T)) msin m T ,0≤t≤s≤T, GM1[m2, T](t, s) =        cos (m s) sin (m(t−T)) mcos m T ,0≤s≤t≤T, cos (m t) sin (m(s−T)) mcos m T ,0≤t≤s≤T, GM2[m2, T](t, s) =        −sin (m s) cos (m(T−t)) mcos m T ,0≤s≤t≤T, −sin (m t) cos (m(T−s)) mcos m T ,0≤t≤s≤T, and GA[m2,2T](t, s) =        −sin (m(s−t+T)) 2mcos m T ,0≤s≤t≤T, −sin (m(−s+t+T)) 2mcos m T ,0≤t≤s≤T. 67 Second Order Equation We observe then that λD 0[T] = π T2. In this case, ΛN[T]=ΛD[T]∪{0}= ΛP[2 T] and ΛM1[T] = ΛM2[T]=ΛA[2 T]. Then, if we represent graphically the discriminant (given in Oscillation Theorem (Theorem 3.1.6)), ˜ ∆(λ) = u1(2 T, λ) + u0 2(2 T, λ), we obtain Figure 3.2.5. λ ˜ ∆(λ) ˜ ∆(λ) = 2 ˜ ∆(λ) = −2 λN 0[T]λD 0[T] = λN 1[T]λD 1[T] = λN 2[T] λM1 0[T] = λM2 0[T]λM1 1[T] = λM2 1[T]λM1 2[T] = λM2 2[T] Figure 3.2.5: Graphic of ˜ ∆(λ)for a(t)=0. Example 3.2.42. If we consider T= 2 and a(t) = (0, t ∈[0,1], 1 10, t ∈[1,2], the eigenvalues can be directly obtained and we can verify that λN 0[2] = λN 0[4] = λP 0[4] ≈ −0.0508, λM2 0[2] = λD 0[4] = λA 0[4] ≈0.5346, λM1 0[2] ≈0.5984 and λD 0[2] ≈2.4170. Graphically, the situation would be represented in Figure 3.2.6. 68 3.2 Hill’s Equation λ ˜ ∆(λ) ˜ ∆(λ) = 2 ˜ ∆(λ) = −2 λN 0[2] λD 0[2] λN 1[2] λD 1[2] λN 2[2] λM2 0[2] λM1 0[2] λM2 1[2] λM1 1[2] λM2 2[2] λM1 2[2] Figure 3.2.6: Graphic of ˜ ∆(λ)for a piecewise constant potential a. Note that the k-th eigenvalue of problem (M2, T )always appears before the one of problem (M1, T). In addition, the order between the eigenvalues of (N, T)and (D, T)is also maintained. Example 3.2.43. Considering T=πand a(t) = cos t, we obtain the following approximations λN 0[π] = λN 0[2 π] = λP 0[2 π] = λP 0[4 π]≈ −0.378, λM1 0[π] = λA 0[2 π]≈ −0.348, λM2 0[π] = λD 0[2 π]≈0.5948 and λD 0[π]≈0.918. Graphically we would obtain Figure 3.2.7. λ ˜ ∆(λ) ˜ ∆(λ) = 2 ˜ ∆(λ) = −2 λN 0[π]λD 0[π]λN 1[π]λD 1[π]λN 2[π] λM1 0[π]λM2 0[π]λM1 1[π]λM2 1[π]λM1 2[π]λM2 2[π] Figure 3.2.7: Graphic of ˜ ∆(λ)for a(t) = cos t. 69 Second Order Equation In this case, the k-th eigenvalue of (M1, T)is smaller than the one of (M2, T). Again, the order between the eigenvalues of (N, T)and (D, T )is maintained. The following example shows that eigenvalues related to problem (N, T)do not necessarily have to alternate with the ones of (D, T ). Example 3.2.44. Considering T=πand a(t) = cos 2 t, we obtain the following approximation for the spectra of the considered problems ΛP[4 π] = {−0.1218,0.0923,0.47065,1.4668,2.34076,3.9792,4.1009, . . . }, ΛP[2 π] = {−0.1218,0.47065,1.4668,3.9792,4.1009, . . . }, ΛM1[π] = ΛM2[π] = ΛA[2 π] = {0.0923,2.34076, . . . }, ΛN[π] = {−0.1218,0.47065,4.1009, . . . } and ΛD[π] = {1.4668,3.9792, . . . }. We observe that in this case λN 0[π]< λN 1[π]< λD 0[π]< λD 1[π]< λN 2[π]. Note that the eigenvalues of mixed problems coincide. This is due to the fact that a(t) = a(π−t)(see Corollary 2.3.1). Consequently, all the eigenvalues of ΛA[2 π] are a double root of ˜ ∆(λ) = −2. Graphically we get Figure 3.2.8. λ ˜ ∆(λ) ˜ ∆(λ)=2 ˜ ∆(λ) = −2 λN 0[π]λN 1[π]λD 0[π]λD 1[π]λN 2[a, π]λN 3[π]λD 2[π] λM1 0[π] = λM2 0[π]λM1 1[π] = λM2 1[π]λM1 2[π] = λM2 2[π] Figure 3.2.8: Graphic of ˜ ∆(λ)for a(t) = cos 2 t. 70 3.2 Hill’s Equation Remark 3.2.45. The numerical results obtained in the considered examples suggest an order of eigenvalues even more precise than the one theoretically proved. It is observed that the eigenvalues of mixed problems alternate, with one eigenvalue of a mixed problem between two consecutive eigenvalues of the other one, and reciprocally. This has been observed in all the considered examples in which the spectra of the two mixed problems are different (Examples 3.2.42 and 3.2.43), independently of which of them appears first. We also appreciate in the examples an alternation between Neumann and Dirichlet eigenvalues except for the case in which the spectrum of the mixed problems is the same (in this case the order of appearance of Dirichlet and Neumann changes between one pair of eigenvalues and the next one, as we can see in Example 3.2.44). This situation suggests the existence of some property justifying this fact. However, up to this moment, this has not been formally proved and these speculations are uniquely based on the numerical results obtained while working with different potentials. 3.2.6. Explicit Criteria to Ensure Constant Sign of Green’s Functions As we have commented before, being able to ensure the constant sign of the Green’s function is important as, among other things, in some cases it allows to warrant the constant sign of the solutions. Moreover, as it has been mentioned at the beginning of this section, the periodic problem related to Hill’s equation has been widely studied. In particular, many characterizations of maximum and antimaximum principles have been proved. All these criteria, by virtue of Theorem 1.1.7, can be used to ensure the constant sign of the Green’s function related to problem (P, T ). We will compile next these criteria and, for the sake of simplicity, we will formulate them in terms of the constant sign of the Green’s function, although most of them are originally proved for maximum and antimaximum principles. First, we will introduce now some notation that we will use in this section: The positive part h+(t) = max{h(t),0}, t ∈I and the negative one h−(t) = −min{h(t),0}, t ∈I, are defined as usual. Given 1≤α≤ ∞ we will denote by α∗its conjugate, that is, the number satisfying the relation 1 α+1 α∗= 1. If α= 1 then α∗=∞and vice-versa. 71 Second Order Equation Moreover, if f∈ AC([α, β]) and g∈ AC([c, d]) is monotone, it is verified that (f◦g)0(t) = f0(g(t)) g0(t)a. e. t∈[c, d]. This result can be seen in [37, Remark 3] and is deduced from [113, Theorems 9.3 and 38.4]. Therefore, for a. e. t∈K, the following equality is satisfied: y0(t) = u0(w−1(t)) (w−1(t))0=u0(w−1(t)) 1 w0(w−1(t)) = (p u0) (w−1(t)). In an analogous way it can be deduced that (p u0)◦w−1∈ AC(K)and y00(t) = ((p u0)◦w−1)0(t)=(p u0)0(w−1(t)) p(w−1(t)) ∈L1(I). Consequently, we have that, for a. e. t∈K, y00(t) + p(w−1(t)) ¯a(w−1(t)) y(t) = p(p u0)0(w−1(t)) + (p¯a) (w−1(t)) u(w−1(t)) = (p¯σ) (w−1(t)), and, moreover, y(0) = u(w−1(0)) = u(0) = u(T) = u(w−1(R)) = y(R), y0(0) = lim t→0+y0(t) = lim t→0+(p u0) (w−1(t)) = lim s→0+(p u0)(s)=(p u0)(0) and y0(R) = lim t→R−y0(t) = lim t→R−(p u0) (w−1(t)) = lim s→T−(p u0)(s)=(p u0)(T). On the other hand, note that ZR 0p(w−1(t)) ¯a(w−1(t))αdt=ZR 0¯a pα−1 α(w−1(t))αp(w−1(t)) d t =ZR 0¯a pα−1 α(w−1(t))α(w−1(t))0dt =Zw−1(R) w−1(0) ¯a pα−1 α(s)αds=ZT 0¯a pα−1 α(s)αds =¯a pα−1 αα Lα(I)<+∞, that is, a(t)≡p(w−1(t)) ¯a(w−1(t)) ∈Lα(K). 78 3.3 General Second Order Equation Remark 3.3.2. The fact that ais measurable is deduced from both p◦w−1and ¯a◦w−1being measurable. Indeed, we will see that if Vis open, then (p◦w−1)−1(V) = w(p−1(V)) is a measurable set. As pis a measurable function, then p−1(V)is measurable. So, it is enough to verify that wtakes measurable sets into measurable sets. Since wis absolutely continuous, the following theorem guarantees that the image by wof any set of measure zero has measure zero. Theorem 3.3.3 (Banach-Zarecki).[75, Theorem 18.25] A function fis absolutely continuous on an interval [c, d]if and only if the two following conditions are verified: (i) fis continuous and of bounded variation on [c, d]. (ii) The image through fof any subset of [c, d]with measure zero is a set with measure zero. Moreover, as a consequence of [136, Chapter 6, Exercise 6] we have that, as wis a continuous function, the image of a set with measure zero has measure zero if and only if the image of any measurable set is measurable. Consequently, w(p−1(V)) is a measurable set and the function p◦w−1is measurable. An analogous reasoning could be considered for ¯a◦w−1. Similarly we obtain that σ(t)≡p(w−1(t)) ¯σ(w−1(t)) ∈Lα(K). As a consequence, y∈W2,1(K)is a solution of (y00(t) + a(t)y(t) = σ(t),a. e. t∈K, y(0) = y(R), y0(0) = y0(R),(3.3.2) with a, σ ∈Lα(K). Reciprocally, let a, σ ∈Lα(K)be arbitrary and let y∈W2,1(K)be a solution of problem (3.3.2). Consider functions pand win the previous conditions. Defining u:I→Ras u(t) := y(w(t)) and using the fact that y∈ AC(K) and w:I→Kis monotone and satisfies that w∈ AC(I), we deduce from Theorem 3.3.1 that u=y◦w∈ AC(I). Therefore, applying again the chain rule, we have that u0(t) = y0(w(t)) w0(t) = y0(w(t)) 1 p(t),a. e. t∈I, 79 Second Order Equation that is, (p u0)(t) = y0(w(t)),a. e. t∈I. Since y0∈ AC(K)and w:I→Jis monotone and satisfies that w∈ AC(I), we deduce again from Theorem 3.3.1 that p u0=y0◦w∈ AC(I), and p(p u0)0(t) = y00(w(t)),a. e. t∈I. Moreover (p u0)0(t) + ¯a(t)u(t) = y00(w(t)) p(t)+ ¯a(t)y(w(t)) = ¯σ(t),a. e. t∈I, with ¯a(t) = a(w(t)) p(t)and ¯σ(t) = σ(w(t)) p(t), t ∈I. Obviously u(0) = y(w(0)) = y(0) = y(R) = y(w(T)) = u(T).(3.3.3) The monotony assumptions on function wlet us affirm that p u0(0) = lim t→0+(p u0)(t) = lim t→0+y0(w(t)) = lim s→0+y0(s) = y0(0) (3.3.4) and p u0(T) = lim t→T−(p u0)(t) = lim t→T−y0(w(t)) = lim s→R−y0(s) = y0(R).(3.3.5) Finally, we observe that ZT 0pα−1 α(t) ¯a(t)αdt=ZT 0a(w(t)) p−1 α(t)αdt =ZT 0|a(w(t))|αdt p(t)=ZT 0|a(w(t))|αw0(t) d t =Zw(T) w(0) |a(s)|αds=ZR 0|a(s)|αds =kakα Lα(K)<+∞. (3.3.6) As a consequence u∈ AC(I), with p u0∈ AC(I), is a solution of problem (3.3.2) with pα−1 α¯a, pα−1 α¯σ∈Lα(I). We have proved that problems (3.3.1) and (3.3.2) are equivalent and the qualitative properties of the solutions of both problems are the same. We will see next the relation between the corresponding Green’s functions. 80 3.3 General Second Order Equation Lemma 3.3.4. Let ¯ GP[T]and GP[R]be the Green’s functions related to problems (3.3.1) and (3.3.2), respectively. It is verified that ¯ GP[T](t, s) = GP[R](w(t), w(s)),∀(t, s)∈I×I. Proof. If uis the unique solution of problem (3.3.1), then, as we have proved before, y(t) = u(w−1(t)) is the unique solution of (3.3.2) for σ(t) = p(w−1(t)) ¯σ(w−1(t)) and satisfies that y(t) = ZR 0 GP[R](t, s)p(w−1(s)) ¯σ(w−1(s)) d s. Conversely, if yis the unique solution of (3.3.2) then, using previous arguments again, u=y(w(t)) is the unique solution of problem (3.3.1). Consequently u(t) = y(w(t)) = ZR 0 GP[R](w(t), s)p(w−1(s)) ¯σ(w−1(s)) d s =ZT 0 GP[R](w(t), w(s)) ¯σ(s) d s, from where we deduce the result. As an immediate consequence we obtain the following corollary. Corollary 3.3.5. The following equivalences hold: 1. ¯ GP[T]≥0on I×Iif and only if GP[R]≥0on K×K. 2. ¯ GP[T]<0on I×Iif and only if GP[R]<0on K×K. The previous result lets us obtain some criteria about the constant sign of the Green’s function of problem (3.3.1) from Lemmas 3.2.47, 3.2.48, 3.2.49 and 3.2.50. These results are deduced by simply taking into account the following facts: As a consequence from (3.3.6) we have that kakLα(K)=pα−1 α¯aLα(I). It is verified that ZR 0 a(t) d t=ZR 0 p(w−1(t)) ¯a(w−1(t)) d t=ZR 0 ¯a(w−1(t)) (w−1(t))0dt =ZT 0 ¯a(s) d s. 81 Second Order Equation Since by hypothesis p > 0a. e. t∈I, condition a0(respectively, ≺0) is equivalent to ¯a0(respectively, ≺0). Moreover, a+(t) = p(w−1(t)) ¯a+(w−1(t)), a−(t) = p(w−1(t)) ¯a−(w−1(t)) and consequently ZR 0 a+(t) d t=ZT 0 ¯a+(t) d tand ZR 0 a−(t) d t=ZT 0 ¯a−(t) d t. Corollary 3.3.6. If ¯a≺0, then ¯ GP[T]<0on I×I. Corollary 3.3.7. If ¯a0and pα−1 α¯aLα(I)≤K(2 α∗, R), then ¯ GP[T]≥0on I×I. Corollary 3.3.8. If RT 0¯a(t) d t≥0,¯a6≡ 0on Iand pα−1 α¯a+Lα(I)≤K(2 α∗, R), then ¯ GP[T]≥0on I×I. Corollary 3.3.9. If ¯a∈L1(I),¯a6≡ 0on Iand ZT 0 ¯a+(s) d s < 4 R,RT 0¯a+(s) d s 1−R 4RT 0¯a+(s) d s≤ZT 0 ¯a−(s) d s, then ¯ GP[T]<0on I×I. Corollary 3.3.10. If ¯ GP[T](t, s)<0for all (t, s)∈I×Ithen RT 0¯a(s) d s < 0. Considering now the discriminant ∆(λ) = u1(R, λ) + u0 2(R, λ)for problem (3.3.2), Lemmas 3.2.51 and 3.2.52 can be rewritten in order to obtain some conditions that assure the constant sign of the Green’s function of problem (3.3.1). Corollary 3.3.11. The following properties hold: (i) ¯ GP[T]<0on I×Iif and only if ∆(λ)>2for all λ≤0. (ii) ¯ GP[T]≥0on I×Iif and only if ∆(λ)>−2for all λ < 0and ∆(0) <2. Corollary 3.3.12. Let ¯a(t) = ˆ ¯a(t) + λ1 p(t), where λ=1 RRT 0¯a(s) d sis T Rtimes the mean value of ¯aand suppose that pα−1 αˆ ¯a+Lα(I)≤K(2 α∗, R). Then 82 3.3 General Second Order Equation (i) ¯ GP[T]<0on I×Iif and only if RT 0¯a(s) d s < 0and ∆(0) >2. (ii) If RT 0¯a(s) d s < 0and ∆(0) <2, then ¯ GP[T]≥0on I×I. (iii) If operator Lunder periodic boundary conditions is nonresonant and 0≤ZT 0 ¯a(s) d s≤π2 R  1−pα−1 αˆ ¯a+Lα(I) K(2 α∗, R)  , then ¯ GP[T]≥0on I×I. Proof. The hypotheses of Lemma 3.2.52 (applied to a(t) = p(w−1(t)) ¯a(w−1(t))) will be rewritten in terms of problem (3.3.1). Indeed, such result considers that kˆa+kLα(K)≤K(2 α∗, R), with a(t) = ˆa(t) + λ, where λ=1 RRR 0a(s) d sis the mean value of a. It is immediate to verify that λ=1 RZT 0 ¯a(s) d s and, clearly, this is T Rtimes the mean value of ¯a. Moreover, kˆa+kα Lα(K)=k(a−λ)+kα Lα(K)=ZR 0 (a(t)−λ)α +dt =ZR 0p(w−1(t)) ¯a(w−1(t)) −λα +dt =ZT 0 (p(s) ¯a(s)−λ)α + 1 p(s)ds =ZT 0pα−1 α¯a(s)−λ p(s)+α ds =pα−1 α¯a−λ p+ α Lα(I) =pα−1 αˆ ¯a+α Lα(I), that is, kˆa+kLα(K)=pα−1 αˆ ¯a+Lα(I). The other changes in this corollary with respect to Lemma 3.2.52 are immediately obtained from the same considerations as in previous results. 83 Second Order Equation 3.3.2. Non-periodic Conditions All the previous reasonings have been done considering periodic boundary conditions. Nevertheless, equalities (3.3.3), (3.3.4) and (3.3.5) guarantee that u(0) = y(0), u(T) = y(R), and (p u0)(0) = y0(0),(p u0)(T) = y0(R). Consequently, periodic conditions in problems (3.3.1) and (3.3.2) can be substituted by any other kind of boundary conditions and this does not affect to the equivalence of the problems. We obtain the same relation between the Green’s functions corresponding to each case, that is, using an analogous notation to the periodic case, we have the following equalities: ¯ GN[T](t, s) = GN[R](w(t), w(s)),∀(t, s)∈I×I, ¯ GD[T](t, s) = GD[R](w(t), w(s)),∀(t, s)∈I×I, ¯ GM1[T](t, s) = GM1[R](w(t), w(s)),∀(t, s)∈I×I, ¯ GM2[T](t, s) = GM2[R](w(t), w(s)) ∀(t, s)∈I×I and ¯ GA[T](t, s) = GA[R](w(t), w(s)),∀(t, s)∈I×I. As a consequence all the results relating different Green’s functions which were obtained for Hill’s equation are still valid in this more general case. In particular, all the corollaries in Subsection 3.3.1 can be rewritten in terms of the Green’s function of other boundary conditions, in an analogous way to what we have made in Subsection 3.2.6. To do that it is enough to consider Lemmas 3.2.47, 3.2.48, 3.2.49 and 3.2.50 for ˜a(the even extension of a) and take into account the following considerations: We have that k˜a+kLα[0,2R]= 21/αka+kLα(K)= 21/α pα−1 α¯a+Lα(I). Condition ˜a0(respectively, ≺0) is equivalent to a0(respectively, ≺0) which, at the same time, is equivalent to ¯a0(respectively, ≺0). 84 3.3 General Second Order Equation The integrals of potentials present the following relation Z2R 0 ˜a(t) d t= 2 ZR 0 a(t) d t= 2 ZT 0 ¯a(t) d t. Analogously, since p > 0a. e. t∈I, Z2R 0 ˜a+(t) d t= 2 ZT 0 ¯a+(t) d tand Z2R 0 ˜a−(t) d t= 2 ZT 0 ¯a−(t) d t. Corollary 3.3.13. (i) If ¯a≺0, then ¯ GN[T]<0on I×I. (ii) If RT 0¯a(t) d t≥0,¯a6≡ 0and pα−1 α¯a+Lα(I)≤2−1/α K(2 α∗,2R), then ¯ GN[T]≥0on I×I. (iii) If ¯a∈L1(I),¯a6≡ 0and ZT 0 ¯a+(s) d s < 1 R,RT 0¯a+(s) d s 1−RRT 0¯a+(s) d s≤ZT 0 ¯a−(s) d s, then ¯ GN[T]<0on I×I. Any of the previous conditions implies that: 1. ¯ GM1[T](t, s)<0for all (t, s)∈[0, T)×[0, T). 2. ¯ GM2[T](t, s)<0for all (t, s)∈(0, T]×(0, T]. 3. ¯ GD[T](t, s)<0for all (t, s)∈(0, T)×(0, T). 4. ¯ GD[2 T](t, s)<0for all (t, s)∈(0,2T)×(0,2T). Corollary 3.3.14. If ¯ GN[T](t, s)<0for all (t, s)∈I×Ithen RT 0¯a(s) d s < 0. Moreover, considering the discriminant ˜ ∆(λ) = u1(2 R, λ) + u0 2(2 R, λ)for the periodic problem (y00(t) + ˜a(t)y(t) = ˜σ(t),a. e. t∈[0,2R], y(0) = y(2 R), y0(0) = y0(2 R), obtained from (3.3.2) by simply considering ˜aand ˜σthe even extensions of aand σ, we can deduce results for Green’s function different from the periodic one by rewriting Lemmas 3.2.51 and 3.2.52. The results are the following ones. 85 Second Order Equation Corollary 3.3.15. The following properties hold: (i) ¯ GN[T]<0on I×Iif and only if ˜ ∆(λ)>2for all λ≤0. (ii) ¯ GN[T]≥0on I×Iif and only if ˜ ∆(λ)>−2for all λ < 0and ˜ ∆(0) <2. Corollary 3.3.16. If ˜ ∆(λ)>−2for all λ < 0then: 1. ¯ GM1[T](t, s)<0for all (t, s)∈[0, T)×[0, T). 2. ¯ GM2[T](t, s)<0for all (t, s)∈(0, T]×(0, T]. 3. ¯ GD[T](t, s)<0for all (t, s)∈(0, T)×(0, T). 4. ¯ GD[2 T](t, s)<0for all (t, s)∈(0,2T)×(0,2T). Corollary 3.3.17. Suppose that pα−1 αˆ ¯a+Lα(I)≤2−1/α K(2 α∗,2R), with ¯a(t) = ˆ ¯a(t) + λ1 p(t), where λ=1 RRT 0¯a(s) d sis T Rtimes the mean value of ¯a. Then (i) ¯ GN[T]<0on I×Iif and only if RT 0¯a(s) d s < 0and ˜ ∆(0) >2. (ii) If RT 0¯a(s) d s < 0and ˜ ∆(0) <2, then ¯ GN[T]≥0on I×I. (iii) If operator Lunder periodic boundary conditions on [0,2T]is nonresonant and 0≤ZT 0 ¯a(s) d s≤π2 4R  1− 21/α pα−1 αˆ ¯a+Lα(I) K(2 α∗,2R)  , then ¯ GN[T]≥0on I×I. Proof. This result is obtained by applying Lemma 3.2.52 to ˜a, the even extension of a(t) = p(w−1(t)) ¯a(w−1(t)). The hypotheses of that lemma consider that ˆ ˜a+Lα[0,2R]≤K(2 α∗,2R), with ˜a(t) = ˆ ˜a(t) + λ, where λ=1 2RR2R 0˜a(s) d sis the mean value of ˜a. From the fact that ˜ais and even function and from the relation between aand ¯awe have that λ=1 RZR 0 a(s) d s=1 RZT 0 ¯a(s) d s, 86 3.3 General Second Order Equation so λis T Rtimes the mean value of ¯a. In addition, using again that ˜ais even (and so ˆ ˜ais even too) and taking into account the reasoning developed in the proof of Corollary 3.3.12, we arrive at the following relation between the norms of ˆ ˜aand ˆ ¯a ˆ ˜a+Lα[0,2R]= 21/α kˆa+kLα(K)= 21/α pα−1 αˆ ¯a+Lα(I). The rest of variations in this corollary with respect to Lemma 3.2.52 are immediately deduced from the considerations used in previous results. Corollary 3.3.18. If either ZT 0 ¯a(s) d s < 0 or 0≤ZT 0 ¯a(s) d s≤π2 4R  1− 21/α pα−1 αˆ ¯a+Lα(I) K(2 α∗,2R)  , then: 1. ¯ GM1[T](t, s)<0for all (t, s)∈[0, T)×[0, T). 2. ¯ GM2[T](t, s)<0for all (t, s)∈(0, T]×(0, T]. 3. ¯ GD[T](t, s)<0for all (t, s)∈(0, T)×(0, T). 4. ¯ GD[2 T](t, s)<0for all (t, s)∈(0,2T)×(0,2T). We will finish this subsection with an example in which we will use the relation between the Green’s functions of problems (3.3.1) and (3.3.2) to obtain the explicit expression of one of them through another one of a problem with constant coefficients. Example 3.3.19. Consider the equation 1 tu0(t)0+λ t u(t)=0, t ∈[0,1],(3.3.7) u(0) = u(1),lim t→0+(t u0(t))(t) = lim t→1−(t u0(t))(t),(3.3.8) which is a periodic problem of the type of (3.3.1) with ¯a(t) = λ t,p(t) = 1 tand [0, T] = [0,1]. 87 Solutions for Even Order Nonlinear BVPs with Constant Sign Green’s Functions We include now some references that the reader can consult to find more information about this theory in a more general framework. First, we refer to the monograph [9], where the authors develop the classical theory of lower and upper solutions. Moreover, some recent results and open problems can be found in the works of Mawhin [109–112] and in the surveys of Cabada [17] and De Coster and Habets [44,45], together with their monograph [46], and the references therein. The novelty in our approach with respect to others presented in the literature is that we are able to ensure the existence of solution of a problem by means of lower and upper solutions of another problem with different boundary conditions. To the best of our knowledge, this approach is new in the literature. All the results in this chapter are collected in [31]. In particular, we will consider nonlinear problems that fulfil the following scheme L u(t) = f(t, u(t)), t ∈I, u ∈X, (4.0.1) being Lthe 2n-th order general linear operator defined in (2.1.1), namely L u(t)≡u(2n)(t) + a2n−1(t)u(2n−1)(t) + ···+a1(t)u0(t) +a0(t)u(t), t ∈I, with ak:I→R, ak∈Lα(I), α ≥1,k= 0,...,2n−1. We will assume that Lis nonresonant on X, where, as in previous chapters, X⊂W2n,1(I)is a Banach space where the boundary conditions are included. Finally, we shall mention that the constant sign of the Green’s function will be a basic assumption to prove the existence of solutions. This chapter is organized in the following way: First, Section 4.1 compiles some basic definitions and conditions that will be used throughout the remaining of the chapter. Section 4.2 includes the results proving the existence of solutions via lower and upper solutions method. Finally, Section 4.3 provides an example in which we prove the existence of solutions of the Dirichlet problem via lower and upper solutions of Neumann problem. 4.1. Preliminaries It is clear that solutions of problem (4.0.1) correspond with the fixed points in X of the following integral operator L−1u(t) = ZT 0 G[T](t, s)f(s, u(s)) d s. In particular, when the Banach space Xis either XN,T ,XD,T ,XM1,T or XM2,T (defined as in Chapter 2), we obtain, respectively, the following nonlinear problems: 94 4.1 Preliminaries Neumann problem: L u(t) = f(t, u(t)), t ∈I, u ∈XN,T ,(4.1.1) Dirichlet problem: L u(t) = f(t, u(t)), t ∈I, u ∈XD,T ,(4.1.2) Mixed problem 1: L u(t) = f(t, u(t)), t ∈I, u ∈XM1,T ,(4.1.3) Mixed problem 2: L u(t) = f(t, u(t)), t ∈I, u ∈XM2,T ;(4.1.4) each of them with its corresponding equivalent integral operator: TNu(t) = ZT 0 GN[T](t, s)f(s, u(s)) d s, TDu(t) = ZT 0 GD[T](t, s)f(s, u(s)) d s, TM1u(t) = ZT 0 GM1[T](t, s)f(s, u(s)) d s, TM2u(t) = ZT 0 GM2[T](t, s)f(s, u(s)) d s. Notation 4.1.1. Note that, as in Chapters 2 and 3, we will use the notation G[T]to refer to the Green’s function related to operator L. Moreover, analogously to what we have done in Chapter 3, we will consider the parametrized operator defined from L, namely L[λ]u(t)≡L u(t) + λ u(t). In this case, to stress also its dependence on λ, we will denote by G[λ, T]the Green’s function related to L[λ]. For the purpose of finding fixed points of the previously defined integral operators, we shall use the following definitions. 95 Solutions for Even Order Nonlinear BVPs with Constant Sign Green’s Functions Definition 4.1.2. We say that a function α∈Xis a lower solution of problem (4.0.1) if L α(t)≥f(t, α(t)) for a. e. t∈I. Analogously, a function β∈Xis said to be an upper solution of problem (4.0.1) if L β(t)≤f(t, β(t)) for a. e. t∈I. Previous definitions are adapted to each of the considered problems by simply changing Xby any of the suitable Banach spaces XN,T ,XD,T ,XM1,T or XM2,T . Before proving existence results for some of the problems, we will consider some conditions that will be used in the remainder of the chapter. First, we will ask the nonlinearity fto satisfy the following property: (L0)The function f:I×R→Ris a L1-Carath´ eodory function, that is, •f(·, x)is measurable for all x∈R. •f(t, ·)is continuous for a. e. t∈I. • For every R > 0there exists ϕR∈L1(I)such that |f(t, x)| ≤ ϕR(t), for all x∈[−R, R]and a. e. t∈I. Moreover, given two continuous functions αand β, we will state the following conditions: (L1)There exists some λ∈Rsuch that for every t∈Iand x∈[α(t), β(t)], it holds that f(t, α(t)) + λ α(t)≥f(t, x) + λ x ≥f(t, β(t)) + λ β(t), and f(t, α(t)) + λ α(t)≥0≥f(t, β(t)) + λ β(t). (L2)There exists some λ∈Rsuch that for every t∈Iand x∈[β(t), α(t)], it holds that f(t, α(t)) + λ α(t)≥f(t, x) + λ x ≥f(t, β(t)) + λ β(t), and f(t, α(t)) + λ α(t)≥0≥f(t, β(t)) + λ β(t). 96 4.2 Results of Existence of Solutions 4.2. Results of Existence of Solutions In this section we will use the lower and upper solutions methods to prove the existence of solutions of the considered problems. Theorem 4.2.1. Assume that condition (L0)holds and let αand βbe lower and upper solutions of the Neumann problem (4.1.1), respectively, such that α(t)≤β(t)for all t∈I. Moreover, assume that there exists some λfor which GP[λ, 2T]≤0on J×J, GD[λ, T]≤0on I×Iand (L1)holds. Then, there exists a solution uof the Dirichlet problem (4.1.2) such that α(t)≤u(t)≤β(t),for all t∈I. Proof. Let λbe such that GP[λ, 2T]≤0on J×J,GD[λ, T ]≤0on I×Iand condition (L1)holds. Consider the problem L[λ]u(t) = f(t, u(t)) + λ u(t), t ∈I, u ∈XD,T ,(4.2.1) with L[λ]u(t)≡L u(t) + λ u(t). As a consequence, the solutions of problem (4.2.1) coincide with the solutions of (4.1.2). Also, these solutions correspond with fixed points of the following integral operator TD[λ]u(t) = ZT 0 GD[λ, T](t, s) (f(s, u(s)) + λ u(s)) d s. We will divide the proof into several steps. In particular, Steps 1 to 3 follow standard techniques but we include them for the sake of completeness. Step 1: TD[λ]: C(I)→ C(I)is well-defined: Let u∈ C(I)and (tn)n∈N⊂Isuch that lim n→∞tn=t0∈I. On the one hand, from property (G2) in the definition of Green’s function (Definition 1.1.1), GD[λ, T](·, s)is uniformly continuous on I. Thus, lim n→∞GD[λ, T](tn, s) (f(s, u(s)) + λ u(s)) = GD[λ, T](t0, s) (f(s, u(s)) + λ u(s)). On the other hand it holds that |GD[λ, T](t, s) (f(s, u(s)) + λ u(s))| ≤ |GD[λ, T](t, s)|ϕkuk(s) + λkuk, 97 Solutions for Even Order Nonlinear BVPs with Constant Sign Green’s Functions where kukdenotes the usual supremum norm. Moreover, from (G2) in Definition 1.1.1, GD[λ, T]is continuous on I×Iand so it is bounded on I×Iby some constant M. Therefore |GD[λ, T](t, s) (f(s, u(s)) + λ u(s))| ≤ Mϕkuk(s) + λkuk,a. e. s∈I and, since the right hand side of previous inequality is in L1(I)by Lebesgue’s Dominated Convergence Theorem we obtain that lim n→∞TD[λ]u(tn) = lim n→∞ZT 0 GD[λ, T](tn, s) (f(s, u(s)) + λ u(s)) d s =ZT 0 lim n→∞GD[λ, T](tn, s) (f(s, u(s)) + λ u(s)) d s =ZT 0 GD[λ, T](t0, s) (f(s, u(s)) + λ u(s)) d s=TD[λ]u(t0). Thus, TD[λ]u∈ C(I). Step 2: TD[λ]is continuous: Let {un}n∈Nbe a sequence which converges to uin C(I). Then, there exists some R∈R+such that kunk ≤ Rfor all n∈N. Now, on the one hand, from (L0), we deduce that lim n→∞f(s, un(s)) + λ un(s) = f(s, u(s)) + λ u(s),for a. e. s∈I. On the other hand, |GD[λ, T](t, s)||f(s, un(s)) + λ un(s)| ≤ M(ϕR(s) + λ R),a. e. s∈I and, since the right hand side of previous inequality is in L1(I)by Lebesgue’s Dominated Convergence Theorem we deduce that lim n→∞TD[λ]un(t) = lim n→∞ZT 0 GD[λ, T](t, s) (f(s, un(s)) + λ un(s)) d s =ZT 0 lim n→∞GD[λ, T](t, s) (f(s, un(s)) + λ un(s)) d s =ZT 0 GD[λ, T](t, s) (f(s, u(s)) + λ u(s)) d s=TD[λ]u(t). Thus we can conclude that operator TD[λ]is continuous. Step 3: TD[λ]is a compact operator: 98 4.2 Results of Existence of Solutions Take B={u∈ C(I); kuk< r}. First, we will prove that TD[λ](B)is uniformly bounded: kTD[λ]uk= sup t∈IZT 0 GD[λ, T](t, s) (f(s, u(s)) + λ u(s)) d s ≤ZT 0 M(ϕr(s) + λ r) d s=MZT 0 ϕr(s) d s+λ r T and, since ϕr∈L1(I), it is clear that TD[λ](B)is uniformly bounded. Now, we will prove that TD[λ](B)is equicontinuous. We have that |(TD[λ]u)(t1)−(TD[λ]u)(t2)| ≤ZT 0|GD[λ, T](t1, s)−GD[λ, T](t2, s)|(|f(s, u(s))|+λ|u(s)|) d s ≤ZT 0|GD[λ, T](t1, s)−GD[λ, T](t2, s)|(ϕr(s) + λ r) d s and, since GD[λ, T]is uniformly continuous on I×I, it occurs that for every ε > 0 there exists δ > 0such that when |t1−t2|< δ, |(TD[λ]u)(t1)−(TD[λ]u)(t2)| ≤ εZT 0 (ϕr(s) + λ r) d s. Thus, the fact that ϕr∈L1(I)lets us conclude that TD[λ](B)is equicontinuous. As a consequence, by Ascoli-Arzel` a’s Theorem (Theorem 1.2.2), we deduce that TD[λ](B)is relatively compact in C(I)and thus TD[λ]is a compact operator. Step 4: α≤TD[λ]αand β≥TD[λ]βon I. From Corollary 2.5.1, we know that GP[λ, 2T]≤0on J×Jimplies that GN[λ, T](t, s)≤ −|GD[λ, T](t, s)|,for all t, s ∈I. (4.2.2) On the other hand, the fact that α∈XN,T ⊂W2n,1(I)and L α(t)≥f(t, α(t)) for a. e. t∈Imeans that there exists a nonnegative function g∈L1(I), such that L α(t) + λ α(t) = f(t, α(t)) + λ α(t) + g(t),for a. e. t∈I. Therefore, since α∈XN,T , it holds that α(t) =ZT 0 GN[λ, T](t, s) (f(s, α(s)) + λ α(s)) d s+ZT 0 GN[λ, T](t, s)g(s) d s. 99 Solutions for Even Order Nonlinear BVPs with Constant Sign Green’s Functions From (4.2.2), it is deduced that GN[λ, T ]is nonpositive. Thus, GN[λ, T ](t, s)g(s) is nonpositive for a. e. t, s ∈Iand so, the second integral in previous expression is less or equal than zero. Moreover, we also deduce from (4.2.2) that GN[λ, T]≤GD[λ, T]on I×I. Therefore, taking into account the fact that (from (L1))f(s, α(s)) + λ α(s)≥0for a. e. s∈I, we obtain the following inequalities for all t∈I: α(t)≤ZT 0 GN[λ, T](t, s) (f(s, α(s)) + λ α(s)) d s ≤ZT 0 GD[λ, T](t, s) (f(s, α(s)) + λ α(s)) d s=TD[λ]α(t). Analogously, from β∈W2n,1(I)and L β(t)≤f(t, β(t)), we deduce that there exists a nonpositive function h∈L1(I), such that L β(t) + λ β(t) = f(t, β(t)) + λ β(t) + h(t),for a. e. t∈I. As a consequence, reasoning analogously to the previous case and taking into account that f(s, β(s)) + λ β(s)≤0for a. e. s∈I, we obtain the following inequalities β(t) = ZT 0 GN[λ, T](t, s) (f(s, β(s)) + λ β(s)) d s+ZT 0 GN[λ, T](t, s)h(s) d s ≥ZT 0 GN[λ, T](t, s) (f(s, β(s)) + λ β(s)) d s ≥ZT 0 GD[λ, T](t, s) (f(s, β(s)) + λ β(s)) d s=TD[λ]β(t). Step 5: TD[λ]([α, β]) ⊂[α, β], where [α, β]≡ {u∈ C(I) : α(t)≤u(t)≤β(t),for all t∈I}. We will decompose operator TD[λ]as a composition of two operators. First, consider the Nemytskii operator N[λ]: C(I)→L1(I)defined in the following way N[λ]u(t) = f(t, u(t)) + λ u(t),for a. e. t∈I. On the other hand, consider operator K[λ]: L1(I)→ C(I)defined as K[λ]σ(t) = ZT 0 GD[λ, T](t, s)σ(s) d s, for all t∈I. 100 4.2 Results of Existence of Solutions It is clear that TD[λ] = K[λ]◦N[λ]. Moreover, let’s see that operator K[λ]is nonincreasing in [α, β]. Indeed, take σ1, σ2∈L1(I)such that σ1(t)≤σ2(t)for a. e. t∈I. Then, since GD[λ, T]is nonpositive, it holds that GD[λ, T](t, s)σ1(s)≥GD[λ, T](t, s)σ2(s),for a. e. t, s ∈I and, therefore, K[λ]σ1(t) = ZT 0 GD[λ, T](t, s)σ1(s) d s≥ZT 0 GD[λ, T](t, s)σ2(s) d s =K[λ]σ2(t),for all t∈I. Now, let u∈[α, β]. From (L1)we have that f(t, α(t)) + λ α(t)≥f(t, u(t)) + λ u(t)≥f(t, β(t)) + λ β(t),a. e. t∈I and so α(t)≤TD[λ]α(t)≤TD[λ]u(t)≤TD[λ]β(t)≤β(t),∀t∈I. We conclude that TD[λ]u∈[α, β]for all u∈[α, β]. Step 6: Operator TD[λ]has a fixed point in XD∩[α, β]. Since the interval [α, β]is a closed, convex, bounded and nonempty subset of the Banach space X,TD[λ]is a compact operator and TD[λ]([α, β]) ⊂[α, β], then we are in the suitable conditions to apply Schauder’s fixed point Theorem (Theorem 1.2.3) which ensures us the existence of a fixed point of TD[λ]on [α, β]. Obviously, this fixed point satisfies Dirichlet boundary conditions and therefore it is a solution of problem (4.1.2). Remark 4.2.2. Note that the functions αand βconsidered in previous theorem are not required to belong to XD,T , that is, they may not be lower and upper solutions of Dirichlet problem, that is, the equalities u(2k)(0) = 0 and u(2k)(T) = 0 may fail for some values of k. In an analogous way, we can prove the following result when GP[λ, 2T]is nonnegative and hypothesis (L2)holds. Theorem 4.2.3. Assume that condition (L0)holds and let αand βbe lower and upper solutions of Neumann problem (4.1.1), respectively, such that α(t)≥β(t)for all t∈I. 101 Solutions for Even Order Nonlinear BVPs with Constant Sign Green’s Functions Moreover, assume that there exists some λfor which GP[λ, 2T]≥0on J×J, GD[λ, T]≥0on I×Iand (L2)is satisfied. Then, there exists a solution uof the Dirichlet problem (4.1.2) such that β(t)≤u(t)≤α(t),for all t∈I. Proof. The proof is analogous to the one of Theorem 4.2.1, so we will only detail the parts of it which present some differences. Let λbe such that GP[λ, 2T]≥0on J×J,GD[λ, T]≥0on I×Iand condition (L2)holds. Consider operator TD[λ]as defined in the proof of Theorem 4.2.1. Step 1: TD[λ]: C(I)→ C(I)is well-defined, continuous and compact: The proof is exactly the same than in Theorem 4.2.1. Step 2: α≥TD[λ]αand β≤TD[λ]β. From Corollary 2.5.1, we know that GP[λ, 2T]≤0on J×Jimplies that GN[λ, T](t, s)≥ |GD[λ, T](t, s)|=GD[λ, T](t, s),for all t, s ∈I. (4.2.3) In this case, there exists a nonnegative function g∈L1(I)such that L α(t) + λ α(t) = f(t, α(t)) + λ α(t) + g(t),for a. e. t∈I and so the following inequalities hold for all t∈I: α(t) = ZT 0 GN[λ, T](t, s) (f(s, α(s)) + λ α(s)) d s+ZT 0 GN[λ, T](t, s)g(s) d s ≥ZT 0 GN[λ, T](t, s) (f(s, α(s)) + λ α(s)) d s ≥ZT 0 GD[λ, T](t, s) (f(s, α(s)) + λ α(s)) d s=TD[λ]α(t). Analogously, there exists a nonpositive function h∈L1(I)such that L β(t) + λ β(t) = f(t, β(t)) + λ β(t) + h(t),for a. e. t∈I and β(t) = ZT 0 GN[λ, T](t, s) (f(s, β(s)) + λ β(s)) d s+ZT 0 GN[λ, T](t, s)h(s) d s ≤ZT 0 GN[λ, T](t, s) (f(s, β(s)) + λ β(s)) d s ≤ZT 0 GD[λ, T](t, s) (f(s, β(s)) + λ β(s)) d s=TD[λ]β(t). 102 4.2 Results of Existence of Solutions Step 3: TD[λ]([β, α]) ⊂[β, α]. In this case, operator K[λ]is nondecreasing. Let u∈[β, α]. From (L2)we have that f(t, α(t)) + λ α(t)≥f(t, u(t)) + λ u(t)≥f(t, β(t)) + λ β(t),a. e. t∈I and so α(t)≥TD[λ]α(t)≥TD[λ]u(t)≥TD[λ]β(t)≥β(t),∀t∈I and we conclude that TD[λ]u∈[β, α]for all u∈[β, α]. Step 4: Operator TD[λ]has a fixed point in XD∩[β, α]. Analogously to the proof of Theorem 4.2.1, this fact is deduced from Schauder’s fixed point Theorem. Remark 4.2.4. We must note that when GD[λ, T]has constant sign, there exist αand β, lower and upper solutions of Dirichlet problem, respectively, and it is satisfied that f(t, α(t)) + λ α(t)≥f(t, x) + λ x ≥f(t, β(t)) + λ β(t) for every t∈Iand x∈[α(t), β(t)], then there exists a solution of the Dirichlet problem (4.1.2) (see, for instance, [25] for the case 2n= 4). In this case, by adding the hypotheses on the sign of f(t, α(t)) + λ α(t)and f(t, β(t)) + λ β(t), we can ensure the existence of a solution for problem (4.1.2) when we have lower and upper solutions of Neumann problem (4.1.1). Now, using the inequalities in Corollary 2.5.3, we can obtain similar results to prove the existence of solutions of Mixed 1 and Dirichlet problems. Theorem 4.2.5. Assume that condition (L0)holds and let αand βbe lower and upper solutions of the Neumann problem (4.1.1), respectively, such that α(t)≤β(t)for all t∈I. Moreover, assume that there exists some λfor which GN[λ, 2T]≤0on J×J, GM1[λ, T]≤0on I×Iand (L1)holds. Then, there exists a solution uof the Mixed problem 1(4.1.3) such that α(t)≤u(t)≤β(t),for all t∈I. 103 Positive Solutions for Second Order BVPs with Sign-Changing Green’s Functions the periodic boundary value problem (with T= 1 in the paper) (u00(t) + a(t)u(t) = g(t)f(u(t)), t ∈(0, T), u(0) = u(T), u0(0) = u0(T), with fand gnonnegative continuous functions and gsatisfying the condition min t∈[0,T]g(t)>0. Moreover, they assumed the Green’s function to be nonnegative and to satisfy the following condition: min 0≤s≤TZT 0 G(t, s) d t > 0.(5.1.1) We note that the method in the aforementioned reference can not be used for Dirichlet and Mixed problems, as their related Green’s functions do not satisfy condition (5.1.1). In [147], Webb considered weaker assumptions to prove the existence of positive solutions of the previous problem, but he still assumed the Green’s function to be nonnegative. Despite our results do not require the Green’s function to be nonnegative, as we will see, they could be applied to this particular case, obtaining positive solutions assuming an integral condition weaker than (5.1.1) (see Remarks 5.3.6 and 5.3.11 in Section 5.3). On the other hand, some existence results for boundary value problems with signchanging Green’s function were considered in [28, 79], where the authors asked for the existence of a subinterval [c, d]⊂[0, T], a function φ∈L1([0, T]) and a constant c∈(0,1] such that the Green’s function Gsatisfies the following condition: |G(t, s)| ≤ φ(s)for all t∈[0, T]and almost every s∈[0, T], G(t, s)≥c φ(s)for all t∈[c, d]and almost every s∈[0, T].(5.1.2) It must be pointed out that, if we consider a periodic problem with constant potential a(t) = ρ2for which the related Green’s function changes its sign (i.e. ρ > π/T, ρ6= 2kπ/T,k= 1,2, . . .), condition (5.1.2) is never fulfilled for any strictly positive function φ. This is due to the fact that in such situation the Green’s function is constant along the straight lines of slope equals to one (as we have seen in Lemma 2.4.3). On the other hand, as we will prove in Section 5.4, our results can be applied without further complications to this case. Moreover, for Dirichlet boundary value problem with constant potential a(t) = ρ2 with sign-changing Green’s function (i.e. ρ > π/T,ρ6=kπ/T,k= 1,2, . . .), as a direct consequence of expression (5.5.1) below, it is immediate to verify that condition (5.1.2) holds if and only if ρ2lies between the first and the second eigenvalues 110 5.1 Introduction of the problem ( π T< ρ < 2π T) but it is never satisfied for ρ > 2π T. However, as we will point out in Section 5.5, our results can be applied for any nonresonant value of ρ > π/T . Despite of this, we must note that the conditions are more restrictive when ρincreases. Furthermore, in [28,79] the authors proved the existence of solutions in the cone K0=u∈ C[0, T],min t∈[c,d]u(t)≥ckuk, that is, they ensured the positivity of the solutions on the subinterval [c, d]but such solutions were allowed to change sign when considering the whole interval [0, T ]. As far as we know, positive solutions for boundary value problems with signchanging Green’s function can be tracked only as back as 2011 in the papers [104, 163]. In the first of these papers, R. Ma considers the following one parameter family of problems: (u00(t) + a(t)u(t) = λ g(t)f(u(t)), t ∈(0, T ), u(0) = u(T), u0(0) = u0(T).(5.1.3) By using the Schauder’s fixed point Theorem, the author obtains the existence of a positive solution for sufficiently small values of λ. These existence results are not comparable with the ones we will obtain in this chapter. In the second paper [163], S. Zhong and Y. An study the following autonomous periodic boundary value problem, with constant potential ρ∈0,3π 2T: (u00(t) + ρ2u(t) = f(u(t)), t ∈(0, T), u(0) = u(T), u0(0) = u0(T).(5.1.4) In this case, it is very well-known that the related Green’s function GP(t, s)≥0for all ρ∈0,π Tand it changes sign for ρ∈(π T,3π 2T](see [16,18]). With this, it can be defined the constant δ=         ∞, ρ ∈0,π Ti, inf t∈IRT 0G+ P(t, s) d s RT 0G− P(t, s) d s, ρ ∈π T,3π 2T, and using the Krasnoselskii’s fixed point Theorem, the authors prove the following existence result. Theorem 5.1.1 ( [163, Theorem 3]).Suppose that the following assumptions are fulfilled: 111 Positive Solutions for Second Order BVPs with Sign-Changing Green’s Functions (J1) f: [0,∞)→[0,∞)is continuous. (J2) 0 ≤m= inf u≥0{f(u)}and M= sup u≥0{f(u)} ≤ M≤ ∞. (J3) M/m ≤δ, with M/m =∞when m= 0. Moreover, if δ=∞assume that lim x→∞ f(x) x< ρ2<lim x→0+ f(x) x. Then problem (5.1.4) has a positive solution on [0, T]. Concerning this specific case, we improve the range of the values ρfor which the result is still valid. Furthermore, we apply our study to nonconstant potentials and nonautonomous nonlinear parts. As we will see, some of the positivity conditions imposed for the periodic boundary value problem cannot be adapted for Dirichlet problem, so the approach that must be used needs to be considerably modified, by using, in this case, a different type of cones. This chapter is divided in the following way: in Section 5.2 we state some preliminary results considering the Hill’s operator, in Section 5.3 some new results concerning the existence of a positive solution for the Hill’s periodic problem in the case that the Green’s function may change sign are proved. Moreover, in this section, such existence results are generalized to other boundary conditions. In Section 5.4 we improve Theorem 5.1.1 for the periodic problem with a constant potential and in Section 5.5 we approach the Dirichlet problem, also in the case of a constant potential, where as far as we know, no results for sign-changing Green’s functions were proved before. All the results in this chapter are compiled in [27]. 5.2. Preliminaries Consider the particular case of operator Ldefined in (2.1.1) for n= 1 and a1≡0, that is, the Hill’s operator related to the potential a L u(t)≡u00(t) + a(t)u(t), t ∈I, where a:I→R,a∈Lα(I),α≥1. As in previous chapters, we denote x0on Iif and only if x≥0on Iand RT 0x(s) d s > 0. 112 5.3 Periodic Boundary Value Problems Since throughout this chapter we will always work on the same interval [0, T], it is not necessary to stress the dependence of the problem on the parameter T. Therefore, when working both with Green’s functions and eigenvalues, and contrary to what we did in the previous chapter, we will skip the indication about the parameter T. This way, we will denote by GP,GN,GD,GM1and GM2the related Green’s functions and λP 0,λN 0,λD 0,λM1 0and λM2 0the corresponding smallest eigenvalue of each of the problems (periodic, Neumann, Dirichlet, Mixed 1 and Mixed 2), all of them considered on the interval I. Analogously, λA 0will be the smallest eigenvalue of the anti-periodic problem. For the reader’s convenience, we rewrite now the following relations which have been proved in previous chapters and will be the key points to show some of the following results. Lemma 5.2.1. 1. GN(t, s)<0on I×Iif and only if λN 0>0. 2. GN(t, s)≥0on I×Iif and only if λN 0<0,λM1 0≥0and λM2 0≥0. 3. GNchanges sign if and only if min{λM1 0, λM2 0}<0. 4. GD(t, s)<0on (0, T)×(0, T)if and only if λD 0>0. 5. GDchanges sign if and only if λD 0<0. 6. GM1(t, s)<0on [0, T)×[0, T)if and only if λM1 0>0. 7. GM1changes sign if and only if λM1 0<0. 8. GM2(t, s)<0on (0, T]×(0, T]if and only if λM2 0>0. 9. GM2changes sign if and only if λM2 0<0. 10. GP(t, s)<0on I×Iif and only if λP 0>0. 11. GP(t, s)≥0on I×Iif and only if λP 0<0,λA 0≥0. 12. GPchanges sign if and only if λA 0<0. 5.3. Periodic Boundary Value Problems Consider now the following nonlinear and nonautonomous periodic boundary value problem: (u00(t) + a(t)u(t) = f(t, u(t)), t ∈I, u(0) = u(T), u0(0) = u0(T).(5.3.1) 113 Positive Solutions for Second Order BVPs with Sign-Changing Green’s Functions We will assume that the Hill’s operator coupled with periodic conditions is nonresonant and λA 0<0. From Lemma 5.2.1, we know that in this case the related Green’s function changes its sign on I×I. On the other hand (as we have seen in Section 3.1), there exists vP, a positive eigenfunction on I, unique up to a constant, related to λP 0, that is, vPis such that (v00 P(t) + a(t)vP(t) = −λP 0vP(t),for a. e. t∈I, vP(0) = vP(T), v0 P(0) = v0 P(T). Therefore, vP(t) = −λP 0ZT 0 GP(t, s)vP(s) d s and, since vPis positive and λP 0< λA 0<0, we have that ZT 0 GP(t, s)vP(s) d s > 0∀t∈I and, consequently, ZT 0 G+ P(t, s)vP(s) d s > ZT 0 G− P(t, s)vP(s) d s∀t∈I, where G+ Pand G− Pare the positive and negative parts of GP. Since the Green’s function changes sign, it makes sense to define γ= inf t∈IRT 0G+ P(t, s)vP(s) d s RT 0G− P(t, s)vP(s) d s(>1).(5.3.2) Moreover, in order to ensure the existence of solutions of problem (5.3.1), we will make the following assumptions: (H1)f:I×[0,∞)→[0,∞)satisfies L1-Carath´ eodory conditions, that is: f(·, u)is measurable for every u∈R. f(t, ·)is continuous for a. e. t∈I. For each r > 0there exists φr∈L1(I)such that f(t, u)≤φr(t)for all u∈[−r, r],a. e. t∈I. (H2)There exist two positive constants mand Msuch that m vP(t)≤f(t, x)≤M vP(t) for every t∈Iand x≥0. Moreover, these constants satisfy that M m≤γ. 114 5.3 Periodic Boundary Value Problems (H3)There exists [c, d]⊂Isuch that Rd cGP(t, s) d t≥0,for all s∈Iand Rd cGP(t, s) d t > 0,for all s∈[c, d]. Remark 5.3.1. Note that condition (H1)is the same as (L0)given in Chapter 4 but, for the reader’s convenience, we have decided to rewrite it so that one can find all the hypotheses used in this chapter together. Remark 5.3.2. We note that condition (H2)includes, as particular cases, hypotheses (J2) and (J3) in Theorem 5.1.1 used in [163]. This is due to the fact that, if a(t) = ρ2, as in problem (5.1.4), we have that λP 0=−ρ2and vP(t)=1for all t∈I. Moreover, as we will point out in Section 5.4, we have that, if a(t) = ρ2, then ZT 0 GP(t, s) d s=1 ρ2, and condition (H3)is trivially fulfilled for [c, d] = I. Moreover, we note that in (H2)we are not considering the possibility of m= 0. Theorem 5.1.1 includes this case, but only when δ= +∞, which only happens when the Green’s function is nonnegative. In [163] the authors consider this possibility because they are assuming that ρ∈0,3π 2Tand, when ρ∈0,π T,GPis nonnegative. As we will see in Corollary 5.3.5, hypothesis (H2)is not necessary in case that the Green’s function is nonnegative, so this is the reason why we do not consider the possibility m= 0. We will consider the Banach space (C(I, R),k·k)coupled with the supremum norm kuk≡kuk∞, and define the cone K=u∈ C(I, R): u≥0on I, ZT 0 u(s) d s≥σkuk, where σ=η max t, s∈I{GP(t, s)}, being η= min s∈[c,d]Zd c GP(t, s) d t>0.(5.3.3) Now, it is clear that uis a solution of the periodic problem (5.3.1) if and only if it is a fixed point of the following operator: Tu(t) = ZT 0 GP(t, s)f(s, u(s)) d s. 115 Positive Solutions for Second Order BVPs with Sign-Changing Green’s Functions Lemma 5.3.3. Assume that λA 0<0and (H1)–(H3)hold. Then T:C(I)→ C(I)is a compact operator which maps the cone Kto itself. Proof. We will divide the proof into several steps. We note that Steps 1 to 3 follow standard techniques but we include them for the sake of completeness. Step 1: T:C(I)→ C(I)is well-defined: Let u∈ C(I)and (tn)n∈N⊂Isuch that lim n→∞tn=t0∈I. On the one hand, from property (G2) in the definition of Green’s function (Definition 1.1.1), GP(·, s)is uniformly continuous on I. Thus, lim n→∞GP(tn, s)f(s, u(s)) = GP(t0, s)f(s, u(s)),a. e. s∈I. On the other hand it holds that |GP(t, s)f(s, u(s))| ≤ |GP(t, s)|φkuk(s),a. e. s∈I. Moreover, from (G2) in Definition 1.1.1, GPis continuous on I×Iand so it is bounded on I×Iby some constant C. Therefore |GP(t, s)f(s, u(s))| ≤ C φkuk(s),a. e. s∈I, and, since the right hand side of previous inequality is in L1(I), by Lebesgue’s Dominated Convergence Theorem we obtain that lim n→∞Tu(tn) = lim n→∞ZT 0 GP(tn, s)f(s, u(s)) d s =ZT 0 lim n→∞GP(tn, s)f(s, u(s)) d s =ZT 0 GP(t0, s)f(s, u(s)) d s=Tu(t0). Thus, Tu∈ C(I). Step 2: Operator Tis continuous: Let {un}n∈N⊂ C(I)be a sequence which converges to uin C(I). Then, there exists some R∈R+such that kunk ≤ Rfor all n∈N. Now, from (H1), we deduce that lim n→∞f(s, un(s)) = f(s, u(s)),for a. e. s∈I. On the other hand, |GP(t, s)|f(s, un(s)) ≤C φR(s),for a. e. s∈I 116 5.3 Periodic Boundary Value Problems and, since the right hand side of previous inequality is in L1(I), by Lebesgue’s Dominated Convergence Theorem we deduce that lim n→∞Tun(t) = lim n→∞ZT 0 GP(t, s)f(s, un(s)) d s =ZT 0 lim n→∞GP(t, s)f(s, un(s)) d s =ZT 0 GP(t, s)f(s, u(s)) d s=Tu(t). Thus we can conclude that operator Tis continuous. Step 3: Tis a compact operator: Take B={u∈ C(I) : kuk< r}. First, we will prove that T(B)is uniformly bounded: kTuk= sup t∈IZT 0 GP(t, s)f(s, u(s)) d s≤ZT 0 C φr(s) d s and, since φr∈L1(I), it is clear that T(B)is uniformly bounded. Now, we will prove that Tis equicontinuous. We have that |(Tu)(t1)−(Tu)(t2)| ≤ ZT 0|GP(t1, s)−GP(t2, s)|f(s, u(s)) d s ≤ZT 0|GP(t1, s)−GP(t2, s)|φr(s) d s and, since GPis uniformly continuous on I×I, it occurs that for every ε > 0there exists δ > 0such that when |t1−t2|< δ, |(Tu)(t1)−(Tu)(t1)| ≤ εZT 0 φr(s) d s. Thus, the fact that φr∈L1(I)lets us conclude that Tis equicontinuous. As a consequence, by Ascoli-Arzel` a’s Theorem (Theorem 1.2.2), we deduce that T(B)is relatively compact in C(I)and thus Tis a compact operator. Step 4: Tmaps the cone to itself. 117 Positive Solutions for Second Order BVPs with Sign-Changing Green’s Functions Considering u∈K, then, from (5.3.2), the following inequalities are fulfilled for all t∈I: Tu(t) = ZT 0 GP(t, s)f(s, u(s)) d s=ZT 0G+ P(t, s)−G− P(t, s)f(s, u(s)) d s ≥ZT 0m vP(s)G+ P(t, s)−M vP(s)G− P(t, s)ds ≥mZT 0 G+ P(t, s)vP(s) d s−γZT 0 G− P(t, s)vP(s) d s≥0. Moreover, ZT 0Tu(t) d t≥Zd cTu(t) d t=Zd cZT 0 GP(t, s)f(s, u(s)) d sdt =ZT 0 f(s, u(s)) Zd c GP(t, s) d tds≥ηZT 0 f(s, u(s)) d s, and, since Tu(t)≤max t, s∈I{GP(t, s)}ZT 0 f(s, u(s)) d s, we deduce that ZT 0Tu(t) d t≥σTu(t)for all t∈I. Thus, ZT 0Tu(t) d t≥σkTuk, and the result is concluded. Now, in order to prove the existence of solutions of problem (5.3.1), we will use some classical results regarding the fixed point index which have been compiled in Lemma 1.2.7. First, we note that, as an immediate consequence of condition (H2), we deduce the following properties: f0= lim x→0+min t∈[c,d] f(t, x) x=∞, f∞= lim x→∞ max t∈I f(t, x) x= 0, where the interval [c, d]is given in (H3). These properties will let us prove the following theorem. 118 5.3 Periodic Boundary Value Problems Theorem 5.3.4. Assume that λA 0<0and hypotheses (H1)–(H3)hold. Then there exists at least one positive solution of problem (5.3.1) in the cone K. Proof. Taking into account the definition of f0, we know that there exists δ1>0 such that when kuk ≤ δ1, then f(t, u(t)) >u(t) η,∀t∈[c, d], with ηdefined in (5.3.3). Let Ω1={u∈K:kuk< δ1} and choose u∈∂Ω1and e∈K\{0}. We will prove that u6=Tu+λ e for every λ > 0. Assume, on the contrary, that there exists some λ > 0such that u=Tu+λ e, that is, u(t) = Tu(t) + λ e(t)≥ Tu(t)∀t∈I. Then Zd c u(t) d t≥Zd cTu(t) d t=Zd cZT 0 GP(t, s)f(s, u(s)) d sdt =ZT 0Zd c GP(t, s) d tf(s, u(s)) d s ≥Zd cZd c GP(t, s) d tf(s, u(s)) d s > Zd c u(s) d s, which is a contradiction. Therefore, we deduce from Lemma 1.2.7 that iK(T, Ω1) = 0. Now, proceeding in an analogous way to [21,62, 63], we define ˜ f(t, u) = max 0≤z≤uf(t, z). Clearly, ˜ f(t, ·)is a nondecreasing function on [0,∞)and ˜ f(t, x)≥f(t, x)for all t∈I,x∈[0,∞). Moreover, since f∞= 0 it is obvious that lim x→∞ (max t∈I ˜ f(t, x) x)= 0. As a consequence, we know that there exists δ2>0such that if kuk ≥ δ2then ˜ f(t, kuk)<σ2 T2ηkuk,∀t∈I. 119 Existence of Solutions of Integral Equations with Asymptotic Conditions Now we deduce, by application of Lebesgue’s Dominated Convergence Theorem, that lim n→∞ ∂jg Tun ∂tj−∂jf Tu ∂tj∞≤lim n→∞Z∞ −∞ Mj(s)|f(s, un(s)) −f(s, u(s))|ds =Z∞ −∞ lim n→∞Mj(s)|f(s, un(s)) −f(s, u(s))|ds= 0. Therefore, we deduce that Tun→Tu in e Cn ϕ. Hence, operator Tis continuous. Step 3: Compactness: Let B⊂e Cn ϕbe a bounded set, that is, there exists some R > 0such that kukϕ≤Rfor all u∈B. First, we will see that T(B)is uniformly bounded. Using the General Leibniz’s Rule (for differentiation), it is clear that ∂jf Tu ∂tj=∂j(Tu/ϕ) ∂tj= j X l=0 j l∂lTu ∂tl ∂j−l ∂tj−l 1 ϕ. Moreover, from Leibniz’s Integral Rule, ∂lTu ∂tl(t) = Z∞ −∞ ∂lk ∂tl(t, s)η(s)f(s, u(s)) d s+∂lp ∂tl(t), t ∈R. Thus,  ∂jf Tu ∂tj∞ = j X l=0 j l∂lTu ∂tl ∂j−l ∂tj−l 1 ϕ∞≤ j X l=0 j l ∂lTu ∂tl ∂j−l ∂tj−l 1 ϕ∞ = j X l=0j l ∂j−l ∂tj−l 1 ϕ(·)Z∞ −∞ ∂lk ∂tl(·, s)η(s)f(s, u(s)) d s+∂lp ∂tl∞ ≤ j X l=0j l ∂j−l ∂tj−l 1 ϕ(·)Z∞ −∞ ∂lk ∂tl(·, s)η(s)f(s, u(s)) d s∞ + ∂j−l ∂tj−l 1 ϕ ∂lp ∂tl∞. (8.4.4) 222 8.4 Fixed Points of Integral Equations It is satisfied that  ∂j−l ∂tj−l 1 ϕ(t)Z∞ −∞ ∂lk ∂tl(t, s)η(s)f(s, u(s)) d s ≤ ∂j−l ∂tj−l 1 ϕ(t)Z∞ −∞  ∂lk ∂tl(t, s)η(s)f(s, u(s)) d s ≤ ∂j−l ∂tj−l 1 ϕ(t)Z∞ −∞  ∂lk ∂tl(t, s)η(s)φR(s) d s, (8.4.5) and so, from (8.4.4) and (8.4.5) and using (C3)and (C4),  ∂jf Tu ∂tj∞ ≤ j X l=0j l ∂j−l ∂tj−l 1 ϕ(·)Z∞ −∞ ∂lk ∂tl(·, s)η(s) φR(s) d s∞ + ∂j−l ∂tj−l 1 ϕ ∂lp ∂tl∞<∞. So, we have found an upper bound which does not depend on u. Therefore it is clear that the set T(B)is uniformly bounded. On the other hand, taking into account the upper bound obtained in (8.4.3), we have that given ε > 0there exists δ > 0such that if t1, t2∈R,|t1−t2|< δ, then, for j= 0, . . . , n,  ∂jf Tu ∂tj(t1)−∂jf Tu ∂tj(t2)≤εZ∞ −∞ wj(s)f(s, u(s)) d s+ 1 ≤εZ∞ −∞ wj(s)φR(s) d s+ 1. Then, since from (C3)wjφR∈L1(R), we can conclude that there exists some constant csuch that  ∂jf Tu ∂tj(t1)−∂jf Tu ∂tj(t2)< ε c, for all u∈B. This implies that T(B)is equicontinuous. In conclusion, we derive, by application of Theorem 8.3.2, that T(B)is relatively compact in e Cn ϕ. Therefore, Tis a compact operator. Step 4: Tmaps Kαto Kα: It is an immediate consequence of conditions (C5)and (C6). Now we will give some conditions under which we can assure that the index of some subsets of Kαis 1 or 0. We will consider the following sets: Kβ, ρ α:= {u∈Kα:β(u)< ρ}, 223 Existence of Solutions of Integral Equations with Asymptotic Conditions Kγ, ρ α:= {u∈Kα:γ(u)< ρ}. Now, hypothesis (C9)implies that either there exists a function b:R+→R given by b(ρ) := sup {β(u): u∈Kα, γ(u)< ρ}, or there exists c:R+→Rsuch that c(ρ) := sup {γ(u): u∈Kα, β(u)< ρ}. With these definitions, Kβ, ρ α⊂Kγ, c(ρ) αand Kγ, ρ α⊂Kβ, b(ρ) α, in case that the aforementioned functions exist. To prove that the index of some of these subsets is 1or 0, we will use the sufficient conditions given in Lemma 1.2.7. Lemma 8.4.2. Assume that conditions (C1)–(C7)hold. Moreover let there exist ρ > 0such that 0< fρZ∞ −∞ β(k(·, s)η(s)) d s+β(p) ρ<1,(I1 ρ) where fρ= sup f(t, u(t)) ρ:t∈R, u ∈Kα, β(u) = ρ. Then iKα(T, Kβ, ρ α)=1. Proof. We will prove that Tu 6=µ u for all u∈∂Kβ, ρ αand for every µ≥1. Suppose, on the contrary, that there exist some u∈∂Kβ, ρ αand µ≥1such that µ u(t) = Z∞ −∞ k(t, s)η(s)f(s, u(s)) d s+p(t). Then, taking βon both sides and using (C7), we get µ ρ =µ β(u) = β(Tu)≤Z∞ −∞ β(k(·, s)η(s)) f(s, u(s)) d s+β(p) ≤ρfρZ∞ −∞ β(k(·, s)η(s)) d s+β(p) ρ< ρ, which is a contradiction. Therefore, from Lemma 1.2.7, we conclude the veracity of the result. 224 8.4 Fixed Points of Integral Equations Lemma 8.4.3. Assume that conditions (C1)–(C6)and (C8)hold. Moreover, let there exist ρ > 0such that Kγ, ρ αis bounded and fρZ∞ −∞ γ(k(·, s)η(s)) d s+γ(p) ρ>1,(I0 ρ) where fρ= inf f(t, u(t)) ρ:t∈R, u ∈Kα, γ(u) = ρ. Then iKα(T, Kγ, ρ α) = 0. Proof. We will prove that there exists e∈Kγ, ρ α\{0}such that u6=T u +λ e for all u∈∂Kγ, ρ αand all λ > 0. Let us take e=ξin (C8)and suppose, on the contrary, that there exist some u∈∂Kγ, ρ αand λ > 0such that u(t) = Z∞ −∞ k(t, s)η(s)f(s, u(s)) d s+p(t) + λ e(t). Now, taking γon both sides and using (C7)and (C8), ρ=γ(u) = γ(Tu +λ e)≥γ(Tu) + λ γ(e)≥γ(Tu) ≥Z∞ −∞ γ(k(·, s)η(s)) f(s, u(s)) d s+γ(p) ≥ρfρZ∞ −∞ γ(k(·, s)η(s)) d s+γ(p) ρ> ρ, which is a contradiction. The result follows from Lemma 1.2.7. From previous lemmas, it is possible to formulate the following theorem. In this case, we establish conditions to ensure the existence of one or two solutions of the integral equation (8.4.1). However, similar results can be formulated to ensure the existence of three or more solutions. Theorem 8.4.4. Assume that conditions (C1)–(C9)hold. The integral equation (8.4.1) has at least one nontrivial solution in Kαif one of the following conditions holds: (S1)There exist bgiven in condition (C9)and ρ1, ρ2∈(0,∞)with ρ2> b(ρ1) such that (I0 ρ1)and (I1 ρ2)hold. (S2)There exist cgiven in condition (C9)and ρ1, ρ2∈(0,∞)with ρ2> c(ρ1) such that (I1 ρ1)and (I0 ρ2)hold. 225 Existence of Solutions of Integral Equations with Asymptotic Conditions The integral equation (8.4.1) has at least two nontrivial solutions in Kαif one of the following conditions holds: (S3)There exist both band cin condition (C9)and ρ1, ρ2, ρ3∈(0,∞)with ρ2> b(ρ1)and ρ3> c(ρ2)such that (I0 ρ1),(I1 ρ2)and (I0 ρ3)hold. (S4)There exist both band cin condition (C9)and ρ1, ρ2, ρ3∈(0,∞)with ρ2> c(ρ1)and ρ3> b(ρ2)such that (I1 ρ1),(I0 ρ2)and (I1 ρ3)hold. The proof of previous theorem is immediate from Lemmas 8.4.2 and 8.4.3, together with the general properties of fixed point index given in Lemma 1.2.7. Remark 8.4.5. We note that the previous results could also be formulated for either e Cn ϕ([a, ∞)) or e Cn ϕ((−∞, a]) for any a∈R. 8.4.1. An Example: Asymptotic Behavior of a Self Propelled Projectile We will finally apply the theory developed in this section to solve a particular case of the problem of the self propelled projectile which has been formulated in Section 8.2, namely      u00(t) = −gR2 (u(t) + R)2+h(t, u(t)) −ρ(u(t)) u0(t), t ∈[0,∞), u(0) = 0, u0(0) = v0. (8.4.6) As stated in Section 8.2, we will ignore the friction term (the term depending on u0) because it is only related to atmospheric drag and therefore does not affect the asymptotic behavior. Hence, we will study the problem (u00(t) = f(t, u(t)), t ∈[0,∞); u(0) = 0, u0(0) = v0,(8.4.7) with f: [0,∞)×[0,∞)→Rdefined as f(t, y) = −g R2 (y+R)2+h(t, y), where h: [0,∞)×[0,∞)→Rrepresents the propulsion of the projectile. Given the domain of fand hand taking into account Remark 8.4.5, we will work on the interval [0,∞). 226 8.4 Fixed Points of Integral Equations Rewriting (8.4.7) as an integral problem, we know that the solutions of (8.4.7) coincide with the fixed points of the following integral operator, Tu(t) = p(t) + Z∞ 0 k(t, s)f(s, u(s)) d s, (8.4.8) where p(t) = v0t and k(t, s) = (t−s, 0≤s≤t, 0,otherwise, is the corresponding Green’s function. We note that in this case the Green’s function is nonnegative on [0,∞)×[0,∞). We will take h(s, y) = g R2 (y+R)2+y e−s for s, y ∈[0,∞). To ensure the constant sign of f, we extend h(and thus f) in the following way: h(s, y) = g R2 (y+R)2for y < 0. We note that this extension does not have a physical meaning, as we know that the variable ywill never be negative in reality, but it is considered to ensure the applicability of the results in this section in order to solve the problem. We will consider ϕ(t) = t+ 1, and work in the space e Cϕ([0,∞)). Our cone Kα=nu∈e Cϕ([0,∞)) : α(u)≥0o will be defined by the functional α:e Cϕ([0,∞)) −→ R u7−→ α(u) = Z∞ 0 u(t) ϕ2(t)dt−kukϕ3, with ϕ2(t) = C etfor some constant C > 0, which will be calculated later, and ϕ3(t) = et. 227 Existence of Solutions of Integral Equations with Asymptotic Conditions The functional αis well-defined because if u∈e Cϕ([0,∞)), then it holds that u(t)=(t+ 1) ˜u(t), with ˜u∈ C([0,∞],R), which implies that ˜uis uniformly bounded for some constant N. Then, Z∞ 0 u(t) C etdt=Z∞ 0 (t+ 1) ˜u(t) C etdt≤Z∞ 0 t|˜u(t)| C etdt+Z∞ 0 |˜u(t)| C etdt ≤NZ∞ 0 t C etdt+Z∞ 0 1 C etdt=2N C, and sup t∈[0,∞) |u(t)| et≤sup t∈[0,∞) |u(t)| t+ 1 =kukϕ, so α(u)∈Rfor all u∈e Cϕ([0,∞)). Moreover, it is easy to check that αsatisfies properties (P1)–(P3)and therefore the cone Kαis well-defined: (P1)For all u, v ∈e Cϕ([0,∞)), it holds that α(u+v) = Z∞ 0 u(t) + v(t) ϕ2(t)dt−ku+vkϕ3 ≥Z∞ 0 u(t) ϕ2(t)dt+Z∞ 0 v(t) ϕ2(t)dt−kukϕ3−kvkϕ3 =α(u) + α(v). (P2)For all u∈e Cϕ([0,∞)) and λ≥0, α(λ u) = Z∞ 0 λ u(t) ϕ2(t)dt−kλ ukϕ3=λ α(u). (P3)If α(u) = Z∞ 0 u(t) ϕ2(t)dt−kukϕ3≥0 and α(−u) = Z∞ 0 −u(t) ϕ2(t)dt−kukϕ3≥0, then −2kukϕ3≥0. This implies that kukϕ3= 0, which is equivalent to u≡0. We will see now that hypotheses (C1)–(C9)for n= 0 are satisfied: 228 8.4 Fixed Points of Integral Equations (C1)In this case η≡1and k(t, ·)η(·)∈L1([0,∞)) for every t∈[0,∞); indeed Z∞ 0|k(t, s)η(s)|ds=Zt 0 (t−s) d s=t2 2. Moreover, k(·, s)η(s)∈e Cϕ([0,∞)) for every s∈[0,∞). This is deduced from the fact that k(·, s)η(s)∈ C([0,∞)) and there exist both limits lim t→∞ k(t, s)η(s) ϕ(t)= lim t→∞ t−s t+ 1 = 1 and lim t→0 k(t, s)η(s) ϕ(t)= 0. Finally, we will see that last condition in (C1)is satisfied for ω0(s) = 1 + s. Fix ε > 0. Since 1 ϕis a uniformly continuous function, there exists δ < ε such that for |t1−t2|< δ,1 t1+1 −1 t2+1< ε. We will compute now the difference k(t1,s) ϕ(t1)−k(t2,s) ϕ(t2). Fix s∈[0,∞), • If t1, t2> s, then  k(t1, s) ϕ(t1)−k(t2, s) ϕ(t2)= t1−s t1+ 1 −t2−s t2+ 1=−1−s t1+ 1 −−1−s t2+ 1  = (1 + s) 1 t1+ 1 −1 t2+ 1< ε ω0(s). • If t1> s and t2< s, then  k(t1, s) ϕ(t1)−k(t2, s) ϕ(t2)= t1−s t1+ 1< t1−t2 t1+ 1 <ε t1+ 1 < ε < ε ω0(s). • If t1, t2< s, then  k(t1, s) ϕ(t1)−k(t2, s) ϕ(t2)= 0. (C2)By definition of h, we have that f(t, y) = 0 for y < 0and f(t, y) = y e−t≥0 for y≥0. Clearly, f(·, y)is measurable for each fixed y∈Rand f(t, ·)is continuous for a. e. t∈[0,∞). Finally, for each r > 0, f(t, y ϕ(t)) = 0,for all y∈[−r, 0], t ∈[0,∞) 229 Existence of Solutions of Integral Equations with Asymptotic Conditions and f(t, y ϕ(t)) = y ϕ(t)e−t≤r ϕ(t)e−t,for all y∈[0, r], t ∈[0,∞). Therefore condition (C2)is satisfied if we take φr(t) = r ϕ(t)e−t. (C3)For a fixed r > 0, we have that 1 ϕ(t)Z∞ 0|k(t, s)η(s)|φr(s) d s=1 t+ 1 Zt 0 (t−s)r(s+ 1) e−sds =r t+ 1 (−3+2t+e−t(3 + t)), so 1 ϕ(t)Z∞ 0|k(t, s)η(s)|φr(s) d s∈L∞([0,∞)). Moreover, Z∞ 0 ω0(s)φr(s) d s=Z∞ 0 r(s+ 1)2e−sds= 5 r, that is, ω0φr∈L1([0,∞)). Finally, from the limits calculated in (C1)and the expression of Green’s function, we have that z+(s)=1,z−(s)=0and M0(s)=1, so it is clear that |z+|φr,|z−|φr, M0φr∈L1([0,∞)). (C4)It is clear that p(t) = v0t∈e Cϕ([0,∞)) since p∈ C([0,∞)) and there exist both lim t→∞ p(t) ϕ(t)=v0 and lim t→0 p(t) ϕ(t)= 0. (C5)We have to prove that α(k(·, s)) = Z∞ 0 k(τ, s) ϕ2(τ)dτ−kk(·, s)kϕ3≥0for a. e. s∈[0,∞). We have that Z∞ 0 k(τ, s) ϕ2(τ)dτ=Z∞ s τ−s C eτdτ=e−s C. 230 8.4 Fixed Points of Integral Equations On the other hand, fixed s, we have that  k(t, s) et= 0, t ≤s, and  k(t, s) et=t−s et=e−st−s et−s≤e−se−1, t ≥s. Therefore, it is enough to take C≤eto ensure that α(k(·, s)) ≥0. On the other hand, α(p) = Z∞ 0 p(t) ϕ2(t)dt−kpkϕ3=Z∞ 0 v0t C etdt−sup t∈[0,∞) v0t et=v0 C−v0e−1. Therefore, α(p)≥0if and only if C≤e. (C6)By definition, α(Tu) = Z∞ 0 Tu(t) ϕ2(t)dt−kTukϕ3. We have that Z∞ 0 Tu(t) ϕ2(t)dt=Z∞ 0Z∞ 0 k(t, s) ϕ2(t)f(s, u(s)) d s+p(t) ϕ2(t)dt =Z∞ 0Z∞ 0 k(t, s) ϕ2(t)dtf(s, u(s)) d s+Z∞ 0 p(t) ϕ2(t)dt, and kTukϕ3=Z∞ 0 k(·, s)f(s, u(s)) d s+pϕ3 ≤Z∞ 0 k(·, s)f(s, u(s)) d sϕ3 +kpkϕ3 ≤Z∞ 0kk(·, s)kϕ3f(s, u(s)) d s+kpkϕ3, and, consequently, α(Tu)≥Z∞ 0Z∞ 0 k(t, s) ϕ2(t)dtf(s, u(s)) d s −Z∞ 0kk(·, s)kϕ3f(s, u(s)) d s+Z∞ 0 p(t) ϕ2(t)dt−kpkϕ3 =Z∞ 0 α(k(·, s)) f(s, u(s)) d s+α(p). 231 Existence of Solutions of Integral Equations with Asymptotic Conditions Given u∈e Cϕ, we will see that L1u∈e Cϕ. From (e C1), (i), given ε∈R+, there exists some δ∈R+such that for t1, t2∈R, |t1−t2|< δ it is satisfied that g L1u(t1)−g L1u(t2)≤Z∞ −∞ |k(t1, s)η(s)| ϕ(t1)−|k(t2, s)η(s)| ϕ(t2)|u(s)|ds ≤εZ∞ −∞ ω0(s)|u(s)|ds =εZ∞ −∞ ω0(s)|u(s)| ϕ(s)ϕ(s) d s ≤εkukϕZ∞ −∞ ω0(s)ϕ(s) d s (8.5.2) and since, by (e C2),ω0ϕ∈L1(R), the previous expression is bounded from above by εkukϕcfor some positive constant c. Hence, g L1uis continuous in R. Now we will prove that there exists lim t→±∞ g L1u(t)∈R. Indeed, lim t→±∞ g L1u(t) = lim t→±∞ L1u(t) ϕ(t)= lim t→±∞ 1 ϕ(t)Z∞ −∞ |k(t, s)η(s)|u(s) d s. Since k(·, s)η(s)∈e Cϕ, then, for all s∈R, there exists lim t→±∞ |k(t, s)η(s)| ϕ(t)=: z(±)(s)∈R. On the other hand, for all t∈Rand a. e. s∈R, |k(t, s)η(s)| ϕ(t)u(s)≤M0(s)|u(s)|=M0(s)|u(s)| ϕ(s)ϕ(s)≤ kukϕM0(s)ϕ(s) and, from (e C2),M0ϕ∈L1(R). Thus, from Lebesgue’s Dominated Convergence Theorem, lim t→±∞ 1 ϕ(t)Z∞ −∞ |k(t, s)η(s)|u(s) d s=Z∞ −∞ lim t→±∞ |k(t, s)η(s)| ϕ(t)u(s) d s =Z∞ −∞ z(±)(s)u(s) d s, and, since, Z∞ −∞ z(±)(s)u(s) d s≤Z∞ −∞ z(±)(s)|u(s)|ds ≤ kukϕZ∞ −∞ z(±)(s)ϕ(s) d s∈R, 238 8.5 Existence of Solutions via Spectral Theory we deduce that z(±)u∈L1(R). Therefore there exists lim t→±∞ L1u(t) ϕ(t)∈R. Consequently, L1u∈e Cϕ. Step 2: Continuity: It is obvious from the linearity and boundedness of operator L1. Step 3: Compactness: Let B⊂e Cϕbe a bounded set, that is, there exists some R > 0such that kukϕ≤Rfor all u∈B. Then, kL1ukϕ=g L1u∞= L1u ϕ∞ = 1 ϕ(·)Z∞ −∞ |k(·, s)η(s)|u(s) d s∞ ≤kukϕ 1 ϕ(·)Z∞ −∞ |k(·, s)η(s)|ϕ(s) d s∞ ≤R 1 ϕ(·)Z∞ −∞ |k(·, s)η(s)|ϕ(s) d s∞ <∞, (8.5.3) and we have obtained an upper bound which does not depend on u. Therefore it is clear that the set L1(B)is uniformly bounded. On the other hand, taking into account the upper bound found in (8.5.2), we have that if t1, t2∈Rare such that |t1−t2|< δ then g L1u(t1)−g L1u(t2)≤εkukϕZ∞ −∞ ω0(s)ϕ(s) d s≤ε R Z∞ −∞ ω0(s)ϕ(s) d s, and, since ω0ϕ∈L1(R), we conclude that L1(B)is equicontinuous. In conclusion, we derive, by application of Ascoli-Arzela’s Theorem (Theorem 8.3.2), that L1(B)is relatively compact in e Cϕand therefore L1is a compact operator. Step 4: L1maps Pto P∩Kα: Since L1has a positive integral kernel, it clearly maps Pinto P. Finally, it maps Pinto P∩Kαas a direct consequence of hypotheses (e C4)and (e C5). CASE II: n6= 0: We note that in this case we have the additional hypothesis that k(·, s)η(s)is nonnegative for all s∈R. As a consequence, we will omit the absolute value in the definition of L1u. As in Case I, we will divide the proof into four steps. Step 1: L1maps (e Cn ϕ,k·kϕ)to (e Cn ϕ,k·kϕ): Let u∈e Cn ϕ. 239 Existence of Solutions of Integral Equations with Asymptotic Conditions Since k(·,s)η(s) ϕ(·)is integrable for every s∈R, we can use Leibniz’s Integral Rule to get ∂jg L1u ∂tj(t) = ∂j(L1u/ϕ) ∂tj(t) = Z∞ −∞ ∂j(k/ϕ) ∂tj(t, s)η(s)u(s) d s. On the other hand, from (e C1), given ε∈R+, there exists some δ∈R+such that for t1, t2∈R,|t1−t2|< δ it is satisfied that  ∂jg L1u ∂tj(t1)−∂jg L1u ∂tj(t2) ≤Z∞ −∞  ∂j(k/ϕ) ∂tj(t1, s)η(s)−∂j(k/ϕ) ∂tj(t2, s)η(s)|u(s)|ds ≤εZ∞ −∞ ωj(s)|u(s)|ds≤εkukϕZ∞ −∞ ωj(s)ϕ(s) d s. (8.5.4) Since ωjϕ∈L1(R), the previous expression is bounded from above by εkukϕcfor some positive constant c. Hence, ∂jg L1u ∂tjis continuous in Rfor j= 0, . . . , n, that is, g L1u∈ Cn(R,R). Analogously to Case I, it can be proved that there exists lim t→±∞ g L1u(t)and, consequently, L1u∈e Cn ϕ. Step 2: Continuity: Again, it is obvious from the linearity and boundedness of operator L1. Step 3: Compactness: Let B⊂e Cn ϕbe a bounded set, that is, there exists R > 0such that kukϕ≤Rfor all u∈B. We will prove that L1(B)is uniformly bounded. Using the General Leibniz’s Rule (for differentiation), it is clear that ∂jg L1u ∂tj=∂j(L1u/ϕ) ∂tj= j X l=0 j l∂lL1u ∂tl ∂j−l ∂tj−l 1 ϕ. Moreover, from Leibniz’s Integral Rule, ∂lL1u ∂tl(t) = Z∞ −∞ ∂lk ∂tl(t, s)η(s)u(s) d s. 240 8.5 Existence of Solutions via Spectral Theory Thus,  ∂jg L1u ∂tj∞ = j X l=0 j l∂lL1u ∂tl ∂j−l ∂tj−l 1 ϕ∞≤ j X l=0 j l ∂lL1u ∂tl ∂j−l ∂tj−l 1 ϕ∞ = j X l=0 j l ∂j−l ∂tj−l 1 ϕ(·)Z∞ −∞ ∂lk ∂tl(·, s)η(s)u(s) d s∞ . It is satisfied that  ∂j−l ∂tj−l 1 ϕ(t)Z∞ −∞ ∂lk ∂tl(t, s)η(s)u(s) d s ≤ ∂j−l ∂tj−l 1 ϕ(t)Z∞ −∞  ∂lk ∂tl(t, s)η(s)|u(s)|ds ≤R ∂j−l ∂tj−l 1 ϕ(t)Z∞ −∞  ∂lk ∂tl(t, s)η(s)ϕ(s) d s, and so, from two previous inequalities and taking into account condition (e C2), we deduce that  ∂jg L1u ∂tj∞≤R j X l=0 j l ∂j−l ∂tj−l 1 ϕ(·)Z∞ −∞  ∂lk ∂tl(·, s)η(s)ϕ(s) d s∞ <∞. The rest of the proof is analogous to Case I but using equation (8.5.4) instead of (8.5.2). Step 4: L1maps Pto P∩Kα: The proof is the same than in Case I. Theorem 8.5.3. If (e C1),(e C2),(e C6),(e C7)and (e C8)hold, then operator L2is continuous, compact and maps Pinto P∩Kα. Proof. We will distinguish two different cases: CASE I: n= 0: Step 1: L2maps (e Cϕ,k·kϕ)to (e Cϕ,k·kϕ): Let u∈e Cϕ. Since k(·, s)η(s)∈e Cϕfor all s∈R, it is clear that k(·, s)η(s) ϕ(·)+ ≡(k(·, s)η(s))+ ϕ(·)∈ C(R)for all s∈R. 241 Existence of Solutions of Integral Equations with Asymptotic Conditions Reasoning analogously to the proof for L1, from (e C1), (ii), given ε∈R+, there exists some δ∈R+such that for t1, t2∈R,|t1−t2|< δ it is satisfied that g L2u(t1)−g L2u(t2)≤εkukϕZA ω0(s)ϕ(s) d s(8.5.5) and, since ω0ϕ∈L1(R), it can be deduced that g L2uis continuous in R. Now we will see that there exists lim t→±∞ g L2u(t)∈R. We have that lim t→±∞ g L2u(t) = lim t→±∞ L2u(t) ϕ(t)= lim t→±∞ 1 ϕ(t)ZA (k(t, s)η(s))+u(s) d s. Reasoning as before, since k(·, s)η(s)∈e Cϕ, then (k(·, s)η(s))+∈e Cϕand so, for all s∈R, it is ensured the existence of 0≤lim t→±∞ (k(t, s)η(s))+ ϕ(t)≤lim t→±∞ |k(t, s)η(s)| ϕ(t)=z(±)(s)∈R. On the other hand,  (k(t, s)η(s))+ ϕ(t)u(s)≤|k(t, s)η(s)| ϕ(t)|u(s)|≤M0(s)|u(s)|=M0(s)|u(s)| ϕ(s)ϕ(s) ≤ kukϕM0(s)ϕ(s) for all t∈R. From (e C2),M0ϕ∈L1(R)and so M0ϕ∈L1(A). Thus, from Lebesgue’s Dominated Convergence Theorem, lim t→±∞ 1 ϕ(t)ZA (k(t, s)η(s))+u(s) d s=ZA lim t→±∞ (k(t, s)η(s))+ ϕ(t)u(s) d s, and since ZA lim t→±∞ (k(t, s)η(s))+ ϕ(t)u(s) d s≤ZA z(±)(s)|u(s)|ds ≤ kukϕZA z(±)(s)ϕ(s) d s∈R, and z(±)ϕ∈L1(A), it can be concluded that there exists lim t→±∞ L2u(t) ϕ(t)and, consequently, L2u∈e Cϕ. Step 2: Continuity: It is obvious from the linearity and boundedness of operator L2. Step 3: Compactness: 242 8.5 Existence of Solutions via Spectral Theory The proof is analogous to the one for operator L1(Theorem 8.5.2) by considering inequalities We have the following inequalities g L2u∞= L2u ϕ∞ = 1 ϕ(·)ZA (k(·, s)η(s))+u(s) d s∞ ≤kukϕ 1 ϕ(·)ZA (k(·, s)η(s))+ϕ(s) d s∞ ≤kukϕ 1 ϕ(·)ZA|k(·, s)η(s)|ϕ(s) d s∞ ≤kukϕ 1 ϕ(·)Z∞ −∞ |k(·, s)η(s)|ϕ(s) d s∞ , and (8.5.5) instead of (8.5.3) and (8.5.2), respectively. Step 4: L2maps Pto P∩Kα: Since L2has a positive integral kernel, it clearly maps Pinto P. Finally, it maps Pinto P∩Kαas a direct consequence of hypothesis (e C7)and (e C8). CASE II: n6= 0: The proof is analogous to the one made for operator L1, with some small changes in the line of those introduced in Case I. Analogously to the two previous theorems, it can be proved that operator Tsatisfies the following properties. Theorem 8.5.4. If (e C1)–(e C3),(e C9)and (e C10)hold, then operator Tis continuous, compact and maps Kαinto Kα. Proof. The proof, except for the continuity, is completely analogous to previous theorems but using the following inequality f(s, u(s)) = fs, u(s) ϕ(s)ϕ(s)≤φkukϕ(s)ϕ(s)≤φkukϕ∞ϕ(s), instead of u(s)≤ kukϕϕ(s). Continuity: Since Tis not a linear operator, continuity can not be deduced from boundedness, contrary to what we did in previous theorems. Therefore, we shall prove that operator Tis continuous in a different way. Let {un}n∈N⊂e Cϕbe a sequence which converges to uin e Cϕ. Then, there exists some R∈Rsuch that kunkϕ≤Rfor all n∈Nand so it holds that f(s, un(s)) = fs, un(s) ϕ(s)ϕ(s)≤φR(s)ϕ(s)≤ kφRk∞ϕ(s), 243 Existence of Solutions of Integral Equations with Asymptotic Conditions where we have used condition (e C3). Moreover, from (e C3), it holds that f(s, un(s)) →f(s, u(s)) for a. e. s∈R. From (8.4.2), it is clear that, for all t∈Rand j∈ {0, . . . , n},  ∂jg Tun ∂tj(t)−∂jf Tu ∂tj(t)≤Z∞ −∞  ∂j(k/ϕ) ∂tj(t, s)η(s)|f(s, un(s))−f(s, u(s))|ds and, using (˜ C2),  ∂jg Tun ∂tj(t)−∂jf Tu ∂tj(t)≤Z∞ −∞ Mj(s)|f(s, un(s)) −f(s, u(s))|ds ≤2kφRk∞Z∞ −∞ Mj(s)ϕ(s) d s < ∞. Now we deduce, by application of Lebesgue’s Dominated Convergence Theorem, that lim n→∞ ∂jg Tun ∂tj−∂jf Tu ∂tj∞≤lim n→∞Z∞ −∞ Mj(s)|f(s, un(s)) −f(s, u(s))|ds =Z∞ −∞ lim n→∞Mj(s)|f(s, un(s)) −f(s, u(s))|ds= 0. Therefore, we deduce that Tun→Tu in e Cn ϕand so Tis continuous. The following theorem proves that the spectral radius (see Definition 1.3.2) of both operators L1and L2are positive and their related eigenfunctions have constant sign. This result is analogous to [79, Theorem 4.5] and is proven using the facts that the considered operators leave Pinvariant and that Pis a total cone (see Definition 1.2.5), combined with Krein-Rutman Theorem (Theorem 1.3.3). Theorem 8.5.5. Assume that conditions (e C1),(e C2)and (e C4)–(e C8)hold. Then, it holds that r(L1)>0is an eigenvalue of L1with an eigenfunction in P\{0}. Analogously, r(L2)>0is an eigenvalue of L2with an eigenfunction in P\{0}. Proof. We will prove the result for L1. Consider v∈Psuch that v≡1in A. Then, for t∈A, L1v(t) = Z∞ −∞ |k(t, s)η(s)|v(s) d s≥ZA|k(t, s)η(s)|v(s) d s =ZA k(t, s)η(s) d s≥1 f M, 244 8.5 Existence of Solutions via Spectral Theory with 1/f Mgiven in (˜ C6). Then, there exists some open and bounded set B, with A⊂Bsuch that when t∈B,ZA|k(t, s)η(s)|ds≥1 2f M. Now, defining u(t)=1for t∈Aand u(t)=0when t /∈B, from Whitney’s Extension Theorem (see [153, Theorem I]), ucan be extended to R(and this extension will be also denoted by u) as a function of class n. Moreover, from the proof of Whitney’s Extension Theorem, it is possible to deduce that this extension will be nonnegative and upperly bounded by 1. Finally, since u(t) = 0 when t /∈Band Bis a bounded set, then it is clear that lim t→±∞u(t) = 0 and u∈e Cn ϕ, with independence of the choice of ϕ. Therefore, for t∈B, it holds that L1u(t) = Z∞ −∞ |k(t, s)η(s)|u(s) d s≥ZA|k(t, s)η(s)|u(s) d s =ZA|k(t, s)η(s)|ds≥1 2f M≥1 2f Mu(t), and for t /∈B, L1u(t) = Z∞ −∞ |k(t, s)η(s)|u(s) d s≥0 = 1 2f Mu(t). Thus, as a consequence of Theorem 1.3.6, we conclude that r(L1)≥1 2f M>0. Finally, since Pis a total cone and L1maps Pinto P, Krein-Rutman Theorem (Theorem 1.3.3) ensures that r(L1)is an eigenvalue with a related eigenfunction φ∈P\{0}. Remark 8.5.6. As a consequence of Theorems 8.5.2 and 8.5.5, we know that the eigenfunctions mentioned above are in P∩Kα. We will define now the following operator on Cn(A, R) ¯ Lu(t) := ZA k(t, s)η(s)u(s) d s, t ∈A, and consider the cone PAof positive functions in Cn(A, R). As with previous operators, we will prove that ¯ Lsatisfies the following properties. 245 Existence of Solutions of Integral Equations with Asymptotic Conditions Theorem 8.5.7. Assume that conditions (e C1),(e C2)and (e C6)–(e C8)hold. Then, the operator ¯ Lis compact and maps PAinto PA. Proof. Let f∈ Cn(A, R)and B⊂Ran open and bounded set such that A⊂B. Define now g(t) = f(t)for t∈Aand g(t)=0for t∈R\B. Then, from Whitney’s Extension Theorem (see [153, Theorem I]), gcan be extended to Ras a function of class n, that is, there exists an extension of fto Ras a function of class nsuch that this extension vanishes for t∈R\B. Obviously, this extension of fbelongs to Cn ϕ(R). Now, denote by ithe function which maps a function in Cn(A, R)to the aforementioned extension in Cn ϕ(R)and by πthe map which takes every function in Cn ϕ(R)to its restriction to the set A(which clearly belongs to Cn(A, R)). We obtain the following diagram: Cn ϕ(R)Cn ϕ(R) Cn(A, R)Cn(A, R) L2 ¯ L iπ Let us show now that it is commutative. Consider f∈ Cn(A, R). It holds that (π◦L2◦i)(f)(t) = πZA (k(t, s)η(s))+i(f)(s) d s =πZA (k(t, s)η(s))+f(s) d s =ZA k(t, s)η(s)f(s) d s=¯ L(f)(t), t ∈A. Now, since L2is compact and both iand πare continuous, we deduce that ¯ Lis a compact operator. Finally, from (e C6)it is clear that ¯ Lmaps PAinto PA. Remark 8.5.8. We point out that, in the previous proof, Whitney’s extension theorem can be used as a consequence of the fact that Ais a finite union of compact intervals. Theorem 8.5.9. It holds that r(¯ L)>0and it is an eigenvalue of ¯ Lwith an eigenfunction in PA. Proof. Let ψbe the eigenfunction related to L2whose existence is proved in Theorem 8.5.5. Then, if we consider its restriction to A,ψ|A, it is clear that for t∈A ¯ Lψ|A(t) = L2ψ(t) = r(L2)ψ(t) = r(L2)ψ|A(t), 246 8.5 Existence of Solutions via Spectral Theory and so from Theorems 1.3.6 and 8.5.5, we deduce that r(¯ L)≥r(L2)>0. We define the following numbers in the extended real line: f0= lim x→0 sup t∈R f(t, x ϕ(t)) ϕ(t) |x|, f0= lim x→0 inf t∈A f(t, x ϕ(t)) ϕ(t) |x|, f∞= lim |x|→+∞ sup t∈R f(t, x ϕ(t)) ϕ(t) |x|, f∞= lim |x|→+∞ inf t∈A f(t, x ϕ(t)) ϕ(t) |x|. Next, we will give a result in which we will prove that, under suitable conditions, the index of some subsets is 1or 0. Before that, we shall give the following definition that will be implicitly used in Theorem 8.5.11. Definition 8.5.10. Let, X, Y, Z be topological spaces, YHausdorff. Let f:X→Y, g:X→Z. Let z0∈g(X)0. We say that Lis the limit of fwhen g(x)tends to z0 if for every neighborhood NYof Lthere exists a neighborhood NZof z0such that fg−1(NZ\{z0})⊂NY. We write lim g(x)→z0 f(x) = L. A particular case of this definition would be the notion of limit in the case of the topology occurring when studying Stieltjes derivatives with respect to a function g (cf. [59,120]). Now, in order to prove the following theorem, we adapt some of the proofs of [151, Theorems 3.2-3.5] to this new context. In particular, to prove that the index of some subsets of a cone is 1or 0, we will use the sufficient conditions given in Lemma 1.2.7. Theorem 8.5.11. Assume that hypotheses (e C1)–(e C10)hold. Assume also that there exists β:Cn ϕ→[0,∞)such that lim β(u)→0kukϕ= 0,lim β(u)→+∞kukϕ= +∞, and β(u)6= 0 ⇒u6≡ 0. Consider Kβ, ρ α:= {u∈Kα:β(u)< ρ}. Then, the following assertions hold: (1) If 0≤f0< µ(L1), then there exists ρ0>0such that iKα(T, Kβ, ρ α) = 1 for each ρ∈(0, ρ0]. 247 Existence of Solutions of Integral Equations with Asymptotic Conditions 8.5.1. An Example Consider the problem Tu(t) = Z∞ −∞ e−|s| 2sin tp|u(s)|sin2sds, that is, k(t, s) = e−|s| 2sin t,η≡1and f(s, y) = p|y|sin2s. We will take ϕ(t) = |t|, and α(u) = min t∈[π 4,3π 4]u(t)−√2 2kuk∞. We will verify that conditions (e C1)–(e C10)are satisfied for the case n= 0: (e C1)First of all, since k(·, s)∈ C(R)and there exist lim t→±∞ k(t, s) ϕ(t)= lim t→±∞ e−|s| 2sin t |t|= 0, it is clear that k(·, s)∈e Cϕfor all s∈R. Moreover, for every ε > 0there exists δ > 0such that when |t1−t2|< δ, (i)  k(t1, s) ϕ(t1)−k(t2, s) ϕ(t2)= e−|s| 2sin t1 |t1|−e−|s| 2sin t2 |t2|≤ε e−|s| 2, and (ii)  (k(t1, s))+ ϕ(t1)−(k(t2, s))+ ϕ(t2)= e−|s| 2(sin t1)+ |t1|−e−|s| 2(sin t2)+ |t2|≤ε e−|s| 2, so we will take ω0(s) = e−|s| 2. 254 8.5 Existence of Solutions via Spectral Theory (e C2)Clearly, it holds that ω0ϕ∈L1(R). Also, 1 ϕ(t)Z∞ −∞ |k(t, s)|ϕ(s) d s=|sin t| |t|Z∞ −∞ e−|s| 2|s|ds= 8 |sin t| |t|∈L∞(R). Moreover, in this case z(±)(s) = lim t→±∞ e−|s| 2|sin t| |t|= 0, M0(s) = sup t∈R e−|s| 2|sin t| |t|=e−|s| 2, and it holds that z(±)ϕ, M0ϕ∈L1(R). (e C3)It is clear that f(·, y)is measurable for each fixed y∈Rand f(t, ·)is continuous for a. e. t∈R. Finally, for each r > 0, there exists φr(t) = √rsin2t p|t|∈L∞(R) such that f(t, x ϕ(t)) ϕ(t)=p|x t|sin2t |t|≤φr(t), for all x∈[−r, r]and a. e. t∈R. (e C4)In this case, α(|k(·, s)|) = min t∈[π 4,3π 4]|k(t, s)|− √2 2kk(·, s)k∞ =e−|s| 2min t∈[π 4,3π 4]|sin t|− √2 2e−|s| 2= 0. (e C5)It is clear that α(|k(·, s)|)ϕ(s)∈L1(R). Moreover, for all u∈P, it holds that α(L1u) = min t∈[π 4,3π 4]Z∞ −∞ |k(t, s)|u(s) d s−√2 2Z∞ −∞ |k(·, s)|u(s) d s∞ ≥Z∞ −∞ min t∈[π 4,3π 4]|k(t, s)|u(s) d s−√2 2Z∞ −∞ kk(·, s)k∞u(s) d s =Z∞ −∞ α(|k(·, s)|)u(s) d s. 255 Existence of Solutions of Integral Equations with Asymptotic Conditions (e C6)We can take A=π 4,3π 4. For such A, we obtain 1 f M(A)= inf t∈AZA e−|s| 2sin tds= inf t∈An2e−3π 8−1 + eπ 4sin to =√2e−3π 8−1 + eπ 4>0. (e C7)It is analogous to (e C4). The same occurs to (e C9). (e C8)It is analogous to (e C5). The same occurs to (e C10). Finally, we obtain the following values for the limits f∞and f0: f∞= lim |x|→+∞ sup t∈Rp|x|sin2t p|t| |x|≤lim |x|→+∞p|x| |x|= 0, and so f∞= 0. Analogously, f0= lim |x|→0 inf t∈Ap|x|sin2t p|t| |x|= lim |x|→0p|x| √3π|x|= +∞. On the other hand, since both r(L1)and r(L2)are positive (as it has been proved in Theorem 8.5.5), it holds that µ(L1)>0and µ(L2)<+∞. Thus, from (T2)in Theorem 8.5.12, we deduce that our problem has at least a nontrivial solution in Kα⊂e Cϕ. Remark 8.5.15. Note that, as it has been indicated before, the results in Section 8.4 are not applicable to this problem as the nonlinearity fdoes not satisfy condition (C2). 256 Chapter 9 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian In this chapter, we will study the existence of unbounded solutions of a singular nonlinear initial value problem with a φ-Laplacian. The same problem has already been considered in [13], where the authors discussed the existence and properties of bounded solutions. Here we will focus our attention on unbounded solutions of the problem and provide sufficient conditions for their existence. This way, this chapter completes the results obtained in [13]. Contrary to previous chapters, since the problem is singular, it is not possible to construct an equivalent integral problem with a kernel given by a related Green’s function. As a consequence, the techniques developed in this chapter are completely different to the ones used previously in this Thesis. All the results in this chapter are collected in [131]. 9.1. Introduction The aim of this chapter is to analyse the singular nonlinear equation (p(t)φ(u0(t)))0+p(t)f(φ(u(t))) = 0, t > 0,(9.1.1) with the initial conditions u(0) = u0, u0(0) = 0,(9.1.2) where u0∈[L0, L]. A special case of equation (9.1.1) with φ(u)≡uand p(t) = tn−1,n∈N,n≥2, tn−1u0(t)0+tn−1f(u(t)) = 0, t > 0, arises in many areas. For example, in the study of phase transition of Van der Waals fluids ( [60]), in population genetics, where it serves as a model for the spatial distribution of the genetic composition of a population ([55]), in the homogeneous nucleation theory ([1]), in the relativistic cosmology for description of particles which 257 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian can be treated as domains in the universe ([100]), or in the nonlinear field theory, in particular, when describing bubbles generated by scalar fields of the Higgs type in the Minkowski spaces ([47]). The above nonlinear equation was replaced with its abstract and more general form p(t)u0(t)0+q(t)f(u(t)) = 0, t > 0, which was investigated for p=qin [122, 123, 125–128] and for p6=qin [12, 14, 130,144]. Other problems without φ-Laplacian close to (9.1.1)–(9.1.2) can be found in [2,8,10,90–92] and those with φ-Laplacian in [49,82,107,108,137]. Before precising what the main objectives of this chapter are, we need to define what we understand by solution of problem (9.1.1)–(9.1.2). Definition 9.1.1. Let [0, b)⊂[0,∞)be a maximal interval such that a function u∈C1([0, b)) with φ(u0)∈C1((0, b)) satisfies equation (9.1.1) for every t∈(0, b) and let usatisfy the initial conditions (9.1.2). Then uis called a solution of problem on [0, b). If uis a solution of problem (9.1.1)–(9.1.2) on [0,∞), then uis called a solution of problem (9.1.1)–(9.1.2). In particular, following the line of [13], we will distinguish three different types of solutions. Definition 9.1.2. Consider a solution of problem (9.1.1)–(9.1.2) with u0∈(L0, L) and denote usup = sup{u(t): t∈[0,∞)}. If usup =L, then uis called a homoclinic solution of problem (9.1.1)–(9.1.2). If usup < L, then uis called a damped solution of problem (9.1.1)–(9.1.2). Solutions from Definition 9.1.2 are bounded. Therefore, we are mostly interested in another type of solutions specified in the next definition. Definition 9.1.3. Let ube a solution of problem (9.1.1)–(9.1.2) on [0, b), where b∈(0,∞]. If there exists some c∈(0, b)such that u(c) = L, u0(c)>0,(9.1.3) then uis called an escape solution of problem (9.1.1)–(9.1.2) on [0, b). The three considered types of solutions can be seen in Figure 9.1.1. As we have mentioned before, analytical properties of the solutions of problem (9.1.1)–(9.1.2) with a φ-Laplacian have been already studied in [13], with a focus on 258 9.1 Introduction 0t L L0 damped solutions homoclinic solution escape solution Figure 9.1.1: Types of solutions of problem (9.1.1)–(9.1.2). the existence of bounded solutions on [0,∞). In particular, the existence of damped solutions was proved for certain values of u0. Some results derived in [13] will also be useful here when the existence and properties of unbounded solutions are of interest. Therefore, we will recapitulate them in Section 9.2. The goal of this chapter is to find conditions which guarantee the existence of escape solutions of problem (9.1.1)–(9.1.2) which are unbounded. Note that the analysis of problem (9.1.1)–(9.1.2) with a general φ-Laplacian includes, for example, φ(x) = |x|αsign x, for α > 1. Let us emphasize that in this case, φ−1(x) = |x|1 αsign xis not locally Lipschitz continuous. Since φ−1is present in the operator form of (9.1.1)–(9.1.2), namely u(t) = u0+Zt 0 φ−1−1 p(s)Zs 0 p(τ)f(φ(u(τ))) d τds, t ≥0, the standard technique based on the Lipschitz property is not applicable here and another approach needs to be developed. Therefore, we will distinguish two cases: In the first case, where functions φ−1and fare Lipschitz continuous, the uniqueness of solution of problem (9.1.1)–(9.1.2) is guaranteed. In this case, we will obtain a sequence of escape solutions with different initial values. In the second case, functions φ−1and fdo not have to be Lipschitz continuous. The lack of uniqueness causes difficulties and therefore is more challenging. The problems are overcome by means of the lower and upper solutions method. Also here sufficient conditions for the existence of escape solutions are derived. 259 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian However, contrary to previous case, it may occur in this one that all the escape solutions have the same initial value L0. Moreover, since, in general, an escape solution does not need to be unbounded, criteria for an escape solution to tend to infinity will be derived. In this manner, we will obtain new existence results for unbounded solutions of problem (9.1.1)–(9.1.2). This chapter is organized in the following way: Preliminary results for an auxiliary problem with a bounded nonlinearity are stated in Section 9.2. Auxiliary lemmas necessary for proofs of the existence of escape solutions of the auxiliary problem are given in Section 9.3. The existence of escape solutions of this problem is further discussed in Section 9.4. Namely, the first existence result in Section 9.4 is derived by an approach based on the Lipschitz property. The other case without the Lipschitz condition is studied by means of the lower and upper solutions method. In Section 9.5, several criteria for escape solutions of the original problem to be unbounded are proved. The main results about the existence of unbounded solutions with examples are given in Section 9.6. 9.2. Preliminaries Throughout this chapter, we will make the following basic assumptions: φ∈C1(R)is a Laplacian, that is, φ(R) = R, φ(0) = 0 and φ0(x)>0for x∈(R\{0}).(B1) L0<0< L and the nonlinearity fsatisfies the following properties f∈C[φ(L0),∞), f(φ(L0)) = f(0) = f(φ(L)) = 0.(B2) Moreover, foscillates in the following way: xf(x)>0for x∈(φ(L0), φ(L)) \{0}, f(x)≤0for x > φ(L).(B3) Finally, p∈C[0,∞)∩C1(0,∞)is an increasing function, that is, p(0) = 0 and p0(t)>0for t∈(0,∞).(B4) Moreover, in order to derive the main existence results about unbounded solutions of problem (9.1.1)–(9.1.2), we introduce the following auxiliary equation with a bounded nonlinearity (p(t)φ(u0(t)))0+p(t)e f(φ(u(t))) = 0, t ∈(0,∞),(9.2.1) 260 9.2 Preliminaries where e f(x) = (f(x)for x∈[φ(L0), φ(L)], 0for x < φ(L0), x > φ(L).(9.2.2) Since e fis bounded on R, the maximal interval of existence for each solution of problem (9.2.1), (9.1.2) is [0,∞). In this section, we collect preliminary results for solutions of problem (9.2.1), (9.1.2) derived in [13]. Properties, asymptotic behaviour and a priori estimates of such solutions are specified in Lemmas 9.2.1–9.2.9. The existence and continuous dependence on initial values of solutions are provided in Theorems 9.2.10 and 9.2.12, respectively. Lemma 9.2.1 ( [13, Lemma 2.1 b) ]).Let (B1)–(B4)hold and let ube a solution of equation (9.2.1). Assume that there exists a≥0such that u(a)∈(L0,0) and u0(a) = 0. Then u0(t)>0for t∈(a, θ], where θis the first zero of uon (a, ∞).If such θdoes not exist, then u0(t)>0for t∈(a, ∞). Lemma 9.2.2 ( [13, Lemma 2.2]).Let (B1)–(B4)hold and let ube a solution of equation (9.2.1). Assume that there exists a≥0such that u(a) = Land u0(a) = 0. a) Let θ > a be the first zero of uon (a, ∞). Then there exists a1∈[a, θ)such that u(a1) = L, u0(a1) = 0,0≤u(t)< L, u0(t)<0, t ∈(a1, θ]. b) Let u > 0on [a, ∞)and u6≡ Lon [a, ∞). Then there exists a1∈[a, ∞)such that u(a1) = L, u0(a1) = 0,0< u(t)< L, u0(t)<0, t ∈(a1,∞). In both cases, u(t) = Lfor t∈[a, a1]. Lemma 9.2.3 ([13, Lemma 2.6]).Assume (B1)–(B4), lim t→∞ p0(t) p(t)= 0,(9.2.3) and ∃¯ B∈(L0,0): e F¯ B=e F(L),where e F(x) = Zx 0e f(φ(s)) d s, x ∈R.(9.2.4) Let ube a solution of equation (9.2.1) and let there exist b≥0and θ > b such that u(b)∈[¯ B, 0), u0(b) = 0, u(θ) = 0, u(t)<0, t ∈[b, θ).(9.2.5) 261 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Then there exists a∈(θ, ∞)such that u0(a) = 0, u0(t)>0, t ∈(b, a), u(a)∈(0, L). Lemma 9.2.4 ( [13, Lemma 2.7]).Assume that hypotheses (B1)–(B4),(9.2.3) and (9.2.4) hold. Let ube a solution of equation (9.2.1) and let a≥0and θ > a be such that u(a)∈(0, L], u0(a) = 0, u(θ) = 0, u(t)>0, t ∈[a, θ).(9.2.6) Then there exists b∈(θ, ∞)such that u0(b) = 0, u0(t)<0, t ∈(a, b), u(b)∈(¯ B, 0). Lemma 9.2.5 ([13, Lemma 2.8]).Assume that (B1)–(B4)and (9.2.3) hold. Let ube a solution of equation (9.2.1) and let b≥0be such that u(b)∈(L0,0), u0(b) = 0, u(t)<0, t ∈[b, ∞). Then lim t→∞u(t)=0,lim t→∞u0(t)=0. Lemma 9.2.6 ([13, Lemma 2.9]).Assume that (B1)–(B4)and (9.2.3) hold. Let ube a solution of equation (9.2.1) and let a≥0be such that u(a)∈(0, L], u0(a) = 0, u(t)>0, t ∈[a, ∞). Then either u(t) = L, t ∈[a, ∞) or lim t→∞u(t)=0,lim t→∞u0(t)=0. Lemma 9.2.7 ( [13, Lemma 3.1]).Assume that hypotheses (B1)–(B4),(9.2.3) and (9.2.4) hold. Let ube a solution of problem (9.2.1),(9.1.2) with u0∈L0,¯ B. Let there exist θ > 0,a > θ such that u(θ) = 0, u(t)<0, t ∈[0, θ) and u0(a) = 0, u0(t)>0, t ∈(θ, a). Then u(a)∈(0, L]and u0(t)>0, t ∈(0, a). 262 9.2 Preliminaries Lemma 9.2.8 ( [13, Lemma 3.2]).Let assumptions (B1)–(B4),(9.2.3) and (9.2.4) hold. Let ube a solution of problem (9.2.1),(9.1.2) with u0∈(L0,0) ∪(0, L). Then u0∈¯ B, 0∪(0, L)implies that ¯ B < u(t)< L, t ∈(0,∞) and u0∈L0,¯ Bimplies that u0< u(t), t ∈(0,∞). For the following result, we introduce a function ϕdefined as ϕ(t) :=      1 p(t)Zt 0 p(s) d s, t ∈(0, T] 0, t = 0. (9.2.7) This function is continuous on [0, T]and satisfies that 0< ϕ(t)≤t, t ∈(0, T]and lim t→0+ϕ(t) = 0.(9.2.8) Moreover, since e fis bounded, there exists some constant f M > 0such that |e f(x)| ≤ f M, ∀x∈R.(9.2.9) Lemma 9.2.9 ([13, Lemma 3.4]).Assume (B1)–(B4). Let ube a solution of problem (9.2.1),(9.1.2) with u0∈[L0, L]. The inequality Zβ 0 p0(t) p(t)φ(u0(t))dt≤f M(β−ϕ(β)) , with ϕgiven in (9.2.7), is valid for every β > 0. If moreover (9.2.3) and (9.2.4) hold, then there exists ec > 0such that |u0(t)| ≤ ec, t ∈[0,∞), for every solution uof (9.2.1),(9.1.2) with u0∈(L0,0) ∪(0, L). The existence of solutions of the auxiliary problem (9.2.1), (9.1.2) is proved in [13] by means of the Schauder’s fixed point Theorem. We state this existence result in the next theorem. Theorem 9.2.10 ([13, Theorem 4.1]).Assume that conditions (B1)–(B4)hold. Then, for each u0∈[L0, L], there exists a solution uof problem (9.2.1),(9.1.2). 263 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian The previous lemma gives a powerful tool to ensure the existence of escape solutions with u0∈(L0, C). However, if φ−1and fare not Lipschitz continuous, then problem (9.2.1), (9.1.2) with u0∈[L0, L]\{0}can have multiple solutions. These solutions may be escape solutions. In particular, more solutions can start at L0, not only the constant solution u≡L0. Therefore, we need to extend the assertions of Lemma 9.3.2, which deal with values greater than L0, to the case u0=L0. For this purpose next two lemmas will be helpful. Lemma 9.3.3. Let (B1)–(B4)hold and let ube a solution of problem (9.2.1),(9.1.2) such that u0=L0, u 6≡ L0, u(t)≥L0for t∈[0,∞).(9.3.13) Then there exists a≥0such that u(t) = L0for t∈[0, a](9.3.14) and u0(t)>0for t∈(a, θ], where θis the first zero of uin (a, ∞). If such θdoes not exist, then u0(t)>0for t∈(a, ∞). If θ∈(a, ∞)and there exist a1> θ such that u0(a1) = 0 and u0(t)>0, t ∈(θ, a1),(9.3.15) then u(a1)∈(0, L]. Proof. By (9.3.13), there exists τ > 0such that L0< u(τ)<0. Put a:= inf {τ > 0 : L0< u(τ)<0}. Then ufulfils (9.3.14) and u0(a) = 0. Put θ:= sup {τ > a :L0< u(τ)<0}. Then p(t)e f(φ(u(t))) <0, t ∈(a, θ).(9.3.16) 270 9.3 Auxiliary Results Integrating equation (9.2.1) over [a, t], we get, by (9.3.16), p(t)φu0(t)=−Zt a p(s)e f(φ(u(s))) d s > 0, t ∈(a, θ)(9.3.17) and, since p(t)>0, necessarily u0(t)>0for t∈(a, θ). If θ=∞, then the proof is finished. On the other hand, if θ < ∞, then θis the first zero of uon (a, ∞)and (9.3.17) yields u0(θ)>0. Let θ∈(a, ∞)and a1> θ such that (9.3.15) holds. Since u(θ) = 0 and u0(t)>0on (θ, a1), then u(a1)>0. Assume that u(a1)> L. Then there exists a0∈(θ, a1)such that u > L on (a0, a1]. Integrating equation (9.2.1) over (a0, a1) and using (9.2.2), we obtain p(a0)φ(u0(a0)) −p(a1)φ(u0(a1)) = Za1 a0 p(s)e f(φ(u(s))) d s= 0, and so, p(a0)φ(u0(a0)) = 0. Consequently, u0(a0) = 0, which contradicts that u0>0on (θ, a1). We have proved that u(a1)≤L, which completes the proof. Lemma 9.3.4. Let (B1)–(B4)and (9.2.3) hold and let ube a solution of (9.2.1), (9.1.2) satisfying that u0=L0, u 6≡ L0, u(t)≥L0for t∈[0,∞). Assume that u(t)<0, t ∈[0,∞). Then lim t→∞u(t)=0,lim t→∞u0(t)=0. Proof. By Lemma 9.3.3, there exists a≥0such that u(t) = L0for t∈[0, a]and u0(t)>0for t∈(a, ∞). Hence, uis increasing on (a, ∞)and so L0< u(t)<0, t ∈(a, ∞) and there exists lim t→∞u(t) =: `∈(L0,0]. Now, if uis a solution of (9.2.1), then φ0(u0(t)) u00(t) + p0(t) p(t)φ(u0(t)) + e f(φ(u(t))) = 0, t ∈(0,∞).(9.3.18) 271 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Multiplying previous equation by u0and integrating it from ato t, we obtain ψ1(t) + ψ2(t) + ψ3(t) = 0, t ∈(a, ∞), where ψ1(t) = Zu0(t) u0(a) x φ0(x) d x, ψ2(t) = Zt a p0(s) p(s)φ(u0(s)) u0(s) d s, ψ3(t) = Zu(t) u(a)e f(φ(x)) d x. It holds that ψ3(t) = e F(u(t)) −e F(u(a)). Since e F(x)is decreasing for x∈(L0,0) and uis increasing on (a, ∞),e F(u(t)) is decreasing for t∈(a, ∞)and so lim t→∞ e F(u(t)) = e F(`). Therefore, lim t→∞ψ3(t) =: Q3∈−e F(L0),0. On the other hand, since φ1is positive on (a, ∞), it occurs that ψ2(t)< ψ3(t) for t∈(a, ∞). This way, since ψ2is continuous, increasing and positive on (a, ∞), it holds that lim t→∞ψ2(t) =: Q2∈(0,−Q3]. As a consequence, we get that lim t→∞ψ1(t) =: Q1∈h0,−e F(L0). Thus, defining Φ(z) := Zz 0 x φ0(x) d x, it occurs that lim t→∞Φ(u0(t)) = Q1. Moreover, since Φis positive, continuous and increasing on (0,∞), its inverse Φ−1is also positive, continuous and increasing. Consequently, lim t→∞u0(t) = lim t→∞Φ−1Φu0(t)= Φ−1(Q1)≥0 272 9.4 Existence of Escape Solutions and, since there exists lim t→∞u(t) =: `∈(L0,0], we conclude that lim t→∞u0(t)=0. Now, assume that `6= 0. Then, taking the limit when tgoes to ∞in (9.3.18) and using (9.2.3), we obtain φ0(0) lim t→∞u00(t) = −e f(φ(l)). Since −e f(φ(l)) ∈(−∞,0), then necessarily lim t→∞u00(t)>0, which is a contradiction with lim t→∞u0(t) = 0. Therefore, `= 0 and the result is proved. Lemma 9.3.5 (Basic Lemma II).Let (B1)–(B4),(9.2.3) and (9.2.4) hold. Choose C∈(L0,¯ B). For each n∈N, let unbe a solution of problem (9.2.1),(9.1.2) with u0=L0and let (an, bn)be the maximal interval such that L0< un(t)< L and u0 n(t)>0, t ∈(an, bn). Finally, let γn∈(an, bn)be such that un(γn) = C. If the sequence {γn}∞ n=1 is unbounded, then the sequence {un}∞ n=1 contains an escape solution of problem (9.2.1),(9.1.2) with u0=L0. Proof. The proof is held in an analogous way to the proof of Lemma 9.3.2 where in Step 1, Lemmas 9.3.3 and 9.3.4 are used instead of Lemmas 9.2.1 and 9.2.5, respectively. 9.4. Existence of Escape Solutions This section is devoted to prove the existence of escape solutions of problem (9.2.1), (9.1.2). First, we will discuss the existence of escape solutions provided the Lipschitz continuity of φ−1and f. For this purpose we choose a sequence of solutions which converges locally uniformly to the constant solution u≡L0. In this manner we obtain an unbounded sequence {γn}∞ n=1 required in the Basic Lemma I (Lemma 9.3.2) for the existence of an escape solution. This approach fails without the assumption on the Lipschitz condition. This situation is subject of investigation in the rest of this section. In particular, we will solve this problem with the lower and upper solutions method. 273 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Theorem 9.4.1 (Existence of escape solutions of problem (9.2.1), (9.1.2) I).Assume that (B1)–(B4),(9.2.3),(9.2.4),(9.2.10) and (9.2.11) hold. Then there exist infinitely many escape solutions of problem (9.2.1),(9.1.2) with different starting values in L0,¯ B. Proof. Choose n∈N,C∈L0,¯ Band Bn∈(L0, C). By Theorems 9.2.10 and 9.2.12, there exists a unique solution unof problem (9.2.1), (9.1.2) with u0=Bn. By Lemma 9.2.1, there exists a maximal an>0such that u0 n>0on (0, an). Since un(0) <0, there exists a maximal ean>0such that un< L on [0,ean). If we put bn= min{an,ean}, then un(t)<0, u0 n(t)>0, t ∈(0, bn). Further, due to Lemmas 9.2.1 and 9.2.5, either lim t→∞un(t) = 0 or unhas a zero θn∈(0, bn). Consequently, there exists γn∈(0, bn)satisfying that un(γn) = C. This way, from the sequence {Bn}∞ n=1 ⊂(L0, C), we get the sequence {un}∞ n=1 of solutions of problem (9.2.1), (9.1.2) with u0=Bn, and the corresponding sequence of {γn}∞ n=1. Assume that lim n→∞Bn=L0. Then, by Theorem 9.2.12, the sequence {un}∞ n=1 converges locally uniformly on [0,∞)to the constant function u≡L0. Therefore, limn→∞ γn=∞and the sequence {γn}∞ n=1 is unbounded. Thus, by Lemma 9.3.2 there exists n0∈Nsuch that un0is an escape solution of problem (9.2.1), (9.1.2). We have un0(0) = Bn0> L0. Now, consider the unbounded sequence {γn}∞ n=n0+1. By Lemma 9.3.2 there exists n1∈Nsuch that un1is an escape solution of problem (9.2.1), (9.1.2) such that un1(0) = Bn1> L0. Repeating this procedure, we obtain the sequence {unk}∞ k=0 of escape solutions of problem (9.2.1), (9.1.2). Remark 9.4.2. We note that the proof of previous theorem does not remain valid if we eliminate hypotheses (9.2.10) and (9.2.11). The reason is that, without these hypotheses, we do not have uniqueness of solution. Then, in the previous proof, lim n→∞Bn=L0implies that the sequence {un}∞ n=1 converges locally uniformly on [0,∞)to a function usuch that u(0) = L0. However, since there is no uniqueness of solution, we can not affirm that u≡L0and so we can not ensure that the sequence {γn}∞ n=1 is unbounded. Since previous method is not valid in case φ−1and fare not Lipschitz continuous, we need to find an alternative approach to investigate the existence of escape solutions in such a case. In order to prove this existence result, we consider the lower 274 9.4 Existence of Escape Solutions and upper solutions method for an auxiliary mixed problem on [0, T]. In particular, we will use this method to find solutions of (9.2.1) which satisfy that u0(0) = 0, u(T) = C, C ∈[L0, L].(9.4.1) Definition 9.4.3. A function u∈ C1([0, T ]) with φ(u0)∈ C1((0, T]) is a solution of problem (9.2.1),(9.4.1) if ufulfills (9.2.1) for t∈(0, T ]and satisfies (9.4.1). Definition 9.4.4. A function σ1∈ C([0, T]) is a lower solution of problem (9.2.1), (9.4.1) if there exists a finite (possibly empty) set Σ1⊂(0, T)such that σ1∈ C2((0, T]\Σ1)and p(t)φ(σ0 1(t))0+p(t)e f(φ(σ1(t))) ≥0, t ∈(0, T]\Σ1,(9.4.2) −∞ < σ0 1(τ−)< σ0 1(τ+)<∞, τ ∈Σ1,(9.4.3) σ0 1(0+)≥0, σ1(T)≤C. (9.4.4) Analogously, Definition 9.4.5. A function σ2∈ C([0, T ]) is an upper solution of problem (9.2.1), (9.4.1) if there exists a finite (possibly empty) set Σ2⊂(0, T)such that σ2∈ C2((0, T]\Σ2)and p(t)φ(σ0 2(t))0+p(t)e f(φ(σ2(t))) ≤0, t ∈(0, T]\Σ2,(9.4.5) −∞ < σ0 2(τ+)< σ0 2(τ−)<∞, τ ∈Σ2,(9.4.6) σ0 2(0+)≤0, σ2(T)≥C. (9.4.7) Theorem 9.4.6 (Lower and upper solutions method).Let (B1)–(B4)hold and let σ1 and σ2be lower and upper solutions of problem (9.2.1),(9.4.1) such that σ1(t)≤σ2(t), t ∈[0, T]. Then problem (9.2.1),(9.4.1) has a solution usuch that σ1(t)≤u(t)≤σ2(t), t ∈[0, T]. Proof. The proof is divided into two steps. Step 1: Construction of an auxiliary problem and its solvability. For t∈[0, T]and x∈Rwe define the following auxiliary nonlinearity f∗(t, x) =         e f(φ(σ1(t))) + σ1(t)−x σ1(t)−x+1, x < σ1(t), e f(φ(x)), σ1(t)≤x≤σ2(t), e f(φ(σ2(t))) −x−σ2(t) x−σ2(t)+1 , x > σ2(t). 275 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Note that f∗is bounded, that is, there exists M∗>0such that |f∗(t, x)| ≤ M∗,∀(t, x)∈[0, T]×R.(9.4.8) Consider the auxiliary equation p(t)φ(u0(t))0+p(t)f∗(t, u(t)) = 0, t ∈(0, T].(9.4.9) Integrating (9.4.9), we get the equivalent form of problem (9.4.9), (9.4.1): u(t) = C−ZT t φ−1−1 p(s)Zs 0 p(τ)f∗(τ, u(τ)) d τds, t ∈[0, T]. Now, consider the Banach space C([0, T ]) with the maximum norm and define an operator F:C([0, T]) → C([0, T]) in the following way: (Fu)(t) := C−ZT t φ−1−1 p(s)Zs 0 p(τ)f∗(τ, u(τ)) d τds. Put Λ := max{|L0|, L}and consider the ball B(0, R) = u∈ C([0, T]): kukC([0,T ]) ≤R, where R:= Λ + T φ−1(M∗T)and M∗is the upper bound given in (9.4.8). Since φ is increasing on R,φ−1is also increasing on Rand, by (9.2.8), φ−1(M∗ϕ(t)) ≤φ−1(M∗T), t ∈[0, T], where ϕis defined in (9.2.7). Then, the norm of Fucan be estimated as follows kFukC([0,T ]) = max t∈[0,T]C−ZT t φ−1−1 p(s)Zs 0 p(τ)f∗(τ, u(τ)) d τds ≤Λ + ZT tφ−1(M∗ϕ(s))ds≤Λ + ZT t φ−1(M∗T) d s ≤Λ + T φ−1(M∗T) = R, which yields that Fmaps B(0, R)to itself. Let us prove that Fis compact on B(0, R). First, we will show that Fis continuous. Choose a sequence {un} ⊂ C([0, T]) such that limn→∞ kun−ukC([0,T]) = 0. We have that (Fun)(t)−(Fu)(t) = −ZT tφ−1−1 p(s)Zs 0 p(τ)f∗(τ, un(τ)) d τ +φ−1−1 p(s)Zs 0 p(τ)f∗(τ, u(τ)) d τds. 276 9.4 Existence of Escape Solutions Since f∗is continuous on [0, T]×R, we get lim n→∞kf∗(·, un(·)) −f∗(·, u(·))kC([0,T ]) = 0. Now, for n∈N, define An(t) :=      −1 p(t)Zt 0 p(τ)f∗(τ, un(τ)) d τ, t ∈(0, T], 0, t = 0 and A(t) :=      −1 p(t)Zt 0 p(τ)f∗(τ, u(τ)) d τ, t ∈(0, T], 0, t = 0. Then, for a fixed n∈N, |An(t)−A(t)|= 1 p(t)Zt 0 p(τ) (f∗(τ, u(τ)) −f∗(τ, un(τ))) d τ, t ∈(0, T] and, by (9.2.8) and (9.4.8), lim t→0+|An(t)−A(t)|= 0. Therefore, An−A∈ C([0, T]) and from |An(t)−A(t)| ≤ kf∗(·, un(·)) −f∗(·, u(·))kC([0,T])  1 p(t)Zt 0 p(τ) d τ ≤ kf∗(·, un(·)) −f∗(·, u(·))kC([0,T]) max {p(τ) : τ∈[0, t]} p(t)t =kf∗(·, un(·)) −f∗(·, u(·))kC([0,T]) t, t ∈[0, T], we deduce that kAn−AkC([0,T]) ≤ kf∗(·, un(·)) −f∗(·, u(·))kC([0,T]) T, n ∈N. This implies that lim n→∞kAn−AkC([0,T ]) = 0. Using the continuity of φ−1on R, we have lim n→∞φ−1(An)−φ−1(A)C([0,T ]) = 0. 277 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Therefore, lim n→∞kFun−FukC([0,T]) = lim n→∞ZT tφ−1(An(s)) −φ−1(A(s))dsC([0,T]) ≤Tlim n→∞φ−1(An)−φ−1(A)C([0,T ]) = 0, that is, operator Fis continuous. On the other hand, choose an arbitrary ε > 0and put δ:= ε φ−1(M∗T). Then, for t1, t2∈[0, T],|t1−t2|< δ, and u∈ B(0, R), it holds that |(Fu) (t1)−(Fu) (t2)|=Zt1 t2 φ−1−1 p(s)Zs 0 p(τ)f∗(τ, u(τ)) d τds ≤Zt1 t2 φ−1(M∗ϕ(s)) d s≤Zt1 t2 φ−1(M∗T) d s =φ−1(M∗T)|t1−t2|< φ−1(M∗T)δ=ε. Hence, functions in F(B(0, R)) are equicontinuous, and, by the Arzel` a–Ascoli’s Theorem (Theorem 1.2.2), the set F(B(0, R)) is relatively compact. Consequently, the operator Fis compact on B(0, R). Then, Schauder’s fixed point Theorem (Theorem 1.2.3) yields the existence of a fixed point u?of Fin B(0, R). Therefore, u?(t) = C−ZT t φ−1−1 p(s)Zs 0 p(τ)f∗(τ, u?(τ)) d τds is a solution of (9.4.9), (9.4.1). Step 2: Solvability of the original problem (9.2.1), (9.4.1). We will prove that any solution uof problem (9.4.9), (9.4.1) satisfies that σ1(t)≤u(t)≤σ2(t), t ∈[0, T], and, therefore, it is a solution of problem (9.2.1), (9.4.1). Put v(t) = u(t)−σ2(t)for t∈[0, T]and assume that max{v(t) : t∈[0, T]}=v(t0)>0.(9.4.10) By (9.4.6), v0(τ−)< v0(τ+)for each τ∈Σ2, so t0/∈Σ2. Moreover, σ2(T)≥Cand u(T) = C, so v(T)≤0and, consequently, t06=T. Therefore, t0∈[0, T)\Σ2. We distinguish two cases: 278 9.4 Existence of Escape Solutions (i) If t0= 0, then (9.4.1) and (9.4.7) yield v0(0+) = u0(0+)−σ0 2(0+) = −σ0 2(0+)≥0. If v0(0+)>0, we get a contradiction with (9.4.10); hence, v0(0+) = 0. (ii) If t0∈(0, T)\Σ2, (9.4.10) also implies that v0(t0) = 0. Since t0∈[0, T)\Σ2, there exists δ > 0such that (t0, t0+δ)⊂(0, T)\Σ2and v(t)>0for t∈(t0, t0+δ). Moreover, for t∈(t0, t0+δ), we have that p(t)φ(u0(t))0−p(t)φ(σ0 2(t))0≥p(t)−f∗(t, u(t)) + e f(φ(σ2(t))) =p(t)v(t) v(t)+1 >0 and integrating the previous expression, we obtain that Zt t0p(s)φ(u0(s))0−p(s)φ(σ0 2(s))0ds=p(t)φ(u0(t)) −φ(σ0 2(t))>0, for t∈(t0, t0+δ). Therefore, since φis increasing, we have that v0(t)>0on (t0, t0+δ), which is a contradiction with (9.4.10). Consequently, we have proved that u(t)≤σ2(t), t ∈[0, T]. Analogously, it can be proved that u(t)≥σ1(t), t ∈[0, T]. We conclude that the solution uof problem (9.4.9), (9.4.1) is a solution of (9.2.1), (9.4.1). The main result of this section (which proves the existence of escape solutions in case that φ−1and fare not Lipschitz continuous) is contained in Theorem 9.4.8. Its proof is based on Lemmas 9.3.2 and 9.3.5, where a suitable sequence {un}∞ n=1 of solutions of problem (9.2.1), (9.1.2) is used. In order to get such a sequence with the starting values equal to L0(see part (ii) in the proof of Theorem 9.4.8), we need the next lemma. 279 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian where cis from (9.5.1). Moreover, if b < ∞, then lim t→b−u(t) = ∞. Proof. Let ube an escape solution of problem (9.1.1)–(9.1.2) on [0, b). Then ∃c∈(0,∞): u(t)∈[L0, L), t ∈[0, c), u(c) = L, u0(c)>0. Assume that there exists c1> c such that u0(c1) = 0, u(t)> L, u0(t)>0for t∈(c, c1). Integrating equation (9.1.1) over [c, c1], dividing by p(t)and using (B1),(B3)and (B4), we get φ(u0(t)) = p(c)φ(u0(c)) p(t)−1 p(t)Zt c p(s)f(φ(u(s))) d s > 0, t ∈[c, c1], which contradicts that u0(c1) = 0. Hence, u(t)> L and u0(t)>0for t∈(c, b). Let b < ∞. Since [0, b)is the maximal interval where the solution uis defined, ucannot be extended behind b. Therefore, since u0(t)>0for t∈(c, b), it holds that lim t→b−u(t) = ∞and thus, the solution uis unbounded. Since all escape solutions of (9.2.1), (9.1.2) on [0, b)which cannot be extended to the half-line [0,∞)are naturally unbounded, we continue our investigation about unboundedness of escape solutions defined on [0,∞). That is, we will assume from now on that [0, b) = [0,∞). Theorem 9.5.2. Assume (B1)–(B4)hold and let lim t→∞p(t)<∞.(9.5.2) Let ube an escape solution of problem (9.1.1)–(9.1.2). Then lim t→∞u(t) = ∞. Proof. Let ube an escape solution of problem (9.1.1)–(9.1.2). Lemma 9.5.1 ensures that u0(t)>0, t ∈(c, ∞), with cfrom (9.5.1), and so, there exists lim t→∞u(t)∈(L, ∞]. Due to (B1),(B4)and (9.5.1), p(c)φ(u0(c)) =: c0∈(0,∞). 286 9.5 Unbounded Solutions Integrating now equation (9.1.1) from cto t>c, we get, by (B3)and (B4), that u(t) = L+Zt c φ−1c0 p(s)−1 p(s)Zs c p(τ)f(φ(u(τ))) d τds >Zt c φ−1c0 p(s)ds, for t∈(c, ∞). Conditions (B4)and (9.5.2) warrant that lim s→∞ c0 p(s)∈(0,∞) and, by (B1),Z∞ 1 φ−1c0 p(s)ds=∞. Therefore, lim t→∞u(t)≥Z∞ c φ−1c0 p(s)ds=∞, which implies that the solution is unbounded. Theorem 9.5.3. Assume (B1)–(B4),(9.2.3) and f(x)<0for x>φ(L).(9.5.3) Let ube an escape solution of problem (9.1.1)–(9.1.2). Then uis unbounded. Proof. Let ube an escape solution of problem (9.1.1)–(9.1.2). Lemma 9.5.1 implies that u0>0on (c, ∞)and hence, there exists limt→∞ u(t)∈(L, ∞]. Assume on the contrary that lim t→∞u(t) =: A∈(L, ∞).(9.5.4) Step 1: We prove that u0is bounded. Assume that u0is unbounded. Then there exists a sequence {tn}∞ n=1 such that lim n→∞tn=∞and lim n→∞u0(tn) = ∞. Equation (9.1.1) has an equivalent form φ0(u0(t)) u00(t) + p0(t) p(t)φ(u0(t)) + f(φ(u(t))) = 0, t ∈(0,∞).(9.5.5) Choose n∈N. Multiplying this equation by u0and integrating it from cto t>c, we obtain for t=tnthat ψ1(tn) + ψ2(tn) + ψ3(tn) = 0, tn∈[c, ∞),(9.5.6) 287 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian where ψ1(tn) = Zu0(tn) u0(c) x φ0(x) d x, ψ2(tn) = Ztn c p0(s) p(s)φ(u0(s)) u0(s) d s, ψ3(tn) = Zu(tn) L f(φ(x)) d x. Then ψ3(tn) = F(u(tn)) −F(L), where F(x) := Zx 0 f(φ(s)) d s, x ∈R. Due to (B1)and (9.5.3), F(x)is decreasing for x > φ(L). Since uis increasing on (c, ∞),F(u(tn)) is decreasing for tn∈(c, ∞)and limn→∞ F(u(tn)) = F(A). According to (9.5.4), lim n→∞ψ3(tn)∈(−∞,0) , and, by (B1)and (B4), lim n→∞ψ1(tn) = ∞and lim n→∞ψ2(tn)>0. Hence, letting n→ ∞in (9.5.6), we obtain 0 = lim n→∞(ψ1(tn) + ψ2(tn) + ψ3(tn)) = ∞, which is a contradiction. So, u0is bounded. Step 2: We will prove that lim t→∞u(t) = ∞. Since u0is bounded, letting t→ ∞ in (9.5.5) and using (9.2.3), (9.5.3) and (9.5.4), we get lim t→∞φ0(u0(t)) u00(t) = −f(φ(A)) >0. Since φ0(u0(t)) >0for t > c, there exists τ > c such that u00(t)>0for t≥τ. Therefore, u0is increasing on [τ, ∞)and there exists lim t→∞u0(t)>0, which contradicts lim t→∞u(t) = A < ∞. Thus, the solution is unbounded. The following corollary can be deduced from the proof of Theorem 9.5.3. 288 9.5 Unbounded Solutions Corollary 9.5.4. Assume conditions (B1)–(B4)and (9.2.3) and let ube a solution of problem (9.1.1)–(9.1.2). If usatisfies that lim t→∞u(t) =: A∈(L, ∞), then f(φ(A)) = 0. Remark 9.5.5. Note that, in previous corollary, f(φ(A)) = 0 is equivalent with the fact that u(t)≡Ais a solution of equation (9.1.1). For f≡0on (φ(L),∞), we are able to find necessary and sufficient condition for the unboundedness of escape solutions of problem (9.1.1)–(9.1.2). Theorem 9.5.6. Assume (B1)–(B4), f(x)≡0for x>φ(L)(9.5.7) and φ(a b) = φ(a)φ(b), a, b ∈(0,∞).(9.5.8) Let ube an escape solution of problem (9.1.1)–(9.1.2). Then uis unbounded if and only if Z∞ 1 φ−11 p(s)ds=∞.(9.5.9) If we replace condition (9.5.8) by φ(a b)≤φ(a)φ(b), a, b ∈(0,∞),(9.5.10) then (9.5.9) implies that uis unbounded. Proof. Let ube an escape solution of problem (9.1.1)–(9.1.2). Then, according to Lemma 9.5.1, u0>0on (c, ∞). Thus there exists t0> c such that u(t0)> L, u0(t)>0for t∈[t0,∞). Therefore, there exists lim t→∞u(t)∈(L, ∞]. Using (9.5.8), we obtain φ−1(a)φ−1(b) = φ−1(φ(φ−1(a)φ−1(b))) = φ−1(φ(φ−1(a)) φ(φ−1(b))) =φ−1(a b), a, b ∈(0,∞).(9.5.11) Due to (B1),(B4)and (9.5.7), p(t0)φ(u0(t0)) =: c0∈(0,∞)and f(φ(u(t))) = 0 for t∈[t0,∞). 289 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Thus, integrating equation (9.1.1) from t0to t>t0and using (9.5.11), we get u(t) = u(t0) + Zt t0 φ−1c0 p(s)ds=u(t0) +φ−1(c0)Zt 1 φ−11 p(s)ds−Zt0 1 φ−11 p(s)ds, t ∈(t0,∞). Letting t→ ∞here, we get the equivalence. Now, let us consider (9.5.10) instead of (9.5.8) and assume that (9.5.9). Then we continue analogously and obtain φ−1(a)φ−1(b) = φ−1φ(φ−1(a)φ−1(b))≤φ−1φ(φ−1(a)) φ(φ−1(b)) =φ−1(a b), with a, b ∈(0,∞), and u(t) = u(t0) + Zt t0 φ−1c0 p(s)ds≥u(t0) +φ−1(c0)Zt 1 φ−11 p(s)ds−Zt0 1 φ−11 p(s)ds, t ∈(t0,∞). We let t→ ∞here and obtain that if (9.5.9), then the solution is unbounded. 9.6. Main Results and Examples In this section, we first present the existence results about unbounded solutions of the original problem (9.1.1)–(9.1.2) in case that φ−1and fare Lipschitz continuous (see Theorems 9.6.1, 9.6.3 and 9.6.5). Each of these theorems is afterwards illustrated by an example which is chosen in such a way that only this theorem is applicable, while none of the remaining two theorems can be used for this example. Then, in Theorems 9.6.7, 9.6.9 and 9.6.11, we present the main existence results about unbounded solutions of the original problem (9.1.1)–(9.1.2) provided φ−1and fdo not need to be Lipschitz continuous. The illustration by examples is done as in the previous case and shows that none of these theorems is included in any of the two remaining ones. In the whole section, we assume that (due to Definition 9.1.1) for each n∈N, [0, bn)⊂[0,∞)is a maximal interval such that a function unsatisfies equation (9.1.1) for every t∈(0, bn). Theorem 9.6.1. Assume that conditions (B1)–(B4),(9.2.3),(9.2.4),(9.2.10),(9.2.11) and (9.5.2) hold. Then there exist infinitely many unbounded solutions unof problem (9.1.1)–(9.1.2) on [0, bn)with different starting values in L0,¯ B,n∈N. 290 9.6 Main Results and Examples Proof. By Theorem 9.4.1, there exist infinitely many escape solutions unof problem (9.2.1), (9.1.2) with starting values in L0,¯ B. Let us choose n∈N. Then ∃cn∈(0,∞): un(t)∈(L0, L), t ∈[0, cn), un(cn) = L, u0 n(cn)>0. Consider the restriction of unto [0, cn]. Then there exists bn> cnsuch that uncan be extended as a solution of problem (9.1.1)–(9.1.2) on [0, bn). If bn<∞, then, due to Lemma 9.5.1, lim t→b− n un(t) = ∞, so unis unbounded. If bn=∞, then Theorem 9.5.2 yields lim t→∞un(t) = ∞, that is unis unbounded, as well. Example 9.6.2. Consider problem (9.1.1)–(9.1.2) with φ(x) = sinh x=ex−e−x 2, x ∈R, f(x) = (x(x+ sinh 4) (sinh 1 −x), x ∈[−sinh 4,sinh 1], cos(x−sinh 1) −1, x > sinh 1, p(t) = arctan tor p(t) = tanh t=et−e−t et+e−t, t ∈[0,∞). Here L0=−4,L= 1,φ−1(x) = arcsinh x= ln x+√x2+ 1. These functions psatisfy (B4),(9.5.2) and lim t→∞ (arctan t)0 arctan t= lim t→∞ 1 t2+1 arctan t= 0,lim t→∞ (tanh t)0 tanh t= lim t→∞ 1 cosh2t tanh t= 0, that is, (9.2.3) holds, as well. Functions φand ffulfil (B1)–(B3). Moreover, 0< L < −L0,φis odd and e F(L0) = Z−4 0 φ(s) (φ(s) + sinh 4) (sinh 1 −φ(s)) d s =Z4 0 φ(s) (sinh 4 −φ(s)) (sinh 1 + φ(s)) d s >Z1 0 φ(s) (sinh 4 −φ(s)) (sinh 1 + φ(s)) d s >Z1 0 φ(s) (φ(s) + sinh 4) (sinh 1 −φ(s)) d s=e F(L), 291 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian thus, (9.2.4) holds. Since fand φ−1are Lipschitz continuous, conditions (9.2.10) and (9.2.11) are valid, too. We have fulfilled all assumptions of Theorem 9.6.1. Since fhas isolated zeros on (sinh 1,∞), we cannot use neither Theorem 9.6.3 nor Theorem 9.6.5 here. In the same way as in the proof of Theorem 9.6.1, we can prove the following Theorems 9.6.3 or 9.6.5, if we use in the proof Theorems 9.5.3 or 9.5.6, respectively, instead of Theorem 9.5.2. Theorem 9.6.3. Let (B1)–(B4),(9.2.3),(9.2.4),(9.2.10),(9.2.11) and (9.5.3) hold. Then there exist infinitely many unbounded solutions unof problem (9.1.1)–(9.1.2) on [0, bn)with different starting values in L0,¯ B,n∈N. Example 9.6.4. Let us consider problem (9.1.1)–(9.1.2) with φ(x) = ln(|x|+ 1) sign x, x ∈R, f(x) = x(x+ ln 4) (ln 2 −x), x ∈[−ln 4,∞), p(t) = tβ, β > 0, t ∈[0,∞). Here L0=−3,L= 1 and φ−1(x) = e|x|−1sign x. We can easily check that φ,fand psatisfy (B1)–(B4),(9.2.3) and (9.5.3). In addition, 0< L < −L0,φis odd and we can show, similarly to Example 9.6.2, that (9.2.4) holds. The Lipschitz continuity of fand φ−1yields (9.2.10) and (9.2.11). Thus, we can apply Theorem 9.6.3 here. Since lim t→∞tβ=∞and f(x)<0for x > ln 2, we can not use neither Theorem 9.6.1 nor Theorem 9.6.5. Theorem 9.6.5. Assume that (B1)–(B4),(9.2.3),(9.2.4),(9.2.10),(9.2.11),(9.5.7), (9.5.10) and (9.5.9) hold. Then there exist infinitely many unbounded solutions unof problem (9.1.1)–(9.1.2) on [0, bn)with different starting values in L0,¯ B,n∈N. Example 9.6.6. Consider problem (9.1.1)–(9.1.2) with φ(x) = x, x ∈R, p(t) = √t , t ∈[0,∞), f(x) = (x3(x−φ(L0))(φ(L)−x), x ∈[φ(L0), φ(L)], 0, x > φ(L),0< L < −L0. 292 9.6 Main Results and Examples Functions φ,f,pand φ−1(x) = xsatisfy (B1)–(B4),(9.2.3),(9.2.10),(9.2.11), (9.5.7),(9.5.8) and consequently, (9.5.10). Since f(φ(x)) = f(x)and L < −L0, we have e F(L)<e F(L0)and (9.2.4) holds. In addition, Z∞ 1 φ−11 p(s)ds=Z∞ 1 1 √sds=∞. We have satisfied all assumptions of Theorem 9.6.5. Since lim t→∞√t=∞and f(x)<0for x > ln 2, we cannot use neither Theorem 9.6.1 nor Theorem 9.6.3. Now, applying Theorem 9.4.8 instead of Theorem 9.4.1, we get as before the existence results about unbounded solutions in each case, where φ−1and fdo not have to be Lipschitz continuous. Theorem 9.6.7. Let (B1)–(B4),(9.2.3),(9.2.4) and (9.5.2) hold. Then there exist infinitely many unbounded solutions unof problem (9.1.1)–(9.1.2) on [0, bn)with not necessarily different starting values in L0,¯ B,n∈N. Example 9.6.8. Let us consider problem (9.1.1)–(9.1.2) with 0< L < −L0 φ(x) = |x|αsign x, α > 1, x ∈R, f(x) =        p|x|sign x(x−φ(L0)) (φ(L)−x), x ∈[φ(L0), φ(L)], (φ(L)−x)(φ(2L)−x), x ∈(φ(L), φ(2L)), 0, x ≥φ(2L), p(t) = arctan tor p(t) = tanh t=et−e−t et+e−t, t ∈[0,∞). According to Example 9.6.2, functions psatisfy (B4),(9.2.3) and (9.5.2). Functions φand ffulfil (B1)–(B3). Since fis continuous, 0< L < −L0and φis a continuous and odd function, (9.2.4) holds, too. We have verified all assumptions of Theorem 9.6.7. The form of fimplies that neither Theorem 9.6.9 nor Theorem 9.6.11 can be applied. Theorem 9.6.9. Assume that (B1)–(B4),(9.2.3),(9.2.4) and (9.5.3) hold. Then there exist infinitely many unbounded solutions unof problem (9.1.1)–(9.1.2) on [0, bn) with not necessarily different starting values in L0,¯ B,n∈N. 293 On Unbounded Solutions of Singular Initial Value Problems with φ-Laplacian Example 9.6.10. Consider problem (9.1.1)–(9.1.2) with φ(x) = x3, x ∈R, f(x) = 3 √x(x+ 8) (1 −x), x ∈[−8,∞), p(t) = tβ, β > 0, t ∈[0,∞). Here L0=−2,L= 1,φ−1(x) = 3 √x. It is easy to see that φ,fand pfulfil (B1)–(B4),(9.2.3) and (9.5.3). Further, e F(L0) = Z−2 0 ss3+ 81−s3ds=144 5 and e F(L) = Z1 0 ss3+ 81−s3ds=99 40 . So, e F(L0)>e F(L)which yields (9.2.4). Therefore, we can apply Theorem 9.6.9 here. Since lim t→∞tβ=∞and f(x)<0for x > 1, we cannot use neither Theorem 9.6.7 nor Theorem 9.6.11. Theorem 9.6.11. Let (B1)–(B4),(9.2.3),(9.2.4),(9.5.7),(9.5.10) and (9.5.9) hold. Then there exist infinitely many unbounded solutions unof problem (9.1.1)–(9.1.2) on [0, bn)with not necessarily different starting values in L0,¯ B,n∈N. Example 9.6.12. Let us consider problem (9.1.1)–(9.1.2) with φ(x) = |x|αsign x, α > 1, x ∈R, p(t) = tβ, β ∈(0, α], t ∈[0,∞), f(x) = (3 √x(x−φ(L0)) (φ(L)−x), x ∈[φ(L0), φ(L)], 0, x > φ(L),0< L < −L0. Functions φ,fand psatisfy (B1)–(B4),(9.2.3),(9.5.7),(9.5.8) and consequently, (9.5.10). Moreover, 0< L < −L0and φis odd function which yields (9.2.4). Furthermore, φ−1(x) = x1 αfor x > 0 and Z∞ 1 φ−11 p(s)ds=Z∞ 1 s−β αds=∞, that is, we have verified all assumptions of Theorem 9.6.11. Since lim t→∞tβ=∞and f(x) = 0 for x > φ(L), neither Theorem 9.6.7 nor Theorem 9.6.9 are applicable. 294 9.6 Main Results and Examples It si clear that every unbounded solution of problem (9.1.1)–(9.1.2) is an escape solution. According to the proofs of above theorems, we can formulate also the reverse assertion. Corollary 9.6.13. Assume all assumptions of Theorem 9.6.1 or 9.6.3 or 9.6.5 or 9.6.7 or 9.6.9 or 9.6.11. Then each escape solution of problem (9.1.1)–(9.1.2) is unbounded. 295 Resumen Se demuestra en esta secci´ on que la funci´ on de Green de cualquier problema de frontera asociado a la ecuaci´ on previa se puede expresar en t´ erminos de la funci´ on de Green asociada al operador de Hill con las mismas condiciones de frontera. Como consecuencia, todos los resultados obtenidos en la secci´ on anterior se pueden adaptar en t´ erminos de este problema. Este cap´ ıtulo recoge resultados de [22] y [23]. Cap´ ıtulo 4: Soluciones para Problemas de Frontera No Lineales de Orden Par con Funciones de Green de Signo Constante En este cap´ ıtulo se consideran por primera vez problemas de frontera no lineales. En particular, se considerar´ an problemas que sigan el siguiente esquema: L u(t) = f(t, u(t)), t ∈I, u ∈X, siendo Lel operador general lineal de orden 2ndefinido en el Cap´ ıtulo 3. Por otra parte, consideraremos X⊂W2n,1(I)como un espacio de Banach que incluye las condiciones de frontera y en el cual Les no resonante. En estas condiciones se tiene que las soluciones del problema de frontera anterior se corresponden con los puntos fijos en Xdel siguiente operador integral L−1u(t) = ZT 0 G[T](t, s)f(s, u(s)) d s, siendo G[T]la funci´ on de Green asociada. El m´ etodo utilizado para garantizar la existencia de puntos fijos de este operador integral es el de sub y sobresoluciones. La novedad principal de nuestra aproximaci´ on frente a referencias previas presentes en la literatura es el hecho de que conseguimos garantizar la existencia de soluci´ on del problema mediante un par de sub y sobresoluciones de otro problema distinto (compuesto por el mismo operador sometido a condiciones de frontera diferentes). Esto ser´ a posible gracias a las relaciones punto a punto entre distintas funciones de Green probadas en los Cap´ ıtulos 2 y 3. Cabe comentar tambi´ en que una de las hip´ otesis b´ asicas de este cap´ ıtulo es la del signo constante de las funciones de Green. Los resultados de este cap´ ıtulo se pueden ver en [31]. 302 Resumen Cap´ ıtulo 5: Soluciones Positivas para Problemas de Frontera No Lineales de Orden Dos con Funciones de Green de Signo No Constante Este cap´ ıtulo est´ a dedicado a estudiar la existencia de soluciones de signo constante de un problema de frontera de orden dos asociado al operador de Hill en el caso en que, al contrario de lo que ocurr´ ıa en el cap´ ıtulo anterior, la funci´ on de Green cambie de signo. La idea b´ asica de este cap´ ıtulo se fundamenta en el hecho de que, pese a que la funci´ on de Green cambie de signo, se puede asegurar que la integral de esta funci´ on multiplicada por la autofunci´ on correspondiente al primer autovalor del problema es positiva. Expondremos el siguiente razonamiento en t´ erminos del problema peri´ odico, aunque resulta igualmente v´ alido para cualquier otra condici´ on de frontera. Consideremos pues el siguiente problema peri´ odico (u00(t) + a(t)u(t) = f(t, u(t)), t ∈I, u(0) = u(T), u0(0) = u0(T), y sean GPsu funci´ on de Green asociada y vPla autofunci´ on correspondiente al primer autovalor. Entonces se tiene que ZT 0 GP(t, s)vP(s) d s > 0,para todo t∈I, lo cual justifica que tiene sentido definir la siguiente constante: γ= ´ınf t∈IRT 0G+ P(t, s)vP(s) d s RT 0G− P(t, s)vP(s) d s(>1). Supongamos que se cumplen las siguientes hip´ otesis: (H1)f:I×[0,∞)→[0,∞)satisface las condiciones de L1-Carath´ eodory. (H2)Existen dos constantes positivas myMtales que m vP(t)≤f(t, x)≤M vP(t) para todo t∈Iyx≥0. Adem´ as, estas constantes deben cumplir que M m≤γ. (H3)Existe un subintervalo [c, d]⊂Ital que Rd cGP(t, s) d t≥0,para todo s∈I yRd cGP(t, s) d t > 0,para todo s∈[c, d]. 303 Resumen Entonces, si la funci´ on de Green cambia de signo, se demuestra que existe una soluci´ on del problema en el cono K=u∈ C(I, R): u≥0en I, ZT 0 u(s) d s≥σkuk, donde σ=η m´ax t, s∈I{GP(t, s)}, y η= m´ın s∈[c,d]Zd c GP(t, s) d t>0. N´ otese que esta soluci´ on es no negativa. Todos los resultados de este cap´ ıtulo se recogen en [27]. Cap´ ıtulo 6: Resultados de Existencia y Multiplicidad de Soluciones para Ecuaciones Generalizadas de Hammerstein con un Par´ ametro En este cap´ ıtulo estudiamos problemas integrales definidos en espacios de Banach que reciben el nombre de ecuaciones generalizadas de Hammerstein. En particular, estudiamos la existencia y multiplicidad de puntos fijos del siguiente operador integral Tu(t) = λZT 0 k(t, s)f(s, u(s), u0(s), . . . , u(m)(s)) d s, t ∈I, donde λ > 0es un par´ ametro positivo, k:I×I→Res una funci´ on n´ ucleo que verificar´ a ciertas propiedades, mun entero positivo y f:I×Rm+1 →[0,+∞)es una funci´ on L1-Carath´ eodory. Este cap´ ıtulo generaliza varios resultados presentes en la literatura al pedir condiciones menos restrictivas de lo habitual sobre el n´ ucleo. En concreto, se pedir´ a que el n´ ucleo y algunas de sus derivadas (no necesariamente todas) sean positivos ´ unicamente en un subintervalo de I. Este subintervalo podr´ ıa incluso llegar a ser degenerado, es decir, podr´ ıa tratarse de un ´ unico punto. Por otra parte, buscaremos n´ ucleos para los cuales algunas de sus derivadas (de nuevo, no necesariamente todas) satisfagan las siguientes desigualdades:  ∂jk ∂tj(t, s)≤φj(s)para todo t∈[cj, dj]y c. t. p. s∈I, 304 Resumen y ∂jk ∂tj(t, s)≥ξjφj(s)para todo t∈[aj, bj]y c. t. p. s∈I, siendo φjfunciones integrables y ξjconstantes. Cabe comentar que los intervalos [aj, bj]y[cj, dj]deben tener intersecci´ on no vac´ ıa pero podr´ ıan ser distintos e, incluso, no comparables. Bajo diversas hip´ otesis (v´ eanse (H1)–(H7)en la Secci´ on 6.2), podemos demostrar entonces la existencia de puntos fijos del operador integral considerado en el cono K=     u∈ Cm(I, R): u(i)(t)≥0, t ∈[mi, ni], i ∈J0; m´ın t∈[aj,bj]u(j)(t)≥ξjku(j)k[cj,dj], j ∈J1    , donde u(j)[cj,dj]:= m´ax t∈[cj,dj]u(j)(t), J≡ {0,1, . . . , m}yJ1⊂J0⊂J,J16=∅. Este tipo de conos, hasta donde la autora tiene conocimiento, es nuevo en la literatura. En cuanto a las t´ ecnicas para demostrar la existencia de puntos fijos, se utilizan dos diferentes. En primer lugar, en la Secci´ on 6.3, se prueba la existencia de un punto fijo utilizando el ´ ındice de punto fijo para conjuntos abiertos arbitrarios (algunos de los cuales son no acotados). Por otra parte, en la Secci´ on 6.4 se dan resultados de existencia y multiplicidad de soluciones. Estos resultados se basan tambi´ en en el ´ ındice de punto fijo, esta vez sobre conjuntos abiertos y acotados. La diferencia principal entre ambas secciones es que las hip´ otesis que se le piden a la no linealidad fson diferentes y, de hecho, en la Secci´ on 6.5 se muestran ejemplos en los que se ve que ambos m´ etodos no son comparables. A continuaci´ on, la Secci´ on 6.6 presenta una aplicaci´ on de los resultados previos para garantizar la existencia de soluci´ on de problemas de Dirichlet de orden par arbitrario    u(2n)(t) = ft, u(t), . . . , u(2n−1)(t), t ∈[0,1], u(2k)(0) = u(2k)(1) = 0, k = 0, . . . , n −1. Este estudio generaliza los existentes en la literatura puesto que en este tipo de problemas se suele considerar que la funci´ on fdepende ´ unicamente de las derivadas de orden par, mientras que en este cap´ ıtulo se admite la dependencia de cualquier derivada hasta orden 2n−1. 305 Resumen Finalmente, la Secci´ on 6.7 considera el caso particular del siguiente problema diferencial de orden tres (−u(3)(t) = λ f(t, u(t), u0(t), u00(t)), t ∈[0,1], u(0) = u0(0) = 0, u0(1) = α u0(η), siendo 0< η < 1y1< α < 1 ηconstantes dadas. Los resultados de este cap´ ıtulo se pueden encontrar en [32] y [102]. Cap´ ıtulo 7: Problemas Multipunto Resonantes en la Semirrecta En este cap´ ıtulo consideraremos por primera vez un problema definido en un dominio no acotado. En particular, probaremos la existencia de soluciones acotadas para el siguiente problema multipunto       u00(t) = f(t, u(t), u0(t)), t ∈[0,∞), u(0) = 0, u0(+∞) = m−1 X i=1 αiu0(ξi), siendo αi>0y0 = ξ1<··· < ξm−1<+∞. Asumiremos que los coeficientes αi cumplen la siguiente condici´ on m−1 X i=1 αi= 1, la cual implica que nos encontramos ante un problema resonante. Para resolver este problema consideraremos otro modificado (el cual se construir´ a a˜ nadiendo nuevos t´ erminos a ambos lados de la ecuaci´ on) que ser´ a equivalente al primero y no resonante. Este problema modificado lo transformaremos en un problema integral cuyos puntos fijos se corresponder´ an con las soluciones del problema inicial. En concreto, el problema integral con el que trabajaremos ser´ a Tu(t) = Z∞ 0 G(t, s)f(s, u(s), u0(s)) + k u0(s) + M u(s)ds, donde Ges la funci´ on de Green del problema        u00(t) + k u0(t) + M u(t)=0, t ∈[0,∞), u(0) = 0, u0(+∞) = m−1 X i=1 αiu0(ξi), 306 Resumen ykyMson dos constantes positivas que cumplen ciertas condiciones. Adem´ as, el problema modificado satisfar´ a otra propiedad importante: su funci´ on de Green estar´ a en el espacio L1[0,∞)∩L∞[0,∞). Esto permitir´ a que el operador integral sea compacto tanto si la no linealidad fsatisface las condiciones L1oL∞- Carath´ eodory. N´ otese que este hecho permite garantizar la existencia de soluci´ on para un mayor n´ umero de casos puesto que, al estar considerando en este cap´ ıtulo un intervalo no acotado, los espacios L1[0,∞)yL∞[0,∞)no son comparables. Para probar la existencia de puntos fijos del operador integral utilizaremos el m´ etodo de sub y sobresoluciones. En particular, para demostrar que el operador integral es compacto utilizaremos el criterio de compacidad dado en el Theorem 1, que involucra una cierta condici´ on de equiconvergencia en infinito. Los resultados de este cap´ ıtulo se recogen en [103]. Cap´ ıtulo 8: Existencia de Soluciones de Ecuaciones Integrales con Condiciones Asint´ oticas En este cap´ ıtulo estudiamos los puntos fijos de un operador integral definido sobre la recta real. En general, la mayor dificultad cuando se intenta probar la existencia de puntos fijos de operadores integrales definidos en intervalos no acotados surge al demostrar que el operador considerado es compacto. Estos problemas se deben principalmente a la imposibilidad de utilizar el Teorema de Ascoli-Arzel` a para probar la compacidad del operador. La forma m´ as habitual de resolver este problema consiste en utilizar un cierto criterio de compacidad (el cual hemos utilizado, precisamente, en el Cap´ ıtulo 7), que se recoge en el Theorem 1, en la p´ agina 181. En este cap´ ıtulo presentamos un m´ etodo alternativo que tendr´ a un doble beneficio: por una parte, nos permitir´ a utilizar el Teorema de Ascoli-Arzel` a para probar la compacidad del operador. Por otra, nos garantizar´ a que las soluciones encontradas tienen un cierto comportamiento asint´ otico. Para ello, definimos un espacio de Banach que incluya esas propiedades asint´ oticas. En particular, para n∈N, consideramos el espacio de las funciones reales de variable real que son de clase ny tienen l´ ımite en ±∞: Cn(R,R) := f:R→R:f|R∈ Cn(R,R),∃l´ım t→±∞f(j)(t)∈R, j = 0, . . . , n, siendo R≡[−∞,∞]. Se tiene que Cn(R,R),n∈Nes un espacio de Banach con la norma kfk(n):= sup nf(k)∞:k= 0, . . . , no. 307 Resumen Entonces, dada una funci´ on ϕ∈ Cn(R,R+), definimos el espacio de las ϕextensiones de clase na infinito como sigue: e Cn ϕ≡e Cn ϕ(R,R) = nf∈ Cn(R,R) : ∃e f∈ Cn(R,R), f =ϕ·e f|Ro. En particular, este espacio es de Banach con la norma inducida kfkϕ:= e f(n), f ∈e Cϕ, de donde se deduce que los espacios Cn(R,R)ye Cn ϕson isomorfos. De la existencia de dicho isomorfismo se deduce que, puesto que el Teorema de Ascoli-Arzel` a se puede aplicar al espacio Cn(R,R)(por ser Rcompacto), entonces este teorema se puede aplicar tambi´ en al espacio e Cn ϕ. Buscaremos pues puntos fijos de operadores integrales de la forma Tu(t) := p(t) + Z∞ −∞ k(t, s)η(s)f(s, u(s)) d s en el espacio de Banach e Cn ϕ, para una cierta funci´ on ϕque representar´ a precisamente el comportamiento asint´ otico de las soluciones. Dicho de otro modo, que los puntos fijos del operador se encuentren en el espacio e Cn ϕimplicar´ a que tales funciones se comporten asint´ oticamente de forma similar a ϕ. En cuanto al m´ etodo empleado para garantizar la existencia de puntos fijos, consideramos en este cap´ ıtulo dos aproximaciones diferentes: la primera de ellas, desarrollada en la Secci´ on 8.4 se basa en el ´ ındice de punto fijo en conos y presenta hip´ otesis bastante restrictivas sobre la funci´ on no lineal f. Por otra parte, la segunda aproximaci´ on, analizada en la Secci´ on 8.5, se basa en definir una serie de operadores lineales auxiliares y estudiar sus propiedades espectrales. En particular, si el radio espectral de estos operadores y ciertos l´ ımites obtenidos a partir de la funci´ on no lineal fsatisfacen ciertas propiedades, ser´ a posible probar la existencia de puntos fijos. En este caso, las restricciones sobre la funci´ on fson mucho menos restrictivas que las impuestas por el m´ etodo anterior, pero a expensas de pedir que el n´ ucleo ksatisfaga condiciones m´ as fuertes. Tal y como se muestra en el cap´ ıtulo con ejemplos de los dos m´ etodos, estos son no comparables. Todos estos resultados se pueden ver en [33] y [34]. Cap´ ıtulo 9: Soluciones no Acotadas de Problemas de Valores Iniciales Singulares con φ-Laplaciano En este ´ ultimo cap´ ıtulo se estudia un problema de valor inicial singular con φLaplaciano, prestando especial inter´ es a la existencia de soluciones no acotadas del 308 Resumen mismo. En este caso, al tratarse de un problema singular, no es posible construir un problema integral equivalente, tal y como se hace en los cap´ ıtulos anteriores. Consecuentemente, las t´ ecnicas utilizadas en este cap´ ıtulo difieren totalmente de las de consideradas hasta el momento. En particular, consideraremos el siguiente problema no lineal: ((p(t)φ(u0(t)))0+p(t)f(φ(u(t))) = 0, t > 0, u(0) = u0, u0(0) = 0, u0∈[L0, L]. Comenzamos el cap´ ıtulo definiendo tres tipos de soluciones posibles que podemos obtener. As´ ı, si denotamos usup = sup{u(t): t∈[0,∞)}, diremos que Una soluci´ on udel problema es “oscilante” (damped) si usup < L. Una soluci´ on udel problema es homocl´ ınica si usup =L. Una soluci´ on udel problema ser´ a “de escape” si usup > L. Puesto que tanto las soluciones oscilantes como las homocl´ ınicas est´ an acotadas, las soluciones no acotadas ser´ an un subconjunto de las de escape. Esto motiva la divisi´ on del cap´ ıtulo en dos partes: 1. B´ usqueda de condiciones para garantizar la existencia de soluciones de escape. 2. B´ usqueda de condiciones necesarias o suficientes para garantizar que una soluci´ on de escape es no acotada. Adem´ as, para la b´ usqueda de condiciones que aseguren la existencia de soluciones de escape tendremos que considerar dos casos diferenciados: el primero de ellos, en el que tanto fcomo φ−1son funciones lipschitzianas, resulta bastante m´ as sencillo puesto que en estas condiciones la unicidad de soluci´ on del problema est´ a garantizada. Por el contrario, el segundo caso (con fyφ−1no lipschitzianas), presenta una serie de complicaciones derivadas de la no unicidad de soluci´ on. Para solventar estos problemas se considera el m´ etodo de sub y sobresoluciones. Estos dos casos presentan adem´ as otra diferencia importante en cuanto a los resultados obtenidos: mientras que en el primero se garantiza la existencia de una sucesi´ on 309 Resumen de soluciones de escape que toman diferentes valores iniciales, en el segundo podr´ ıa ocurrir que todas las soluciones tuvieran el mismo valor inicial L0. 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