Existence and Multiplicity of Solutions of Functional Differential Equations

TESIS DOCTORAL: Existence and Multiplicity of Solutions of Functional Differential Equations F. ADRIÁN FDEZ. TOJO Director de Tesis: ALBERTO CABADA FERNÁNDEZ Existence and Multiplicity of Solutions of Functional Differential Equations INFORME DEL DIRECTOR DE TESIS: La presente memoria fue realizada por D. Fernando Adrián Fernández Tojo, bajo la dirección de D. Alberto Cabada Fernández, catedrático del Departamento de Análise Matemática de la Universidade de Santiago de Compostela, para optar al grado de Doctor en Matematicas por la Universidade de Santiago de Compostela. El director de Tesis declara que no incurre en ninguna de las causas de abstención recogidas en la ley 30/1992 (artículos 28 y 29). Santiago de Compostela, 5 de abril de 2015. El director de la Tesis Fdo.: Alberto Cabada Fernández El doctorando Fdo.: F. Adrián Fernández Tojo A mi familia, a mis padres Cristina y Fernando y a mi hermano Jacobo, por estar siempre ahí Acknowledgments I would not dare to start this work without profiting the opportunity to give thanks to all those who, without their support and affection, I could not have taken to a deserving ending this exiting, but yet exhausting, work I lavish the reader on. I would like to express my gratitude towards my supervisor, Alberto Cabada, for his unbreakable tenacity and unconditional drive even when the work seemed overwhelming. Thanks as well to all the other researchers and Professors who so willingly and selflessly provided advice and greatly contributed to my work, specially to Gennaro Infante and Paola Pietramala, with whom I undertook part of the investigation and who made possible the second part of this Thesis, but also to J. Ángel Cid and Pedro Torres –whom I visited during my research–, who opened the doors to new opportunities to extend my analysis. Thanks to all of the other co-organizers, present and past, of the Introduction to Research Seminar I have worked with these years in an attempt to show other young scientists like us the work our community of starting researchers is doing. I would like to thank my family, my mum, dad and brother, who were there, are there, and will be there whenever I need them. And last my friends –Pablo, Santi (to whom I owe more than one good idea), Sandra, David, Dani, Alberto…–, debate comrades –Atenea, Iago…– and Kung Fu brothers and sisters –Manchi, Julia, Vins…– who have taught me those things in live you cannot learn otherwise –to survive to my work and even enjoy it– and made me laugh when I most needed it. F. Adrián F. Tojo April 9, 2015 Contents Notation 13 Preface 15 I Green’s functions 17 1 Involutions and differential equations 21 1.1 The straight line problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 1.2 Involutions and their properties . . . . . . . . . . . . . . . . . . . . . . . . . 23 1.2.1 The concept of involution . . . . . . . . . . . . . . . . . . . . . . . . 23 1.2.2 Properties of involutions . . . . . . . . . . . . . . . . . . . . . . . . . 24 1.2.3 Characterization of involutions . . . . . . . . . . . . . . . . . . . . . . 26 1.3 Differential Operators with Involutions . . . . . . . . . . . . . . . . . . . . . . 28 1.3.1 Algebraic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 1.3.2 Differential equations with involutions . . . . . . . . . . . . . . . . . . 29 2 General results for differential equations with involutions 33 2.1 The bases of the study . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2.2 Differential equations with reflection . . . . . . . . . . . . . . . . . . . . . . . 36 3 Order one problems with constant coefficients 39 3.1 Reduction of differential equations with involutions . . . . . . . . . . . . . . . 39 3.2 Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) ............... 44 3.2.1 Constant sign of function 𝐺....................... 49 3.2.2 Lower and upper solutions method . . . . . . . . . . . . . . . . . . . 54 3.2.3 Existence of solutions via Krasnosel’skiĭ’s Fixed Point Theorem . . . . . 55 3.2.4 Examples ................................. 58 3.3 The antiperiodic case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3.3.1 The general case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 3.4 Examples ..................................... 61 3.5 Solutions of the initial value problem . . . . . . . . . . . . . . . . . . . . . . . 66 certain values depending on the variables and these ripples will cause, precisely, the existence of many solutions. This situation is similar to what happens to a bucket of water when we shake it. If we mark a line a little bit above the water level and rock the bucket, ripples start to appear and, when they get high enough, they reach the line we have marked. The more ripples there are, the more times that level is reached. Simple as it may sound, the conditions that have to be satisfied in order to apply this method can, as we will see, get really convoluted with the increasing generality of the problems studied. All these discoveries appear in several publications the author has written during the preparation of the Thesis. The reader may consult [34,35,39–44,96,165,166]. Part I Green’s functions 19 Involutions have been an interesting subject of research at least since Rothe first computed the number of different involutions on finite sets in 1800 [152]. After that, Babbage published in 1815 [7] the foundational paper in which functional equations are first considered, in particular those of the form 𝑓(𝑓(𝑡))=𝑡which are called involutions†. Despite of the progresses on the theory of functional equations, we have to wait for Silberstein who, in 1940 [156], solved the first functional differential equation with an involution. The interest on differential equations with involutions is retaken by Wiener in 1969 [186]. Wiener, together with Watkins, will lead the discoveries in this direction in the following decades [1, 155,173,174,186–189]. Quite a lot of work has been done ever since by several authors. We will make a brief review on this in Chapter 2. In 2013 the first Green’s function for a differential equation with an involution was computed [39] and the field rapidly expanded [40,41,43,44]. This first part goes through those discoveries related to Green’s functions. In order to do that, first we recall some general results concerning involutions which will help us understand their remarkable analytic and algebraic properties. Chapter 1 will deal about this subject while Chapter 2 will give a brief overview on differential equations with involutions to set the reader in the appropriate research framework. We recommend the reader to go through the monograph [187] which has a whole chapter on the subject and, although it was written more than twenty years ago, it contains most of what is worth knowing on the matter. In Chapter 3 we start working on the theory of Green’s functions for functional differential equations with involutions in the most simple cases: order one problems with constant coefficients and reflection. Here we solve the problem with different boundary conditions, studying the specific characteristics which appear when considering periodic, anti-periodic, initial or arbitrary linear boundary conditions. We also apply some very well known techniques (lower and upper solutions method or Krasnosel’skiĭ’s Fixed Point Theorem, for instance) in order to further derive results. Computing explicitly the Green’s function for a problem with nonconstant coefficients is not simple, not even in the case of ordinary differential equations. We face these obstacles in Chapter 4, where we reduce a new, more general problem containing nonconstant coefficients and arbitrary differentiable involutions, to the one studied in Chapter 3. In order to do this we use a double trick. First, we reduce the case of a general involution to the case of the reflection using some of the knowledge gathered in Chapter 1 and then we use a special change of variable (only valid in some cases) that allows the obtaining of the Green’s function of problems with nonconstant coefficients from the Green’s functions of constant-coefficient analogs. To end this part of the work, we have Chapter 5, in which we deepen in the algebraic nature of reflections and extrapolate these properties to other algebras. In this way, we do not only generalize the results of Chapter 3 to the case of 𝑛-th order problems and general twopoint boundary conditions, but also solve functional differential problems in which the Hilbert transform or other adequate operators are involved. The last chapters of this part are about applying the results we have proved so far to some related problems. First, in Chapter 6, setting again the spotlight on some interesting relation be- †Babbage, in the preface to his work [7], described very well the importance of involutions: «Many of the calculations with which we are familiar, consist of two parts, a direct, and an inverse; thus, when we consider an exponent of a quantity: to raise any number to a given power, is the direct operation: to extract a given root of any number, is the inverse method […] In all these cases the inverse method is by far de most difficult, and it might perhaps be added, the most useful». 20 tween an equation with reflection and an equation with a 𝜑-Laplacian, we obtain some results concerning the periodicity of solutions of that first problem with reflection. Chapter 7 moves to a more practical setting. It is of the greatest interest to have adequate computer programs in order to derive the Green’s functions obtained in Chapter 5 for, in general, the computations involved are very convoluted. Being so, we present in this chapter such an algorithm, implemented in Mathematica. We also add some considerations which could lead to simplifying the computations and therefore the time necessary to run the program. The reader can find in the appendix the exact code of the program. 1. Involutions and differential equations 1.1 The straight line problem Before moving to the study of involutions, we will motivate it with a simple problem derived from some considerations on the straight line. Let us assume that 𝑥(𝑡) = 𝑎𝑡+𝑏, where 𝑎,𝑏 ∈ ℝ, is a straight line on the real plane. Then, using the formula of the slope between two points (−𝑡,𝑥(−𝑡))and (𝑡,𝑥(𝑡))we have that 𝑥′(𝑡)= 𝑥(𝑡)−𝑥(−𝑡) 2𝑡 .(1.1.1) Every straight line satisfies this equation. Nevertheless, observe that we are not asking for the slope to be constant and therefore we may ask the following questions in a natural way: Are all of the solutions of equation (1.1.1) straight lines? (see here the spirit of Babbage’s words concerning inverse problems), How can we solve differential equations of this kind? How can we guarantee the existence of solution?, How do the solutions of the equation depend on the fact that 𝑥′varies depending on both 𝑡as well as of the image of 𝑡by a symmetry (in this case the reflection), or, more generally of an involution? In order to answer the first question, we will study the even and odd functions – each one with a different symmetry property– and how does the derivative operator act on them. We do this study in its most basic form, on groups, and then apply it to the real case (a group with the sum). Definition 1.1.1. Let 𝐺and 𝐻be groups, 𝐴 ⊂ 𝐺and define 𝐴−1 ∶= {𝑥−1 |𝑥 ∈ 𝐺}. Assume that 𝐴−1 ⊂𝐴. We will say that 𝑓∶𝐴→𝐻is a symmetric or even function if 𝑓(𝑥−1)= 𝑓(𝑥) ∀𝑥∈𝐴. We will say that 𝑓is an antisymmetric or odd function if 𝑓(𝑥−1)=𝑓(𝑥)−1 ∀𝑥∈ 𝐴. Remark 1.1.2. If 𝑓is a homomorphism, 𝑓is odd. That is because, first if 𝑓is an homeomorphism, 𝐴is a subgroup of 𝐺and 𝑓(𝐴)a subgroup of 𝐻. Now if 𝑒represents the identity element of 𝐺,𝑒′that of 𝐻, and 𝑥∈𝐴,𝑒′= 𝑓(𝑒) = 𝑓(𝑥𝑥−1) = 𝑓(𝑥)𝑓(𝑥−1), so 𝑓(𝑥−1) = 𝑓(𝑥)−1. On the other hand, if 𝑓is an even homomorphism, all of the elements of 𝑓(𝐴)satisfy 𝑦2=𝑒′for every 𝑦∈𝑓(𝐴). For this reason, the only even and odd function with real values, that is, with values in the abelian group (ℝ,+), is the 0function. Remark 1.1.3. The set of even – respectively odd– functions of a subset 𝐴⊂𝐺to a commutative group 𝐻is a group with the point-wise operation induced by the operation of 𝐻, that is, (𝑓𝑔)(𝑥)∶=𝑓(𝑥)𝑔(𝑥)for every 𝑥∈𝐴,𝑓,𝑔∶𝐴→𝐻both even or odd. Proposition 1.1.4. Let 𝐺be a group, 𝐴⊂𝐺such that 𝐴−1 ⊂𝐴,𝑉is a vector space on a field 𝔽of characteristic not equal to 2†. Then there exist two maps 𝑓u�∶𝐴→(𝑉,+)and 𝑓u�∶𝐴→ (𝑉,+), even and odd respectively, such that 𝑓 =𝑓u�+𝑓u�. Furthermore, this decomposition is unique. †This condition is taken in order to be allowed to divide by 2in the vector space 𝑉. 22 1.1. The straight line problem Proof. It is enough to define 𝑓u�(𝑥)∶= 𝑓(𝑥)+𝑓(𝑥−1) 2, 𝑓u�(𝑥)∶= 𝑓(𝑥)−𝑓(𝑥−1) 2. It is clear that 𝑓u�and 𝑓u�are even and odd respectively and that 𝑓 =𝑓u�+𝑓u�. Assume know that there exist two such decompositions: 𝑓 = 𝑓u�+𝑓u�= 𝑓u�+ 𝑓u�. Then, 𝑓u�− 𝑓u�= 𝑓u�−𝑓u�, but 𝑓u�− 𝑓u�is even and  𝑓u�−𝑓u�odd, hence 𝑓u�− 𝑓u�= 𝑓u�−𝑓u�= 0and the decomposition is unique.  From now on, given a function 𝑓 ∶𝐴→𝑉,𝑓u�will stand for its even part and 𝑓u�for its odd part. Corollary 1.1.5. In the conditions of Proposition 1.1.4, the vector space ℱ(𝐺,𝑉)∶={𝑓 ∶𝐺→ 𝑉}can be decomposed in the direct sum of vector spaces ℱu�(𝐺,𝑉)∶={𝑓 ∶𝐺→𝑉|𝑓 even } and ℱu�(𝐺,𝑉)∶={𝑓 ∶𝐺→𝑉|𝑓 odd }, that is, ℱ(𝐺,𝑉)=ℱu�(𝐺,𝑉)⊕ℱu�(𝐺,𝑉). For rest of the section, let 𝐴⊂ℝbe such that −𝐴⊂𝐴. Given the expression of 𝑓u�and 𝑓u� in the decomposition we can claim that u�(𝐴,ℝ)=u�u�(𝐴,ℝ)⊕u�u�(𝐴,ℝ)where u�(𝐴,ℝ) are the differentiable functions from 𝐴to ℝand u�u�(𝐴,ℝ)and u�u�(𝐴,ℝ)the sets of those functions which are, respectively, even differentiable and odd differentiable functions. The following Proposition is an elemental result in differential calculus. Proposition 1.1.6. The derivative operator acts in the following way: u�u�(𝐴,ℝ)⊕u�u�(𝐴,ℝ) u�u�(𝐴,ℝ)⊕u�u�(𝐴,ℝ) (𝑔,ℎ) (0 u� u� 0 )( u� ℎ)=(ℎ′,𝑔′) u� Corollary 1.1.7. For every 𝑓 ∈u�(𝐴,ℝ)we have that (1) (𝑓′)u�=𝑓′⟺ 𝑓 =𝑓u�+𝑐, 𝑐∈ℝ, (2) (𝑓′)u�=𝑓′⟺ 𝑓 =𝑓u�. Now we can solve the “straight line problem” as follows: equation (1.1.1) can be written as 𝑥′(𝑡)= 𝑥(𝑡)−𝑥(−𝑡) 2𝑡 =𝑥u�(𝑡) 𝑡, and since u�u�(u�) u�is symmetric, taking into account Proposition 1.1.6, we arrive at the equivalent system of differential equations (𝑥u�)′(𝑡)=0, (𝑥u�)′(𝑡)= 𝑥u�(𝑡) 𝑡. Hence, 𝑥u�(𝑡)=𝑐,𝑥u�(𝑡)=𝑘𝑡with 𝑐,𝑘∈ℝ, that is, 𝑥is the straight line 𝑥(𝑡)=𝑘𝑡+𝑐, which answers the first question we posed. 1. Involutions and their properties 23 Further on we will use this decomposition method in order to obtain solutions of more complex differential equations with reflection. Involutions, as we will see, have very special properties. This is due to their double nature, analytic and algebraic. This chapter is therefore divided in two sections that will explore the two kinds of properties, arriving at last to some parallelism between involutions and complex numbers for its capability to decompose certain polynomials (see Remark 1.3.6). In this chapter we recall results from [39,41,46,132,187,189,196]. 1.2 Involutions and their properties 1.2.1 The concept of involution The concept of involution is fundamental for the theory of groups and algebras, but, at the same time, being an object in mathematical analysis, their analytical properties allow the obtaining of further information concerning this object. In order to be clear in this respect, let us define what we understand by involution in this analytical context. We follow the definitions of [187,189]. Definition 1.2.1. Let 𝐴⊂ℝbe a set containing more that one point and 𝑓 ∶𝐴→𝐴a function such that 𝑓is not the identity Id. Then 𝑓is an involution if 𝑓2≡𝑓 ∘𝑓 =Id or, equivalently, if 𝑓 =𝑓−1. If 𝐴=ℝ, we say that 𝑓is a strong involution [187]. Involutions are also known as Carleman functions in the literature [46,148]. Example 1.2.2. The following involutions are the most common examples: (1) 𝑓 ∶ℝ→ℝ, 𝑓(𝑥)=−𝑥is an involution known as reflection. (2) 𝑓 ∶ℝ\{0}→ℝ\{0}, 𝑓(𝑥)= 1 u�known as inversion. (3) Let 𝑎,𝑏,𝑐∈ℝ,𝑐𝑏+𝑎2≠0,𝑐≠0, 𝑓 ∶ℝ\{𝑎 𝑐}→ℝ\{𝑎 𝑐}, 𝑓(𝑥)= 𝑎𝑥+𝑏 𝑐𝑥−𝑎 is a family of functions known as bilinear involutions. If 𝑎2+𝑏𝑐> 0, the involution is said hyperbolic and has two fixed points in its domain. The definition of involution can be extended in a natural way to arbitrary sets (not necessarily of real numbers) or, in the following way, to order 𝑛involutions. Definition 1.2.3. Let 𝐴⊂ℝ,𝑓 ∶𝐴→𝐴,𝑛∈ℕ,𝑛≥2. We say that 𝑓is an order 𝑛involution if (1) 𝑓u�≡𝑓∘ u� ⌣ ⋯∘𝑓 =Id, 24 1.2. Involutions and their properties (2) 𝑓u�≠Id ∀𝑘=1,…,𝑛−1. Example 1.2.4. The following is an example of an involution defined on a set which is not a subset of ℝ: 𝑓 ∶ℂ→ℂ,𝑓(𝑧)=𝑒2u� u�u�is an order 𝑛involution on the complex plane. Example 1.2.5. 𝑓(𝑥)=⎧ { { ⎨ { { ⎩ 𝑥, 𝑥∈(−∞,0)∪(𝑛,+∞), 𝑥+1, 𝑥∈(0,1)∪(1,2)∪⋯∪(𝑛−2,𝑛−1), 𝑥−(𝑛−1), 𝑥∈(𝑛−1,𝑛) is an order 𝑛involution in ℝ\{0,1,…,𝑛}. Observe that 𝑓is not defined on a connected set of ℝ, neither admits a continuous extension to a connected set. This fact is related to the statement of Theorem 1.2.9. 1.2.2 Properties of involutions Now we will establish a series of results useful when it comes to study involutions. Proposition 1.2.6. Let 𝐴⊂ℝ,𝑓 ∶𝐴→𝐴be an order 𝑛involution, then 𝑓is invertible. Proof. If ℎ∘𝑔is bijective, then ℎis surjective, since (ℎ∘𝑔)(𝐴)⊂ℎ(𝐴), and 𝑔injective, since 𝑔(𝑥)=𝑔(𝑦)implies (ℎ∘𝑔)(𝑥)=(ℎ∘𝑔)(𝑦). Hence, since 𝑓 ∘𝑓u�−1 =𝑓u�−1 ∘𝑓 =Id,𝑓is bijective (invertible).  The following proposition [189] is a classical result regarding involutions. Here we present it for connected subsets of ℝ. Proposition 1.2.7. Let 𝐴⊂ℝbe connected and 𝑓 ∶𝐴→𝐴an order two continuous involution. Then, (1) 𝑓is strictly decreasing, and (2) 𝑓has a unique fixed point. Proof. (1). Since 𝑓is invertible, it is strictly monotone. 𝑓 ≠ Id, so there exists 𝑥0∈ 𝐴 such that 𝑓(𝑥0)≠𝑥0. Let us assume that 𝑓is increasing. If 𝑥0<𝑓(𝑥0), since 𝐴is connected, 𝑓(𝑥0)<𝑓2(𝑥0)=𝑥0(contradiction) and the same occurs if 𝑓(𝑥0)<𝑥0. Thus, 𝑓is decreasing. (2). Since 𝐴is connected, 𝐴is an interval. Let 𝑎∈𝐴, then 𝑓(𝑎)∈𝐴. Let us assume, without lost of generality, that 𝑓(𝑎) >𝑎. Then, [𝑎,𝑓(𝑎)]⊂𝐴and 𝑓([𝑎,𝑓(𝑎)])=[𝑎,𝑓(𝑎)]. Let 𝑔=𝑓−Id,𝑔is continuous and 𝑔(𝑎)=𝑓(𝑎)−𝑎>0,𝑔(𝑓(𝑎))=𝑎−𝑓(𝑎)<0, therefore, by Bolzano’s Theorem, there exists 𝛼∈(𝑎,𝑓(𝑎))such that 𝑔(𝛼)=0, i.e. 𝑓(𝛼)=𝛼. Since 𝑓is strictly decreasing, such point is unique.  Remark 1.2.8. If 𝐴is not connected, point (2)of Proposition 1.2.7 may not be satisfied. For instance, bilinear involutions have 0or 2fixed points. 1. Involutions and their properties 25 Now we will prove a theorem that illustrates the importance of order two involutions. Similar proofs can be found in [39,132,148]. Theorem 1.2.9. The only continuous involutions defined in connected sets of ℝare of order 2. Proof. Let 𝐴be a connected subset of ℝand 𝑓∶𝐴→𝐴a continuous involution of order 𝑛. Let us prove in several steps that 𝑛=2. (a) 𝑛is even. We will prove first that 𝑓is decreasing. Since 𝑓 ≠Id, there exists 𝑥0∈𝐴such that 𝑓(𝑥0)≠𝑥0. Let us assume that 𝑓is increasing. If 𝑥0<𝑓(𝑥0), using that 𝐴is connected, 𝑥0<𝑓(𝑥0)<𝑓2(𝑥0)<⋯<𝑓u�−1(𝑥0)<𝑓u�(𝑥0)=𝑥0, which is a contradiction. The same happens if 𝑓(𝑥0)<𝑥0. Therefore 𝑓is decreasing. The composition of two functions, both increasing or decreasing is increasing. If one is increasing and the other decreasing, then the composition is a decreasing function. Therefore, if 𝑛is odd and 𝑓is decreasing, 𝑓u�is decreasing, which is absurd since 𝑓u�=Id. (b) 𝑛=2𝑚with 𝑚odd. Otherwise, 𝑛=4𝑘for some 𝑘∈ℕ. Then, if 𝑔=𝑓2u�,𝑔≠Id and 𝑔2=Idand, using Proposition 1.2.7, 𝑔is decreasing, but this is a contradiction since 2𝑘 is even. (c) 𝑛=2. If 𝑛 = 2𝑚with 𝑚odd, 𝑚 ≥3, take 𝑔 = 𝑓2. Then 𝑔 ≠Idand 𝑔u�= Id, so 𝑔is an involution of order 𝑘≤𝑚. But, by part (𝑎), this implies that 𝑔is decreasing, which is impossible since 𝑔=𝑓2. From now on, if we do not specify the order of the involution, we will assume it is of order two. The proof of Proposition 1.2.7 suggests a way of constructing an iterative method convergent to the fixed point of the involution. This is illustrated in the following theorem. Theorem 1.2.10. Let 𝐴⊂ℝbe a connected set, 𝑓 ∶𝐴→𝐴a continuous involution, 𝛼is the unique fixed point of 𝑓and 𝑓of class two in a neighborhood of 𝛼. Then, the iterative method {𝑥0∈𝐴, 𝑥u�+1 =𝑔(𝑥u�), 𝑘=0,1,…, where 𝑔∶= u�+Id 2, is globally convergent to 𝛼and of order at least 2. Proof. Let us consider the closed interval of extremal points 𝑥u�and 𝑓(𝑥u�)that we will denote in this proof by [𝑥u�,𝑓(𝑥u�)]. Since 𝑥u�+1 is the middle point of the interval [𝑥u�,𝑓(𝑥u�)],𝑥u�+1 ∈ [𝑥u�,𝑓(𝑥u�)]and, furthermore, since 𝑓([𝑥u�,𝑓(𝑥u�)])=[𝑥u�,𝑓(𝑥u�)], we have that 𝑓(𝑥u�+1)∈ [𝑥u�,𝑓(𝑥u�)]. Therefore, |𝑓(𝑥u�+1)−𝑥u�+1|≤ 1 2|𝑓(𝑥u�)−𝑥u�|≤⋯≤ 1 2u�+1|𝑓(𝑥0)−𝑥0|. Hence, |𝑥u�+1 −𝑥u�|=∣𝑓(𝑥u�)+𝑥u� 2−𝑥u�∣=1 2|𝑓(𝑥u�)−𝑥u�|≤ 1 2u�|𝑓(𝑥0)−𝑥0|. 32 1.3. Differential Operators with Involutions =(𝑇∗−(𝜑(𝑇))∗)𝐿𝑥−(𝑇∗−(𝜑(𝑇))∗)𝑚𝜑∗𝑥 =ℎ(𝑇)−ℎ(𝜑(𝑇))−𝑚(𝑇)𝑥(𝜑(𝑇))+𝑚(𝜑(𝑇))𝑥(𝜑(𝜑(𝑇))) =ℎ(𝑇)−ℎ(𝜑(𝑇))−𝑚(𝑇)𝑥(𝜑(𝑇))+𝑚(𝑇)𝑥(𝑇) =ℎ(𝑇)−ℎ(𝜑(𝑇)). Hence, any solution of problem (1.3.8) is a solution of problem 𝑅𝐿𝑥=𝑅ℎ, 𝑥(𝜑(𝑇))=𝑥(𝑇), 𝑥′(𝑇)−𝑥′(𝜑(𝑇))=ℎ(𝑇)−ℎ(𝜑(𝑇)). Rewriting this expression, 𝑥″(𝑡)−𝑚′(𝑡) 𝑚(𝑡)𝑥′(𝑡)−𝜑′(𝑡)𝑚(𝜑(𝑡))𝑚(𝑡)𝑥(𝑡) =ℎ′(𝑡)−𝑚(𝑡)𝜑′(𝑡)ℎ(𝜑(𝑡))−𝑚′(𝑡) 𝑚(𝑡)ℎ(𝑡), 𝑥(𝜑(𝑇))=𝑥(𝑇), 𝑥′(𝑇)−𝑥′(𝜑(𝑇))=ℎ(𝑇)−ℎ(𝜑(𝑇)), which is a system of ordinary differential equations with nonhomogeneous boundary conditions. The reverse problem, determining whetherthe solution of this system is a solution of (1.3.8), is more difficult and it is not always the case. We will deepen in this factfurther on and compute the Green’s function in those cases there is a unique solution. 2. General results for differential equations with involutions As mentioned in the Introduction, this chapter is devoted to those results related to differential equations with involution not directly associated with Green’s functions. The proofs of the results can be found in the bibliography cited for each case. We will not deepen into these results, but we summarize their nature for the convenience of the reader. The reader may consult as well the book by Wiener [187] as a good starting point for general results in this direction. It is interesting to observe the progression and different kinds of results collected in this Chapter with those related to Green’s functions that we will show latter on. 2.1 The bases of the study As was pointed out in the introduction, the study of differential equations with reflection starts with the solving of the Siberstein equation in 1940 [156]. Theorem 2.1.1. The equation 𝑥′(𝑡)=𝑥(1 𝑡), 𝑡∈ℝ+, has exactly the following solutions: 𝑥(𝑡)=𝑐√𝑡cos(√3 2ln𝑡−𝜋 6), 𝑐∈ℝ. In Silberstein’s article it was written u� 3instead of u� 6, which appears corrected in [186,187]. Wiener provides a more general result in this line. Theorem 2.1.2 ( [187]).Let 𝑛∈ℝ. The equation 𝑡u�𝑥′(𝑡)=𝑥(1 𝑡), 𝑡∈ℝ+ has exactly the following solutions: 𝑥(𝑡)= ⎧ { { { { { ⎨ { { { { { ⎩ 𝑐𝑡, 𝑛=−1, 𝑐𝑡(1−2ln𝑡), 𝑛=3, 𝑐(𝑡u�1+𝜆1𝑡u�2), 𝑛<−1or 𝑛>3, 𝑐𝑡1−u� 2⎡ ⎢ ⎣cos(𝛼ln𝑡)+√𝑛+1 3−𝑛sin(𝛼ln𝑡)⎤ ⎥ ⎦, 𝑛∈(−1,3), 34 2.1. The bases of the study where 𝑐∈ℝ,𝜆1and 𝜆2are the roots of the polynomial 𝜆2+(𝑛−1)𝜆+1and 𝛼= √(𝑛+1)(3−𝑛) 2. It is also Wiener [186,187] who formalizes the concept of differential equation with involutions. Definition 2.1.3 ([186]).An expression of the form 𝑓(𝑡,𝑥(𝜑1(𝑡),…,𝑥(𝜑u�(𝑡)),…,𝑥u�)(𝜑1(𝑡)),…,𝑥u�)(𝜑u�(𝑡)))=0, 𝑡∈ℝ where 𝜑1,…,𝜑u�are involutions and 𝑓is a real function of 𝑛𝑘+1real variables is called differential equation with involutions. The first objective in the research concerning this kind of equations was to find a way of reducing them to ordinary differential equations of systems of ordinary differential equations. In this sense, we have the following reduction results for the existence of solutions [186,187]. Theorem 2.1.4. Consider the equation 𝑥′(𝑡)=𝑓(𝑡,𝑥(𝑡),𝑥(𝜑(𝑡))), 𝑡∈ℝ (2.1.1) and assume the following hypotheses are satisfied: • The function 𝜑is a continuously differentiable strong involution with fixed point 𝑡0. • The function 𝑓(𝑡,𝑦,𝑧)is defined and is continuously differentiable in the space where its arguments take values. • The equation (2.1.1) is uniquely solvable with respect to 𝑥(𝜑(𝑡)), i.e. there exists a unique function 𝑔(𝑡,𝑥(𝑡),𝑥′(𝑡))such that 𝑥(𝜑(𝑡))=𝑔(𝑡,𝑥(𝑡),𝑥′(𝑡)). Then, the solution of the ordinary differential equation 𝑥″(𝑡)= [𝜕𝑓 𝜕𝑡 +𝑥′(𝑡)𝜕𝑓 𝜕𝑦+𝜑′(𝑡)𝑓(𝜑(𝑡),𝑔(𝑡,𝑥(𝑡),𝑥′(𝑡)),𝑥(𝑡))𝜕𝑓 𝜕𝑧](𝑡,𝑥(𝑡),𝑔(𝑡,𝑥(𝑡),𝑥′(𝑡))), with initial conditions 𝑥(𝑡0)=𝑥0, 𝑥′(𝑡0)=𝑓(𝑡0,𝑥0,𝑥0), is a solution of the equation (2.1.1) with initial conditions 𝑥(𝑡0)=𝑥0. Corollary 2.1.5 ( [186]).Let us assume that in the equation 𝑥′(𝑡)=𝑓(𝑥(𝜑(𝑡))) (2.1.2) 2. The bases of the study 35 the function 𝜑is a continuously differentiable function with a fixed point 𝑡0and the function 𝑓 is monotone and continuously differentiable in ℝ. Then, the solution of the equations 𝑥″(𝑡)=𝑓′(𝑓−1(𝑥′(𝑡)))𝑓(𝑥(𝑡))𝜑′(𝑡), 𝑥(𝜑(𝑡))=𝑓−1(𝑥′(𝑡)), with initial conditions 𝑥(𝑡0)=𝑥0, 𝑥′(𝑡0)=𝑓(𝑥0), is a solution of the equation (2.1.2) with initial condition 𝑥(𝑡0)=𝑥0. In Lemma 3.1.1 (page 39) we prove a result more general than Corollary 2.1.5. There we show the equivalence of 𝑥′(𝑡)=𝑓(𝑥(𝜑(𝑡)))and 𝑥″(𝑡)=𝑓′(𝑓−1(𝑥′(𝑡)))𝑓(𝑥(𝑡))𝜑′(𝑡). Lucić has extended these results to more general ones which include higher order derivatives or different involutions. We refer the reader to [128,129,187]. On the other hand, Šarkovskiĭ [169] studies the equation 𝑥′(𝑡) = 𝑓(𝑥(𝑡),𝑥(−𝑡))and, noting 𝑦(𝑡)∶=𝑥(−𝑡), arrives to the conclusion that the solutions of such equation are solutions of the system 𝑥′(𝑡)=𝑓(𝑥,𝑦), 𝑦′(𝑡)=−𝑓(𝑦,𝑥), with the condition 𝑥(0)=𝑦(0). Then he applies this expression to the stability of differentialdifference equations. We will arrive to this expression by other means in Proposition 3.1.7 (see page 43). The traditional study of differential equations with involutions has been done for the case of connected domains. Watkins [173] extends these results (in particular Theorem 2.1.4) to the case of nonconnected domains, as it is the case of the inversion 1/𝑡in ℝ\{0}. The asymptotic behavior of equations with involutions has also been studied. Theorem 2.1.6 ([174]).Let 𝑎>0. Assume 𝜑∶[0,+∞)→[0,+∞)is a continuously differentiable involution such that 𝜑(𝑥)−𝜑(𝑏)< 1 𝑥−1 𝑏,for all 𝑥,𝑏∈(𝑎,+∞), 𝑥>𝑏. Then the equation 𝑦′(𝑡)=𝑦(𝜑(𝑡))has an oscillatory solution. We will deepen in the fact that such a type of equations oscillate and compute the period later on (see page 124). Related to this oscillatory behavior is the fact, pointed out by Zampieri [196], that involutions are related to a potential of some second order differential equations. Definition 2.1.7. An equilibrium point of a planar vector field is called a (local) center if all orbits in a neighborhood are periodic and enclose it. The center is called isochronous if all periodic orbits have the same period in a neighborhood of the center. 36 2.2. Differential equations with reflection Theorem 2.1.8 ([196]).Let 𝜑∈u�1(𝐽)be an involution, 𝜔>0, and define 𝑉(𝑥)= 𝜔2 8(𝑥−𝜑(𝑥))2, 𝑥∈𝐽. Then the origin is an isochronous center for 𝑥″(𝑡) = −𝑉′(𝑥(𝑡)). Namely, all orbits which intersect 𝐽and the interval of the 𝑥-axis in the 𝑥,𝑥′-plane, are periodic and have the same period 2𝜋/𝜔. On the other hand, if 𝑔is a continuous function defined on a neighborhood of 0∈ℝ, such that 𝑔(0) = 0, there exists 𝑔′(0) > 0and the origin is an isochronous center for 𝑥″(𝑡) = 𝑔(𝑥(𝑡)), then there exist an open interval 𝐽,0∈𝐽, which is a subset of the domain of 𝑔, and an involution 𝜑∶𝐽→𝐽such that ∫u� 0𝑔(𝑦)d𝑦= 𝜔2 8(𝑥−𝜑(𝑥))2, 𝑥∈𝐽, where 𝜔=√𝑔′(0). 2.2 Differential equations with reflection The particular field of differential equations with reflection has been subject to much study motivated by the simplicity of this particular involution and its good algebraic properties. O’Regan [136] studies the existence of solutions for problems of the form 𝑦(u�)(𝑡)=𝑓(𝑡,𝑦(𝑡),𝑦(−𝑡),…,𝑦(u�−1)(𝑡),𝑦(u�−1)(−𝑡)), −𝑇≤𝑡≤𝑇, 𝑦∈ℬ, where ℬrepresents some initial or boundary value conditions, using a nonlinear alternative result. On the same line, existence and uniqueness results are proven by Hai [84] for problems of the kind 𝑥″(𝑡)+𝑐𝑥′(𝑡)+𝑔(𝑡,𝑥(𝑡),𝑥(−𝑡))=ℎ(𝑡), 𝑡∈[−1,1], 𝑥(−1)=𝑎𝑥′(−1), 𝑥(1)=−𝑏𝑥′(1), with 𝑐∈ℝ,𝑎,𝑏≥0. Wiener and Watkins study in [189] the solution of the equation 𝑥′(𝑡)−𝑎𝑥(−𝑡) = 0 with initial conditions. Equation 𝑥′(𝑡)+𝑎𝑥(𝑡)+𝑏𝑥(−𝑡)= 𝑔(𝑡)has been treated by Piao in [141,142]. For the equation 𝑥′(𝑡)+𝑎𝑥(𝑡)+𝑏𝑥(−𝑡)=𝑓(𝑡,𝑥(𝑡),𝑥(−𝑡)), 𝑏≠0, 𝑡∈ℝ, Piao [141] obtains existence results concerning periodic and almost periodic solutions using topological degree techniques (in particular Leray-Schauder Theorem). In [122,155, 173, 187, 189] some results are introduced to transform this kind of problems with involutions and initial conditions into second order ordinary differential equations with initial conditions or first order two dimensional systems, granting that the solution of the last will be a solution to the first. 2. Differential equations with reflection 37 Beyond existence, in all its particular forms, the spectral properties of equations with reflection have also been studied. In [117], the focus is set on the eigenvalue problem 𝑢′(−𝑡)+𝛼𝑢(𝑡)=𝜆𝑢(𝑡), 𝑡∈[−1,1], 𝑢(−1)=𝛾𝑢(1). If 𝛼2∈(−1,1)and 𝛾≠𝛼±√1−𝛼2, the eigenvalues are given by 𝜆u�=√1−𝛼2⎡ ⎢ ⎣𝑘𝜋+arctan⎛ ⎜ ⎝1−𝛾 1+𝛾√1+𝛼 1−𝛼⎞ ⎟ ⎠⎤ ⎥ ⎦, 𝑘∈ℤ, and the related eigenfunctions by 𝑢u�(𝑡)∶=√1+𝛼cos⎡ ⎢ ⎣𝑘𝜋+arctan⎛ ⎜ ⎝1−𝛾 1+𝛾√1+𝛼 1−𝛼⎞ ⎟ ⎠⎤ ⎥ ⎦𝑡 +√1−𝛼sin⎡ ⎢ ⎣𝑘𝜋+arctan⎛ ⎜ ⎝1−𝛾 1+𝛾√1+𝛼 1−𝛼⎞ ⎟ ⎠⎤ ⎥ ⎦𝑡, 𝑘∈ℤ. The study of equations with reflection extends also to partial differential equations. See for instance [23,187]. Furthermore, asymptotic properties and boundedness of the solutions of initial first order problems are studied in [174] and [1] respectively. Second order boundary value problems have been considered in [82, 83, 137, 187] for Dirichlet and Sturm-Liouville boundary value conditions, higher order equations has been studied in [136]. Other techniques applied to problems with reflection of the argument can be found in [5,131,188]. 3. Order one problems with constant coefficients In this chapter we recall some results in [39,40,43]. We start studying the first order operator 𝑥′(𝑡)+𝑚𝑥(−𝑡)coupled with periodic boundary value conditions. We describe the eigenvalues of the operator and obtain the expression of its related Green’s function in the nonresonant case. We also obtain the range of the values of the real parameter 𝑚for which the integral kernel, which provides the unique solution, has constant sign. In this way, we automatically establish maximum and anti-maximum principles for the equation. In the last part of the chapter we generalize these results to the case of antiperiodic and general conditions and study the different maximum and anti-maximum principles derived illustrating them with some examples. Also, we put special attention in the case of initial conditions, in which we obtain the Green’s function in a particular way and undertake a study of its sign in different circumstances. 3.1 Reduction of differential equations with involutions Let us consider the problems 𝑥′(𝑡)=𝑓(𝑥(𝜑(𝑡))), 𝑥(𝑐)=𝑥u�(3.1.1) and 𝑥″(𝑡)=𝑓′(𝑓−1(𝑥′(𝑡)))𝑓(𝑥(𝑡))𝜑′(𝑡), 𝑥(𝑐)=𝑥u�, 𝑥′(𝑐)=𝑓(𝑥u�). (3.1.2) Then we have the following Lemma: Lemma 3.1.1. Let (𝑎,𝑏)⊂ℝand let 𝑓 ∶ℝ→ℝbe a diffeomorphism. Let 𝜑∈u�1((𝑎,𝑏)) be an involution. Let 𝑐be a fixed point of 𝜑. Then 𝑥is a solution of the first order differential equation with involution (3.1.1) if and only if 𝑥is a solution of the second order ordinary differential equation (3.1.2). We note that this lemma improves Corollary 2.1.5. Remark 3.1.2. This result is still valid for 𝑓 ∶𝐽1→𝐽2, being 𝐽1,𝐽2two real intervals as long as the values of the solution 𝑥stay in 𝐽2. We will detail more on this subject in Chapter 6. Proof. That those solutions of (3.1.1) are solutions of (3.1.2) is almost trivial. The boundary conditions are justified by the fact that 𝜑(𝑐)=𝑐. Differentiating (3.1.1) we get 𝑥″(𝑡)=𝑓′(𝑥(𝜑(𝑡)))𝑥′(𝜑(𝑡))𝜑′(𝑡) and, taking into account that 𝑥′(𝜑(𝑡))=𝑓(𝑥(𝑡))by (3.1.1), we obtain (3.1.2). 40 3.1. Reduction of differential equations with involutions Conversely, let 𝑥be a solution of (3.1.2). The equation implies that (𝑓−1)′(𝑥′(𝑡))𝑥″(𝑡)=𝑓(𝑥(𝑡))𝜑′(𝑡). (3.1.3) Integrating from 𝑐to 𝑡in (3.1.3), 𝑓−1(𝑥′(𝑡))−𝑥u�=𝑓−1(𝑥′(𝑡))−𝑓−1(𝑥′(𝑐))=∫u� u�𝑓(𝑥(𝑠))𝜑′(𝑠)d𝑠 (3.1.4) and thus, defining 𝑔(𝑠)∶=𝑓(𝑥(𝜑(𝑠)))−𝑥′(𝑠), we conclude from (3.1.4) that 𝑥′(𝑡)=𝑓(𝑥u�+∫u� u�𝑓(𝑥(𝑠))𝜑′(𝑠)d𝑠) =𝑓(𝑥(𝜑(𝑡))+∫u� u�(𝑓(𝑥(𝑠))−𝑥′(𝜑(𝑠)))𝜑′(𝑠)d𝑠) =𝑓(𝑥(𝜑(𝑡))+∫u�(u�) u�(𝑓(𝑥(𝜑(𝑠)))−𝑥′(𝑠))d𝑠) =𝑓(𝑥(𝜑(𝑡))+∫u�(u�) u�𝑔(𝑠)d𝑠). Let us fix 𝑡 > 𝑐where 𝑥(𝑡)is defined. We will prove that (3.1.1) is satisfied in [𝑐,𝑡](the proof is done analogously for 𝑡<𝑐). Recall that 𝜑has to be decreasing, so 𝜑(𝑡) <𝑐. Also, since 𝑓is a diffeomorphism, the derivative of 𝑓is bounded on [𝑐,𝑡], so 𝑓is Lipschitz on [𝑐,𝑡]. Since 𝑓,𝑥,𝑥′and 𝜑′are continuous, we can define 𝐾1∶=inf{𝛼∈ℝ+∶∣𝑓(𝑥(𝜑(𝑟))+∫u�(u�) u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝜑(𝑟)))∣ ≤𝛼∣∫u�(u�) u�𝑔(𝑠)d𝑠∣∀𝑟∈[𝑐,𝑡]}, and 𝐾2∶=inf{𝛼∈ℝ+∶∣𝑓(𝑥(𝑟)+∫u� u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝑟))∣ ≤𝛼∣∫u� u�𝑔(𝑠)d𝑠∣∀𝑟∈[𝑐,𝑡]}. Let 𝐾 =max{𝐾1,𝐾2}. Now, |𝑔(𝑡)|=∣𝑓(𝑥(𝜑(𝑡))+∫u�(u�) u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝜑(𝑡)))∣≤𝐾∣∫u�(u�) u�𝑔(𝑠)d𝑠∣ ≤−𝐾∫u�(u�) u�|𝑔(𝑠)|d𝑠=−𝐾∫u� u�|𝑔(𝜑(𝑠))|𝜑′(𝑠)d𝑠. Applying this inequality at 𝑟=𝜑(𝑠)inside the integral we deduce that |𝑔(𝑡)|≤−𝐾∫u� u�𝐾∣∫u� u�𝑔(𝑟)d𝑟∣𝜑′(𝑠)d𝑠≤−𝐾2∫u� u�∫u� u�|𝑔(𝑟)|d𝑟 𝜑′(𝑠)d𝑠 =𝐾2|𝜑(𝑡)−𝜑(𝑐)|∫u� u�|𝑔(𝑟)|d𝑟≤𝐾2(𝑐−𝑎)∫u� u�|𝑔(𝑟)|d𝑟. Thus, by Grönwall’s Lemma, 𝑔(𝑡) = 0and hence (3.1.1) is satisfied for all 𝑡 < 𝑏where 𝑥is defined.  Notice that, as an immediate consequence of this result, we have that the unique solution of the equation 𝑥″(𝑡)=−√1+(𝑥′(𝑡))2sinh𝑥(𝑡), 𝑥(0)=𝑥0, 𝑥′(0)=sinh𝑥0, 3. Reduction of differential equations with involutions 41 coincide with the unique solution of 𝑥′(𝑡)=sinh𝑥(−𝑡), 𝑥(0)=𝑥0. Furthermore, Lemma 3.1.1 can be extended, with a very similar proof, to the case with periodic boundary value conditions. Let us consider the equations 𝑥′(𝑡)=𝑓(𝑥(𝜑(𝑡))), 𝑥(𝑎)=𝑥(𝑏) (3.1.5) and 𝑥″(𝑡)=𝑓′(𝑓−1(𝑥′(𝑡)))𝑓(𝑥(𝑡))𝜑′(𝑡), 𝑥(𝑎)=𝑥(𝑏)=𝑓−1(𝑥′(𝑎)). (3.1.6) Lemma 3.1.3. Let [𝑎,𝑏]⊂ℝand let 𝑓 ∶ℝ→ℝbe a diffeomorphism. Let 𝜑∈u�1([𝑎,𝑏]) be an involution such that 𝜑([𝑎,𝑏]) = [𝑎,𝑏]. Then 𝑥is a solution of the first order differential equation with involution (3.1.5) if and only if 𝑥is a solution of the second order ordinary differential equation (3.1.6). Proof. Let 𝑥be a solution of (3.1.5). Since 𝜑(𝑎) = 𝑏we trivially get that 𝑥is a solution of (3.1.6). Let 𝑥be a solution of (3.1.6). As in the proof of the previous lemma, we have that 𝑥′(𝑡)=𝑓(𝑥(𝜑(𝑡))+∫u�(u�) u�𝑔(𝑠)d𝑠), where 𝑔(𝑠)∶=𝑓(𝑥(𝜑(𝑠)))−𝑥′(𝑠). Let 𝐾1∶=inf{𝛼∈ℝ+∶∣𝑓(𝑥(𝜑(𝑟))+∫u�(u�) u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝜑(𝑟)))∣ ≤𝛼∣∫u�(u�) u�𝑔(𝑠)d𝑠∣∀𝑟∈[𝑎,𝑏]}, 𝐾2∶=inf{𝛼∈ℝ+∶∣𝑓(𝑥(𝑟)+∫u� u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝑟))∣ ≤𝛼∣∫u� u�𝑔(𝑠)d𝑠∣∀𝑟∈[𝑎,𝑏]}. 𝐾′ 1∶=inf{𝛼∈ℝ+∶∣𝑓(𝑥(𝜑(𝑟))+∫u�(u�) u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝜑(𝑟)))∣ ≤𝛼∣∫u�(u�) u�𝑔(𝑠)d𝑠∣∀𝑟∈[𝑎,𝑏]}, 𝐾′ 2∶=inf{𝛼∈ℝ+∶∣𝑓(𝑥(𝑟)+∫u� u�𝑔(𝑠)d𝑠)−𝑓(𝑥(𝑟))∣ ≤𝛼∣∫u� u�𝑔(𝑠)d𝑠∣∀𝑟∈[𝑎,𝑏]} 𝐾1,𝐾2be as in the proof of Lemma 3.1.1but changing 𝑐by 𝑎and [𝑐,𝑡]by [𝑎,𝑏]. Let 𝐾′ 1, 𝐾′ 2be as 𝐾1,𝐾2but changing 𝑐by 𝑏. Let 𝐾 =max{𝐾1,𝐾2,𝐾′ 1,𝐾′ 2}. Then, for 𝑡in [𝑎,𝑏], |𝑔(𝑡)|≤𝐾∣∫u�(u�) u�𝑔(𝑠)d𝑠∣≤−𝐾∫u� u�|𝑔(𝜑(𝑠))|𝜑′(𝑠)d𝑠 48 3.2. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) Using (𝐼𝐼𝐼), we deduce that this last expression is equal to ℎ(𝑡)+∫u� −u� [𝑚𝜕𝐺 𝜕𝑡(𝑡,−𝑠)− 𝜕2𝐺 𝜕𝑡𝜕𝑠(𝑡,𝑠)+𝑚2𝐺(−𝑡,−𝑠)−𝑚𝜕𝐺 𝜕𝑠(−𝑡,𝑠)]ℎ(𝑠)d𝑠. which is, by (𝐼𝑉),(𝑉𝐼𝐼),(𝐼𝑋)and (𝑋), equal to ℎ(𝑡)+∫u� −u� (𝑚[𝜕𝐺 𝜕𝑡(𝑡,−𝑠)−𝜕𝐺 𝜕𝑠(−𝑡,𝑠)]+𝜕2𝐺 𝜕𝑡2(𝑡,𝑠)+𝑚2𝐺(𝑡,𝑠))ℎ(𝑠)d𝑠=ℎ(𝑡). Therefore, (3.2.1a) is satisfied. Condition (𝑉)allows us to verify the boundary condition: 𝑢(𝑇)−𝑢(−𝑇) =∫u� −u� [𝑚𝐺(𝑇,−𝑠)−𝜕𝐺 𝜕𝑠(𝑇,𝑠)−𝑚𝐺(−𝑇,−𝑠)+𝜕𝐺 𝜕𝑠(−𝑇,𝑠)]ℎ(𝑠)=0.  As the original Green’s function, 𝐺satisfies several properties. Proposition 3.2.3. 𝐺satisfies the following properties: (𝐼′)u�u� u�u� exists and is continuous in {(𝑡,𝑠)∈𝐼2|𝑠≠𝑡}, (𝐼𝐼′) 𝐺(𝑡,𝑡−)and 𝐺(𝑡,𝑡+)exist for all 𝑡∈𝐼and satisfy 𝐺(𝑡,𝑡−)−𝐺(𝑡,𝑡+)=1 ∀𝑡∈𝐼, (𝐼𝐼𝐼′)u�u� u�u� (𝑡,𝑠)+𝑚𝐺(−𝑡,𝑠)=0for a. e. 𝑡,𝑠∈𝐼, 𝑠≠𝑡, (𝐼𝑉′) 𝐺(𝑇,𝑠)=𝐺(−𝑇,𝑠) ∀𝑠∈(−𝑇,𝑇), (𝑉′) 𝐺(𝑡,𝑠)=𝐺(−𝑠,−𝑡) ∀𝑡,𝑠∈𝐼. Proof. Properties (𝐼′),(𝐼𝐼′)and (𝐼𝑉′)are straightforward from the analogous properties for function 𝐺. (𝐼𝐼𝐼′). In the proof of Proposition 3.2.2 we implicitely showed that function 𝑢defined in (3.2.3), and thus the unique solution of (3.2.1), satisfies 𝑢′(𝑡)=ℎ(𝑡)+∫u� −u� 𝜕𝐺 𝜕𝑡(𝑡,𝑠)ℎ(𝑠)d𝑠. Hence, since 𝑢′(𝑡)−ℎ(𝑡)+𝑚𝑢(−𝑡)=0, ∫u� −u� 𝜕𝐺 𝜕𝑡(𝑡,𝑠)ℎ(𝑠)d𝑠+𝑚∫u� −u� 𝐺(−𝑡,𝑠)ℎ(𝑠)d𝑠=0, 3. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) 49 this is, ∫u� −u� ⎡ ⎢ ⎣𝜕𝐺 𝜕𝑡(𝑡,𝑠)+𝑚𝐺(−𝑡,𝑠)⎤ ⎥ ⎦ℎ(𝑠)d𝑠=0for all ℎ∈L1(𝐼), and thus 𝜕𝐺 𝜕𝑡(𝑡,𝑠)+𝑚𝐺(−𝑡,𝑠)=0for a. e. 𝑡,𝑠∈𝐼, 𝑠≠𝑡. (𝑉′). This result is proven using properties (𝑉𝐼)−(𝑋): 𝐺(−𝑠,−𝑡)=𝑚𝐺(−𝑠,𝑡)−𝜕𝐺 𝜕𝑠(−𝑠,−𝑡)=𝑚𝐺(𝑡,−𝑠)+𝜕𝐺 𝜕𝑡(−𝑠,−𝑡) =𝑚𝐺(𝑡,−𝑠)−𝜕𝐺 𝜕𝑡(𝑠,𝑡)=𝑚𝐺(𝑡,−𝑠)−𝜕𝐺 𝜕𝑠(𝑡,𝑠)=𝐺(𝑡,𝑠).  Remark 3.2.4. Due to the expression of 𝐺given in next section, properties (𝐼𝐼)and (𝐼′)can be improved by adding that 𝐺and 𝐺are analytic on {(𝑡,𝑠) ∈ 𝐼2|𝑠 ≠ 𝑡}and {(𝑡,𝑠) ∈ 𝐼2||𝑠|≠|𝑡|}respectively. Using properties (𝐼𝐼′)−(𝑉′)we obtain the following corollary of Proposition 3.2.2. Corollary 3.2.5. Suppose that 𝑚≠𝑘𝜋/𝑇,𝑘∈ℤ. Then the problem 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡), 𝑡∈𝐼∶=[−𝑇,𝑇], 𝑥(−𝑇)−𝑥(𝑇)=𝜆, with 𝜆∈ℝhas a unique solution given by the expression 𝑢(𝑡)∶=∫u� −u� 𝐺(𝑡,𝑠)ℎ(𝑠)d𝑠+𝜆𝐺(𝑡,−𝑇). 3.2.1 Constant sign of function 𝐺 We will now give a result on the positivity or negativity of the Green’s function for problem (3.2.1). In order to achieve this, we need a new lemma and the explicit expression of the function 𝐺. Let 𝛼∶=𝑚𝑇and 𝐺u�be the Green’s function for problem (3.2.1) for a particular value of the parameter 𝛼. Note that sign(𝛼)=sign(𝑚)because 𝑇is always positive. Lemma 3.2.6. 𝐺u�(𝑡,𝑠)=−𝐺−u�(−𝑡,−𝑠) ∀𝑡,𝑠∈𝐼. Proof. Let 𝑢(𝑡)∶=∫u� −u� 𝐺u�(𝑡,𝑠)ℎ(𝑠)d𝑠be a solution to (3.2.1). Let 𝑣(𝑡)∶=−𝑢(−𝑡). Then 𝑣′(𝑡)−𝑚𝑣(−𝑡)=𝑢′(−𝑡)+𝑚𝑢(𝑡)=ℎ(−𝑡), and therefore 𝑣(𝑡)=∫u� −u� 𝐺−u�(𝑡,𝑠)ℎ(−𝑠)d𝑠. On the other hand, by definition of 𝑣, 𝑣(𝑡)=−∫u� −u� 𝐺u�(−𝑡,𝑠)ℎ(𝑠)d𝑠=−∫u� −u� 𝐺u�(−𝑡,−𝑠)ℎ(−𝑠)d𝑠, therefore we can conclude that 𝐺u�(𝑡,𝑠)=−𝐺−u�(−𝑡,−𝑠)for all 𝑡, 𝑠∈𝐼. 50 3.2. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) Corollary 3.2.7. 𝐺u�is positive if and only if 𝐺−u� is negative on 𝐼2. With this corollary, to make a complete study of the positivity and negativity of the Green’s function, it is enough to find out for what values 𝛼 = 𝑚𝑇 ∈ ℝ+function 𝐺is positive and for which is not. This will be very useful to state maximum and anti-maximum principles for (3.2.1) due to the way we express its solution as an integral operator with kernel 𝐺. Using the algorithm described in [31] we can obtain the explicit expression of 𝐺: 2𝑚sin(𝑚𝑇)𝐺(𝑡,𝑠)=⎧ { ⎨ { ⎩cos𝑚(𝑇+𝑠−𝑡) if 𝑠≤𝑡, cos𝑚(𝑇−𝑠+𝑡) if 𝑠>𝑡. Therefore, 2sin(𝑚𝑇)𝐺(𝑡,𝑠)=⎧ { { { ⎨ { { { ⎩ cos𝑚(𝑇−𝑠−𝑡)+sin𝑚(𝑇+𝑠−𝑡) if −𝑡≤𝑠<𝑡, cos𝑚(𝑇−𝑠−𝑡)−sin𝑚(𝑇−𝑠+𝑡) if −𝑠≤𝑡<𝑠, cos𝑚(𝑇+𝑠+𝑡)+sin𝑚(𝑇+𝑠−𝑡) if −|𝑡|>𝑠, cos𝑚(𝑇+𝑠+𝑡)−sin𝑚(𝑇−𝑠+𝑡) if 𝑡<−|𝑠|. Realize that 𝐺is continuous in {(𝑡,𝑠)∈𝐼2|𝑡≠𝑠}. Making the change of variables 𝑡=𝑇𝑧, 𝑠=𝑇𝑦, we can simplify this expression to 2sin(𝛼)𝐺(𝑧,𝑦)=⎧ { { { ⎨ { { { ⎩ cos𝛼(1−𝑦−𝑧)+sin𝛼(1+𝑦−𝑧) if −𝑧≤𝑦<𝑧, cos𝛼(1−𝑦−𝑧)−sin𝛼(1−𝑦+𝑧) if −𝑦≤𝑧<𝑦, cos𝛼(1+𝑦+𝑧)+sin𝛼(1+𝑦−𝑧) if −|𝑧|>𝑦, cos𝛼(1+𝑦+𝑧)−sin𝛼(1−𝑦+𝑧) if 𝑧<−|𝑦|. Using the trigonometric identity cos(𝑎−𝑏)±sin(𝑎+𝑏)=(cos𝑎±sin𝑎)(cos𝑏±sin𝑏), we can factorise this expression as follows: 2sin(𝛼)𝐺(𝑧,𝑦)=⎧ { { { ⎨ { { { ⎩ [cos𝛼(1−𝑧)+sin𝛼(1−𝑧)][sin𝛼𝑦+cos𝛼𝑦] if −𝑧≤𝑦<𝑧, [cos𝛼𝑧−sin𝛼𝑧][sin𝛼(𝑦−1)+cos𝛼(𝑦−1)] if −𝑦≤𝑧<𝑦, [cos𝛼(1+𝑦)+sin𝛼(1+𝑦)][cos𝛼𝑧−sin𝛼𝑧] if −|𝑧|>𝑦, [cos𝛼𝑦+sin𝛼𝑦][cos𝛼(𝑧+1)−sin𝛼(𝑧+1)] if 𝑧<−|𝑦|. (3.2.5) Note that cos𝜉+sin𝜉>0 ∀𝜉∈(2𝑘𝜋−𝜋 4, 2𝑘𝜋+3𝜋 4), 𝑘∈ℤ cos𝜉+sin𝜉<0 ∀𝜉∈(2𝑘𝜋+3𝜋 4, 2𝑘𝜋+7𝜋 4), 𝑘∈ℤ cos𝜉−sin𝜉>0 ∀𝜉∈(2𝑘𝜋−3𝜋 4, 2𝑘𝜋+𝜋 4), 𝑘∈ℤ cos𝜉−sin𝜉<0 ∀𝜉∈(2𝑘𝜋+𝜋 4, 2𝑘𝜋+5𝜋 4), 𝑘∈ℤ (3.2.6) 3. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) 51 Figure 3.2.1: Plot of the function 𝐺(𝑧,𝑦)for 𝛼= u� 4. As we have seen, the Green’s function 𝐺is not defined on the diagonal of 𝐼2. For easier manipulation, we will define it in the diagonal as follows: 𝐺(𝑡,𝑡)=⎧ { ⎨ { ⎩lim u�→u�+𝐺(𝑡,𝑠) if 𝑚>0 lim u�→u�−𝐺(𝑡,𝑠) if 𝑚<0 for 𝑡∈(−𝑇,𝑇); 𝐺(𝑇,𝑇)= lim u�→u�−𝐺(𝑠,𝑠), 𝐺(−𝑇,−𝑇)= lim u�→−u�+𝐺(𝑠,𝑠) Using expression (3.2.5) and formulae (3.2.6) we can prove the following theorem. Theorem 3.2.8. (1) If 𝛼∈(0,u� 4)then 𝐺is strictly positive on 𝐼2. (2) If 𝛼∈(−u� 4,0)then 𝐺is strictly negative on 𝐼2. (3) If 𝛼= u� 4then 𝐺vanishes on 𝑃∶={(−𝑇,−𝑇),(0,0),(𝑇,𝑇),(𝑇,−𝑇)}and is strictly positive on (𝐼2)\𝑃. (4) If 𝛼=−u� 4then 𝐺vanishes on 𝑃and is strictly negative on (𝐼2)\𝑃. (5) If 𝛼∈ℝ\[−u� 4,u� 4]then 𝐺is not positive nor negative on 𝐼2. Proof. Lemma 3.2.6 allows us to restrict the proof to the positive values of 𝛼. We study here the positive values of 𝐺(𝑧,𝑦)in 𝐴∶={(𝑧,𝑦)∈[−1,1]2|𝑧≥|𝑦|}. The rest of cases are done in an analogous fashion. Let 𝐵1∶=⋃ u�1∈ℤ (1−𝜋 𝛼(2𝑘1+3 4),1−𝜋 𝛼(2𝑘1−1 4)), 52 3.2. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) 𝐵2∶=⋃ u�2∈ℤ 𝜋 𝛼(2𝑘2−1 4,2𝑘2+3 4), 𝐶1∶=⋃ u�1∈ℤ (1−𝜋 𝛼(2𝑘1+7 4),1−𝜋 𝛼(2𝑘1+3 4)), 𝐶2∶=⋃ u�2∈ℤ 𝜋 𝛼(2𝑘2+3 4,2𝑘2+7 4), 𝐵∶={(𝑧,𝑦)∈𝐵1×𝐵2|𝑧>|𝑦|}, and 𝐶∶={(𝑧,𝑦)∈𝐶1×𝐶2|𝑧>|𝑦|}. Realize that 𝐵∩𝐶=∅. Moreover, we have that 𝐺(𝑧,𝑦)>0on 𝐴if and only if 𝐴⊂𝐵∪𝐶. To prove the case 𝐴⊂𝐵, it is a necessary and sufficient condition that [−1,1]⊂𝐵2and [0,1]⊂𝐵1. [−1,1]⊂𝐵2if and only if 𝑘2∈1 2(u� u�−3 4,1 4−u� u�)for some 𝑘2∈ℤ, but, since 𝛼>0, this only happens if 𝑘2=0. In such a case [−1,1]⊂ u� 4u�(−1,3), which implies 𝛼< u� 4. Hence, u� u�>4, so [0,1]⊂(1−3 4u� u�,1+1 4u� u�)=(1−u� u�(2𝑘1+3 4),1−u� u�(2𝑘1−1 4))for 𝑘1=0. Therefore 𝐴⊂𝐵. We repeat this study for the case 𝐴⊂𝐶and all the other subdivisions of the domain of 𝐺, proving the statement.  The following definitions [25] lead to a direct corollary of Theorem 3.2.8. Definition 3.2.9. Let ℱu�(𝐼)be the set of real differentiable functions 𝑓defined on 𝐼such that 𝑓(−𝑇)−𝑓(𝑇)=𝜆. A linear operator 𝑅∶ℱu�(𝐼)→L1(𝐼)is said to be (1) strongly inverse positive on ℱu�(𝐼)if 𝑅𝑥≻0on I ⇒𝑥>0on I ∀𝑥∈ℱu�(𝐼), (2) strongly inverse negative on ℱu�(𝐼)if 𝑅𝑥≻0on I ⇒𝑥<0on I ∀𝑥∈ℱu�(𝐼), where 𝑥≻0stands for 𝑥≥0and ∫u� −u� 𝑥(𝑡)d𝑡>0. Respectively, 𝑥≺0stand for stands for 𝑥≤0and ∫u� −u� 𝑥(𝑡)d𝑡<0. Corollary 3.2.10. The operator 𝑅u�∶ℱu�(𝐼) → L1(𝐼)defined as 𝑅u�(𝑥(𝑡)) = 𝑥′(𝑡)+ 𝑚𝑥(−𝑡), with 𝑚∈ℝ\{0}, satisfies (1) 𝑅u�is strongly inverse positive on ℱu�(𝐼)if and only if 𝑚∈(0, u� 4u�]and 𝜆≥0, (2) 𝑅u�is strongly inverse negative on ℱu�(𝐼)if and only if 𝑚∈[−u� 4u�,0)and 𝜆≥0. This last corollary establishes a maximum and anti-maximum principle (cf. [25, Lemma 2.5, Remark 2.3]). The function 𝐺has a fairly convoluted expression which does not allow us to see in a straightforward way its dependence on 𝑚(see Figure 3.2.1). This dependency can be analyzed, without computing and evaluating the derivative with respect to 𝑚, just using the properties of equation (3.2.1a) in those regions where the operator 𝑅u�is inverse positive or inverse negative. A different method to the one used here but pursuing a similar purpose can be found in [30, Lemma 2.8] for the Green’s function related to the second order Hill’s equation. In [28, Section 1.8] the reader can find a weaker result for 𝑛-th order equations. 3. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) 53 Proposition 3.2.11. Let 𝐺u�u�∶𝐼→ℝbe the Green’s function and 𝑢u�the solution to the problem (3.2.1) with constant 𝑚=𝑚u�, 𝑖=1,2respectively. Then the following assertions hold. (1) If 0<𝑚1<𝑚2≤u� 4u� then 𝑢1>𝑢2>0on 𝐼for every ℎ≻0on 𝐼and 𝐺u�1>𝐺u�2>0 on 𝐼2. (2) If −u� 4u� ≤ 𝑚1< 𝑚2< 0then 0>𝑢1> 𝑢2> 0on 𝐼for every ℎ≻0on 𝐼and 0>𝐺u�1>𝐺u�2on 𝐼2. Proof. (1). Let ℎ ≻ 0in equation (3.2.1a). Then, by Corollary 3.2.10, 𝑢u�> 0on 𝐼,𝑖 = 1,2. We have that 𝑢′ u�(𝑡)+𝑚u�𝑢u�(−𝑡)=ℎ(𝑡) 𝑖=1,2. Therefore, for a. e. 𝑡∈𝐼, 0=(𝑢2−𝑢1)′(𝑡)+𝑚2𝑢2(−𝑡)−𝑚1𝑢1(−𝑡)>(𝑢2−𝑢1)′(𝑡)+𝑚1(𝑢2−𝑢1)(−𝑡), and 0=(𝑢2−𝑢1)(𝑇)−(𝑢2−𝑢1)(−𝑇). Hence, from Corollary 3.2.10, 𝑢2<𝑢1on 𝐼. On the other hand, for all 𝑡∈𝐼, it is satisfied that 0>(𝑢2−𝑢1)(𝑡)=∫u� −u�(𝐺u�2(𝑡,𝑠)−𝐺u�1(𝑡,𝑠))ℎ(𝑠)d𝑠 ∀ℎ≻0. (3.2.7) This makes clear that 0≺𝐺u�2≺𝐺u�1a. e. on 𝐼2. To prove that 𝐺u�2<𝐺u�1on 𝐼2, let 𝑠∈𝐼be fixed, and define 𝑣u�∶ℝ→ℝas the 2𝑇– periodic extension to the whole real line of 𝐺u�u�(⋅,𝑠). Using (𝐼′)–(𝐼𝑉′), we have that 𝑣2−𝑣1is a continuosly differentiable function on 𝐼u�≡ (𝑠,𝑠+2𝑇). Futhermore, it is clear that (𝑣2−𝑣1)′is absolutely continuous on 𝐼u�. Using (𝐼𝐼𝐼′), we have that (𝑣2−𝑣1)′(𝑡)+𝑚2𝑣2(−𝑡)−𝑚1𝑣1(−𝑡)=0 on 𝐼u�. As consequence, 𝑣″ u�(𝑡)+𝑚2 u�𝑣u�(𝑡)=0a. e. on 𝐼u�. Moreover, using (𝐼𝐼′)and (𝐼𝑉′)we know that (𝑣2−𝑣1)(𝑠)=(𝑣2−𝑣1)(𝑠+2𝑇), (𝑣2−𝑣1)′(𝑠)=(𝑣2−𝑣1)′(𝑠+2𝑇). Hence, for all 𝑡∈𝐼u�, we have that 0=(𝑣2−𝑣1)″(𝑡)+𝑚2 2𝑣2(𝑡)−𝑚2 1𝑣1(𝑡)>(𝑣2−𝑣1)″(𝑡)+𝑚2 1(𝑣2−𝑣1)(𝑡). The periodic boundary value conditions, together the fact that for this range of values of 𝑚1, operator 𝑣″+𝑚2 1𝑣is strongly inverse positive (see Corollary 3.2.10), we conclude that 𝑣2<𝑣1on 𝐼u�, this is, 𝐺u�2(𝑡,𝑠)<𝐺u�1(𝑡,𝑠)for all 𝑡,𝑠∈𝐼. (2). This is straightforward using part (1), Lemma 3.2.6 and Theorem 3.2.8: 𝐺u�2(𝑡,𝑠)=−𝐺−u�2(−𝑡,−𝑠)<−𝐺−u�1(−𝑡,−𝑠)=𝐺u�1(𝑡,𝑠)<0 ∀𝑡,𝑠∈𝐼. By equation (3.2.7), 𝑢2<𝑢1on 𝐼.  54 3.2. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) Remark 3.2.12. In (1) and (2) we could have added that 𝑢1< 𝑢2∀ℎ ≺ 0. These are straightforward consequences of the rest of the proposition. The next subsection is devoted to point out some applications of the given results to the existence of solutions of nonlinear periodic boundary value problems. Due to the fact that the proofs follow similar steps to the ones given in some previous papers (see [25, 167]), we omit them. 3.2.2 Lower and upper solutions method Lower and upper solutions methods are a variety of widespread techniques that supply information about the existence –and sometimes construction– of solutions of differential equations. Depending on the particular type of differential equation and the involved boundary value conditions, it is subject to these techniques change but are in general suitable –with proper modifications– to other cases. For this application we will follow the steps in [25] and use Corollary 3.2.10 to establish conditions under which the more general problem 𝑥′(𝑡)=𝑓(𝑡,𝑥(−𝑡)) ∀𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇), (3.2.8) has a solution. Here 𝑓 ∶ 𝐼×ℝ → ℝis an Lp-Carathéodory function, that is, 𝑓(⋅,𝑥)is measurable for all 𝑥∈ℝ,𝑓(𝑡,⋅)is continuous for a. e. 𝑡 ∈ 𝐼, and for every 𝑅>0, there exists ℎu�∈Lp(𝐼)such that, if with |𝑥|<𝑅then |𝑓(𝑡,𝑥)|≤ℎu�(𝑡) for a. e. 𝑡∈𝐼. Definition 3.2.13. We say 𝑢 ∈ u�(𝐼)is an absolutely continuous function in 𝐼if there exists 𝑓 ∈L1(𝐼)such that for all 𝑎∈𝐼, 𝑢(𝑡)=𝑢(𝑎)+∫u� u�𝑓(𝑠)d𝑠, 𝑡∈𝐼. We denote by 𝐴𝐶(𝐼)the set of absolutely continuous functions defined on 𝐼. Definition 3.2.14. We say that 𝛼∈𝐴𝐶(𝐼)is a lower solution of (3.2.8) if 𝛼satisfies 𝛼′(𝑡)≥𝑓(𝑡,𝛼(−𝑡)) for a. e. 𝑡∈𝐼, 𝛼(−𝑇)−𝛼(𝑇)≥0. Definition 3.2.15. We say that 𝛽∈𝐴𝐶(𝐼)is an upper solution of (3.2.8) if 𝛽satisfies 𝛽′(𝑡)≤𝑓(𝑡,𝛽(−𝑡)) for a. e. 𝑡∈𝐼, 𝛽(−𝑇)−𝛽(𝑇)≤0. We establish now a theorem that proves the existence of solutions of (3.2.8) under some conditions. The proof follows the same steps of [25, Theorem 3.1] and we omit it here. Theorem 3.2.16. Let 𝑓 ∶𝐼×ℝ→ℝbe a L1-Carathéodory function. If there exist 𝛼≥𝛽lower and upper solutions of (3.2.8) respectively and 𝑚∈(0, u� 4u�]such that 𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)≥−𝑚(𝑥−𝑦) for a. e. 𝑡∈𝐼with 𝛽(𝑡)≤𝑦≤𝑥≤𝛼(𝑡), 3. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) 55 then there exist two monotone sequences (𝛼u�)u�∈ℕ,(𝛽u�)u�∈ℕ, nonincreasing and nondecreasing respectively, with 𝛼0=𝛼,𝛽0=𝛽, which converge uniformly to the extremal solutions in [𝛽,𝛼]of (3.2.8). Furthermore, the estimate 𝑚= u� 4u� is best possible in the sense that, for every fixed 𝑚> u� 4u�, there are problems with its unique solution outside of the interval [𝛽,𝛼]. In an analogous way we can prove the following theorem. Theorem 3.2.17. Let 𝑓 ∶𝐼×ℝ→ℝbe a 𝐿1-Carathéodory function. If there exist 𝛼≤𝛽lower and upper solutions of (3.2.8) respectively and 𝑚∈[−u� 4u�,0)such that 𝑓(𝑡,𝑥)−𝑓(𝑡,𝑦)≤−𝑚(𝑥−𝑦) for a. e. 𝑡∈𝐼with 𝛼(𝑡)≤𝑦≤𝑥≤𝛽(𝑡), then there exist two monotone sequences (𝛼u�)u�∈ℕ,(𝛽u�)u�∈ℕ, nonincreasing and nondecreasing respectively, with 𝛼0=𝛼,𝛽0=𝛽, which converge uniformly to the extremal solutions in [𝛼,𝛽]of (3.2.8). Furthermore, the estimate 𝑚=−u� 4u� is best possible in the sense that, for every fixed 𝑚<−u� 4u�, there are problems with its unique solution outside of the interval [𝛼,𝛽]. 3.2.3 Existence of solutions via Krasnosel’skiĭ’s Fixed Point Theorem In this section we implement the methods used in [120] for the existence of solutions of second order differential equations to prove new existence results for problem 𝑥′(𝑡)=𝑓(𝑡,𝑥(−𝑡),𝑥(𝑡)) ∀𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇), (3.2.9) where 𝑓 ∶ 𝐼×ℝ×ℝ → ℝis 2𝑇-periodic on 𝑡and an L1-Carathéodory function, that is, 𝑓(⋅,𝑢,𝑣)is measurable for each fixed 𝑢and 𝑣and 𝑓(𝑡,⋅,⋅)is continuous for a. e. 𝑡∈[−𝑇,𝑇], and for each 𝑟>0, there exists 𝜑u�∈L1([−𝑇,𝑇])such that 𝑓(𝑡,𝑢,𝑣)≤𝜑u�(𝑡) for all (𝑢,𝑣)∈[−𝑟,𝑟]×[−𝑟,𝑟], and a. e. 𝑡∈[−𝑇,𝑇]. . Let us first establish the fixed point theorem we are going to use [120]. Definition 3.2.18. Let u�be a real topological vector space. A cone 𝐾in u�is closed set such that is closed under the sum (that is, 𝑥+𝑦∈𝐾for all 𝑥,𝑦∈𝐾), closed under the multiplication by nonnegative scalars (that is 𝜆𝑥∈𝐾for all 𝜆∈[0,+∞),𝑥∈𝐾) and such that 𝐾∩(−𝐾)= {0}(that is, if 𝑥,−𝑥∈𝐾, then 𝑥=0). Theorem 3.2.19 (Krasnosel’skiĭ).Let ℬbe a Banach space, and let u�⊂ℬbe a cone in ℬ. Assume Ω1,Ω2are open subsets of ℬwith 0∈Ω1,Ω1⊂Ω2and let 𝐴∶u�∩(Ω2\Ω1)→u� be a compact and continuous operator such that one of the following conditions is satisfied: (1) ‖𝐴𝑢‖≤‖𝑢‖if 𝑢∈u�∩𝜕Ω1and ‖𝐴𝑢‖≥‖𝑢‖if 𝑢∈u�∩𝜕Ω2, (2) ‖𝐴𝑢‖≥‖𝑢‖if 𝑢∈u�∩𝜕Ω1and ‖𝐴𝑢‖≤‖𝑢‖if 𝑢∈u�∩𝜕Ω2. 56 3.2. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) Then, 𝐴has at least one fixed point in u�∩(Ω2\Ω1). In the following, let 𝑚∈ℝ\{0}and 𝐺be the Green function for problem 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡), 𝑥(−𝑇)=𝑥(𝑇). Let 𝑀=sup{𝐺(𝑡,𝑠) ∶ 𝑡,𝑠∈𝐼},𝐿=inf{𝐺(𝑡,𝑠) ∶ 𝑡,𝑠∈𝐼}. Theorem 3.2.20. Let 𝑚∈(0, u� 4u�). Assume there exist 𝑟,𝑅∈ℝ+,𝑟<𝑅such that 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥0 ∀𝑥,𝑦∈[𝐿 𝑀𝑟,𝑀 𝐿𝑅], a. e. 𝑡∈𝐼. Then, if one of the following conditions holds, (1) 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥ 𝑀 2𝑇𝐿2𝑥 ∀𝑥,𝑦∈[𝐿 𝑀𝑟,𝑟], a. e. 𝑡∈𝐼, 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤ 1 2𝑇𝑀𝑥 ∀𝑥,𝑦∈[𝑅,𝑀 𝐿𝑅], a. e. 𝑡∈𝐼; (2) 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤ 1 2𝑇𝑀𝑥 ∀𝑥,𝑦∈[𝐿 𝑀𝑟,𝑟], a. e. 𝑡∈𝐼, 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥ 𝑀 2𝑇𝐿2𝑥 ∀𝑥,𝑦∈[𝑅,𝑀 𝐿𝑅], a. e. 𝑡∈𝐼; problem (3.2.9) has a positive solution. If ℬ=(u�(𝐼),‖⋅‖∞), by defining the absolutely continuous operator 𝐴∶ℬ→ℬsuch that (𝐴𝑥)(𝑡)∶=∫u� −u� 𝐺(𝑡,𝑠)[𝑓(𝑠,𝑥(−𝑠),𝑥(𝑠))+𝑚𝑥(−𝑠)],d𝑠 we deduce the result following the same steps as in [167]. We present now two corollaries (analogous to the ones in [167]). The first one is obtained by strengthening the hypothesis and making them easier to check. Corollary 3.2.21. Let 𝑚∈(0, u� 4u�),𝑓(𝑡,𝑥,𝑦)≥0for all 𝑥,𝑦∈ℝ+and a. e. 𝑡∈𝐼. Then, if one of the following condition holds: (1) lim u�,u�→0+𝑓(𝑡,𝑥,𝑦) 𝑥=+∞, lim u�,u�→+∞ 𝑓(𝑡,𝑥,𝑦) 𝑥=0, (2) lim u�,u�→0+𝑓(𝑡,𝑥,𝑦) 𝑥=0, lim u�,u�→+∞ 𝑓(𝑡,𝑥,𝑦) 𝑥=+∞ uniformly for a. e. 𝑡∈𝐼, then problem (3.2.9) has a positive solution. 3. Solution of the equation 𝑥′(𝑡)+𝑚𝑥(−𝑡)=ℎ(𝑡) 57 Corollary 3.2.22. Let 𝑚∈(0, u� 4u�). Assume there exist 𝑟,𝑅∈ℝ+,𝑟<𝑅such that 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤0 ∀𝑥,𝑦∈[−𝑀 𝐿𝑅,−𝐿 𝑀𝑟], a. e. 𝑡∈𝐼. Then, if one of the following conditions holds, (1) 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤ 𝑀 2𝑇𝐿2𝑥 ∀𝑥,𝑦∈[−𝑟,−𝐿 𝑀𝑟], a. e. 𝑡∈𝐼, 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥ 1 2𝑇𝑀𝑥 ∀𝑥,𝑦∈[−𝑀 𝐿𝑅,−𝑅], a. e. 𝑡∈𝐼; (2) 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥ 1 2𝑇𝑀𝑥 ∀𝑥,𝑦∈[−𝑟,−𝐿 𝑀𝑟], a. e. 𝑡∈𝐼, 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤ 𝑀 2𝑇𝐿2𝑥 ∀𝑥,𝑦∈[−𝑀 𝐿𝑅,−𝑅], a. e. 𝑡∈𝐼; problem (3.2.9) has a negative solution. Similar results to these –with analogous proofs– can be given when the Green’s function is negative. Theorem 3.2.23. Let 𝑚∈(−u� 4u�,0). Assume there exist 𝑟,𝑅∈ℝ+,𝑟<𝑅such that 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤0 ∀𝑥,𝑦∈[𝑀 𝐿𝑟, 𝐿 𝑀𝑅], a. e. 𝑡∈𝐼. Then, if one of the following conditions holds, (1) 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤ 𝐿 2𝑇𝑀2𝑥 ∀𝑥,𝑦∈[𝑀 𝐿𝑟,𝑟], a. e. 𝑡∈𝐼, 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥ 1 2𝑇𝐿𝑥 ∀𝑥,𝑦∈[𝑅, 𝐿 𝑀𝑅], a. e. 𝑡∈𝐼; (2) 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥ 1 2𝑇𝐿𝑥 ∀𝑥,𝑦∈[𝑀 𝐿𝑟,𝑟], a. e. 𝑡∈𝐼, 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≤ 𝐿 2𝑇𝑀2𝑥 ∀𝑥,𝑦∈[𝑅, 𝐿 𝑀𝑅], a. e. 𝑡∈𝐼; problem (3.2.9) has a positive solution. Corollary 3.2.24. Let 𝑚∈(−u� 4u�,0). Assume there exist 𝑟,𝑅∈ℝ+,𝑟<𝑅such that 𝑓(𝑡,𝑥,𝑦)+𝑚𝑥≥0 ∀𝑥,𝑦∈[−𝐿 𝑀𝑅,−𝑀 𝐿𝑟], a. e. 𝑡∈𝐼. Then, if one of the following conditions holds, 64 3.4. Examples =1 𝑚∫u� −u� ℎ(𝑠)d𝑠− 1 𝑚∫u� −u� 𝐺(𝑡,𝑠)ℎ(𝑠)d𝑠 𝐺(𝑡,−𝑇) ≤1 𝑚∫u� −u� |ℎ(𝑠)|d𝑠+ 1 𝑚∫u� −u� 𝐺(𝑡,𝑠)|ℎ(𝑠)|d𝑠 𝐺(𝑡,−𝑇) ≤⎛ ⎜ ⎜ ⎝1+max u�∈u� maxu�∈u� 𝐺(𝑡,𝑠) 𝐺(𝑡,−𝑇) ⎞ ⎟ ⎟ ⎠‖ℎ‖1 𝑚≤(1+ 𝑀 minu�∈u� 𝐺(𝑡,−𝑇))‖ℎ‖1 𝑚 =(1+ 2𝑀 cot𝛼−1)‖ℎ‖1 𝑚. This provides a new sufficient condition to ensure that 𝑢>0. 𝑐>(1+ 2𝑀 cot𝛼−1)‖ℎ‖1 𝑚=∶𝑘2. Observe that 𝑘2 𝑘1=1+(2𝑀−1)tan𝛼 4𝑀𝛼 . In order to quantify the improvement of the estimate, we have to know the value of 𝑀. Lemma 3.4.6. 𝑀= 1 2(1+csc𝛼). Proof. By [34, Lemma 5.9] we know that, after the change of variable 𝑡=𝑇𝑧,𝑦=𝑇𝑠, (sin𝛼)Φ(𝑦)= max u�∈[−1,1]𝐺(𝑧,𝑦)=⎧ { ⎨ { ⎩cos[𝛼(𝑦−1)+u� 4]cos(𝛼𝑦−u� 4)if 𝑦∈[0,1], cos(𝛼𝑦+u� 4)cos[𝛼(𝑦+1)−u� 4]if 𝑦∈[−1,0). Observe that Φis symmetric, hence, it is enough to study it on [0,1]. Differentiating and equalizing to zero it is easy to check that the maximum is reached at 𝑧= 1 2. Thus, 𝑓(𝛼)∶= 𝑘2 𝑘1=1 2𝛼⋅1+sec𝛼 1+csc𝛼. 𝑓is strictly decreasing on (0,u� 4),𝑓(0+)=1and 𝑓(u� 4−)=2 u�. Example 3.4.7. We give now an example for which we compute the optimal constant 𝑐that ensures the solution is positive and compare it to the aforementioned estimate. Consider the problem 𝑥′(𝑡)+ 𝑥(−𝑡)=𝑒u�, 𝑡∈[−1 2,1 2],∫1 2 −1 2𝑥(𝑠)d𝑠=𝑐. (3.4.3) For this specific case, 𝑘2=cos𝛼+1 cos𝛼−sin𝛼‖ℎ‖1 𝑚=2cot1 4sinh1 2 cot1 4−1 =4.91464… 3. Examples 65 Figure 3.4.1: u�2 u�1 as a function of 𝛼. Figure 3.4.2: Solution of problem (3.4.3) for 𝑐=0.850502… Now, using the expression of 𝐺, it is clear that 𝑢(𝑡)=sinh𝑡+ 𝑐 2sin1 2(cos𝑡−sin𝑡) is the unique solution of problem (3.4.3). It is easy to check that the minimum of the solution is reached at −1for 𝑐∈[0,1]. Also that the solution is positive for 𝑐>2sin1 2sinh1/(cos1+ sin1)=0.850502…, which illustrates that the estimate is far from being optimal. 66 3.5. Solutions of the initial value problem 3.5 Solutions of the initial value problem In this section we analyze a particular case for the boundary conditions in the previous section: the initial – or, better said, middle point– problem. We will show that this specific case admits an interesting way of constructing the Green’s function. The results of the Section follow [43]. 3.5.1 The 𝑛-th order problem Consider the following 𝑛-th order differential equation with involution with involution 𝐿𝑢∶= u� ∑ u�=0[𝑎u�𝑢(u�)(−𝑡)+𝑏u�𝑢(u�)(𝑡)]=ℎ(𝑡), 𝑡∈ℝ; 𝑢(𝑡0)=𝑐, (3.5.1) where ℎ ∈ 𝐿1 loc(ℝ),𝑡0,𝑐,𝑎u�,𝑏u�∈ ℝfor 𝑘 = 0,…𝑛−1;𝑎u�= 0;𝑏u�= 1. A solution to this problem will be a function 𝑢∈𝑊u�,1 loc (ℝ), that is, 𝑢is 𝑘times differentiable in the sense of distributions and each of the derivatives satisfies 𝑢u�)|u�∈ L1(𝐾)for every compact set 𝐾 ⊂ℝand 𝑘=0,…,𝑛. Theorem 3.5.1. Assume that there exist 𝑢and 𝑣, functions such that satisfy u�−u� ∑ u�=0 (𝑖+𝑗 𝑗)[(−1)u�+u�−1𝑎u�+u� 𝑢(u�)(−𝑡)+𝑏u�+u� 𝑢(u�)(𝑡)]=0, 𝑡∈ℝ; 𝑗=0,…,𝑛−1, (3.5.2) u�−u� ∑ u�=0 (𝑖+𝑗 𝑗)[(−1)u�+u�𝑎u�+u� 𝑣(u�)(−𝑡)+𝑏u�+u� 𝑣(u�)(𝑡)]=0, 𝑡∈ℝ; 𝑗=0,…,𝑛−1, (3.5.3) ( 𝑢u�𝑣u�− 𝑢u�𝑣u�)(𝑡)≠0, 𝑡∈ℝ. (3.5.4) and also one of the following (ℎ1) 𝐿 𝑢=0and 𝑢(𝑡0)≠0, (ℎ2) 𝐿 𝑣=0and 𝑣(𝑡0)≠0, (ℎ3) 𝑎0+𝑏0≠0and (𝑎0+𝑏0)∫u�0 0(𝑡0−𝑠)u�−1 𝑣(𝑡0) 𝑢u�(𝑠)− 𝑢(𝑡0) 𝑣u�(𝑠) ( 𝑢u�𝑣u�− 𝑢u�𝑣u�)(𝑠) d𝑠≠1. Then problem (3.5.1) has a solution. Proof. Define 𝜑∶= ℎu�𝑣u�−ℎu�𝑣u� 𝑢u�𝑣u�− 𝑢u�𝑣u�,and 𝜓∶= ℎu�𝑢u�−ℎu�𝑢0 𝑢u�𝑣u�− 𝑢u�𝑣u�. Observe that 𝜑is odd, 𝜓is even and ℎ=𝜑 𝑢+𝜓 𝑣. So, in order to ensure the existence of solution of problem (3.5.1) it is enough to find 𝑦and 𝑧such that 𝐿𝑦=𝜑 𝑢and 𝐿𝑧=𝜓 𝑣for, in that case, defining 𝑢= 𝑦+𝑧, we can conclude that 𝐿𝑢=ℎ. We will deal with the initial condition later on. Take 𝑦= 𝜑 𝑢, where 𝜑(𝑡)∶=∫u� 0∫u�u� 0⋯∫u�2 0𝜑(𝑠1)d𝑠1⋯d𝑠u�=1 (𝑛−1)!∫u� 0(𝑡−𝑠)u�−1𝜑(𝑠)d𝑠. 3. Solutions of the initial value problem 67 Observe that 𝜑is even if 𝑛is odd and vice-versa. In particular, we have that 𝜑(u�)(𝑡)=(−1)u�+u�−1 𝜑(u�)(−𝑡), 𝑗=0,…,𝑛. Thus, 𝐿𝑦(𝑡)= u� ∑ u�=0[𝑎u�( 𝜑 𝑢)(u�)(−𝑡)+𝑏u�( 𝜑 𝑢)(u�)(𝑡)] =u� ∑ u�=0 u� ∑ u�=0 (𝑘 𝑗)[(−1)u�𝑎u�𝜑(u�)(−𝑡) 𝑢(u�−u�)(−𝑡)+𝑏u�𝜑(u�)(𝑡) 𝑢(u�−u�)(𝑡)] =u� ∑ u�=0 u� ∑ u�=0 (𝑘 𝑗)𝜑(u�)(𝑡)[(−1)u�+u�+u�−1𝑎u�𝑢(u�−u�)(−𝑡)+𝑏u�𝑢(u�−u�)(𝑡)] =u� ∑ u�=0 𝜑(u�)(𝑡) u� ∑ u�=u� (𝑘 𝑗)[(−1)u�+u�+u�−1𝑎u�𝑢(u�−u�)(−𝑡)+𝑏u�𝑢(u�−u�)(𝑡)] =u� ∑ u�=0 𝜑(u�)(𝑡)u�−u� ∑ u�=0 (𝑖+𝑗 𝑗)[(−1)u�+u�−1𝑎u�+u� 𝑢(u�)(−𝑡)+𝑏u�+u� 𝑢(u�)(𝑡)]= 𝜑(u�)(𝑡) 𝑢(𝑡) =𝜑(𝑡) 𝑢(𝑡). Hence, 𝐿𝑦=𝜑 𝑢. All the same, by taking 𝑧=  𝜓 𝑣with  𝜓(𝑡)∶= 1 (u�−1)! ∫u� 0(𝑡−𝑠)u�−1𝜓(𝑠)d𝑠, we have that 𝐿𝑧=𝜓 𝑣. Hence, defining 𝑢∶=𝑦+𝑧= 𝜑 𝑢+  𝜓 𝑣we have that 𝑢satisfies 𝐿 𝑢=ℎand 𝑢(0)=0. If we assume (ℎ1), 𝑤= 𝑢+𝑐− 𝑢(𝑡0) 𝑢(𝑡0)𝑢 is clearly a solution of problem (3.5.1). When (ℎ2)is fulfilled a solution of problem (3.5.1) is given by 𝑤= 𝑢+𝑐− 𝑢(𝑡0) 𝑣(𝑡0)𝑣. If (ℎ3)holds, using the aforementioned construction we can find 𝑤1such that 𝐿𝑤1=1 and 𝑤1(0)=0. Now, 𝑤2∶=𝑤1−1/(𝑎0+𝑏0)satisfies 𝐿𝑤2=0. Observe that the second part of condition (ℎ3)is precisely 𝑤2(𝑡0)≠0, and hence, defining 𝑤= 𝑢+𝑐− 𝑢(𝑡0) 𝑤2(𝑡0)𝑤2 we have that 𝑤is a solution of problem (3.5.1).  Remark 3.5.2. Having in mind condition (ℎ1)in Theorem 3.5.1, it is immediate to verify that 𝐿 𝑢=0provided that 68 3.5. Solutions of the initial value problem 𝑎u�=0for all 𝑖∈{0,…,𝑛−1}such that 𝑛+𝑖is even. In an analogous way, for (ℎ2), one can show that 𝐿 𝑣=0when 𝑎u�=0for all 𝑖∈{0,…,𝑛−1}such that 𝑛+𝑖is odd. 3.5.2 The first order problem After proving the general result for the 𝑛-th order case, we concentrate our work in the first order problem 𝑢′(𝑡)+𝑎𝑢(−𝑡)+𝑏𝑢(𝑡)=ℎ(𝑡), for a. e. 𝑡∈ℝ; 𝑢(𝑡0)=𝑐, (3.5.5) with ℎ∈L1loc(ℝ)and 𝑡0,𝑎,𝑏,𝑐∈ℝ. A solution of this problem will be 𝑢∈𝑊1,1 loc (ℝ). In order to do so, we first study the homogeneous equation 𝑢′(𝑡)+𝑎𝑢(−𝑡)+𝑏𝑢(𝑡)=0, 𝑡∈ℝ. (3.5.6) By differentiating and making the proper substitutions we arrive to the equation 𝑢″(𝑡)+(𝑎2−𝑏2)𝑢(𝑡)=0, 𝑡∈ℝ. (3.5.7) Let 𝜔∶=√|𝑎2−𝑏2|. Equation (3.5.7) presents three different cases: (C1). 𝑎2>𝑏2.In such a case, 𝑢(𝑡)=𝛼cos𝜔𝑡+𝛽sin𝜔𝑡is a solution of (3.5.7) for every 𝛼,𝛽∈ℝ. If we impose equation (3.5.6) to this expression we arrive to the general solution 𝑢(𝑡)=𝛼(cos𝜔𝑡−𝑎+𝑏 𝜔sin𝜔𝑡) of equation (3.5.6) with 𝛼∈ℝ. (C2). 𝑎2< 𝑏2.Now, 𝑢(𝑡) = 𝛼cosh𝜔𝑡+𝛽sinh𝜔𝑡is a solution of (3.5.7) for every 𝛼,𝛽∈ℝ. To get equation (3.5.6) we arrive to the general solution 𝑢(𝑡)=𝛼(cosh𝜔𝑡−𝑎+𝑏 𝜔sinh𝜔𝑡) of equation (3.5.6) with 𝛼∈ℝ. (C3). 𝑎2=𝑏2.In this a case, 𝑢(𝑡)=𝛼𝑡+𝛽is a solution of (3.5.7) for every 𝛼,𝛽 ∈ℝ. So, equation (3.5.6) holds provided that one of the two following cases is fulfilled: (C3.1). 𝑎=𝑏,where 𝑢(𝑡)=𝛼(1−2𝑎𝑡) is the general solution of equation (3.5.6) with 𝛼∈ℝ, and (C3.2). 𝑎=−𝑏,where 𝑢(𝑡)=𝛼 is the general solution of equation (3.5.6) with 𝛼∈ℝ. 3. Solutions of the initial value problem 69 Now, according to Theorem 3.5.1, we denote 𝑢,𝑣satisfying 𝑢′(𝑡)+𝑎 𝑢(−𝑡)+𝑏 𝑢(𝑡)=0, 𝑢(0)=1, (3.5.8) 𝑣′(𝑡)−𝑎 𝑣(−𝑡)+𝑏 𝑣(𝑡)=0, 𝑣(0)=1. (3.5.9) Observe that 𝑢and 𝑣can be obtained from the explicit expressions of the cases (C1)–(C3) by taking 𝛼=1. Remark 3.5.3. Note that if 𝑢is in the case (C3.1), 𝑣is in the case (C3.2) and vice-versa. We have now the following properties of functions 𝑢and 𝑣. Lemma 3.5.4. For every 𝑡,𝑠∈ℝ, the following properties hold. (1) 𝑢u�≡ 𝑣u�,𝑢u�≡𝑘 𝑣u�for some real constant 𝑘almost everywhere, (2) 𝑢u�(𝑠) 𝑣u�(𝑡)= 𝑢u�(𝑡) 𝑣u�(𝑠),𝑢u�(𝑠) 𝑣u�(𝑡)= 𝑢u�(𝑡) 𝑣u�(𝑠), (3) 𝑢u�𝑣u�− 𝑢u�𝑣u�≡1. (4) 𝑢(𝑠) 𝑣(−𝑠)+ 𝑢(−𝑠) 𝑣(𝑠)=2[ 𝑢u�(𝑠) 𝑣u�(𝑠)− 𝑢u�(𝑠) 𝑣u�(𝑠)]=2. Proof. (1)and (3)can be checked by inspection of the different cases. (2)is a direct consequence of (1).(4)is obtained from the definition of even and odd parts and (3). Now, Theorem 3.5.1 has the following corollary. Corollary 3.5.5. Problem (3.5.5) has a unique solution if and only if 𝑢(𝑡0)≠0. Proof. Considering Lemma 3.5.4 (3), 𝑢and 𝑣, defined as in (3.5.8) and (3.5.9) respectively, satisfy the hypothesis of Theorem 3.5.1, (ℎ1), therefore a solution exists. Now, assume 𝑤1and 𝑤2are two solutions of (3.5.5). Then 𝑤2−𝑤1is a solution of (3.5.6). Hence, 𝑤2−𝑤1is of one of the forms covered in the cases (C1)–(C3) and, in any case, a multiple of 𝑢, that is 𝑤2−𝑤1=𝜆 𝑢for some 𝜆∈ℝ. Also, it is clear that (𝑤2−𝑤1)(𝑡0)=0, but we have 𝑢(𝑡0)≠0as a hypothesis, therefore 𝜆=0and 𝑤1=𝑤2. This is, problem (3.5.5) has a unique solution. Assume now that 𝑤is a solution of (3.5.5) and 𝑢(𝑡0)=0. Then 𝑤+𝜆 𝑢is also a solution of (3.5.5) for every 𝜆∈ℝ, which proves the result.  This last Theorem raises an obvious question: In which circumstances 𝑢(𝑡0)≠0? In order to answer this question, it is enough to study the cases (C1)–(C3). We summarize this study in the following Lemma which can be checked easily. Lemma 3.5.6. 𝑢(𝑡0)=0only in the following cases, • if 𝑎2>𝑏2and 𝑡0=1 u�(arctan u� u�+u� +𝑘𝜋)for some 𝑘∈ℤ, • if 𝑎2<𝑏2,𝑎𝑏>0†and 𝑡0=1 u�arctanh u� u�+u�, †𝑎𝑏>0is equivalent to |𝑏−𝑎|<|𝑏+𝑎|. 70 3.5. Solutions of the initial value problem • if 𝑎=𝑏and 𝑡0=1 2u�. Definition 3.5.7. Let 𝑡1,𝑡2∈ ℝ. We define the oriented characteristic function of the pair (𝑡1,𝑡2)as 𝜒u�2 u�1(𝑡)∶=⎧ { { ⎨ { { ⎩ 1, 𝑡1≤𝑡≤𝑡2, −1, 𝑡2≤𝑡<𝑡1, 0, otherwise. Remark 3.5.8. The previous definition implies that, for any given integrable function 𝑓 ∶ℝ→ ℝ,∫u�2 u�1𝑓(𝑠)d𝑠=∫∞ −∞ 𝜒u�2 u�1(𝑠)𝑓(𝑠)d𝑠. Also, 𝜒u�2 u�1=−𝜒u�1 u�2. The following corollary gives us the expression of the Green’s function for problem (3.5.5). Corollary 3.5.9. Suppose 𝑢(𝑡0)≠0. Then the unique solution of problem (3.5.5) is given by 𝑢(𝑡)∶=∫∞ −∞ 𝐺(𝑡,𝑠)ℎ(𝑠)d𝑠+𝑐− 𝑢(𝑡0) 𝑢(𝑡0)𝑢(𝑡), 𝑡∈ℝ, where 𝐺(𝑡,𝑠)∶= 1 2([ 𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)]𝜒u� 0(𝑠)+[ 𝑢(−𝑠) 𝑣(𝑡)− 𝑣(−𝑠) 𝑢(𝑡)]𝜒0 −u�(𝑠)), (3.5.10) for every 𝑡,𝑠∈ℝ. Proof. First observe that 𝐺(𝑡,⋅)is bounded and of compact support for every fixed 𝑡∈ℝ, so the integral ∫∞ −∞ 𝐺(𝑡,𝑠)ℎ(𝑠)d𝑠is well defined. It is not difficult to verify, for any 𝑡∈ℝ, the following equalities: 𝑢′(𝑡)−𝑐− 𝑢(𝑡0) 𝑢(𝑡0)𝑢′(𝑡)=1 2(d d𝑡∫u� 0[𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)]ℎ(𝑠)d𝑠 +d d𝑡∫0 −u� [𝑢(−𝑠) 𝑣(𝑡)− 𝑣(−𝑠) 𝑢(𝑡)]ℎ(𝑠)d𝑠) =1 2(d d𝑡∫u� 0[𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)]ℎ(𝑠)d𝑠 +d d𝑡∫u� 0[𝑢(𝑠) 𝑣(𝑡)− 𝑣(𝑠) 𝑢(𝑡)]ℎ(−𝑠)d𝑠) =ℎ(𝑡)+1 2(∫u� 0[𝑢(−𝑠) 𝑣′(𝑡)+ 𝑣(−𝑠) 𝑢′(𝑡)]ℎ(𝑠)d𝑠 +∫u� 0[𝑢(𝑠) 𝑣′(𝑡)− 𝑣(𝑠) 𝑢′(𝑡)]ℎ(−𝑠)d𝑠). (3.5.11) On the other hand, 𝑎[𝑢(−𝑡)−𝑐− 𝑢(𝑡0) 𝑢(𝑡0)𝑢(−𝑡)]+𝑏[𝑢(𝑡)−𝑐− 𝑢(𝑡0) 𝑢(𝑡0)𝑢(𝑡)] =1 2𝑎∫−u� 0([ 𝑢(−𝑠) 𝑣(−𝑡)+ 𝑣(−𝑠) 𝑢(−𝑡)]ℎ(𝑠) 3. Solutions of the initial value problem 71 +[ 𝑢(𝑠) 𝑣(−𝑡)− 𝑣(𝑠) 𝑢(−𝑡)]ℎ(−𝑠))d𝑠 +1 2𝑏∫u� 0([ 𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)]ℎ(𝑠)+[ 𝑢(𝑠) 𝑣(𝑡)− 𝑣(𝑠) 𝑢(𝑡)]ℎ(−𝑠))d𝑠 =−1 2𝑎∫u� 0([ 𝑢(𝑠) 𝑣(−𝑡)+ 𝑣(𝑠) 𝑢(−𝑡)]ℎ(−𝑠) +[ 𝑢(−𝑠) 𝑣(−𝑡)− 𝑣(−𝑠) 𝑢(−𝑡)]ℎ(𝑠))d𝑠 +1 2𝑏∫u� 0([ 𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)]ℎ(𝑠)+[ 𝑢(𝑠) 𝑣(𝑡)− 𝑣(𝑠) 𝑢(𝑡)]ℎ(−𝑠))d𝑠 =1 2∫u� 0(−𝑎[ 𝑢(−𝑠) 𝑣(−𝑡)− 𝑣(−𝑠) 𝑢(−𝑡)]+𝑏[ 𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)])ℎ(𝑠)d𝑠 +1 2∫u� 0(−𝑎[ 𝑢(𝑠) 𝑣(−𝑡)+ 𝑣(𝑠) 𝑢(−𝑡)]+𝑏[ 𝑢(𝑠) 𝑣(𝑡)− 𝑣(𝑠) 𝑢(𝑡)])ℎ(−𝑠)d𝑠 =1 2∫u� 0( 𝑢(−𝑠)[−𝑎 𝑣(−𝑡)+𝑏 𝑣(𝑡)]+ 𝑣(−𝑠)[𝑎 𝑢(−𝑡)+𝑏 𝑢(𝑡)]ℎ(𝑠)d𝑠 +1 2∫u� 0( 𝑢(𝑠)[−𝑎 𝑣(−𝑡)+𝑏 𝑣(𝑡)]− 𝑣(𝑠)[𝑎 𝑢(−𝑡)+𝑏 𝑢(𝑡)])ℎ(−𝑠)d𝑠 =−1 2(∫u� 0( 𝑢(−𝑠) 𝑣′(𝑡)+ 𝑣(−𝑠) 𝑢′(𝑡))ℎ(𝑠)d𝑠 +∫u� 0( 𝑢(𝑠) 𝑣′(𝑡)− 𝑣(𝑠) 𝑢′(𝑡))ℎ(−𝑠)d𝑠).(3.5.12) Thus, adding (3.5.11) and (3.5.12), it is clear that 𝑢′(𝑡)+𝑎𝑢(−𝑡)+𝑏𝑢(𝑡)=ℎ(𝑡). We now check the initial condition. 𝑢(𝑡0)=𝑐− 𝑢(𝑡0)+ 1 2∫u�0 0([ 𝑢(−𝑠) 𝑣(𝑡0)+ 𝑣(−𝑠) 𝑢(𝑡0)]ℎ(𝑠)+[ 𝑢(𝑠) 𝑣(𝑡0)− 𝑣(𝑠) 𝑢(𝑡0)]ℎ(−𝑠))d𝑠. It can be directly checked that, for all 𝑡∈ℝ, 𝑢(𝑡)= 1 2∫u� 0([ 𝑢(−𝑠) 𝑣(𝑡)+ 𝑣(−𝑠) 𝑢(𝑡)]ℎ(𝑠)+[ 𝑢(𝑠) 𝑣(𝑡)− 𝑣(𝑠) 𝑢(𝑡)]ℎ(−𝑠))d𝑠, is a solution of problem (3.5.5), which proves the result.  Denote now 𝐺u�,u� the Green’s function for problem (3.5.5) with constant coefficients 𝑎and 𝑏. The following Lemma is analogous to Lemma 3.2.6. Lemma 3.5.10. 𝐺u�,u�(𝑡,𝑠)=−𝐺−u�,−u�(−𝑡,−𝑠), for all 𝑡,𝑠∈𝐼. Proof. Let 𝑢(𝑡)∶=∫∞ −∞ 𝐺u�,u�(𝑡,𝑠)ℎ(𝑠)d𝑠be the solution to 𝑢′(𝑡)+𝑎𝑢(−𝑡)+𝑏𝑢(𝑡)=ℎ(𝑡), 𝑢(0)=0. Let 𝑣(𝑡) ∶= −𝑢(−𝑡). Then 𝑣′(𝑡)−𝑎𝑣(−𝑡)−𝑏𝑣(𝑡) = ℎ(−𝑡), and therefore 𝑣(𝑡) = ∫∞ −∞ 𝐺−u�,−u�(𝑡,𝑠)ℎ(−𝑠)d𝑠. On the other hand, by definition of 𝑣, 𝑣(𝑡)=−∫∞ −∞ 𝐺u�,u�(−𝑡,𝑠)ℎ(𝑠)d𝑠=−∫∞ −∞ 𝐺u�,u�(−𝑡,−𝑠)ℎ(−𝑠)d𝑠, therefore we can conclude that 𝐺u�,u�(𝑡,𝑠)=−𝐺−u�,−u�(−𝑡,−𝑠)for all 𝑡, 𝑠∈𝐼. 72 3.6. Sign of the Green’s Function As a consequence of the previous result, we arrive at the following immediate conclusion. Corollary 3.5.11. 𝐺u�,u� is positive in 𝐼2if and only if 𝐺−u�,−u� is negative on 𝐼2. 3.6 Sign of the Green’s Function In this section we use the above obtained expressions to obtain the explicit expression of the Green’s function, depending on the values of the constants 𝑎and 𝑏. Moreover we study the sign of the function and deduce suitable comparison results. We separate the study in three cases, taking into consideration the expression of the general solution of equation (3.5.6). 3.6.1 The case (C1) Now, assume the case (𝐶1), i.e., 𝑎2> 𝑏2. Using equation (3.5.10), we get the following expression of 𝐺for this situation: 𝐺(𝑡,𝑠)=[cos(𝜔(𝑠−𝑡))+ 𝑏 𝜔sin(𝜔(𝑠−𝑡))]𝜒u� 0(𝑠)+ 𝑎 𝜔sin(𝜔(𝑠+𝑡))𝜒0 −u�(𝑠), which we can rewrite as 𝐺(𝑡,𝑠)= ⎧ { { { { { { { ⎨ { { { { { { { ⎩ cos𝜔(𝑠−𝑡)+ 𝑏 𝜔sin𝜔(𝑠−𝑡), 0≤𝑠≤𝑡, −cos𝜔(𝑠−𝑡)− 𝑏 𝜔sin𝜔(𝑠−𝑡), 𝑡≤𝑠≤0, 𝑎 𝜔sin𝜔(𝑠+𝑡), −𝑡≤𝑠<0, −𝑎 𝜔sin𝜔(𝑠+𝑡), 0<𝑠≤−𝑡, 0, otherwise. (3.6.1a) (3.6.1b) (3.6.1c) (3.6.1d) (3.6.1e) Studying the expression of 𝐺we can obtain maximum and antimaximum principles. In order to do this, we will be interested in those maximal strips (in the sense of inclusion) of the kind [𝛼,𝛽]×ℝwhere 𝐺does not change sign depending on the parameters. So, we are in a position to study the sign of the Green’s function in the different triangles of definition. The result is the following: Lemma 3.6.1. Assume 𝑎2>𝑏2and define 𝜂(𝑎,𝑏)∶= ⎧ { { { { { ⎨ { { { { { ⎩ 1 √𝑎2−𝑏2arctan√𝑎2−𝑏2 𝑏, 𝑏>0, 𝜋 2|𝑎|, 𝑏=0, 1 √𝑎2−𝑏2(arctan√𝑎2−𝑏2 𝑏+𝜋), 𝑏<0. 3. Sign of the Green’s Function 73 Then, the Green’s function of problem (3.5.5) is • positive on {(𝑡,𝑠), 0<𝑠<𝑡}if and only if 𝑡∈(0,𝜂(𝑎,𝑏)), • negative on {(𝑡,𝑠), 𝑡<𝑠<0}if and only if 𝑡∈(−𝜂(𝑎,−𝑏),0). If 𝑎>0, the Green’s function of problem (3.5.5) is • positive on {(𝑡,𝑠), −𝑡<𝑠<0}if and only if 𝑡∈(0,𝜋/√𝑎2−𝑏2), • positive on {(𝑡,𝑠), 0<𝑠<−𝑡}if and only if 𝑡∈(−𝜋/√𝑎2−𝑏2,0), and, if 𝑎<0, the Green’s function of problem (3.5.5) is • negative on {(𝑡,𝑠), −𝑡<𝑠<0}if and only if 𝑡∈(0,𝜋/√𝑎2−𝑏2), • negative on {(𝑡,𝑠), 0<𝑠<−𝑡}if and only if 𝑡∈(−𝜋/√𝑎2−𝑏2,0). Proof. For 0<𝑏<𝑎, the argument of the sinin (3.6.1c) is positive, so (3.6.1c) is positive for 𝑡<𝜋/𝜔. On the other hand, it is easy to check that (3.6.1a) is positive as long as 𝑡<𝜂(𝑎,𝑏). The rest of the proof continues similarly.  As a corollary of the previous result we obtain the following one: Lemma 3.6.2. Assume 𝑎2>𝑏2. Then, • if 𝑎>0, the Green’s function of problem (3.5.5) is nonnegative on [0,𝜂(𝑎,𝑏)]×ℝ, • if 𝑎<0, the Green’s function of problem (3.5.5) is nonpositive on [−𝜂(𝑎,−𝑏),0]×ℝ, • the Green’s function of problem (3.5.5) changes sign in any other strip not a subset of the aforementioned. Proof. The proof follows from the previous result together with the fact that 𝜂(𝑎,𝑏)≤ 𝜋 2𝜔 <𝜋 𝜔.  Remark 3.6.3. Realize that the strips defined in the previous Lemma are optimal in the sense that 𝐺changes sign in a bigger rectangle. The same observation applies to the similar results we will prove for the other cases. This fact implies that we cannot have maximum or anti-maximum principles on bigger intervals for the solution, something that is widely known and which the following results, together with Example 3.6.12, illustrate. 80 3.6. Sign of the Green’s Function Example 3.6.16. Consider the problem 𝑥′(𝑡)+𝜆(𝑥(−𝑡)−𝑥(𝑡))= 𝜆𝑡2−2𝑡+𝜆 (1+𝑡2)2, 𝑥(0)=𝜆 for 𝜆∈ℝ. We can apply the theory in order to get the solution 𝑢(𝑡)= 1 1+𝑡2+𝜆(1+2𝜆𝑡)arctan𝑡−𝜆2ln(1+𝑡2)+𝜆−1 where 𝑢(𝑡)= 1 1+u�2+𝜆(1+2𝜆𝑡)arctan𝑡−𝜆2ln(1+𝑡2)−1. Observe that the real function ℎ(𝑡)∶= 𝜆𝑡2−2𝑡+𝜆 (1+𝑡2)2 is positive on ℝif 𝜆>1and negative on ℝfor all 𝜆<−1. Therefore, Lemma 3.6.14 guarantees that 𝑢will be positive on (0,∞)for 𝜆>1and in (−∞,0)when 𝜆<−1. 4. The nonconstant case In the previous chapter we dealt with order one differential equations with reflection, constant coefficients and different boundary conditions. Now, following [41] we reduce a new, more general problem containing nonconstant coefficients and arbitrary differentiable involutions, to the one studied in Chapter 3. As we will see, we will do this in three steps. First weadd a term depending on 𝑥(𝑡)which does not change much with respect to the previous situations. Then, moving from the reflection to a general involution is fairly simple using some of the knowledge gathered in Chapter 1. The last step, changing from constant to nonconstant coefficients, is another matter. In the nonconstant case computing the Green’s function gets trickier and it is only possible in some situations. We use a special change of variable (only valid in some cases) that allows the obtaining the Green’s function of problems with nonconstant coefficients from the Green’s functions of constant-coefficient analogs. 4.1 Order one linear problems with involutions Assume 𝜑is a differentiable involution on a compact interval 𝐽1⊂ℝ. Let 𝑎,𝑏,𝑐,𝑑∈L1(𝐽1) and consider the following problem 𝑑(𝑡)𝑥′(𝑡)+𝑐(𝑡)𝑥′(𝜑(𝑡))+𝑏(𝑡)𝑥(𝑡)+𝑎(𝑡)𝑥(𝜑(𝑡))=ℎ(𝑡), 𝑥(inf𝐽1)=𝑥(sup𝐽1). (4.1.1) It would be interesting to know under what circumstances problem (4.1.1) is equivalent to another problem of the same kind but with a different involution, in particular the reflection. The following corollary of Lemma 1.2.14 will help us to clarify this situation. Corollary 4.1.1 (CHANGE OF INVOLUTION).Under the hypothesis of Lemma 1.2.14, problem (4.1.1) is equivalent to 𝑑(𝑓(𝑠)) 𝑓′(𝑠) 𝑦′(𝑠)+ 𝑐(𝑓(𝑠)) 𝑓′(𝜓(𝑠))𝑦′(𝜓(𝑠))+𝑏(𝑓(𝑠))𝑦(𝑠)+𝑎(𝑓(𝑠))𝑦(𝜓(𝑠))=ℎ(𝑓(𝑠)), 𝑦(inf𝐽2)=𝑦(sup𝐽2). (4.1.2) Proof. Consider the change of variable 𝑡 = 𝑓(𝑠)and 𝑦(𝑠) ∶= 𝑥(𝑡) = 𝑥(𝑓(𝑠)). Then, using Lemma 1.2.14, it is clear that d𝑦 d𝑠(𝑠)= d𝑥 d𝑡(𝑓(𝑠))d𝑓 d𝑠(𝑠) and d𝑦 d𝑠(𝜓(𝑠))= d𝑥 d𝑡(𝜑(𝑓(𝑠)))d𝑓 d𝑠(𝜓(𝑠)). Making the proper substitutions in problem (4.1.1) we get problem (4.1.2) and vice-versa.  82 4.2. Study of the homogeneous equation This last results allows us to restrict our study of problem (4.1.1) to the case where 𝜑is the reflection 𝜑(𝑡)=−𝑡. Now, take 𝑇 ∈ ℝ+,𝐼 ∶= [−𝑇,𝑇]. Equation (4.1.1), for the case 𝜑(𝑡) = −𝑡, can be reduced to the following system Λ(𝑥′ u� 𝑥′ u�)=(𝑎u�−𝑏u�−𝑎u�−𝑏u� 𝑎u�−𝑏u�−𝑎u�−𝑏u�)(𝑥u� 𝑥u�)+(ℎu� ℎu�), where Λ=(𝑐u�+𝑑u�𝑑u�−𝑐u� 𝑐u�+𝑑u�𝑑u�−𝑐u�). To see this, just compute the even and odd parts of both sides of the equation taking into account Corollary 1.1.7. Now, if det(Λ(𝑡))=𝑑(𝑡)𝑑(−𝑡)−𝑐(𝑡)𝑐(−𝑡)≠0for a. e. 𝑡∈𝐼,Λ(𝑡)is invertible a. e. and (𝑥′ u� 𝑥′ u�)=Λ−1 (𝑎u�−𝑏u�−𝑎u�−𝑏u� 𝑎u�−𝑏u�−𝑎u�−𝑏u�)(𝑥u� 𝑥u�)+Λ−1 (ℎu� ℎu�). So the general case where 𝑐  ≡0is reduced to the case 𝑐=0, taking Λ−1 (𝑎u�−𝑏u�−𝑎u�−𝑏u� 𝑎u�−𝑏u�−𝑎u�−𝑏u�) as coefficient matrix. Hence, in the following section we will further restrict our assumptions to the case where 𝑐≡0in problem (4.1.1). 4.2 Study of the homogeneous equation In this section we will study some different cases for the homogeneous equation 𝑥′(𝑡)+𝑎(𝑡)𝑥(−𝑡)+𝑏(𝑡)𝑥(𝑡)=0, 𝑡∈𝐼, (4.2.1) where 𝑎,𝑏∈L1(𝐼). The solutions of equation (4.2.1) satisfy (𝑥′ u� 𝑥′ u�)=(𝑎u�−𝑏u�−𝑎u�−𝑏u� 𝑎u�−𝑏u�−𝑎u�−𝑏u�)(𝑥u� 𝑥u�).(4.2.2) Realize that, a priori, solutions of system (4.2.2) need not to be pairs of even and odd functions, nor provide solutions of (4.2.1). In order to solvethis system, we will restrict problem (4.2.2) to those cases where the matrix 𝑀(𝑡)=(𝑎u�−𝑏u�−𝑎u�−𝑏u� 𝑎u�−𝑏u�−𝑎u�−𝑏u�)(𝑡) satisfies that [𝑀(𝑡),𝑀(𝑠)] ∶= 𝑀(𝑡)𝑀(𝑠)−𝑀(𝑠)𝑀(𝑡) = 0 ∀𝑡,𝑠 ∈ 𝐼, for in that case, the solution of the system (4.2.2) is given by the exponential of the integral of 𝑀. To see this, we have to present a definition and a related result [119]. 4. Study of the homogeneous equation 83 Definition 4.2.1. Let 𝑆 ⊂ ℝbe an interval. Define ℳ⊂u�1(ℝ,ℳu�×u�(ℝ))such that for every 𝑀∈ℳ, • there exists 𝑃∈u�1(ℝ,ℳu�×u�(ℝ))such that 𝑀(𝑡)=𝑃−1(𝑡)𝐽(𝑡)𝑃(𝑡)for every 𝑡∈ 𝑆where 𝑃−1(𝑡)𝐽(𝑡)𝑃(𝑡)is a Jordan decomposition of 𝑀(𝑡); • the superdiagonal elements of 𝐽are independent of 𝑡, as well as the dimensions of the Jordan boxes associated to the different eigenvalues of 𝑀; • two different Jordan boxes of 𝐽correspond to different eigenvalues; • if two eigenvalues of 𝑀are ever equal, they are identical in the whole interval 𝑆. Theorem 4.2.2 ( [119]).Let 𝑀∈ℳ. Then, the following statements are equivalent. •𝑀commutes with its derivative. •𝑀commutes with its integral. •𝑀commutes functionally, that is 𝑀(𝑡)𝑀(𝑠)=𝑀(𝑠)𝑀(𝑡)for all 𝑡,𝑠∈𝑆. •𝑀=∑u� u�=0 𝛾u�(𝑡)𝐶u�for some 𝐶∈ℳu�×u�(ℝ)and 𝛾u�∈u�1(𝑆,ℝ),𝑘=1,…,𝑟. Furthermore, any of the last properties imply that 𝑀(𝑡)has a set of constant eigenvectors, i.e. a Jordan decomposition 𝑃−1𝐽(𝑡)𝑃where 𝑃is constant. Even though the coefficients 𝑎and 𝑏may in general not have enough regularity to apply Theorem 4.2.2, we will see that we can obtain a basis of constant eigenvectors whenever the matrix 𝑀functionally commutes. That, as we will see, is enough for the solution of the system (4.2.2) to be given by the exponential of the integral of 𝑀. Observe that, [𝑀(𝑡),𝑀(𝑠)]=2(u�u�(u�)u�u�(u�)−u�u�(u�)u�u�(u�) u�u�(u�)[u�u�(u�)+u�u�(u�)]−u�u�(u�)[u�u�(u�)+u�u�(u�)] u�u�(u�)[u�u�(u�)+u�u�(u�)]−u�u�(u�)[u�u�(u�)+u�u�(u�)] u�u�(u�)u�u�(u�)−u�u�(u�)u�u�(u�) ). Let 𝐴(𝑡)∶=∫u� 0𝑎(𝑠)d𝑠,𝐵(𝑡)∶=∫u� 0𝑏(𝑠)d𝑠. Let 𝑀be a primitive (save possibly a constant matrix) of 𝑀,that is, the matrix, 𝑀=(𝐴u�−𝐵u�−𝐴u�−𝐵u� 𝐴u�−𝐵u�−𝐴u�−𝐵u�). We study now the different cases where [𝑀(𝑡),𝑀(𝑠)]=0 ∀𝑡,𝑠∈𝐼. We will always assume 𝑎  ≡0, since the case 𝑎≡0is the well-known case of an ordinary differential equation. Let us see the different possible cases. (D1).𝑏u�=𝑘𝑎, 𝑘∈ℝ, |𝑘|<1. In this case, 𝑎u�=0and 𝑀has the form 𝑀=(𝐵u�−(1+𝑘)𝐴u� (1−𝑘)𝐴u�−𝐵u�). 84 4.2. Study of the homogeneous equation 𝑀has two complex conjugate eigenvalues. What is more, both 𝑀and 𝑀functionally commute, and they have a basis of constant eigenvectors given by the constant matrix 𝑌 ∶=(𝑖√1−𝑘2−𝑖√1−𝑘2 𝑘−1 𝑘−1 ). We have that 𝑌−1𝑀(𝑡)𝑌 =𝑍(𝑡)∶=⎛ ⎜ ⎝−𝐵u�−𝑖√1−𝑘2𝐴u�0 0 −𝐵u�+𝑖√1−𝑘2𝐴u�⎞ ⎟ ⎠. Hence, 𝑒u�(u�) =𝑒u�u�(u�)u�−1 =𝑌𝑒u�(u�)𝑌−1 =𝑒−u�u�(u�) ⎛ ⎜ ⎜ ⎝cos(√1−𝑘2𝐴(𝑡))−1+u� √1−u�2sin(√1−𝑘2𝐴(𝑡)) √1−u�2 1+u� sin(√1−𝑘2𝐴(𝑡))cos(√1−𝑘2𝐴(𝑡))⎞ ⎟ ⎟ ⎠. Therefore, if a solution to equation (4.2.1) exists, it has to be of the form 𝑢(𝑡)=𝛼𝑒−u�u�(u�) cos(√1−𝑘2𝐴(𝑡))+𝛽𝑒−u�u�(u�) 1+𝑘 √1−𝑘2sin(√1−𝑘2𝐴(𝑡)). with 𝛼,𝛽∈ℝ. It is easy to check that all the solutions of equation (4.2.1) are of this form with 𝛽=−𝛼. (D2).𝑏u�=𝑘𝑎, 𝑘∈ℝ, |𝑘|>1. This case is much similar to (D1) In this case 𝑀has again the form 𝑀=(𝐵u�−(1+𝑘)𝐴u� (1−𝑘)𝐴u�−𝐵u�). 𝑀has two real eigenvalues and a basis of constant eigenvectors given by the constant matrix 𝑌 ∶=(√𝑘2−1 −√𝑘2−1 𝑘−1 𝑘−1 ). We have that 𝑌−1𝑀(𝑡)𝑌 =𝑍(𝑡)∶=⎛ ⎜ ⎝−𝐵u�−√𝑘2−1𝐴u�0 0 −𝐵u�+√𝑘2−1𝐴u�⎞ ⎟ ⎠. And so, 𝑒u�(u�) =𝑒u�u�(u�)u�−1 =𝑌𝑒u�(u�)𝑌−1 =𝑒−u�u�(u�) ⎛ ⎜ ⎜ ⎝cosh(√1−𝑘2𝐴(𝑡))−1+u� √u�2−1 sinh(√1−𝑘2𝐴(𝑡)) √1−u�2 1+u� sinh(√𝑘2−1𝐴(𝑡))cosh(√1−𝑘2𝐴(𝑡))⎞ ⎟ ⎟ ⎠. Therefore, it yields solutions of system (4.2.2) of the form 𝑢(𝑡)=𝛼𝑒−u�u�(u�) cosh(√𝑘2−1𝐴(𝑡))+𝛽𝑒−u�u�(u�) 1+𝑘 √𝑘2−1sinh(√𝑘2−1𝐴(𝑡)), 4. The cases (D1)–(D3) for the complete problem 85 which are solutions of equation (4.2.1) when 𝛽=−𝛼. (D3).𝑏u�=𝑎. 𝑀=(𝐵u�−(1+𝑘)𝐴u� 0 −𝐵u�). Since the matrix is triangular, we can solve sequentially for 𝑥u�and 𝑥u�. In this case the solutions of system (4.2.2) are of the form 𝑢(𝑡)=𝛼𝑒−u�u�(u�) +2𝛽𝑒−u�u�(u�)𝐴(𝑡) (4.2.3) which are solutions of equation (4.2.1) when 𝛽=−𝛼. (D4).𝑏u�=−𝑎. 𝑀=(𝐵u�0 (1−𝑘)𝐴u�−𝐵u�). We can solve sequentially for 𝑥u�and 𝑥u�and the solutions of system (4.2.2) are the same as in case (D3), but they are solutions of equation (4.2.1) when 𝛽=0. (D5).𝑏u�=𝑎u�=0. 𝑀=(𝐴u�−𝐵u�0 0 −𝐴u�−𝐵u�) In this case the solutions of system (4.2.2) are of the form 𝑢(𝑡)=𝛼𝑒u�(u�)−u�(u�) +𝛽𝑒−u�(u�)−u�(u�), which are solutions of equation (4.2.1) when 𝛼=0. Remark 4.2.3. Observe that functional matrices appearing in cases (D1)–(D5) belong to ℳ. 4.3 The cases (D1)–(D3) for the complete problem In the more complicated setting of the following nonhomogeneous problem 𝑥′(𝑡)+𝑎(𝑡)𝑥(−𝑡)+𝑏(𝑡)𝑥(𝑡)=ℎ(𝑡), 𝑎.𝑒.𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇), (4.3.1) we have still that, in the cases (D1)–(D3), it can be sorted out very easily. In fact, we get the expression of the Green’s function for the operator. We remark that in the three considered cases along this section the function 𝑎must be even on 𝐼. We note also that 𝑎is allowed to change its sign on 𝐼. First, we are going to prove a generalization of Proposition 3.2.2. Consider problem (4.3.1) with 𝑎and 𝑏constants. 𝑥′(𝑡)+𝑎𝑥(−𝑡)+𝑏𝑥(𝑡)=ℎ(𝑡), 𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇). (4.3.2) Considering the homogeneous case (ℎ=0), differentiating and making proper substitutions, we arrive to the problem. 𝑥″(𝑡)+(𝑎2−𝑏2)𝑥(𝑡)=0, 𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇), 𝑥′(−𝑇)=𝑥′(𝑇). (4.3.3) 86 4.3. The cases (D1)–(D3) for the complete problem Which, for 𝑏2<𝑎2, is the problem of the harmonic oscillator. It was shown in Section 3.2 that, under uniqueness conditions, the Green’s function 𝐺for problem (4.3.3) (that is, problem (3.2.2) satisfies the following properties in the case 𝑏2<𝑎2), but they can be extended almost automatically to the case 𝑏2>𝑎2. Lemma 4.3.1. The Green’s function 𝐺related to problem (4.3.3), satisfies the following properties. (𝐼) 𝐺∈u�(𝐼2,ℝ), (𝐼𝐼) u�u� u�u� and u�2u� u�u�2exist and are continuous in {(𝑡,𝑠)∈𝐼2|𝑠≠𝑡}, (𝐼𝐼𝐼) u�u� u�u� (𝑡,𝑡−)and u�u� u�u� (𝑡,𝑡+)exist for all 𝑡∈𝐼and satisfy 𝜕𝐺 𝜕𝑡(𝑡,𝑡−)−𝜕𝐺 𝜕𝑡(𝑡,𝑡+)=1 ∀𝑡∈𝐼, (𝐼𝑉) u�2u� u�u�2+(𝑎2−𝑏2)𝐺=0in {(𝑡,𝑠)∈𝐼2|𝑠≠𝑡}, (𝑉) (𝑎) 𝐺(𝑇,𝑠)=𝐺(−𝑇,𝑠) ∀𝑠∈𝐼, (𝑏) u�u� u�u� (𝑇,𝑠)= u�u� u�u� (−𝑇,𝑠) ∀𝑠∈(−𝑇,𝑇). (𝑉𝐼) 𝐺(𝑡,𝑠)=𝐺(𝑠,𝑡), (𝑉𝐼𝐼) 𝐺(𝑡,𝑠)=𝐺(−𝑡,−𝑠), (𝑉𝐼𝐼𝐼) u�u� u�u� (𝑡,𝑠)= u�u� u�u� (𝑠,𝑡), (𝐼𝑋) u�u� u�u� (𝑡,𝑠)=−u�u� u�u� (−𝑡,−𝑠), (𝑋) u�u� u�u� (𝑡,𝑠)=−u�u� u�u� (𝑡,𝑠). With these properties, we can prove the following Theorem in the same way we proved Theorem 3.2.2. Theorem 4.3.2. Suppose that 𝑎2−𝑏2≠𝑛2(𝜋/𝑇)2,𝑛=0,1,… Then problem (4.3.2) has a unique solution given by the expression 𝑢(𝑡)∶=∫u� −u� 𝐺(𝑡,𝑠)ℎ(𝑠)d𝑠, where 𝐺(𝑡,𝑠)∶=𝑎𝐺(𝑡,−𝑠)−𝑏𝐺(𝑡,𝑠)+𝜕𝐺 𝜕𝑡(𝑡,𝑠) is called the Green’s function related to problem (4.3.2). This last theorem leads us to the question “Which is the Green’s function for the case (D3) with 𝑎,𝑏constants?”. The following Lemma answers that question. 4. The cases (D1)–(D3) for the complete problem 87 Lemma 4.3.3. Let 𝑎≠0be a constant and let 𝐺u�3 be a real function defined as 𝐺u�3(𝑡,𝑠)∶= 𝑡−𝑠 2−𝑎𝑠𝑡+⎧ { { { ⎨ { { { ⎩ −1 2+𝑎𝑠 if |𝑠|<𝑡, 1 2−𝑎𝑠 if |𝑠|<−𝑡, 1 2+𝑎𝑡 if |𝑡|<𝑠, −1 2−𝑎𝑡 if |𝑡|<−𝑠. Then the following properties hold. •u�u�u�3 u�u� (𝑡,𝑠)+𝑎(𝐺u�3(𝑡,𝑠)+𝐺u�3(−𝑡,𝑠))=0for a. e. 𝑡,𝑠∈(−1,1). •u�u�u�3 u�u� (𝑡,𝑡+)−u�u�u�3 u�u� (𝑡,𝑡−)=1 ∀𝑡∈(−1,1). •𝐺u�3(−1,𝑠)=𝐺u�3(1,𝑠) ∀𝑠∈(−1,1). These properties are straightforward to check. Consider the following problem 𝑥′(𝑡)+𝑎[𝑥(𝑡)+𝑥(−𝑡)]=ℎ(𝑡),𝑡∈[−1,1]; 𝑥(1)=𝑥(−1). (4.3.4) In case of having a solution, it is unique, for if 𝑢,𝑣are solutions, 𝑢−𝑣is in the case (𝐷3)for equation (4.2.1), that is, (𝑢−𝑣)(𝑡)=𝛼(1−2𝑎𝑡). Since (𝑢−𝑣)(−𝑇)=(𝑢−𝑣)(𝑇),𝑢=𝑣. With this and Lemma 4.3.3 in mind, 𝐺u�3 is the Green’s function for the problem (4.3.4), that is, the Green’s function for the case (D3) with 𝑎,𝑏constants and 𝑇=1. For other values of 𝑇, it is enough to make a change of variables  𝑡=𝑇𝑡,𝑠=𝑇𝑠. Remark 4.3.4. The function 𝐺u�3 can be obtained from the Green’s functions for the case (D1) with 𝑎constant, 𝑏u�≡0and 𝑇=1taking the limit 𝑘→1−for 𝑇=1. The following theorem shows how to obtain a Green’s function for non constant coefficients of the equation using the Green’s function for constant coefficients. We can find the same principle, that is, to compose a Green’s function with some other function in order to obtain a new Green’s function, in [29, Theorem 5.1, Remark 5.1] and also in [74, Section 2]. But first, we need to know how the Green’s function should be defined in such a case. Theorem 4.3.2 gives us the expression of the Green’s function for problem (4.3.2), 𝐺(𝑡,𝑠)∶= 𝑎𝐺(𝑡,−𝑠)−𝑏𝐺(𝑡,𝑠)+u�u� u�u� (𝑡,𝑠). For instance, in the case (D1), if 𝜔=√|𝑎2−𝑏2|, we have that 2𝜔sin(𝜔𝑇)𝐺(𝑡,𝑠) ∶=⎧ { { { ⎨ { { { ⎩ 𝑎cos[𝜔(𝑠+𝑡−𝑇)]+𝑏cos[𝜔(𝑠−𝑡+𝑇)]+𝜔sin[𝜔(𝑠−𝑡+𝑇)], 𝑡>|𝑠|, 𝑎cos[𝜔(𝑠+𝑡−𝑇)]+𝑏cos[𝜔(−𝑠+𝑡+𝑇)]−𝜔sin[𝜔(−𝑠+𝑡+𝑇)], 𝑠>|𝑡|, 𝑎cos[𝜔(𝑠+𝑡+𝑇)]+𝑏cos[𝜔(−𝑠+𝑡+𝑇)]−𝜔sin[𝜔(−𝑠+𝑡+𝑇)], −𝑡>|𝑠|, 𝑎cos[𝜔(𝑠+𝑡+𝑇)]+𝑏cos[𝜔(𝑠−𝑡+𝑇)]+𝜔sin[𝜔(𝑠−𝑡+𝑇)], −𝑠>|𝑡|. Also, observe that 𝐺is continuous except at the diagonal, where 𝐺(𝑡,𝑡−)−𝐺(𝑡,𝑡+)=1. 88 4.3. The cases (D1)–(D3) for the complete problem Similarly, we can obtain the explicit expression of the Green’s function 𝐺for the case (D2). Taking again 𝜔=√|𝑎2−𝑏2|, 𝜔2(𝑒2u�u�2−1)𝐺(𝑡,𝑠) ∶= ⎧ { { { { { { { { { ⎨ { { { { { { { { { ⎩ 𝑒u�2(u�−u�) (𝑒u�2(u�+u�) −1)[𝑏(𝑒u�2(u�−u�) −1)+𝜔2] −4𝑎𝑒1 2u�2(u�+u�+2u�) sinh(1 2𝜔2[𝑠−𝑇])sinh(1 2𝜔2[𝑡−𝑇]), |𝑠|<𝑡, 𝑒−u�u�2(𝑒u�2(u�+u�) −1) ⋅[𝑎(𝑒u�u�2−𝑒u�2(u�+u�+u�))+(𝜔2−𝑏)𝑒u�u�2+𝑏𝑒u�u�2], −𝑠>|𝑡|, 𝑒−u�u�2(𝑒u�u�2−𝑒u�u�2) ⋅[𝑎(−𝑒u�2(u�+u�))+𝑎𝑒u�2(u�+u�) +(𝜔2−𝑏)𝑒u�2(u�+u�) +𝑏], 𝑠>|𝑡|, −𝑎(𝑒u�2(u�+u�) −1)(𝑒u�2(u�+u�) −1) +(𝜔2−𝑏)(𝑒u�2(u�+u�) −𝑒u�2(−u�+u�+2u�))+𝑏(−𝑒u�2(u�−u�))+𝑏, |𝑠|<−𝑡. In any case, we have that the Green’s function for problem (4.3.2) can be expressed as 𝐺(𝑡,𝑠)∶=⎧ { { { ⎨ { { { ⎩ 𝑘1(𝑡,𝑠), 𝑡>|𝑠|, 𝑘2(𝑡,𝑠), 𝑠>|𝑡|, 𝑘3(𝑡,𝑠), −𝑡>|𝑠|, 𝑘4(𝑡,𝑠), −𝑠>|𝑡|, were the 𝑘u�,𝑗=1,…,4are analytic functions defined on ℝ2. In order to simplify the statement of the following Theorem, consider the following conditions. (𝐃𝟏∗). (D1) is satisfied, (1−𝑘2)𝐴(𝑇)2≠(𝑛𝜋)2for all 𝑛=0,1,… (𝐃𝟐∗). (D2) is satisfied and 𝐴(𝑇)≠0. (𝐃𝟑∗). (D3) is satisfied and 𝐴(𝑇)≠0. Assume one of (𝐷1∗)–(𝐷3∗). In that case, by Theorem 4.3.2 and Lemma 4.3.3, we are under uniqueness conditions for the solution for the following problem [39]. 𝑥′(𝑡)+𝑥(−𝑡)+𝑘𝑥(𝑡)=ℎ(𝑡), 𝑡∈[−|𝐴(𝑇)|,|𝐴(𝑇)|], 𝑥(𝐴(𝑇))=𝑥(−𝐴(𝑇)). (4.3.5) The Green’s function 𝐺2for problem (4.3.5) is just an specific case of 𝐺and can be expressed as 𝐺2(𝑡,𝑠)∶=⎧ { { { ⎨ { { { ⎩ 𝑘1(𝑡,𝑠), 𝑡>|𝑠|, 𝑘2(𝑡,𝑠), 𝑠>|𝑡|, 𝑘3(𝑡,𝑠), −𝑡>|𝑠|, 𝑘4(𝑡,𝑠), −𝑠>|𝑡|. Define now 𝐺1(𝑡,𝑠)∶=𝑒u�u�(u�)−u�u�(u�)𝐻(𝑡,𝑠)=𝑒u�u�(u�)−u�u�(u�) ⎧ { { { ⎨ { { { ⎩ 𝑘1(𝐴(𝑡),𝐴(𝑠)), 𝑡>|𝑠|, 𝑘2(𝐴(𝑡),𝐴(𝑠)), 𝑠>|𝑡|, 𝑘3(𝐴(𝑡),𝐴(𝑠)), −𝑡>|𝑠|, 𝑘4(𝐴(𝑡),𝐴(𝑠)), −𝑠>|𝑡|. (4.3.6) 4. The cases (D1)–(D3) for the complete problem 89 Defined this way, 𝐺1is continuous except at the diagonal, where 𝐺1(𝑡,𝑡−)−𝐺1(𝑡,𝑡+)=1. Now we can state the following Theorem. Theorem 4.3.5. Assume one of (𝐷1∗)–(𝐷3∗). Let 𝐺1be defined as in (4.3.6). Assume 𝐺1(𝑡,⋅)ℎ(⋅)∈L1(𝐼)for every 𝑡∈𝐼. Then problem (4.3.1) has a unique solution given by 𝑢(𝑡)=∫u� −u� 𝐺1(𝑡,𝑠)ℎ(𝑠)d𝑠. Proof. First realize that, since 𝑎is even, 𝐴is odd, so 𝐴(−𝑡)=−𝐴(𝑡). It is important to note that if 𝑎has not constant sign in 𝐼, then 𝐴may be not injective on 𝐼. From the properties of 𝐺2as a Green’s function, it is clear that 𝜕𝐺2 𝜕𝑡 (𝑡,𝑠)+𝐺2(−𝑡,𝑠)+𝑘𝐺2(𝑡,𝑠)=0 for a. e. 𝑡,𝑠∈𝐴(𝐼), and so, 𝜕𝐻 𝜕𝑡 (𝑡,𝑠)+𝑎(𝑡)𝐻(−𝑡,𝑠)+𝑘𝑎(𝑡)𝐻(𝑡,𝑠)=0 for a. e. 𝑡,𝑠∈𝐼, Hence 𝑢′(𝑡)+𝑎(𝑡)𝑢(−𝑡)+(𝑏u�(𝑡)+𝑘𝑎(𝑡))𝑢(𝑡) =d d𝑡∫u� −u� 𝐺1(𝑡,𝑠)ℎ(𝑠)d𝑠+𝑎(𝑡)∫u� −u� 𝐺1(−𝑡,𝑠)ℎ(𝑠)d𝑠+(𝑏u�(𝑡) +𝑘𝑎(𝑡))∫u� −u� 𝐺1(𝑡,𝑠)ℎ(𝑠)d𝑠 =d d𝑡∫u� −u� 𝑒u�u�(u�)−u�u�(u�)𝐻(𝑡,𝑠)ℎ(𝑠)d𝑠+ d d𝑡∫u� u�𝑒u�u�(u�)−u�u�(u�)𝐻(𝑡,𝑠)ℎ(𝑠)d𝑠 +𝑎(𝑡)∫u� −u� 𝑒u�u�(u�)−u�u�(u�)𝐻(−𝑡,𝑠)ℎ(𝑠)d𝑠 +(𝑏u�(𝑡)+𝑘𝑎(𝑡))∫u� −u� 𝑒u�u�(u�)−u�u�(u�)𝐻(𝑡,𝑠)ℎ(𝑠)d𝑠 =[𝐻(𝑡,𝑡−)−𝐻(𝑡,𝑡+)]ℎ(𝑡)+𝑎(𝑡)𝑒−u�u�(u�) ∫u� −u� 𝑒u�u�(u�)𝜕𝐻 𝜕𝑡 (𝑡,𝑠)ℎ(𝑠)d𝑠 −𝑏u�(𝑡)𝑒−u�u�(u�) ∫u� −u� 𝑒u�u�(u�)𝐻(𝑡,𝑠)ℎ(𝑠)d𝑠+𝑎(𝑡)𝑒−u�u�(u�) ∫u� −u� 𝑒u�u�(u�)𝐻(−𝑡,𝑠)ℎ(𝑠)d𝑠 +(𝑏u�(𝑡)+𝑘𝑎(𝑡))𝑒−u�u�(u�) ∫u� −u� 𝑒u�u�(u�)𝐻(𝑡,𝑠)ℎ(𝑠)d𝑠 =ℎ(𝑡)+ 𝑎(𝑡)𝑒−u�u�(u�) ∫u� −u� 𝑒u�u�(u�) [𝜕𝐻 𝜕𝑡 (𝑡,𝑠)+𝑎(𝑡)𝐻(−𝑡,𝑠)+𝑘𝑎(𝑡)𝐻(𝑡,𝑠)]ℎ(𝑠)d𝑠 =ℎ(𝑡). The boundary conditions are also satisfied. 𝑢(𝑇)−𝑢(−𝑇)=𝑒−u�u�(u�) ∫u� −u� 𝑒u�u�(u�)[𝐻(𝑇,𝑠)−𝐻(−𝑇,𝑠)]ℎ(𝑠)d𝑠=0. In order to check the uniqueness of solution, let 𝑢and 𝑣be solutions of problem (4.3.5). Then 𝑢−𝑣satisfies equation (4.2.1) and so is of the form given in Section 4.2. Also, (𝑢−𝑣)(𝑇)− (𝑢−𝑣)(−𝑇)=2(𝑢−𝑣)u�(𝑇)=0, but this can only happen, by what has been imposed by conditions (𝐷1∗)–(𝐷3∗), if 𝑢−𝑣≡0, thus proving the uniqueness of solution.  96 4.4. The cases (D4) and (D5) 𝑥u�(𝑡)=𝑒−u�u�(u�) [𝑐+∫u� 0(𝑒u�u�(u�)ℎu�(𝑠)+2𝑎u�(𝑠)[𝑐+∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟])d𝑠], where 𝑐,𝑐∈ℝ. 𝑥u�is even independently of the value of 𝑐. Nevertheless, 𝑥u�is odd only when 𝑐=0. Hence, a solution of (4.3.1), if it exists, it has the form (4.4.2). To show the other implication it is enough to check that 𝑢u�is a solution of the problem (4.3.1). 𝑢′ u�(𝑡)=−𝑏u�(𝑡)𝑒−u�u�(u�) [𝑐+∫u� 0(𝑒u�u�(u�)ℎ(𝑠)+2𝑎u�(𝑠)∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟)d𝑠] +𝑒−u�u�(u�) (𝑒u�u�(u�)ℎ(𝑡)+2𝑎u�(𝑡)∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟) =ℎ(𝑡)−𝑏u�(𝑡)𝑢(𝑡)+2𝑎u�(𝑡)𝑒−u�u�(u�) ∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟. Now, 𝑎u�(𝑡)(𝑢u�(−𝑡)−𝑢u�(𝑡))+2𝑎u�(𝑡)𝑒−u�u�(u�) ∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟 =𝑎u�(𝑡)𝑒−u�u�(u�) [𝑐−∫u� 0(𝑒u�u�(u�)ℎ(−𝑠)−2𝑎u�(𝑠)∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟)d𝑠] −𝑎u�(𝑡)𝑒−u�u�(u�) [𝑐+∫u� 0(𝑒u�u�(u�)ℎ(𝑠)+2𝑎u�(𝑠)∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟)d𝑠] +2𝑎u�(𝑡)𝑒−u�u�(u�) ∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟 =−2𝑎u�(𝑡)𝑒−u�u�(u�) ∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑠+2𝑎u�(𝑡)𝑒−u�u�(u�) ∫u� 0𝑒u�u�(u�)ℎu�(𝑟)d𝑟=0. Hence, 𝑢′ u�(𝑡)+𝑎u�(𝑡)𝑢u�(−𝑡)+(−𝑎u�(𝑡)+𝑏u�(𝑡))𝑢u�(𝑡)=ℎ(𝑡), 𝑎.𝑒.𝑡∈𝐼. The boundary condition 𝑢u�(−𝑇)−𝑢u�(𝑇)=0is equivalent to (𝑢u�)u�(𝑇)=0, this is, ∫u� 0𝑒u�u�(u�)ℎu�(𝑠)d𝑠=0 and the result is concluded.  Theorem 4.4.2. If condition (D5) holds, then problem (4.3.1) has solution if and only if ∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠=0, (4.4.4) and in that case the solutions of (4.3.1) are given by 𝑢u�(𝑡)=𝑒u�(u�)−u�(u�) ∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠+𝑒−u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)+u�(u�)ℎu�(𝑠)d𝑠](4.4.5) for every 𝑐∈ℝ. Proof. In the case (D5), 𝑏u�=𝑏and 𝑎u�=𝑎. Also, the matrix in (4.4.1) is diagonal (𝑥′ u� 𝑥′ u�)=(𝑎u�−𝑏u�0 0 −𝑎u�−𝑏u�)(𝑥u� 𝑥u�)+(ℎu� ℎu�).(4.4.6) and the solutions of (4.4.6) are given by 4. The other cases 97 𝑥u�(𝑡)=𝑒u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠], 𝑥u�(𝑡)=𝑒−u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)+u�(u�)ℎu�(𝑠)d𝑠], where 𝑐,𝑐 ∈ ℝ. Since 𝑎and 𝑏are odd, 𝐴and 𝐵are even. So, 𝑥u�is even independently of the value of 𝑐. Nevertheless, 𝑥u�is odd only when 𝑐=0. In such a case, since we need, as in Theorem 4.4.1, that 𝑥u�(𝑇)=0, we get condition (4.4.4), which allows us to deduce the first implication of the Theorem. Any solution 𝑢u�of (4.3.1) has the expression (4.4.5). To show the second implication, it is enough to check that 𝑢is a solution of the problem (4.3.1). 𝑢′ u�(𝑡)=(𝑎(𝑡)−𝑏(𝑡))𝑒u�(u�)−u�(u�) ∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠 −(𝑎(𝑡)+𝑏(𝑡))𝑒−u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)+u�(u�)ℎu�(𝑠)d𝑠]+ℎ(𝑡). Now, 𝑎(𝑡)𝑢u�(−𝑡)+𝑏(𝑡)𝑢u�(𝑡) =𝑎(𝑡)(−𝑒u�(u�)−u�(u�) ∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠+𝑒−u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)+u�(u�)ℎu�(𝑠)d𝑠]) +𝑏(𝑡)(𝑒u�(u�)−u�(u�) ∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠+𝑒−u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)+u�(u�)ℎu�(𝑠)d𝑠]) =−(𝑎(𝑡)−𝑏(𝑡))𝑒u�(u�)−u�(u�) ∫u� 0𝑒u�(u�)−u�(u�)ℎu�(𝑠)d𝑠 +(𝑎(𝑡)+𝑏(𝑡))𝑒−u�(u�)−u�(u�) [𝑐+∫u� 0𝑒u�(u�)+u�(u�)ℎu�(𝑠)d𝑠]. So clearly, 𝑢′ u�(𝑡)+𝑎(𝑡)𝑢u�(−𝑡)+𝑏(𝑡)𝑢u�(𝑡)=ℎ(𝑡) for a.e. 𝑡∈𝐼. which ends the proof.  4.5 The other cases When we are not on the cases (D1)-(D5), since the fundamental matrix of 𝑀is not given by its exponential matrix, it is more difficult to precise when problem (4.3.1) has a solution. Here we present some partial results. Consider the following ordinary differential equation 𝑥′(𝑡)+[𝑎(𝑡)+𝑏(𝑡)]𝑥(𝑡)=ℎ(𝑡), 𝑥(−𝑇)=𝑥(𝑇). (4.5.1) The following lemma gives us the explicit Green’s function for this problem. Let 𝜐=𝑎+𝑏. Lemma 4.5.1. Let ℎ,𝑎,𝑏in problem (4.5.1) be in L1(𝐼)and assume ∫u� −u� 𝜐(𝑡)d𝑡≠0. Then problem (4.5.1) has a unique solution given by 𝑢(𝑡)=∫u� −u� 𝐺3(𝑡,𝑠)ℎ(𝑡)d𝑠, 98 4.5. The other cases where 𝐺3(𝑡,𝑠)=⎧ { ⎨ { ⎩𝜏𝑒∫u� u�u�(u�)du�, 𝑠≤𝑡, (𝜏−1)𝑒∫u� u�u�(u�)du�, 𝑠>𝑡, and 𝜏= 1 1−𝑒−∫u� −u� u�(u�)du�.(4.5.2) Proof. 𝜕𝐺3 𝜕𝑡 (𝑡,𝑠)=⎧ { ⎨ { ⎩−𝜏𝜐(𝑡)𝑒∫u� u�u�(u�)du�, 𝑠≤𝑡, −(𝜏−1)𝜐(𝑡)𝑒∫u� u�u�(u�)du�, 𝑠>𝑡, =−𝜐(𝑡)𝐺3(𝑡,𝑠). Therefore, 𝜕𝐺3 𝜕𝑡 (𝑡,𝑠)+𝜐(𝑡)𝐺3(𝑡,𝑠)=0, 𝑠≠𝑡. Hence, 𝑢′(𝑡)+𝜐(𝑡)𝑢(𝑡) =d d𝑡∫u� −u� 𝐺3(𝑡,𝑠)ℎ(𝑠)d𝑠+ d d𝑡∫u� u�𝐺3(𝑡,𝑠)ℎ(𝑠)d𝑠+𝜐(𝑡)∫u� −u� 𝐺3(𝑡,𝑠)ℎ(𝑠)d𝑠 =[𝐺3(𝑡,𝑡−)−𝐺3(𝑡,𝑡+)]ℎ(𝑡)+∫u� −u� [𝜕𝐺3 𝜕𝑡 (𝑡,𝑠)+𝜐(𝑡)𝐺3(𝑡,𝑠)]ℎ(𝑡)d𝑠 =ℎ(𝑡) a. e. 𝑡∈𝐼. The boundary conditions are also satisfied. 𝑢(𝑇)−𝑢(−𝑇)=∫u� −u� [𝜏𝑒∫u� u�u�(u�)du� −(𝜏−1)𝑒∫u� −u� u�(u�)du�]ℎ(𝑠)d𝑠 =∫u� −u� [𝑒∫u� u�u�(u�)du� 1−𝑒−∫u� −u� u�(u�)du� −𝑒−∫u� −u� u�(u�)du� 𝑒∫u� −u� u�(u�)du� 1−𝑒−∫u� −u� u�(u�)du� ]ℎ(𝑠)d𝑠 =∫u� −u� [𝑒∫u� u�u�(u�)du� 1−𝑒−∫u� −u� u�(u�)du� −𝑒∫u� u�u�(u�)du� 1−𝑒−∫u� −u� u�(u�)du�]ℎ(𝑠)d𝑠=0.  Lemma 4.5.2. |𝐺3(𝑡,𝑠)|≤𝐹(𝜐)∶= 𝑒‖u�‖1 |𝑒‖u�+‖1−𝑒‖u�−‖1|.(4.5.3) Proof. Observe that 𝜏= 1 1−𝑒‖u�−‖1−‖u�+‖1=𝑒‖u�+‖1 𝑒‖u�+‖1−𝑒‖u�−‖1. Hence, 𝜏−1= 𝑒‖u�−‖1 𝑒‖u�+‖1−𝑒‖u�−‖1. On the other hand, 𝑒∫u� u�u�(u�)du� ≤⎧ { ⎨ { ⎩𝑒‖u�−‖1, 𝑠≤𝑡, 𝑒‖u�+‖1, 𝑠>𝑡, which ends the proof.  4. The other cases 99 The next result proves the existence and uniqueness of solution of (4.3.1) when 𝜐is ‘sufficiently small’. Theorem 4.5.3. Let ℎ,𝑎,𝑏in problem (4.3.1) be in L1(𝐼)and assume ∫u� −u� 𝜐(𝑡)d𝑡≠0. Let 𝑊∶={(2𝑇)1 u�(‖𝑎‖u�∗+‖𝑏‖u�∗)}u�∈[1,+∞] where 𝑝−1 +(𝑝∗)−1 =1. If 𝐹(𝜐)‖𝑎‖1(inf𝑊)<1, where 𝐹(𝜐)is defined as in (4.5.3), then problem (4.3.1) has a unique solution. Proof. With some manipulation we get ℎ(𝑡)=𝑥′(𝑡)+𝑎(𝑡)(∫−u� u�𝑥′(𝑠)d𝑠+𝑥(𝑡))+𝑏(𝑡)𝑥(𝑡) =𝑥′(𝑡)+𝜐(𝑡)𝑥(𝑡)+𝑎(𝑡)∫−u� u�(ℎ(𝑠)−𝑎(𝑠)𝑥(−𝑠)−𝑏(𝑠)𝑥(𝑠))d𝑠. Hence, 𝑥′(𝑡)+𝜐(𝑡)𝑥(𝑡)=𝑎(𝑡)∫−u� u�(𝑎(𝑠)𝑥(−𝑠)+𝑏(𝑠)𝑥(𝑠))d𝑠+𝑎(𝑡)∫u� −u� ℎ(𝑠)d𝑠+ℎ(𝑡). Using 𝐺3defined as in (4.5.2) and Lemma 4.5.1, it is clear that 𝑥(𝑡)=∫u� −u� 𝐺3(𝑡,𝑠)𝑎(𝑠)∫−u� u�(𝑎(𝑟)𝑥(−𝑟)+𝑏(𝑟)𝑥(𝑟))d𝑟d𝑠 +∫u� −u� 𝐺3(𝑡,𝑠)[𝑎(𝑠)∫u� −u� ℎ(𝑟)d𝑟+ℎ(𝑠)]d𝑠, this is, 𝑥is a fixed point of an operator of the form 𝐻𝑥(𝑡)+𝛽(𝑡), so, by Banach contraction Theorem, it is enough to prove that ‖𝐻‖<1for some compatible norm of 𝐻. Using Fubini’s Theorem, 𝐻𝑥(𝑡)=−∫u� −u� 𝜌(𝑡,𝑟)(𝑎(𝑟)𝑥(−𝑟)+𝑏(𝑟)𝑥(𝑟))d𝑟, where 𝜌(𝑡,𝑟)=[∫u� |u�| −∫−|u�| −u� ]𝐺3(𝑡,𝑠)𝑎(𝑠)d𝑠. If ∫u� −u� 𝜐(𝑡)d𝑡=‖𝜐+‖1−‖𝜐−‖1>0then 𝐺3is positive and 𝜌(𝑡,𝑟)≤∫u� −u� 𝐺3(𝑡,𝑠)|𝑎(𝑠)|d𝑠≤𝐹(𝜐)‖𝑎‖1. We have the same estimate for −𝜌(𝑡,𝑟). If ∫u� −u� 𝜐(𝑡)d𝑡<0we proceed with an analogous argument and arrive to the conclusion that 𝐺3is negative and |𝜌(𝑡,𝑠)|<𝐹(𝜐)‖𝑎‖1. Hence, |𝐻𝑥(𝑡)|≤𝐹(𝜐)‖𝑎‖1∫u� −u� |𝑎(𝑟)𝑥(−𝑟)+𝑏(𝑟)𝑥(𝑟)|d𝑟 =𝐹(𝜐)‖𝑎‖1‖𝑎(𝑟)𝑥(−𝑟)+𝑏(𝑟)𝑥(𝑟)‖1. Thus, it is clear that ‖𝐻𝑥‖u�≤(2𝑇)1 u�𝐹(𝜐)‖𝑎‖1(‖𝑎‖u�∗+‖𝑏‖u�∗)‖𝑥‖u�, 𝑝∈[1,∞], which ends the proof.  100 4.5. The other cases Remark 4.5.4. In the hypothesis of Theorem 4.5.3, realize that 𝐹(𝜐)≥1. The following result will let us obtain some information on the sign of the solution of problem (4.3.1). In order to prove it, we will use a Theorem from Chapter 8 –Theorem 8.4.11– which is demonstrated independently. Consider an interval [𝑤,𝑑]⊂𝐼, the cone 𝐾 ={𝑢∈u�(𝐼)∶ min u�∈[u�,u�]𝑢(𝑡)≥𝑐‖𝑢‖}, and the following problem 𝑥′(𝑡)=ℎ(𝑡,𝑥(𝑡),𝑥(−𝑡)),𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇), (4.5.4) where ℎis an L1-Carathéodory function. Consider the following conditions. (I1 u�,u�)There exist 𝜌>0and 𝜔∈(0, u� 4u�]such that 𝑓−u�,u� u�<𝜔where 𝑓−u�,u� u�∶=sup{ℎ(𝑡,𝑢,𝑣)+𝜔𝑣 𝜌∶ (𝑡,𝑢,𝑣)∈[−𝑇,𝑇]×[−𝜌,𝜌]×[−𝜌,𝜌]}. (I0 u�,u�)There exists 𝜌>0such that 𝑓u� (u�,u�/u�) ⋅ inf u�∈[u�,u�]∫u� u�𝐺(𝑡,𝑠)𝑑𝑠>1, where 𝑓u� (u�,u�/u�) =inf{ℎ(𝑡,𝑢,𝑣)+𝜔𝑣 𝜌∶ (𝑡,𝑢,𝑣)∈[𝑤,𝑑]×[𝜌,𝜌/𝑐]×[−𝜌/𝑐,𝜌/𝑐]}. Theorem 4.5.5 (Part of Theorem 8.4.11).Let 𝜔 ∈ (0,u� 2𝑇]. Let [𝑤,𝑑] ⊂ 𝐼such that 𝑤 = 𝑇−𝑑∈(max{0,𝑇− u� 4u�},u� 2). Let 𝑐= [1−tan(𝜔𝑑)][1−tan(𝜔𝑤)] [1+tan(𝜔𝑑)][1+tan(𝜔𝑤)].(4.5.5) Problem (4.5.4) has at least one nonzero solution in 𝐾if either of the following conditions hold. (𝑆1)There exist 𝜌1,𝜌2∈(0,∞)with 𝜌1/𝑐<𝜌2such that (I0 u�1,u�)and (I1 u�2,u�)hold. (𝑆2)There exist 𝜌1,𝜌2∈(0,∞)with 𝜌1<𝜌2such that (I1 u�1,u�)and (I0 u�2,u�)hold. Theorem 4.5.6. Let ℎ∈L∞(𝐼),𝑎,𝑏∈L1(𝐼)be such that 0<|𝑏(𝑡)|<𝑎(𝑡)<𝜔< u� 2𝑇 for a. e. 𝑡 ∈ 𝐼and infℎ > 0. Then there exists a solution 𝑢of (4.3.1) such that, 𝑢 > 0in (max{0,𝑇− u� 4u�},min{𝑇, u� 4u�}). 4. The other cases 101 Proof. Problem (4.3.1) can be rewritten as 𝑥′(𝑡)=ℎ(𝑡)−𝑏(𝑡)𝑥(𝑡)−𝑎(𝑡)𝑥(−𝑡), 𝑡∈𝐼, 𝑥(−𝑇)=𝑥(𝑇). With this formulation, we can apply Theorem 4.5.5. Since 0 < 𝑎(𝑡)−|𝑏(𝑡)| < 𝜔for a. e. 𝑡∈𝐼, take 𝜌2∈ℝ+large enough such that ℎ(𝑡)<(𝑎(𝑡)−|𝑏(𝑡)|)𝜌2for a. e. 𝑡∈𝐼. Hence, ℎ(𝑡)<(𝑎(𝑡)−𝜔)𝜌2−|𝑏(𝑡)|𝜌2+𝜌2𝜔for a. e. 𝑡∈𝐼, in particular, ℎ(𝑡)<(𝑎(𝑡)−𝜔)𝑣−|𝑏(𝑡)|𝑢+𝜌2𝜔≤(𝑎(𝑡)−𝜔)𝑣+𝑏(𝑡)𝑢+𝜌2𝜔 for a. e. 𝑡∈𝐼; 𝑢,𝑣∈[−𝜌2,𝜌2]. Therefore, sup{ℎ(𝑡)−𝑏(𝑡)𝑢−𝑎(𝑡)𝑣+𝜔𝑣 𝜌2∶ (𝑡,𝑣)∈[−𝑇,𝑇]×[−𝜌2,𝜌2]}<𝜔, and thus, (I1 u�2,u�)is satisfied. Let [𝑤,𝑑] ⊂ 𝐼be such that [𝑤,𝑑] ⊂ (𝑇− u� 4u�,u� 4u�). Let 𝑐be defined as in (4.5.5) and 𝜖=𝜔∫u� u�𝐺(𝑡,𝑠)d𝑠. Choose 𝛿∈(0,1)such that ℎ(𝑡)>[(1+u� u�)𝜔−(𝑎(𝑡)−|𝑏(𝑡)|)]𝜌2𝛿for a. e. 𝑡∈𝐼 and define 𝜌1∶= 𝛿𝑐𝜌2. Therefore, ℎ > [(𝑎(𝑡)−𝜔)𝑣+𝑏(𝑡)𝑢(𝑡)]u� u�𝜌1for a. e. 𝑡 ∈ 𝐼, 𝑢∈[𝜌1,u�1 u�]and 𝑣∈[−u�1 u�,u�1 u�]. Thus, inf{ℎ(𝑡)−𝑏(𝑡)𝑢−𝑎(𝑡)𝑣+𝜔𝑣 𝜌1∶ (𝑡,𝑣)∈[𝑤,𝑑]×[−𝜌1/𝑐,𝜌1/𝑐]}>𝜔 𝜖, and hence, (I0 u�1,u�)is satisfied. Finally, (𝑆1)in Theorem 4.5.5 is satisfied and we get the desired result.  Remark 4.5.7. In the hypothesis of Theorem 4.5.6, if 𝜔< u� 4𝑇, we can take [𝑤,𝑑]=[−𝑇,𝑇] and continue with the proof of Theorem 4.5.6 as done above. This guarantees that 𝑢is positive in [−𝑇,𝑇]. 5. General linear equations In this chapter we study differential problems in which the reflection operator and the Hilbert transform are involved. We reduce these problems to ordinary differential equations in order to solve them. Also, we describe a general method for obtaining the Green’s function of reducible functional differential equations and illustrate it with the case of homogeneous boundary value problems with reflection and several specific examples. It is important to point out that these transformations, necessary to reduce the problem to an ordinary one, are of a purely algebraic nature. It is, in this sense, similar to the algebraic analysis theory which, through the study of Ore algebras and modules, obtains important information about some functional problems, including explicit solutions [21,50]. Nevertheless, the algebraic structures we deal with here are somewhat different, e. g., they are not in general Ore algebras † Among the reducible functional differential equations, those with reflection have gathered great interest, some of it due to their applications to supersymmetric quantum mechanics [73, 147,153] or to other areas of analysis like topological methods [34]. In this chapter, following [44] we put special emphasis in two operators appearing in the equations: the reflection operator and the Hilbert transform. Both of them have exceptional algebraic properties which make them fit for our approach. 5.1 Differential operators with reflection In this Section we will study a particular family of operators, those that are combinations of the differential operator 𝐷, the pullback operator of the reflection 𝜑(𝑡) = −𝑡, denoted by 𝜑∗(𝑓)(𝑡) = 𝑓(−𝑡), and the identity operator, Id. In order to freely apply the operator 𝐷 without worrying too much about it’s domain of definition, we will consider that 𝐷acts on the set of functions locally of bounded variation on ℝ,BVloc(ℝ)‡. Given a compact interval 𝐾, the space BV(𝐾)is defined as the set {𝑓 ∶ 𝐼 → ℝ|𝑉(𝑓) < +∞}where 𝑉(𝑓) = sup u�∈u�u�∑u�u�−1 u�=0 |𝑓(𝑥u�+1)−𝑓(𝑥u�)|,𝑃={𝑥0,…,𝑥u�u�},min𝐾 =𝑥0<𝑥1<⋯<𝑥u�u�−1 <𝑥u�u�= max𝐾and u�u�is the set of partitions of 𝐾.BVloc(ℝ)is the set {𝑓 ∶ℝ→ℝ|𝑓|u�∈BV(𝐾), for all 𝐾 ⊂ℝcompact}. It is well known that any function locally of bounded variation 𝑓 ∈ BVloc(ℝ)can be expressed as 𝑓(𝑥)=𝑓(𝑥0)+∫u� u�0𝑔(𝑦)d𝑦+ℎ(𝑥), †We refer the reader to [118, 149–151] for an algebraic approach to the abstract theory of boundary value problems and its applications to symbolic computation. ‡Since we will be working with ℝas a domain throughout this chapter, it will be in our interest to take the local versions of the classical function spaces. By local version we mean that, if we restrict the function to a compact set, the restriction belongs to the classical space defined with that compact set as domain for its functions. 104 5.1. Differential operators with reflection for any 𝑥0∈ℝ, where 𝑔∈L1(ℝ), and ℎis the function which is constant except for a countable number of discontinuities (cf. [37,116]). This implies that the distributional derivative (we will call it weak derivative as shorthand) of 𝑓is 𝑓′=𝑔+∑ u�∈ℕℎu�𝛿u�u�,(5.1.1) where 𝛿u�is the Dirac distribution at 𝑥, the 𝑥u�are the points at which ℎhas discontinuities and ℎu�is the magnitude of the discontinuity at 𝑥u�. In this way, we will define 𝐷𝑓 ∶= 𝑔(we will restate this definition in a more general way further on). As we did in Section 2.2, we now consider the real abelian group ℝ[𝐷,𝜑∗]of generators {𝐷u�,𝜑∗𝐷u�}∞ u�=0. If we take the usual composition for operators in ℝ[𝐷,𝜑∗], we observe that 𝐷𝜑∗= −𝜑∗𝐷, so composition is closed in ℝ[𝐷,𝜑∗], which makes it a non commutative algebra. In general, 𝐷u�𝜑∗=(−1)u�𝜑∗𝐷u�for 𝑘=0,1,… The elements of ℝ[𝐷,𝜑∗]are of the form 𝐿=∑ u�𝑎u�𝜑∗𝐷u�+∑ u�𝑏u�𝐷u�∈ℝ[𝐷,𝜑∗]. (5.1.2) For convenience, we consider the sums on 𝑖and 𝑗such that 𝑖,𝑗 ∈ {0,1,…}, but taking into account that the real coefficients 𝑎u�,𝑏u�are zero for big enough indices – that is, we are dealing with finite sums. Despite the non commutativity of the composition in ℝ[𝐷,𝜑∗]there are interesting relations in this algebra. First, notice that ℝ[𝐷,𝜑∗]is not a unique factorization domain. Take a polynomial 𝑃 = 𝐷2+𝛽𝐷+𝛼where 𝛼,𝛽∈ℝ, and define the following operators. If 𝛽2−4𝛼≥0, 𝐿1∶=𝐷+1 2(𝛽−√𝛽2−4𝛼), 𝑅1∶=𝐷+1 2(𝛽+√𝛽2−4𝛼), 𝐿2∶=𝜑∗𝐷−√2𝐷+1 2(𝛽−√𝛽2−4𝛼)𝜑∗+(−𝛽+√𝛽2−4𝛼) √2, 𝑅2∶=𝜑∗𝐷−√2𝐷−1 2(𝛽+√𝛽2−4𝛼)𝜑∗−(𝛽+√𝛽2−4𝛼) √2, 𝐿3∶=𝜑∗𝐷−√2𝐷+1 2(𝛽+√𝛽2−4𝛼)𝜑∗−(𝛽+√𝛽2−4𝛼) √2, 𝑅3∶=𝜑∗𝐷−√2𝐷+1 2(−𝛽+√𝛽2−4𝛼)𝜑∗+(−𝛽+√𝛽2−4𝛼) √2, 𝐿4∶=𝜑∗𝐷+√2𝐷+1 2(𝛽−√𝛽2−4𝛼)𝜑∗+(𝛽−√𝛽2−4𝛼) √2, 5. Differential operators with reflection 105 𝑅4∶=𝜑∗𝐷+√2𝐷−1 2(𝛽+√𝛽2−4𝛼)𝜑∗+(𝛽+√𝛽2−4𝛼) √2, 𝐿5∶=𝜑∗𝐷+√2𝐷+1 2(𝛽+√𝛽2−4𝛼)𝜑∗+(𝛽+√𝛽2−4𝛼) √2, 𝑅5∶=𝜑∗𝐷+√2𝐷+1 2(−𝛽+√𝛽2−4𝛼)𝜑∗+(𝛽−√𝛽2−4𝛼) √2. If 𝛽=0and 𝛼≤0, 𝐿6∶=𝜑∗𝐷+√−𝛼𝜑∗, 𝐿7∶=𝜑∗𝐷−√−𝛼𝜑∗. If 𝛽=0and 𝛼≥0, 𝐿8∶=𝐷+√𝛼𝜑∗, 𝐿9∶=𝐷−√𝛼𝜑∗. If 𝛽=0and 𝛼≤1, 𝐿10 ∶=𝜑∗𝐷−√1−𝛼𝜑∗+1, 𝑅10 ∶=−𝜑∗𝐷+√1−𝛼𝜑∗+1, 𝐿11 ∶=𝜑∗𝐷+√1−𝛼𝜑∗+1, 𝑅11 ∶=−𝜑∗𝐷−√1−𝛼𝜑∗+1. If 𝛽=0,𝛼≠0and 𝛼≤1, 𝐿12 ∶=𝜑∗𝐷−√1−𝛼𝐷+𝛼, 𝑅12 ∶=−1 𝛼𝜑∗𝐷+√1−𝛼 𝛼𝐷+1, 𝐿13 ∶=𝜑∗𝐷+√1−𝛼𝐷+𝛼, 𝑅13 ∶=−1 𝛼𝜑∗𝐷−√1−𝛼 𝛼𝐷+1. Then, 𝑃=𝐿1𝑅1=𝑅1𝐿1=𝑅2𝐿2=𝑅3𝐿3=𝑅4𝐿4=𝑅5𝐿5, and, when 𝛽=0, 𝑃=−𝐿2 6=−𝐿2 7=𝐿2 8=𝐿2 9=𝑅10𝐿10 =𝐿10𝑅10 =𝑅11𝐿11 112 5.3. The reduced problem goal now is to consider abstractly these properties in order to apply them in a more general context with different kinds of operators. Let 𝑋be a vector subspace of L1 loc(ℝ), and (ℝ,𝜏)the real line with its usual topology. Define 𝑋u�∶={𝑓|u�∶ 𝑓 ∈𝑋}for every 𝑈 ∈𝜏(observe that 𝑋u�is a vector space as well). Assume that 𝑋satisfies the following property. (P) For every partition of ℝ,{𝑆u�}u�∈u� ∪{𝑁}, consisting of measurable sets where 𝑁has no accumulation points and the 𝑆u�are open, if 𝑓u�∈𝑋u�u�for every 𝑗∈𝐽, then there exists 𝑓 ∈𝑋 such that 𝑓|u�u�=𝑓u�for every 𝑗∈𝐽. Example 5.3.1. The set of locally absolutely continuous functions ACloc(ℝ)⊂L1 loc(ℝ)does not satisfy (P). To see this just take the following partition of ℝ:𝑆1= (−∞,0),𝑆2= (0,+∞),𝑁 = {0}and consider 𝑓1≡ 0,𝑓2≡ 1.𝑓u�∈ AC(ℝ)u�u�for 𝑗 = 1,2, but any function 𝑓such that 𝑓|u�u�= 𝑓u�,𝑗 = 1,2has a discontinuity at 0, so it cannot be absolutely continuous. That is, (P) is not satisfied. Example 5.3.2. 𝑋 =BVloc(ℝ)satisfies (P). Take a partition of ℝ,{𝑆u�}u�∈u� ∪{𝑁}, with the properties of (P) and a family of functions (𝑓u�)u�∈u� such that 𝑓u�∈ 𝑋u�u�for every 𝑗 ∈ 𝐽. We can further assume, without lost of generality, that the 𝑆u�are connected. Define a function 𝑓 such that 𝑓|u�u�∶=𝑓u�and 𝑓|u�=0. Take a compact set 𝐾 ⊂ℝ. Then, by Bolzano-Weierstrass’ and Heine-Borel’s Theorems, 𝐾 ∩𝑁 is finite for 𝑁has no accumulation points. Therefore, 𝐽u�∶={𝑗∈𝐽∶𝑆u�∩𝐾 ≠∅}is finite as well. To see this denote by 𝜕𝑆the boundary of a set 𝑆and observe that 𝑁∩𝐾 =∪u�∈u�𝜕(𝑆u�∪𝐾)and that the sets 𝜕(𝑆u�∩𝐾)∩𝜕(𝑆u�∩𝐾)are finite for every 𝑗,𝑘∈𝐽. Thus, the variation of 𝑓in 𝐾is 𝑉u�(𝑓) ≤ ∑u�∈u�u�𝑉u�u�(𝑓) < ∞since 𝑓is of bounded variation on each 𝑆u�. Hence, 𝑋satisfies (P). Throughout this section we will consider a function space 𝑋satisfying (P) and two families of linear operators 𝐿={𝐿u�}u�∈u� and 𝑅={𝑅u�}u�∈u� that satisfy Locality: 𝐿u�∈ℒ(𝑋u�,L1 loc(𝑈)),𝑅u�∈ℒ(im(𝐿u�),L1 loc(𝑈)), Restriction: 𝐿u�(𝑓|u�)=𝐿u�(𝑓)|u�,𝑅u�(𝑓|u�)=𝑅u�(𝑓)|u�for every 𝑈,𝑉 ∈ 𝜏 such that 𝑉⊂𝑈†. The following definition allows us to give an example of an space that satisfies the properties of locality and restriction. Definition 5.3.3. Let 𝑓 ∶ ℝ → ℝand assume there exists a partition {𝑆u�}u�∈u� ∪{𝑁}of ℝ consisting of measurable sets where 𝑁is of zero Lebesgue measure satisfying that the weak derivative 𝑔u�exists for every 𝑓|u�u�, then a function 𝑔such that 𝑔|u�u�=𝑔u�is called the very weak derivative (vw-derivative) of 𝑓. Remark 5.3.4. The vw-derivative is uniquely defined save for a zero measure set and is equivalent to the weak derivative for absolutely continuous functions. †The definitions here presented of 𝐿and 𝑅are deeply related to Sheaf Theory. Since the authors want to make this work as self-contained as possible, we will not deepen into that fact. 5. The reduced problem 113 Nevertheless, the vw-derivative is different from the derivative of distributions. For instance, the derivative of the Heavyside function in the distributional sense is de Dirac delta at 0, whereas its vw-derivative is zero. What is more, the kernel of the vw-derivative is the set of functions which are constant on a family of open sets {𝑆u�}u�∈u� such that ℝ\(∪u�∈u�𝑆u�)has Lebesgue measure zero. Example 5.3.5. Take 𝑋 = BVloc(ℝ)and 𝐿=𝐷to be the very weak derivative. Then 𝐿 satisfies the locality and restriction hypotheses. Remark 5.3.6. The vw-derivative, as defined here, is the 𝐷operator defined in (5.1.1) for functions of bounded variation. In other words, the vw-derivative ignores the jumps and considers only those parts with enough regularity. Remark 5.3.7. The locality property allows us to treat the maps 𝐿and 𝑅as if they were just linear operators in ℒ(𝑋,L1 loc(ℝ))and ℒ(im(𝐿),L1 loc(ℝ))respectively, although we must not forget their more complex structure. Assume 𝑋u�⊂im(𝐿u�)⊂im(𝑅u�)for every 𝑈∈𝜏.𝐵u�∈ℒ(im(𝑅ℝ),ℝ),𝑖=1,…,𝑚 and ℎ∈im(𝐿ℝ). Consider now the following problem 𝐿𝑢=ℎ, 𝐵u�𝑢=0, 𝑖=1,…,𝑚. (5.3.1) Let 𝑍∶={𝐺∶ℝ2→ℝ|𝐺(𝑡,⋅)∈𝑋∩ (ℝ)and supp{𝐺(𝑡,⋅)}is compact, 𝑠∈ℝ}. 𝑍is a vector space. Let 𝑓 ∈im(𝐿ℝ)and consider the problem 𝑅𝐿𝑣=𝑓, 𝐵u�𝑣=0, 𝐵u�𝑅𝑣=0, 𝑖=1,…,𝑚. (5.3.2) Let 𝐺 ∈𝑍and define the operator 𝐻u�such that 𝐻u�(ℎ)|u�∶=∫ℝ𝐺(𝑡,𝑠)ℎ(𝑠)d𝑠. We have now the following theorem relating problems (5.3.1) and (5.3.2). Recall that, by definition, ℒ⊢𝐺(𝑡,𝑠)∶=ℒ(𝐺(⋅,𝑠))|u�. Theorem 5.3.8. Assume 𝐿and 𝑅are the aforementioned operators with the locality and restriction properties and let ℎ∈Dom(𝑅ℝ). Assume 𝐿commutes with 𝑅and that there exists 𝐺∈𝑍such that (𝐼) (𝑅𝐿)⊢𝐺=0, (𝐼𝐼) 𝐵u� ⊢𝐺=0, 𝑖=1,…,𝑚, (𝐼𝐼𝐼) (𝐵u�𝑅)⊢𝐺=0, 𝑖=1,…,𝑚, (𝐼𝑉) 𝑅𝐿𝐻u�ℎ=𝐻(u�u�)⊢u�ℎ+ℎ, (𝑉) 𝐿𝐻u�⊢u�ℎ=𝐻u�⊢u�⊢u�ℎ+ℎ. (𝑉𝐼) 𝐵u�𝐻u�=𝐻u�u� ⊢u�, 𝑖=1,…,𝑚, (𝑉𝐼𝐼) 𝐵u�𝑅𝐻u�=𝐵u�𝐻u�⊢u�=𝐻(u�u�u�)⊢u�, 𝑖=1,…,𝑚, Then, 𝑣 ∶= 𝐻u�(ℎ)is a solution of problem (5.3.2) and 𝑢 ∶= 𝐻u�⊢u�(ℎ)is a solution of problem (5.3.1). 114 5.3. The reduced problem Proof. (𝐼)and (𝐼𝑉)imply that 𝑅𝐿𝑣=𝑅𝐿𝐻u�ℎ=𝐻(u�u�)⊢u�ℎ+ℎ=𝐻0ℎ+ℎ=ℎ. On the other hand, (𝐼𝐼𝐼)and (𝑉𝐼𝐼)imply that, for every 𝑖=1,…,𝑚, 𝐵u�𝑅𝑣=𝐵u�𝑅𝐻u�ℎ=𝐻(u�u�u�)⊢u�ℎ=0. All the same, by (𝐼𝐼)and (𝑉𝐼), 𝐵u�𝑣=𝐵u�𝐻u�ℎ=𝐻u�u� ⊢u�ℎ=0. Therefore, 𝑣is a solution to problem (5.3.2). Now, using (𝐼)and (𝑉)and the fact that 𝐿𝑅=𝑅𝐿, we have that 𝐿𝑢=𝐿𝐻u�⊢u�ℎ=𝐻u�⊢u�⊢u�ℎ+ℎ=𝐻(u�u�)⊢u�ℎ+ℎ=𝐻(u�u�)⊢u�ℎ+ℎ=ℎ. Taking into account (𝐼𝐼𝐼)and (𝑉𝐼𝐼), 𝐵u�𝑢=𝐵u�𝐻u�⊢u�(ℎ)=𝐻(u�u�u�)⊢u�ℎ=0, 𝑖=1,…,𝑚. Hence, 𝑢is a solution of problem (5.3.1).  The following Corollary is proved in the same way as the previous Theorem. Corollary 5.3.9. Assume 𝐺∈𝑍satisfies (1) 𝐿⊢𝐺=0, (2) 𝐵u�⊢𝐺=0, 𝑖=1,…,𝑚, (3) 𝐿𝐻u�ℎ=𝐻u�⊢u�ℎ+ℎ, (4) 𝐵u�𝐻u�ℎ=𝐻u�u� ⊢u�ℎ. Then 𝑢=𝐻u�ℎis a solution of problem (5.3.1). Proof of Theorem 5.2.3. Originally, we would need to take ℎ ∈ Dom(𝑅), but by a simple density argument –u�∞(𝐼)is dense in L1(𝐼)– we can take ℎ ∈ L1(𝐼). If we prove that the hypothesis of Theorem 5.3.8 are satisfied, then the existence of solution will be proved. First, Theorem 5.1.1 guarantees the commutativity of 𝐿and 𝑅. Now, Theorem 5.2.1 implies hypothesis (𝐼)−(𝑉𝐼𝐼)of Theorem 5.3.8 in terms of the vw-derivative. Indeed, (𝐼)is straightforward from (𝐺5).(𝐼𝐼)and (𝐼𝐼𝐼)are satisfied because (𝐺1)−−(𝐺6)hold and 𝐵u�𝑢,𝐵u�𝑅𝑢 = 0.(𝐺2)and (𝐺4)imply (𝐼𝑉)and (𝑉).(𝑉𝐼)and (𝑉𝐼𝐼)hold because of (𝐺2),(𝐺5)and the fact that the boundary conditions commute with the integral. On the other hand, the solution to problem (5.2.2) must be unique for, otherwise, the reduced problem 𝑆𝑢 = 0,𝐵u�𝑅𝑢 = 0,𝐵u�𝑢=0,𝑖 = 1,…,𝑛would have several solutions, contradicting the hypotheses.  The following Lemma, in the line of Theorem 4.3.5, extends the application of Theorem 5.2.3 to the case of nonconstant coefficients with some restrictions for problems similar to the one in Example 5.2.4. 5. The reduced problem 115 Lemma 5.3.10. Consider the problem 𝑢″(𝑡)+𝑎(𝑡)𝑢(−𝑡)+𝑏(𝑡)𝑢(𝑡)=ℎ(𝑡),𝑢(−𝑇)=𝑢(𝑇), (5.3.3) where 𝑎∈𝑊2,1 loc (ℝ)is nonnegative and even, 𝑏=𝑘𝑎+ 𝑎″ 4𝑎−5 16(𝑎′ 𝑎)2, for some constant 𝑘∈ℝ,𝑘2≠1and 𝑏is integrable. Define 𝐴(𝑡)∶=∫u� 0√𝑎(𝑠)d𝑠, consider 𝑢″(𝑡)+𝑢(−𝑡)+𝑘𝑢(𝑡)=ℎ(𝑡), 𝑢(−𝐴(𝑇))=𝑢(𝐴(𝑇)) and assume it has a Green’s function 𝐺. Then 𝑢(𝑡)=∫u� −u� 𝐻(𝑡,𝑠)ℎ(𝑠)d𝑠 is a solution of problem (5.3.3) where 𝐻(𝑡,𝑠)∶=4 √𝑎(𝑠) 𝑎(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠)), And 𝐻(𝑡,⋅)ℎ(⋅)is assumed to be integrable in [−𝑇,𝑇]. Proof. Let 𝐺be the Green’s function of the problem 𝑢″(𝑡)+𝑢(−𝑡)+𝑘𝑢(𝑡)=ℎ(𝑡), 𝑢(−𝐴(𝑇))=𝑢(𝐴(𝑇)), 𝑢∈W2,1 loc(ℝ). Observe that, since |𝑘| ≠ 1, we are in the cases (𝐷1)−(𝐷2)in Chapter 4. Now, we show that 𝐻satisfies the equation, that is, 𝜕2𝐻 𝜕𝑡2(𝑡,𝑠)+𝑎(𝑡)𝐻(−𝑡,𝑠)+𝑏(𝑡)𝐻(𝑡,𝑠)=0for a. e. 𝑡,𝑠∈ℝ. 𝜕2𝐻 𝜕𝑡2(𝑡,𝑠)=𝜕2 𝜕𝑡2⎡ ⎢ ⎣ 4 √𝑎(𝑠) 𝑎(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠))⎤ ⎥ ⎦ =𝜕 𝜕𝑡⎡ ⎢ ⎣−𝑎′(𝑡) 4 4 √𝑎(𝑠) 𝑎5(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠))+ 4 √𝑎(𝑠)𝑎(𝑡)𝜕𝐺 𝜕𝑡(𝐴(𝑡),𝐴(𝑠))⎤ ⎥ ⎦ =−𝑎″(𝑡) 4 4 √𝑎(𝑠) 𝑎5(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠))+ 5 16(𝑎′(𝑡))24 √𝑎(𝑠) 𝑎9(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠)) +4 √𝑎(𝑠)𝑎3(𝑡)𝜕2𝐺 𝜕𝑡2(𝐴(𝑡),𝐴(𝑠)). Therefore, 𝜕2𝐻 𝜕𝑡2(𝑡,𝑠)+𝑎(𝑡)𝐻(−𝑡,𝑠)+𝑏(𝑡)𝐻(𝑡,𝑠) 116 5.3. The reduced problem =4 √𝑎(𝑠)𝑎3(𝑡)𝜕2𝐺 𝜕𝑡2(𝐴(𝑡),𝐴(𝑠))+𝑎(𝑡)4 √𝑎(𝑠) 𝑎(𝑡)𝐺(−𝐴(𝑡),𝐴(𝑠)) +𝑘𝑎(𝑡)4 √𝑎(𝑠) 𝑎(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠)) =4 √𝑎(𝑠)𝑎3(𝑡)(𝜕2𝐺 𝜕𝑡2(𝐴(𝑡),𝐴(𝑠))+𝐺(−𝐴(𝑡),𝐴(𝑠))+𝑘𝐺(𝐴(𝑡),𝐴(𝑠)))=0. The boundary conditions are satisfied as well.  The same construction of Lemma 5.3.10 is valid for the case of the initial value problem. We illustrate this in the following example. Example 5.3.11. Let 𝑎(𝑡)=|𝑡|u�,𝑘>1. Taking 𝑏as in Lemma 5.3.10, 𝑏(𝑡)=𝑘|𝑡|u�−𝑝(𝑝+4) 16𝑡2, consider problems 𝑢″(𝑡)+𝑎(𝑡)𝑢(−𝑡)+𝑏(𝑡)𝑢(𝑡)=ℎ(𝑡), 𝑢(0)=𝑢′(0)=0 (5.3.4) and 𝑢″(𝑡)+𝑢(−𝑡)+𝑘𝑢(𝑡)=ℎ(𝑡), 𝑢(0)=𝑢′(0)=0. (5.3.5) Using an argument similar as the one in Example 5.2.4 and considering 𝑅=𝐷2−𝜑∗+𝑘, we can reduce problem (5.3.5) to 𝑢(4)(𝑡)+2𝑘𝑢″(𝑡)+(𝑘2−1)𝑢(𝑡)=𝑓(𝑡), 𝑢(u�)(0)=0, 𝑗=0,…,3, (5.3.6) which can be decomposed in 𝑢″(𝑡)+(𝑘+1)𝑢(𝑡)=𝑣(𝑡), 𝑡∈𝐼, 𝑢(0)=𝑢′(0)=0, 𝑣″(𝑡)+(𝑘−1)𝑣(𝑡)=𝑓(𝑡), 𝑡∈𝐼, 𝑣(0)=𝑣′(0)=0, which have as Green’s functions, respectively,  𝐺1(𝑡,𝑠)= sin(√𝑘+1(𝑡−𝑠)) √𝑘+1 𝜒u� 0(𝑠), 𝑡∈ℝ,  𝐺2(𝑡,𝑠)= sin(√𝑘−1(𝑡−𝑠)) √𝑘−1 𝜒u� 0(𝑠), 𝑡∈ℝ. Then, the Green’s function for problem (5.3.6) is 𝐺(𝑡,𝑠)=∫u� u� 𝐺1(𝑡,𝑟)  𝐺2(𝑟,𝑠)d𝑟 =1 2√𝑘2−1[√𝑘−1sin(√𝑘+1(𝑠−𝑡))−√𝑘+1sin(√𝑘−1(𝑠−𝑡))]𝜒u� 0(𝑠). 5. The Hilbert transform and other algebras 117 Observe that 𝑅⊢𝐺(𝑡,𝑠)=−⎡ ⎢ ⎣sin(√𝑘−1(𝑠−𝑡)) 2√𝑘−1 +sin(√𝑘+1(𝑠−𝑡)) 2√𝑘+1 ⎤ ⎥ ⎦𝜒u� 0(𝑠). Hence, considering 𝐴(𝑡)∶= 2 𝑝+2|𝑡|u� 2𝑡, the Green’s function of problem (5.3.4) follows the expression 𝐻(𝑡,𝑠)∶= 4 √𝑎(𝑠) 𝑎(𝑡)𝐺(𝐴(𝑡),𝐴(𝑠)), This is, 𝐻(𝑡,𝑠)=−∣𝑠 𝑡∣u� 4⎡ ⎢ ⎢ ⎢ ⎣ sin(2√u�−1(u�|u�|u�/2−u�|u�|u�/2) u�+2 ) 2√𝑘−1 +sin(2√u�+1(u�|u�|u�/2−u�|u�|u�/2) u�+2 ) 2√𝑘+1 ⎤ ⎥ ⎥ ⎥ ⎦𝜒u� 0(𝑠). 5.4 The Hilbert transform and other algebras In this section we devote our attention to new algebras to which we can apply the previous results. To achieve this goal we recall the definition and remarkable properties of the Hilbert transform [114]. We define the Hilbert transform 𝖧of a function 𝑓as 𝖧𝑓(𝑡)∶= 1 𝜋lim u�→∞∫u� −u� 𝑓(𝑠) 𝑡−𝑠d𝑠≡ 1 𝜋∫∞ −∞ 𝑓(𝑠) 𝑡−𝑠d𝑠, where the last integral is to be understood as the Cauchy principal value. Among its properties, we would like to point out the following. •𝖧∶Lp(ℝ)→Lp(ℝ)is a linear bounded operator for every 𝑝∈(1,+∞)and ‖𝖧‖u�=⎧ { ⎨ { ⎩tan u� 2u�, 𝑝∈(1,2], cot u� 2u�, 𝑝∈[2,+∞), in particular ‖𝖧‖2=1. •𝖧is an anti-involution: 𝖧2=−Id. • Let 𝜎(𝑡) = 𝑎𝑡+𝑏for 𝑎,𝑏 ∈ ℝ. Then 𝖧𝜎∗= sign(𝑎)𝜎∗𝖧(in particular, 𝖧𝜑∗= −𝜑∗𝖧). Furthermore, if a linear bounded operator 𝖮∶Lp(ℝ)→Lp(ℝ)satisfies this property, 𝖮=𝛽𝐻where 𝛽∈ℝ. •𝖧commutes with the derivative: 𝖧𝐷=𝐷𝖧. 118 5.4. The Hilbert transform and other algebras •𝖧(𝑓 ∗𝑔)=𝑓 ∗𝖧𝑔=𝖧𝑓 ∗𝑔where ∗denotes the convolution. •𝖧is an isometry in L2(ℝ):⟨𝖧𝑓,𝖧𝑔⟩=⟨𝑓,𝑔⟩where ⟨,⟩is the scalar product in L2(ℝ). In particular ‖𝖧𝑓‖2=‖𝑓‖2. Consider now the same construction we did for ℝ[𝐷,𝜑∗]changing 𝜑∗by 𝖧and denote this algebra as ℝ[𝐷,𝖧]. In this case we are dealing with a commutative algebra. Actually, this algebra is isomorphic to the complex polynomials ℂ[𝐷]. Just consider the isomorphism ℝ[𝐷,𝖧] ℂ[𝐷] ∑ u�(𝑎u�𝖧+𝑏u�)𝐷u�∑ u�(𝑎u�𝑖+𝑏u�)𝐷u� Ξ Observe that Ξ|ℝ[u�] =Id|ℝ[u�]. We now state a result analogous to Theorem 5.1.1. Theorem 5.4.1. Take 𝐿=∑ u�(𝑎u�𝖧+𝑏u�)𝐷u�∈ℝ[𝐷,𝖧] and define 𝑅=∑ u�(𝑎u�𝖧−𝑏u�)𝐷u�. Then 𝐿𝑅=𝑅𝐿∈ℝ[𝐷]. Remark 5.4.2. Theorem 5.4.1 is clear from the point of view of ℂ[𝐷]. Since Ξ(𝑅)=−Ξ(𝐿), 𝑅𝐿=Ξ−1(−Ξ(𝐿)Ξ(𝐿))=Ξ−1(−|Ξ(𝐿)|2). Therefore, |Ξ(𝐿)|2∈ℝ[𝐷], implies 𝑅𝐿∈ℝ[𝐷]. Remark 5.4.3. Since ℝ[𝐷,𝖧]is isomorphic to ℂ[𝐷], the Fundamental Theorem of Algebra also applies to ℝ[𝐷,𝖧], which shows a clear classification of the decompositions of an element of ℝ[𝐷,𝖧]in contrast with those of ℝ[𝐷,𝜑∗]which, in page 89, was shown not to be a unique factorization domain. In the following example we will use some properties of the Hilbert transform [114]: 𝖧cos=sin, 𝖧sin=−cos, 𝖧(𝑡𝑓(𝑡))(𝑡)=𝑡𝖧𝑓(𝑡)− 1 𝜋∫∞ −∞ 𝑓(𝑠)d𝑠, where, as we have noted before, the integral is considered as the principal value. 5. The Hilbert transform and other algebras 119 Example 5.4.4. Consider the problem 𝐿𝑢(𝑡)≡𝑢′(𝑡)+𝑎𝖧𝑢(𝑡)=ℎ(𝑡)∶=sin𝑎𝑡, 𝑢(0)=0, (5.4.1) where 𝑎>0. Composing the operator 𝐿=𝐷+𝑎𝖧with the operator 𝑅=𝐷−𝑎𝖧we obtain 𝑆=𝑅𝐿=𝐷2+𝑎2, the harmonic oscillator operator. The extra boundary conditions obtained applying 𝑅are 𝑢′(0)−𝑎𝖧𝑢(0)=0. The general solution to the problem 𝑢″(𝑡)+𝑎2𝑢(𝑡)= 𝑅ℎ(𝑡)=2𝑎cos𝑎𝑡,𝑢(0)=0is given by 𝑣(𝑡)=∫u� 0sin(𝑎[𝑡−𝑠]) 𝑎𝑅ℎ(𝑠)d𝑠+𝛼sin𝑎𝑡=(𝑡+𝛼)sin𝑎𝑡, where 𝛼is a real constant. Hence, 𝖧𝑣(𝑡)=−(𝑡+𝛼)cos𝑎𝑡. If we impose the boundary conditions 𝑣′(0)−𝑎𝖧𝑣(0)=0then we get 𝛼=0. Hence, the unique solution of problem (5.4.1) is 𝑢(𝑡)=𝑡sin𝑎𝑡. Remark 5.4.5. It can be checked that the kernel of 𝐷+𝑎𝖧(𝑎>0) is spanned by sin𝑡and cos𝑡and, also, the kernel of 𝐷−𝑎𝖧is just 0. This defies, in the line of Remark 5.2.6, the usual relation between the degree of the operator and the dimension of the kernel which is held for ordinary differential equations, that is, the operator of a linear ordinary differential equation of order 𝑛has a kernel of dimension 𝑛. In this case we have the order one operator 𝐷+𝑎𝖧 with a dimension two kernel and the injective order one operator 𝐷−𝑎𝖧. Now, we consider operators with reflection and Hilbert transforms, and denote the algebra as ℝ[𝐷,𝖧,𝜑∗]. We can again state a reduction Theorem. Theorem 5.4.6. Take 𝐿=∑ u�𝑎u�𝜑∗𝖧𝐷u�+∑ u�𝑏u�𝖧𝐷u�+∑ u�𝑐u�𝜑∗𝐷u�+∑ u�𝑑u�𝐷u�∈ℝ[𝐷,𝖧,𝜑∗] and define 𝑅=∑ u�𝑎u�𝜑∗𝖧𝐷u�+∑ u�(−1)u�𝑏u�𝖧𝐷u�+∑ u�𝑐u�𝜑∗𝐷u�−∑ u�(−1)u�𝑑u�𝐷u�. Then 𝐿𝑅=𝑅𝐿∈ℝ[𝐷]. 5.4.1 Hyperbolic numbers as operators Finally, we use the same idea behind the isomorphism Ξto construct an operator algebra isomorphic to the algebra of polynomials on the hyperbolic numbers. The hyperbolic numbers†are defined, in a similar way to the complex numbers, as follows, 𝔻={𝑥+𝑗𝑦 ∶ 𝑥,𝑦∈ℝ, 𝑗  ∈ℝ, 𝑗2=1}. †See [6,166] for an introduction to hyperbolic numbers and some of their properties and applications. 120 5.4. The Hilbert transform and other algebras The arithmetic in 𝔻is that obtained assuming the commutative, associative and distributive properties for the sum and product. In a parallel fashion to the complex numbers, if 𝑤∈𝔻, with 𝑤=𝑥+𝑗𝑦, we can define 𝑤∶=𝑥−𝑗𝑦, ℜ(𝑤)∶=𝑥, ℑ(𝑤)∶=𝑦, and, since 𝑤𝑤=𝑥2−𝑦2∈ℝ, we set |𝑤|∶=√|𝑤𝑤|, which is called the Minkowski norm. It is clear that |𝑤1𝑤2|=|𝑤1||𝑤2|for every 𝑤1,𝑤2∈𝔻 and, if |𝑤|≠0, then 𝑤−1 =𝑤/|𝑤|2. If we add the norm ‖𝑤‖=√2(𝑥2+𝑦2), we have that (𝔻,‖⋅‖)is a Banach algebra, so the exponential and the hyperbolic trigonometric functions are well defined. Although, unlike ℂ,𝔻is not a division algebra (not every nonzero element has an inverse), we can derive calculus (differentiation, integration, holomorphic functions…) for 𝔻as well [6]. In this setting, we want to derive an operator 𝐽defined on a suitable space of functions such that satisfies the same algebraic properties as the hyperbolic imaginary unity 𝑗. In other words, we want the map ℝ[𝐷,𝐽] 𝔻[𝐷] ∑ u�(𝑎u�𝐽+𝑏u�)𝐷u�∑ u�(𝑎u�𝑗+𝑏u�)𝐷u� Θ to be an algebra isomorphism. This implies: •𝐽is a linear operator, •𝐽  ∈ℝ[𝐷]. •𝐽2=Id, that is, 𝐽is an involution, •𝐽𝐷=𝐷𝐽. Thereis a simple characterization of linear involutions on a vector space: every linear involution 𝐽is of the form 𝐽=±(2𝑃−Id) where 𝑃is a projection operator, that is, 𝑃2= 𝑃. It is clear that ±(2𝑃−Id)is, indeed a linear operator and an involution. On the other hand, it is simple to check that, if 𝐽is a linear involution, 𝑃∶=(±𝐽+Id)/2is a projection, so 𝐽=±(2𝑃−Id). Hence, it is sufficient to look for a projection 𝑃commuting with de derivative. 5. The Hilbert transform and other algebras 121 Example 5.4.7. Consider the space 𝑊=L2([−𝜋,𝜋])and define 𝑃𝑓(𝑡)∶= ∑ u�∈ℕ∫u� −u� 𝑓(𝑠)cos(2𝑛𝑠)d𝑠cos(2𝑛𝑡)for every𝑓 ∈𝑊, that is, take only the sum over the even coefficients of the Fourier series of 𝑓. Clearly 𝑃𝐷=𝐷𝑃. 𝐽∶=2𝑃−Idsatisfies the aforementioned properties. The algebra ℝ[𝐷,𝐽], being isomorphic to 𝔻[𝐷], satisfies also very good algebraic properties (see, for instance, [146]). In order to get an analogous theorem to Theorem 5.1.1 for the algebra ℝ[𝐷,𝐽]it is enough to take, as in the case of ℝ[𝐷,𝐽],𝑅=Θ−1(Θ(𝐿)). 128 6.1. General solutions Remark 6.1.5. In the hypotheses of Theorem 6.1.3, if instead of 𝑔(0)=𝑓(0)=0we have that 𝑔(𝑠0)=𝑓(𝑠0)=0, define  𝑓(𝑥)∶=𝑓(𝑥+𝑠0), 𝑔(𝑥)∶=𝑔(𝑥+𝑠0). Then  𝑓(0)=  𝑔(0)=0 and problem (6.1.1) is equivalent to ( 𝑔∘𝑣′)′(𝑡)+  𝑓(𝑣(𝑡))=0, 𝑣(𝑎)=𝑐1−𝑠0, 𝑣(𝑎)=𝑐2, with 𝑣(𝑡)=𝑥(𝑡)−𝑠0. Hence, we can apply Theorem 6.1.3 to this case. Remark 6.1.6. Using the notation of Theorem 6.1.3, the explicit form of the solution of problem (6.1.1) is given by 𝑥(𝑡)=⎧ { { ⎨ { { ⎩ 𝐻−1 +(𝑡−⌊𝑡−𝑎 𝑇⌋𝑇), 𝑡∈[𝑎+2𝑇𝑘,𝑎+(2𝑘+1)𝑇], 𝑘∈ℤ, 𝐻−1 −(𝑡−⌊𝑡−𝑎 𝑇⌋𝑇), 𝑡∈[𝑎+(2𝑘−1)𝑇,𝑎+2𝑘𝑇], 𝑘∈ℤ, Remark 6.1.7. Consider the following particular case of problem (6.1.1) with 𝑓(0)=0,𝑔(0)= 0,𝑓and 𝑔increasing and the hypothesis for a unique global solution of the following problem are satisfied in Theorem 6.1.3. (𝑔∘𝑥′)′(𝑡)+ 𝑓(𝑥(𝑡))=0, 𝑥(0)=0, 𝑥′(0)=1. (6.1.6) It is clear that, in the case 𝑔(𝑥)=𝑓(𝑥)=𝑥, the unique solution of problem (6.1.6) is sin(𝑡), which suggests the definition of the sinu�,u� function as the unique solution of problem (6.1.6) for general 𝑔and 𝑓. Correspondingly, arcsin+ u�,u�(𝑟)∶=𝐻+(𝑟). This function, defined as such, coincides with the arcsinu�function defined in [24, 115] for the 𝑝-Laplacian 𝑓(𝑥)=𝑔(𝑥)=|𝑥|u�−2𝑥, the function arcsinu�,u� defined in [14,65,108] for the 𝑝-𝑞-Laplacian 𝑓(𝑥)=|𝑥|u�−2𝑥,𝑔(𝑥)=|𝑥|u�−2𝑥, which first appeared with a slightly different definition in [64], and the hyperbolic version of this function, also in [14,108], which corresponds to the case 𝑓(𝑥)=|𝑥|u�−2𝑥,𝑔(𝑥)=−|𝑥|u�−2𝑥. [164] derives generalized Jacobian functions in a similar way, defining arcsnu�,u�(𝑡,𝑘)∶=∫u� 01 u� √(1−𝑠u�)(1−𝑘u�𝑠u�)d𝑠, of which the inverse (see [164, Proposition 3.2]) is precisely a solution of (𝑓u�∘𝑥′(𝑡))′+𝑞 𝑝∗𝑓u�(𝑥(𝑡))(1+𝑘u�−2𝑘u�|𝑥(𝑡)|u�)=0, where 𝑓u�is the 𝑟-Laplacian for 𝑟 = 𝑝,𝑞and 𝑝∗𝑝=𝑝∗+𝑝. Observe that this case is also covered by our definition. In all of the aforementioned works they are interested on the inverse of the arcsinu�,u� function, the sinu�,u� function, which they extend to the whole real line by symmetry and periodicity. Observe that in our case 𝑓and 𝑔need not to be odd functions, contrary to the above examples, but we can still give the definition of the sinu�,u� function in the whole real line. Also, this lack of symmetry gives rise to a richer set of right inverses of sinu�,u�, for instance, 6. General solutions 129 arcsin− u�,u�(𝑟)∶=𝐻−(𝑟). In general, if we have a problem of the kind Φ((𝑔∘𝑥′)′,𝑥(𝑡))=0; 𝑥(0)=0, 𝑥′(0)=1, and we know it has a unique solution in a neighborhood of 0, then we can define sinu�,Φ as such unique solution and its inverse, in a neighborhood of 0,arcsinu�,Φ. 6.1.1 A particular case Having in mind problem (6.0.2), we now consider a particular case of problem (6.1.1) for the rest of this section. Assume 𝑓is invertible and both 𝑓and 𝑓−1 are continuous. For convenience, assume also that 𝑓is increasing and 𝑓(0)=0. Consider the following problem. (𝑓−1 ∘𝑥′)′(𝑡)+𝜆𝑓(𝑥(𝑡))=0, 𝑥(𝑎)=𝑐, 𝑥′(𝑎)=𝑓(𝑐), (6.1.7) where 𝜆∈ℝ+. The following corollary is just the restatement of Theorem 6.1.3 for this particular case. Corollary 6.1.8. Let 𝑓 ∶ (𝜏1,𝜏2) → (𝜎1,𝜎2)be an invertible function such that 𝑓is continuous and assume 0 ∈ (𝜏1,𝜏2),𝑓(0) = 0and 𝑓increasing, 𝜆>0,(1+𝜆)𝐹(𝑐) < min{𝐹(𝜏1),𝐹(𝜏2)}. Then there exists a unique local solution of problem (6.1.7). Furthermore, if (1+𝜆−1)𝐹(𝑐)<min{𝐹(𝜏1),𝐹(𝜏2)}, then such solution is defined on ℝand is periodic of first period 𝑇∶=∫u�−1 +((1+u�−1)u�(u�)) u�−1 −((1+u�−1)u�(u�))[1 𝑓(𝐹−1 +((1+𝜆)𝐹(𝑐)−𝜆𝐹(𝑟))) −1 𝑓(𝐹−1 −((1+𝜆)𝐹(𝑐)−𝜆𝐹(𝑟)))]d𝑟. (6.1.8) There are some particular cases where the formula (6.1.8) can be simplified. If 𝑓is odd then 𝐹is even and, with the change of variables 𝑟=|𝑐|𝑠, we have that expression (6.1.8) becomes 𝑇=∫u�−1 +((1+u�−1)u�(u�)) |u�| 04|𝑐|d𝑟 𝑓(𝐹−1 +((1+𝜆)𝐹(𝑐)−𝜆𝐹(|𝑐|𝑟))). Also, if we further assume that 𝑓is defined in ℝand that 𝑓(𝑟𝑡) = ℎ(𝑟)𝑓(𝑡)for every 𝑟,𝑡∈ℝ(see Remark 6.1.10 for a classification of such functions) and some function ℎ, then 𝐹(𝑟𝑡)=∫u�u� 0𝑓(𝑠)d𝑠=∫u� 0𝑓(𝑟𝑠)𝑟d𝑠=𝑟ℎ(𝑟)∫u� 0𝑓(𝑠)d𝑠= 𝑟ℎ(𝑟)𝐹(𝑡), so 𝐹satisfies the same kind of property for  ℎ(𝑟)=𝑟ℎ(𝑟). Clearly, for 𝑡>0, 𝐹−1 −( ℎ(𝑟)𝑡)=𝑟𝐹−1 −(𝑡), 𝐹−1 +( ℎ(𝑟)𝑡)=𝑟𝐹−1 +(𝑡). 130 6.1. General solutions Observethat  ℎ(𝑟)=𝐹(𝑟)/𝐹(1), and therefore  ℎ|(−∞,0], ℎ|[0,+∞) areinvertible. Also,  ℎ−1 +(𝑡)= 𝐹−1 +(𝑡𝐹(1))for 𝑡>0. Hence, 𝐹−1 +((1+𝜆−1)𝐹(𝑐)) |𝑐| =𝐹−1 +( ℎ(  ℎ−1 +(1+𝜆−1))𝐹(𝑐)) |𝑐| = ℎ−1 +(1+𝜆−1)𝐹−1 +(𝐹(𝑐)) |𝑐| = ℎ−1 +(1+𝜆−1)=𝐹−1 +((1+𝜆−1)𝐹(1)). All the same, 𝐹−1 −((1+𝜆−1)𝐹(𝑐))/|𝑐|=−𝐹−1 −((1+𝜆−1)𝐹(1)). Also, 𝑓(𝐹−1 +((1+𝜆)𝐹(𝑐)−𝜆𝐹(|𝑐|𝑟)))=𝑓(𝐹−1 +((1+𝜆)  ℎ(|𝑐|)𝐹(1)−𝜆  ℎ(|𝑐|)𝐹(𝑟))) =𝑓(𝐹−1 +( ℎ(|𝑐|)[(1+𝜆)𝐹(1)−𝜆𝐹(𝑟))])=𝑓(|𝑐|𝐹−1 +((1+𝜆)𝐹(1)−𝜆𝐹(𝑟))) =ℎ(|𝑐|)𝑓(𝐹−1 +((1+𝜆)𝐹(1)−𝜆𝐹(𝑟))) =(𝑓(|𝑐|)/𝑓(1))𝑓(𝐹−1 +((1+𝜆)𝐹(1)−𝜆𝐹(𝑟))). With these considerations in mind, we have that we can further reduce expression (6.1.8) to 𝑇(𝑐,𝜆)= 4|𝑐|𝑓(1) 𝑓(|𝑐|) ∫u�−1 +((1+u�−1)u�(1)) 0d𝑟 𝑓(𝐹−1 +((1+𝜆)𝐹(1)−𝜆𝐹(𝑟))). Example 6.1.9. Let 𝑓(𝑡)∶=|𝑡|u�−2𝑡,𝑝>1. Then 𝑇(𝑐,𝜆,𝑝)=4|𝑐|2−u� ∫(1+u�−1)1 u� 0[1+𝜆−𝜆𝑟u�]1−u� u�d𝑟. Observe that with the change of variable 𝑟=(1+𝜆−1)1 u�𝑠we have that 𝑇(𝑐,𝜆,𝑝)=4|𝑐|2−u� ∫1 0(1+𝜆−1)1 u�[(1+𝜆)(1− 𝑠u�)]1−u� u�d𝑠 =4|𝑐|2−u�𝜆−1 u�(1+𝜆)2 u�−1 ∫1 0(1− 𝑠u�)1−u� u�d𝑠 =4|𝑐|2−u�𝜆−1 u�(1+𝜆)2 u�−1 Γ(1 u�)2 𝑝Γ(2 u�). 𝑇is increasing on |𝑐|if 𝑝 ∈ (1,2)and decreasing on |𝑐|if 𝑝>2and independent of |𝑐|if 𝑝=2. If we take 𝜆=1, 𝑇(𝑐,1,𝑝)=22 u�+1 |𝑐|2−u� Γ(1 u�)2 𝑝Γ(2 u�). In particular, 𝑇(𝑐,1,2)=2𝜋(independently of 𝑐). We can also consider the dependence of 𝑇on 𝜆. We do this study for this particular example and in the following section we develop a general theory. 𝜕𝑇 𝜕𝜆(𝑐,𝜆,𝑝)=−4|𝑐|2−u� 𝑝𝜆 (1+ 1 𝜆)1 u�(1+𝜆)1−2u� u�(1+(𝑝−1)𝜆)∫1 0(1−𝑠u�)1−u� u�d𝑠<0. Therefore the period 𝑇is decreasing on 𝜆. 6. General solutions 131 Remark 6.1.10. If a continuous function 𝑓satisfies that 𝑓(𝑟𝑡) = ℎ(𝑟)𝑓(𝑡), we can obtain the explicit expression of 𝑓. Let 𝑐 = 𝑓(1),𝑔(𝑡) ∶= 𝑓(𝑡)/𝑓(1)and 𝛼 = ln𝑔(𝑒). Then 𝑔(𝑡𝑠) = 𝑔(𝑡)𝑔(𝑠). Also, for 𝑡 ≠ 0,1 = 𝑔(1) = 𝑔(𝑡/𝑡) = 𝑔(𝑡)𝑔(1/𝑡)and therefore 𝑔(𝑡−1) = 𝑔(𝑡)−1. If 𝑛∈ℕ,𝑔(𝑡u�) = 𝑔(𝑡)u�, so, for 𝑡 ≥ 0,𝑔(𝑡) = 𝑔(𝑡u� u�) = 𝑔(𝑡1 u�)u�and 𝑔(𝑡1 u�)=𝑔(𝑡)1 u�. Hence, 𝑔(𝑡u� u�)=𝑔(𝑡)u� u�for every 𝑝,𝑞∈ℕ,𝑞≠0and, by the density of ℚ in ℝand the continuity of 𝑓,𝑔(𝑡u�)=𝑔(𝑡)u�for all 𝑡≥0,𝑟∈ℝ+. Now, for 𝑡>0,𝑔(𝑡)=𝑔(𝑒lnu�)=𝑔(𝑒)ln u� =𝑒lnu�(u�)lnu� =𝑡lnu�(u�) =𝑡u�. Hence, 𝑓(𝑡)= 𝛽𝑡u�for 𝑡≥0. On the other hand, 1=𝑔(1)=(𝑔(−1))2, so 𝑔(−1)=±1. Also, 𝑓(−𝑡)= 𝑔(−1)𝑓(𝑡)and thus, 𝑓(−𝑡)=±𝛽𝑡u�for 𝑡>0. In summary, 𝑓(𝑡)=⎧ { ⎨ { ⎩𝛽𝑡u�if 𝑡≥0, ±𝛽(−𝑡)u�if 𝑡<0. If we further ask for 𝑓to be injective, 𝑓(𝑡)=𝛽|𝑡|u�−1𝑡, that is, 𝑓is an 𝛼-laplacian. 6.1.2 Dependence of T on 𝜆and 𝑐 Based on the approach used in Example 6.1.9, we study now the dependence of 𝑇on 𝜆and 𝑐 in a general way. For simplicity, we will assume 𝑐>0. For the case 𝑐<0, just do the change of variable 𝑦(𝑡)=−𝑥(𝑡). We continue to assume the hypotheses for (6.1.7) and further assume that 𝑓is a differentiable function. Let us divide the interval of integration in equation (6.1.2) in [𝐹−1 −((1+ 𝜆−1)𝐹(𝑐)),0]and [0,𝐹−1 +((1+𝜆−1)𝐹(𝑐))]. Observe that 𝐹is injective restricted to any of the two intervals. For the nonnegative interval, taking the change of variables 𝑟=𝐹−1 +((1+𝜆−1)𝐹(𝑐𝑠)), we have that ∫u�−1 +((1+u�−1)u�(u�)) 0[1 𝑓(𝐹−1 +((1+𝜆)𝐹(𝑐)−𝜆𝐹(𝑟))) −1 𝑓(𝐹−1 −((1+𝜆)𝐹(𝑐)−𝜆𝐹(𝑟)))]d𝑟 =∫1 0[1 𝑓(𝐹−1 +((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])−1 𝑓(𝐹−1 −((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])] ⋅[1+𝜆−1]𝑐𝑓(𝑐𝑠) 𝑓(𝐹−1 +((1+𝜆−1)𝐹(𝑐𝑠)))d𝑠. All the same, with the change of variables 𝑟=𝐹−1 −((1+𝜆−1)𝐹(𝑐𝑠)), ∫0 u�−1 −((1+u�−1)u�(u�)) [1 𝑓(𝐹−1 +((1+𝜆)𝐹(𝑐)−𝜆𝐹(𝑟))) 132 6.1. General solutions −1 𝑓(𝐹−1 −((1+𝜆)𝐹(𝑐)−𝜆𝐹(𝑟)))]d𝑟 =∫0 1[1 𝑓(𝐹−1 +((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])−1 𝑓(𝐹−1 −((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])] ⋅[1+𝜆−1]𝑐𝑓(𝑐𝑠) 𝑓(𝐹−1 −((1+𝜆−1)𝐹(𝑐𝑠)))d𝑠. Now let, for 𝜆∈ℝ+and 𝑠∈[0,1], 𝛼(𝜆,𝑠,𝑐)∶=(1+𝜆−1)𝑐𝑓(𝑐𝑠), 𝜕𝛼 𝜕𝜆(𝜆,𝑠,𝑐)=−𝜆−2𝑐𝑓(𝑐𝑠), 𝛽±(𝜆,𝑠,𝑐)∶=𝑓(𝐹−1 ±((1+𝜆−1)𝐹(𝑐𝑠))), 𝜕𝛽± 𝜕𝜆(𝜆,𝑠,𝑐)=−𝜆−2𝐹(𝑐𝑠)𝑓′(𝐹−1 ±((1+𝜆−1)𝐹(𝑐𝑠))) 𝑓(𝐹−1 ±((1+𝜆−1)𝐹(𝑐𝑠))), 𝛾±(𝜆,𝑠,𝑐)∶=𝑓(𝐹−1 ±((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])), 𝜕𝛾± 𝜕𝜆(𝜆,𝑠,𝑐)=[𝐹(𝑐)− 𝐹(𝑐𝑠)]𝑓′(𝐹−1 ±((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])) 𝑓(𝐹−1 ±((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])). Then 𝑇(𝜆,𝑐)=∫1 0𝛼(𝜆,𝑠,𝑐)[1 𝛽+(𝜆,𝑠,𝑐)−1 𝛽−(𝜆,𝑠,𝑐)][ 1 𝛾+(𝜆,𝑠,𝑐)−1 𝛾−(𝜆,𝑠,𝑐)]d𝑠. (6.1.9) Therefore, 𝜕𝑇 𝜕𝜆(𝜆,𝑐)=∫1 0{𝜕𝛼 𝜕𝜆(𝜆,𝑠,𝑐)[1 𝛽+(𝜆,𝑠,𝑐)−1 𝛽−(𝜆,𝑠,𝑐)][ 1 𝛾+(𝜆,𝑠,𝑐)−1 𝛾−(𝜆,𝑠,𝑐)] +𝛼(𝜆,𝑠,𝑐)⎡ ⎢ ⎢ ⎣ u�u�− u�u� (𝜆,𝑠,𝑐) 𝛽−(𝜆,𝑠,𝑐)2−u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛽+(𝜆,𝑠,𝑐)2⎤ ⎥ ⎥ ⎦[1 𝛾+(𝜆,𝑠,𝑐)−1 𝛾−(𝜆,𝑠,𝑐)] +𝛼(𝜆,𝑠,𝑐)[1 𝛽+(𝜆,𝑠,𝑐)−1 𝛽−(𝜆,𝑠,𝑐)]⎡ ⎢ ⎣ u�u�− u�u� (𝜆,𝑠,𝑐) 𝛾−(𝜆,𝑠,𝑐)2−u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛾+(𝜆,𝑠,𝑐)2⎤ ⎥ ⎦ ⎫ } ⎬ } ⎭d𝑠. Observe that 𝛼,𝑓|[0,1],𝑓′,𝐹,𝐹−1 +,𝛽+,u�u�− u�u� ,𝛾+,u�u�+ u�u� are nonnegative, while u�u� u�u�,𝐹−1 −,𝛽−,u�u�+ u�u� , 𝛾−,u�u�− u�u� are nonpositive. In general we cannot tell the sign of 𝑇(𝜆,𝑐)from this expression, but making certain assumptions we can simplify it to derive information. Assume now 𝑓is and odd function. Then 𝐹−1 −=−𝐹−1 +,𝛽−=−𝛽+and 𝛾−=−𝛾+, so 𝜕𝑇 𝜕𝜆(𝜆,𝑐)=4∫1 01 𝛽+(𝜆,𝑠,𝑐)𝛾+(𝜆,𝑠,𝑐)[𝜕𝛼 𝜕𝜆(𝜆,𝑠,𝑐) −𝛼(𝜆,𝑠,𝑐)⎛ ⎜ ⎜ ⎜ ⎝ u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛽+(𝜆,𝑠,𝑐) +u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛾+(𝜆,𝑠,𝑐)⎞ ⎟ ⎟ ⎟ ⎠⎤ ⎥ ⎥ ⎦d𝑠. 6. General solutions 133 Now, if we differentiate equation (6.1.9) with respect to 𝑐, 𝜕𝑇 𝜕𝑐(𝜆,𝑐)=∫1 0{𝜕𝛼 𝜕𝑐(𝜆,𝑠,𝑐)[1 𝛽+(𝜆,𝑠,𝑐)−1 𝛽−(𝜆,𝑠,𝑐)][ 1 𝛾+(𝜆,𝑠,𝑐)−1 𝛾−(𝜆,𝑠,𝑐)] +𝛼(𝜆,𝑠,𝑐)⎡ ⎢ ⎢ ⎣ u�u�− u�u� (𝜆,𝑠,𝑐) 𝛽−(𝜆,𝑠,𝑐)2−u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛽+(𝜆,𝑠,𝑐)2⎤ ⎥ ⎥ ⎦[1 𝛾+(𝜆,𝑠,𝑐)−1 𝛾−(𝜆,𝑠,𝑐)] +𝛼(𝜆,𝑠,𝑐)[1 𝛽+(𝜆,𝑠,𝑐)−1 𝛽−(𝜆,𝑠,𝑐)]⎡ ⎢ ⎣ u�u�− u�u� (𝜆,𝑠,𝑐) 𝛾−(𝜆,𝑠,𝑐)2−u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛾+(𝜆,𝑠,𝑐)2⎤ ⎥ ⎦ ⎫ } ⎬ } ⎭d𝑠. Observe that 𝜕𝛼 𝜕𝑐(𝜆,𝑠,𝑐)=(1+𝜆−1)[𝑓(𝑐𝑠)+𝑐𝑠𝑓′(𝑐𝑠)], 𝜕𝛽± 𝜕𝑐 (𝜆,𝑠,𝑐)=(1+𝜆−1)𝑠𝑓(𝑐𝑠)𝑓′(𝐹−1 ±((1+𝜆−1)𝐹(𝑐𝑠))) 𝑓(𝐹−1 ±((1+𝜆−1)𝐹(𝑐𝑠))), 𝜕𝛾± 𝜕𝑐 (𝜆,𝑠,𝑐)=(1+𝜆)[𝑓(𝑐)−𝑠𝑓(𝑐𝑠)]𝑓′(𝐹−1 ±((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])) 𝑓(𝐹−1 ±((1+𝜆)[𝐹(𝑐)− 𝐹(𝑐𝑠)])). Hence, u�u� u�u� ,u�u�+ u�u� is positive and u�u�− u�u� negative for 𝑐≥0. Assume now 𝑓is an odd function. 𝜕𝑇 𝜕𝑐(𝜆,𝑐)=4∫1 01 𝛽+(𝜆,𝑠,𝑐)𝛾+(𝜆,𝑠,𝑐)[𝜕𝛼 𝜕𝑐(𝜆,𝑠,𝑐) −𝛼(𝜆,𝑠,𝑐)⎛ ⎜ ⎜ ⎜ ⎝ u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛽+(𝜆,𝑠,𝑐) +u�u�+ u�u� (𝜆,𝑠,𝑐) 𝛾+(𝜆,𝑠,𝑐)⎞ ⎟ ⎟ ⎟ ⎠⎤ ⎥ ⎥ ⎦d𝑠. Example 6.1.11. Let 𝑓 ∶ (−1,1) → ℝ,𝑓(𝑥) ∶= 𝑥/√1−𝑥2, 𝑥 ∈ ℝand consider problem (6.1.7)†. Then 𝐹(𝑥)=1−√1−𝑥2, 𝐹−1 +(𝑥)=√2𝑥−𝑥2. In order for the conditions in Corollary 6.1.8 to be satisfied we need (1+𝜆)𝐹(𝑐)<1, (1+𝜆−1)𝐹(𝑐)<1, that is |𝑐|<min⎧ { ⎨ { ⎩√𝜆(𝜆+2) 𝜆+1 ,√2𝜆+1 𝜆+1 ⎫ } ⎬ } ⎭. In Figure 6.1.1 we plot how the period varies as a function of 𝑐and 𝜆. Observe how the period is decreasing in both parameters and limu�,u�→0 𝑇(𝜆,𝑐)=+∞. †The diffeomorphisms 𝑓in this example has been widely studied by Bereanu and Mawhin (see, for instance, [8,10]) and is a type of singular 𝜑-Laplacian known as the mean curvature operator of the Minkowski space. Its inverse, the mean curvature operator of the Euclidean space, also studied in [8], appears in Example 6.1.12. 134 6.1. General solutions Figure 6.1.1: Graph of the period 𝑇function of 𝑐and 𝜆. Example 6.1.12. Let 𝑓be the bounded 𝜑-Laplacian [8] given by 𝑓 ∶ ℝ → (−1,1),𝑓(𝑥) ∶= 𝑥/√1+𝑥2, 𝑥 ∈ ℝand consider problem (6.1.7). 𝑓is effectively the inverse function of the one in the previous example. Then 𝐹(𝑥)=√1+𝑥2−1, 𝐹−1 +(𝑥)=√2𝑥+𝑥2. The conditions in Corollary 6.1.8 are satisfied without any further restrictions. In Figure 6.1.2 we plot how the period varies as a function of 𝑐and 𝜆. Observe in this plot how the period is decreasing in 𝜆, increasing in 𝑐and limu�→0 𝑇(𝑐,𝜆)=limu�→+∞ 𝑇(𝑐,𝜆)=+∞. Figure 6.1.2: Graph of the period 𝑇function of 𝑐and 𝜆. 6. Problems with reflection 135 6.2 Problems with reflection Let us consider again the problem that motivated this chapter, the obtaining of solutions of problem (3.1.5) in the case 𝜑(𝑡)=−𝑡. Hence, consider again the problems (3.1.1) and (3.1.2) in the case 𝜑(𝑡)=−𝑡. Observe that Lemma 3.1.1 (following Remark 3.1.5) can be trivially extended to the following lemma. Lemma 6.2.1. Let 𝑓 ∶(𝜏1,𝜏2)→(𝜎1,𝜎2)an locally Lipschitz a. c. function with a. c. inverse. Then 𝑥is a solution of the first order differential equation with involution (3.1.5) if and only if 𝑥 is a solution of the second order ordinary differential equation (3.1.6). As was previously shown, problem (3.1.6) is equivalent to problem (6.0.1). We can now state the following corollary of Theorem 6.1.3 regarding the periodicity of problem (3.1.5) as foreseen at the beginning of the chapter. Corollary 6.2.2. Let 𝑓 ∶(𝜏1,𝜏2)→(𝜎1,𝜎2)an increasing locally Lipschitz a. c. function with a. c. inverse such that 0∈(𝜏1,𝜏2),𝑓(0)=0and 𝑐>0. Assume 2𝐹(𝑐)<min{𝐹(𝜏1),𝐹(𝜏2)}. Then, if 𝑥u�(𝑡)is a solution of problem (6.0.2) and we assume there exist 𝑐1,𝑐2∈ℝ,𝑐1<𝑐2, such that 2max{𝐹(𝑐1),𝐹(𝑐2)}<min{𝐹(𝜏1),𝐹(𝜏2)}and (𝑥u�1(𝑏)−𝑐1)(𝑥u�2(𝑏)−𝑐2)< 0, then problem (3.1.5) must have at least a solution. We now give an example in which there is no need to find 𝑐1,𝑐2∈ℝin the conditions of Corollary 6.2.2 because the function determining the period has a simple inverse. Example 6.2.3. Take again 𝑓(𝑡)∶=|𝑡|u�−2𝑡,𝑝>1, 𝑐>0and consider the problem 𝑥′(𝑡)=|𝑥(−𝑡)|u�−2𝑥(−𝑡), 𝑡∈ℝ, 𝑥(0)=𝑐. (6.2.1) By Corollaries 6.1.8 and 6.2.2 and Example 6.1.9, we have that the solutions of are periodic for every 𝑐≠0and 𝑇(𝑐,1,𝑝)=22 u�+1 𝑐2−u� Γ(1 u�)2 𝑝Γ(2 u�). Consider now the problem 𝑥′(𝑡)=|𝑥(−𝑡)|u�−2𝑥(−𝑡), 𝑡∈ℝ, 𝑥(𝑎)=𝑥(𝑏). (6.2.2) There is a unique solution for problem (6.2.2) for 𝑝∈(2,+∞). Just take the unique solution of problem (6.2.1) with 𝑐=⎛ ⎜ ⎜ ⎜ ⎜ ⎝𝑏−𝑎 22 u�+1 𝑝Γ(2 u�) Γ(1 u�)2⎞ ⎟ ⎟ ⎟ ⎟ ⎠ 1 2−u� . Observe that for 𝑝 ∈ (0,2)the function 𝑓is not locally Lipschitz, and therefore we cannot apply Lemma 6.2.1. 7. A Mathematica implementation In this chapter we develop an algorithm implemented in Mathematica which allows the obtaining of the Green’s function associated to a differential equation with constant coefficients, reflection and boundary conditions. We also point out possible ways to improve the computational time of the algorithm based on particular decompositions of the problem. The results in this chapter were sent for publication [165]. In order to establish a useful framework to work with these equations, we go back to the notation in Chapter 5. We consider the differential operator 𝐷, the pullback operator of the reflection 𝜑(𝑡)=−𝑡, denoted by 𝜑∗(𝑢)(𝑡)=𝑢(−𝑡), and the identity operator, Id. Let 𝑇∈ℝ+and 𝐼∶=[−𝑇,𝑇]. We consider again the algebra ℝ[𝐷,𝜑∗]. 7.1 The algorithm Theorem 5.2.3 gives a way of computing the Green’s function of a problem with reflection via reduction of the problem. The possibility of computing the Green’s function relies entirely on whether the reduced problem has a unique solution or not. Once we have reduced the problem, we check whether it has a unique solution and, in that case, we use part of the algorithm described in Chapter 6 to derive its Green’s function. Then it is left to compute the function 𝑅⊢𝐺as expressed in Theorem 5.2.3 which will be the Green’s function to our problem. Figure 7.1.1 shows the flow diagram of the algorithm. 7.1.1 Characteristics of the Mathematica notebook We work with the following input variables: •Coefficients 𝐚𝐤: The coefficients associated to the terms 𝑢u�)(𝑡). •Coefficients 𝐛𝐤: The coefficients associated to the terms 𝑢u�)(−𝑡). •𝐓: A positive number, half of the length of the interval on which the solution is defined. •Boundary conditions: A vector in Mathematica notation which specifies the boundary conditions. The input variables may be numbers or abstract symbols. The vectors of coefficents must be introduced in Mathematica notation (there is a default example when the program starts so to get an idea, see Figure 7.1.2). Furthermore, there is a checkbox which allows Mathematica to consider the numbers in the input variables as numerical approximations, which greatly reduces the computation time. While running, the steps of the computation will be shown in the ‘Progress’ frame. These messages will be, in order, ‘Processing data...’, ‘Solving homogeneous equation...’, ‘Computing 144 7.1. The algorithm =(𝑎2 u�−𝑏2 u�)⎡ ⎢ ⎣2u�−1 ∑ u�=0 (−1)u�(𝛼u�𝛼2u�−u�)+(−1)u�𝛼2 u�⎤ ⎥ ⎦, for 𝑘 = 0,…,𝑛where 𝑎u�,𝑏u�,𝛼u�= 0if 𝑘  ∈{0,…,𝑛}and 𝛼u�= 1. These are 𝑛equations with 𝑛unknowns: 𝛼0,…,𝛼u�. We present here the case of 𝑛=2to illustrate the solution of these equations. Example 7.1.4. For 𝑛=2, we have that 𝑅𝐿=(𝑎2 2−𝑏2 2)𝐷4+(−𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2)𝐷2+𝑎2 0−𝑏2 0, (𝑎2 2−𝑏2 2)𝑞(𝐷)𝑞−(𝐷)=(𝑎2 2−𝑏2 2)𝐷4+(2𝛼0−𝛼2 1)(𝑎2 2−𝑏2 2)𝐷2+𝛼2 0(𝑎2 2−𝑏2 2), and the system of equations is 𝑎2 0−𝑏2 0=(𝑎2 2−𝑏2 2)𝛼2 0, −𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2=(𝑎2 2−𝑏2 2)(2𝛼0−𝛼2 1).(7.1.1) Before computing the solutions let us state explicitly de limitations that the fact that 𝑅𝐿, considered as an order 2 polynomial on 𝐷2, that is 𝑅𝐿(𝑥)=𝑎𝑥2+𝑏𝑥+𝑐, has no negative roots implies. There are two options: (1) There are two complex roots, that is, Δ = 𝑏2−4𝑎𝑐 < 0. This is equivalent to 𝑎𝑐 > 0∧|𝑏|<2√𝑎𝑐. Expressed in terms of the coefficients of 𝑅𝐿: (𝑏2 0−𝑎2 0)(𝑏2 2−𝑎2 2)>0, and |−𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2|<2√(𝑏2 0−𝑎2 0)(𝑏2 2−𝑎2 2). (2) There are two nonnegative roots, that is Δ=𝑏2−4𝑎𝑐≥0and (−𝑏+√𝑏2−4𝑎𝑐)/(2𝑎)≤0. This is equivalent to (𝑎,𝑐≥0∧−𝑏≥2√𝑎𝑐)∨(𝑎,𝑐≤0∧𝑏≥2√𝑎𝑐). Expressed in terms of the coefficients of 𝑅𝐿: [(𝑏2 0−𝑎2 0),(𝑏2 2−𝑎2 2)≥0∧−(−𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2)≥2√(𝑏2 0−𝑎2 0)(𝑏2 2−𝑎2 2)] OR [(𝑏2 0−𝑎2 0),(𝑏2 2−𝑎2 2)≤0∧−(−𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2)≥2√(𝑏2 0−𝑎2 0)(𝑏2 2−𝑎2 2)]. Now, with these conditions, the solutions the system of equations (7.1.1) are: Case (I). We have two solutions: 𝛼0=√ √ √ ⎷𝑏2 0−𝑎2 0 𝑏2 2−𝑎2 2, 7. The algorithm 145 𝛼1=±√ √ √ ⎷2sign(𝑎2 2−𝑏2 2)√(𝑏2 0−𝑎2 0)(𝑏2 2−𝑎2 2)−(−𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2) 𝑎2 2−𝑏2 2. Case (II). We have four solutions depending on whether we choose 𝜉=1or 𝜉=−1: 𝛼0=𝜉√ √ √ ⎷𝑏2 0−𝑎2 0 𝑏2 2−𝑎2 2, 𝛼1=±√ √ √ ⎷2𝜉sign(𝑎2 2−𝑏2 2)√(𝑏2 0−𝑎2 0)(𝑏2 2−𝑎2 2)−(−𝑎2 1+2𝑎0𝑎2+𝑏2 1−2𝑏0𝑏2) 𝑎2 2−𝑏2 2. These solution provide well defined real numbers by conditions (I) and (II). Now we could consider those cases where the problem can be decomposed easily. Consider that the reduced problem given by Theorem 5.2.3, 𝑆𝑢=𝑅ℎ,𝐵u�𝑅𝑢=0,𝐵u�𝑢=0,𝑗=1,…,𝑛 can be expressed as an equivalent factored problem 𝐿1𝑢=𝑦, 𝑉u�𝑢=0,𝑗=1,…,𝑛, 𝐿2𝑦=𝑅ℎ,  𝑉u�𝑦=0,𝑗=1,…,𝑛, where the conditions 𝑉u�𝑢 = 0, 𝑉u�𝐿1𝑢 = 0,𝑗 = 1,…,𝑛are equivalent to the conditions 𝐵u�𝑅𝑢=0,𝐵u�𝑢=0,𝑗=1,…,𝑛. Then the Green’s function of problem 𝑆𝑢=𝑅ℎ,𝐵u�𝑅𝑢=0, 𝐵u�𝑢=0,𝑗=1,…,𝑛can be expressed as 𝐺(𝑡,𝑠)=∫u� −u� 𝐺1(𝑡,𝑟)𝐺2(𝑟,𝑠)d𝑟, where 𝐺1is the Green’s function associated to the problem 𝐿1𝑢=𝑦, 𝑉u�𝑢=0,𝑗=1,…,𝑛, and 𝐺2the one associated to the problem 𝐿2𝑦=𝑅ℎ,  𝑉u�𝑦=0,𝑗=1,…,𝑛,in the case both Green’s functions exist. This procedure was already illustrated in Example 5.2.4. Computationally, this procedure poses a big advantage: it is always easier to obtain the Green’s function two order 𝑛problems than to do so for one order 2𝑛problem. Furthermore, if the hypothesis of Lemma 7.1.1 are satisfied and we are able to obtain a factorization of the aforementioned kind using 𝑞and 𝑞−in the place of 𝐿1and 𝐿2, we have an extra advantage: the differential equation given by 𝑞−is the adjoint equation of the one given by 𝑞multiplied by the factor (−1)u�. This fact, together with the following result (which can be found, although not stated as in this work, in [28]), illustrates that in this case it may be possible to solve problem (5.2.2) just computing the Green’s function of one order 𝑛problem. Theorem 7.1.5. Consider an interval 𝐽=[𝑎,𝑏]⊂ℝ, functions 𝜎,𝑎u�∈L1(𝐽),𝑖=1,…,𝑛, real numbers 𝛼u�u�,𝛽u�u�,ℎu�,𝑖=1,…,𝑛,𝑗=0,…,𝑛−1,𝐷(𝐿u�)⊂𝑊u�,1(𝐽)a vector subspace, the operator 𝐿u�𝑢(𝑡)=𝑎0𝑢(u�)(𝑡)+𝑎1(𝑡)𝑢(u�−1)(𝑡)+⋯+𝑎u�−1(𝑡)𝑢′(𝑡)+𝑎u�(𝑡)𝑢(𝑡),𝑡∈𝐽,𝑢∈𝐷(𝐿u�), with 𝑎0=1and the problem 𝐿u�𝑢(𝑡)=𝜎(𝑡), 𝑡∈𝐽, 𝑈u�(𝑢)=ℎu�, 𝑖=1,…,𝑛, (7.1.2) 146 7.1. The algorithm where 𝑈u�(𝑢)∶=u�−1 ∑ u�=0 (𝛼u�u�𝑢(u�)(𝑎)+𝛽u�u�𝑢(u�)(𝑏)), 𝑖=1,…,𝑛. Then, the associated adjoint problem is 𝐿† u�𝑣(𝑡)= u� ∑ u�=0(−1)u�𝑎u�−u�(𝑡)𝑢(u�)(𝑡), 𝑡∈𝐽, 𝑣∈𝐷(𝐿† u�), (7.1.3) where 𝐷(𝐿† u�)= ⎧ { ⎨ { ⎩𝑣∈𝑊u�,2(𝐽) ∶ (𝑏∗−𝑎∗)⎛ ⎜ ⎜ ⎝ u� ∑ u�=1 u�−1 ∑ u�=0(−1)(u�−u�−1)(𝑎u�−u�𝑣)u�−u�−1𝑢(u�)⎞ ⎟ ⎟ ⎠=0, 𝑢∈𝐷(𝐿u�)⎫ } ⎬ } ⎭ Furthermore, if 𝐺(𝑡,𝑠)is the Green’s function of problem (7.1.2), then the one associated to problem (7.1.3) is 𝐺(𝑠,𝑡). Hence, if we can decompose problem (5.2.2) in two adjoint problems, its Green’s function will be 𝐺(𝑡,𝑠)=∫u� −u� 𝐺1(𝑡,𝑟)𝐺2(𝑟,𝑠)d𝑟=∫u� −u� 𝐺1(𝑡,𝑟)𝐺1(𝑠,𝑟)d𝑟. We note though, that unless the operator 𝑞−is the adjoint equation times (−1)u�, the boundary conditions may be not the adjoint ones. Part II Topological Methods 149 We have so far studied differential equations with reflection finding, when possible, the Green’s function in order to derive the solution in the case of uniqueness. Still, many situations, in which nonlinearities are involved, escape the direct construction of solutions and different methods become necessary. Topological methods come handy in these situations, in particular those related to the fixed point index. These tools permit to guarantee the existence and multiplicity of fixed points of continuous maps through an index which counts them with sign. We have already used in Subsection 3.2.3 the celebrated cone contraction-expansion fixed point theorem of Krasnosel’skiĭ. Here we avoid its limitations using an approach developed by Infante and Webb [97] and used in several publications [34,35,87–95,98–100,175–184]. In the following four chapters we will use this method to solve four different kinds of problems increasing in complexity: a problem with reflection, a problem with deviated arguments (applied to a thermostat model), a problem with nonlinear Neumann boundary conditions and a problem with functional nonlinearities in both the equation and the boundary conditions. The structure of the method is fairly consistent and is developed as follows. (1) State the nature of the problem being studied and its specific characteristics. (2) Elaborate a list of properties, of the elements involved in the problem, which is necessary to ask for so we can grant that the existence / multiplicity / nonexistence results can be applied. For instance, the operator 𝐹of which its fixed points will be solutions for our problem has to be continuous. (3) Define an appropriate cone 𝐾in which we will localize the solutions of our problem. Here we have to take an important decision: large cones allow the finding of more solutions but, at the same time, they do not provide good localization results. (4) Prove that the operator 𝐹is compact,continuous and maps 𝐾to 𝐾. (5) Find sufficient conditions for which the fixed point index of the operator 𝐹is 0and ±1 respectively in (at least) two nested subsets of the cone. If we find 𝑛nested subsets for which the index alternates from 0to ±1we can guarantee the existence of at least 𝑛−1 different nontrivial solutions (cf. [123]). By making the cone smaller, we trade solutions for simpler conditions. Also, we may use conditions for the index related to the eigenvalues of the operators involved (see Chapters 10 and 11). (6) Finally, we can apply the results derived to a vast variety of problems and illustrate its usefulness with some examples. As we will see, the particularities of each problem make it impossible to take a common approach to all of the problems studied. Still, there will be important similarities in the different cases which will lead to comparable results. The results in Chapters 8, 9 and 10 have been published in [34], [34] and [96] respectively. Those in Chapter 11 are ready to be sent for publication soon. Due to the bast amount of notation necessary to develop this theory, we will consider it only valid for the chapter in question, so we can use the same symbols for similar (but different) purposes. 8. A cone approximation to a problem with reflection We have studied previously (see Chapter 3), the first order operator 𝑢′(𝑡)+𝜔𝑢(−𝑡)coupled with periodic boundary value conditions, describing the eigenvalues of the operator and providing the expression of the associated Green’s function in the nonresonant case. We provide the range of values of the real parameter 𝜔for which the Green’s function has constant sign and apply these results to prove the existence of constant sign solutions for the nonlinear periodic problem with reflection of the argument (see page 55) 𝑢′(𝑡)=ℎ(𝑡,𝑢(𝑡),𝑢(−𝑡)),𝑡∈[−𝑇,𝑇], 𝑢(−𝑇)=𝑢(𝑇). (8.0.1) The methodology, analogous to the one used by Torres [167] in the case of ordinary differential equations, is to rewrite the problem (8.0.1) as an Hammerstein integral equation with reflections of the type 𝑢(𝑡)=∫u� −u� 𝑘(𝑡,𝑠)[ℎ(𝑠,𝑢(𝑠),𝑢(−𝑠))+𝜔𝑢(−𝑠)]d𝑠, 𝑡∈[−𝑇,𝑇], where the kernel 𝑘has constant sign, and to make use of the well-known Guo-Krasnosel’skiĭ theorem on cone compression-expansion (see Theorem 3.2.19). In this chapter we continue this study and we prove new results regarding the existence of nontrivial solutions of Hammerstein integral equations with reflections of the form 𝑢(𝑡)=∫u� −u� 𝑘(𝑡,𝑠)𝑔(𝑠)𝑓(𝑠,𝑢(𝑠),𝑢(−𝑠))d𝑠, 𝑡∈[−𝑇,𝑇], where the kernel 𝑘is allowed to be not of constant sign. In order to do this, we extend the results of [98], valid for Hammerstein integral equations without reflections, to the new context. We make use of a cone of functions that are allowed to change sign combined with the classical fixed point index for compact maps (we refer to [4] or [81] for further information). As an application of our theory we prove the existence of nontrivial solutions of the periodic problem with reflections (8.0.1). The results of this chapter were published in [34] 8.1 The case of kernels that change sign We begin with the case of kernels that are allowed to change sign. We impose the following conditions on 𝑘,𝑓,𝑔that occur in the integral equation 𝑢(𝑡)=∫u� −u� 𝑘(𝑡,𝑠)𝑔(𝑠)𝑓(𝑠,𝑢(𝑠),𝑢(−𝑠))d𝑠=∶𝐹𝑢(𝑡), (8.1.1) where 𝑇is fixed in (0,∞). (𝐶1)The kernel 𝑘is measurable, and for every 𝜏∈[−𝑇,𝑇]we have lim u�→u� |𝑘(𝑡,𝑠)−𝑘(𝜏,𝑠)|=0 for almost every (a. e.) 𝑠∈[−𝑇,𝑇]. 152 8.1. The case of kernels that change sign (𝐶2)There exist a subinterval [𝑎,𝑏]⊆[−𝑇,𝑇], a measurable function Φwith Φ≥0a. e. and a constant 𝑐=𝑐(𝑎,𝑏)∈(0,1]such that |𝑘(𝑡,𝑠)|≤Φ(𝑠)for all 𝑡∈[−𝑇,𝑇]and a. e. 𝑠∈[−𝑇,𝑇], 𝑘(𝑡,𝑠)≥𝑐Φ(𝑠)for all 𝑡∈[𝑎,𝑏]and a. e. 𝑠∈[−𝑇,𝑇]. (𝐶3)The function 𝑔is measurable and satisfies that 𝑔Φ ∈ L1([−𝑇,𝑇]),𝑔(𝑡) ≥ 0a. e. 𝑡∈[−𝑇,𝑇]and ∫u� u�Φ(𝑠)𝑔(𝑠)d𝑠>0. (𝐶4)The nonlinearity 𝑓 ∶[−𝑇,𝑇]×ℝ×ℝ→[0,∞)satisfies the L∞-Carathéodory conditions, that is, 𝑓(⋅,𝑢,𝑣)is measurable for each fixed 𝑢and 𝑣and 𝑓(𝑡,⋅,⋅)is continuous for a. e. 𝑡∈[−𝑇,𝑇], and for each 𝑟>0, there exists 𝜙u�∈L∞([−𝑇,𝑇])such that 𝑓(𝑡,𝑢,𝑣)≤𝜙u�(𝑡) for all (𝑢,𝑣)∈[−𝑟,𝑟]×[−𝑟,𝑟], and a. e. 𝑡∈[−𝑇,𝑇]. We recall the following definition. Definition 8.1.1. Let 𝑋be a Banach Space. A cone on 𝑋is a closed, convex subset of 𝑋such that 𝜆𝑥∈𝐾for 𝑥∈𝐾and 𝜆≥0and 𝐾∩(−𝐾)={0}. Here we work in the space 𝐶[−𝑇,𝑇], endowed with the usual supremum norm, and we use the cone 𝐾 ={𝑢∈𝐶[−𝑇,𝑇]∶ min u�∈[u�,u�]𝑢(𝑡)≥𝑐‖𝑢‖}, (8.1.2) where 𝑐and [𝑎,𝑏]are defined in (𝐶2). Note that 𝐾 ≠{0}. The cone 𝐾has been essentially introduced by Infante and Webb in [98] and later used in [34,66,69,70,87,93,94,97,99,100,134]. 𝐾is similar to a type of cone of nonnegative functions first used by Krasnosel’skiĭ, see e.g. [121], and D. Guo, see e.g. [81]. Note that functions in 𝐾 are positive on the subset [𝑎,𝑏]but are allowed to change sign in [−𝑇,𝑇]. We require some knowledge of the classical fixed point index for compact maps, see for example [4] or [81] for further information. If Ωis a bounded open subset of 𝐾(in the relative topology) we denote by Ωand 𝜕Ωthe closure and the boundary relative to 𝐾. When 𝐷is an open bounded subset of 𝑋we write 𝐷u�=𝐷∩𝐾, an open subset of 𝐾. Next Lemma is a direct consequence of classical results from degree theory [81]. Lemma 8.1.2. Let Ωbe an open bounded set with 0 ∈ Ωu�and Ωu�≠ 𝐾. Assume that 𝐹∶Ωu�→𝐾is a continuous compact map such that 𝑥≠𝐹𝑥for all 𝑥∈𝜕Ωu�. Then the fixed point index 𝑖u�(𝐹,Ωu�)has the following properties. (1) If there exists 𝑒∈𝐾\{0}such that 𝑥≠𝐹𝑥+𝜆𝑒for all 𝑥∈𝜕Ωu�and all 𝜆>0, then 𝑖u�(𝐹,Ωu�)=0. (2) If 𝜇𝑥≠𝐹𝑥for all 𝑥∈𝜕Ωu�and for every 𝜇≥1, then 𝑖u�(𝐹,Ωu�)=1. (3) If 𝑖u�(𝐹,Ωu�)≠0, then 𝐹has a fixed point in Ωu�. (4) Let Ω1be open in 𝑋with Ω1⊂Ωu�. If 𝑖u�(𝐹,Ωu�)=1and 𝑖u�(𝐹,Ω1 u�)=0, then 𝐹has a fixed point in Ωu�\Ω1 u�. The same result holds if 𝑖u�(𝐹,Ωu�)=0and 𝑖u�(𝐹,Ω1 u�)=1. 8. The case of kernels that change sign 153 Definition 8.1.3. We use the following sets: 𝐾u�={𝑢∈𝐾 ∶‖𝑢‖<𝜌}, 𝑉u�={𝑢∈𝐾 ∶ min u�∈[u�,u�]𝑢(𝑡)<𝜌}. The set 𝑉u�was introduced in [100] and is equal to the set called Ωu�/u� in [97]. The notation 𝑉u� makes shows that choosing 𝑐as large as possible yields a weaker condition to be satisfied by 𝑓 in Lemma 8.1.10. A key feature of these sets is that they can be nested, that is 𝐾u�⊂𝑉u�⊂𝐾u�/u�. Lemma 8.1.4. The operator 𝑁u�(𝑢,𝑣)(𝑡)=∫1 0𝑘(𝑡,𝑠)𝑔(𝑠)𝑓(𝑠,𝑢(𝑠),𝑣(𝑠))d𝑠maps 𝐶(𝐼)× L∞(𝐼)to 𝐶(𝐼)and is compact and continuous. Proof. Fix (𝑢,𝑣)∈𝐶(𝐼)×L∞(𝐼)and let (𝑡u�)u�∈ℕ ⊂𝐼be such that lim u�→∞(𝑡u�)=𝑡∈𝐼. Take 𝑟=‖(𝑢,𝑣)‖∶=‖𝑢‖+‖𝑣‖and consider ℎu�(𝑠)∶=𝑘(𝑡u�,𝑠)𝑔(𝑠)𝑓(𝑠,𝑢(𝑠),𝑣(𝑠)), for a.e. 𝑠∈𝐼. We have, by (𝐶1), that lim u�→∞ℎu�(𝑠)=ℎ(𝑠)∶=𝑘(𝑡,𝑠)𝑔(𝑠)𝑓(𝑠,𝑢(𝑠),𝑣(𝑠)), for a.e. 𝑠∈𝐼. On the other hand, |ℎu�| ≤ Φ𝑔‖𝜙u�‖ ∈ L1(𝐼)so, by the Dominated Convergence Theorem, we have lim u�→∞𝑁u�(𝑢,𝑣)(𝑡u�)=𝑁u�(𝑢,𝑣)(𝑡)and therefore 𝑁u�(𝑢,𝑣)∈𝐶(𝐼). Now let’s see that 𝑁u�is compact, indeed, let (𝑢u�,𝑣u�)u�∈ℕ ⊂𝐶(𝐼)×L∞(𝐼)be such that ‖(𝑢u�,𝑣u�)‖≤𝑅∈ℝ+for all 𝑛∈ℕ. Define 𝑦u�(𝑠) = 𝑓(𝑠,𝑢u�(𝑠),𝑣u�(𝑠)). By Condition (𝐶4)we know that ‖𝑦u�‖ ≤ ‖𝜙u�‖ ∈ L∞(𝐼), therefore (𝑦u�(𝑠))u�∈ℕ is a bounded sequence in ℝand by the Bolzano-Weierstrass Theorem it has a convergent subsequence (𝑦u�u�(𝑠))u�∈ℕ. Take 𝑦(𝑠)∶= lim u�→∞𝑦u�u�(𝑠). Now, since ‖𝑘(𝑡,⋅)𝑔(⋅)𝑦u�u�(⋅)‖≤Φ(⋅)𝑔(⋅)‖𝜙u�‖, for all 𝑡∈𝐼, we can apply the Dominated Convergence Theorem and therefore lim u�→∞𝑁u�(𝑢u�u�,𝑣u�u�)(𝑡)=∫1 0𝑘(𝑡,𝑠)𝑔(𝑠)𝑦(𝑠)d𝑠, for all 𝑡∈𝐼. So we have proved that there exists the point-wise limit on 𝐼. To conclude the assertion of compactness we verify that such convergence is uniform in 𝐼. To this end, we take into account that for all 𝑡∈𝐼it is verified that |𝑁u�(𝑢u�u�,𝑣u�u�)(𝑡)−𝑁u�(𝑢,𝑣)(𝑡)|≤∫1 0|𝑘(𝑡,𝑠)|𝑔(𝑠)|𝑦u�u�(𝑠)−𝑦(𝑠)|d𝑠 ≤∫1 0Φ(𝑠)𝑔(𝑠)|𝑦u�u�(𝑠)−𝑦(𝑠)|d𝑠. Since the last expression on the right is independent of 𝑡we have that such convergence is uniform in 𝐼, and the assertion holds. The continuity is proved in a similar manner.  256 C.2. Segunda Parte El hecho de haber contribuido con la presencia de un funcional en las condiciones de contorno hace que en el Capítulo 10 se estudie otra vez el problema integral de Hammerstein, pero en este caso con la peculiaridad de estar sometido a dos funcionales lineales distintos en las condiciones de contorno que, por otra parte, son de tipo Neumann. A mayores se ofrecen por primera vez resultados para el cálculo del índice de punto fijo relacionados con el radio espectral de los operadores asociados lo cual, en muchos casos, resulta ventajoso a la hora de obtener resultados sin realizar demasiados cálculos. Finalmente, corona la segunda parte de esta memoria el Capítulo 11. Este destaca sobre los anteriores en tanto a que la complejidad del problema estudiado es muy superior. Esto se debe a la presencia de funcionales y operadores no lineales, tanto en la ecuación como en las condiciones de contorno. Tal generalidad obliga a la aparición de una gran profusión de condiciones a ser satisfechas y resultados muy interesantes. En particular, se aplica la generalización de la definición del radio espectral a operadores acotados para poder obtener resultados de índice de punto fijo sencillos. Más allá de las dos partes que constituyen el núcleo del trabajo realizado, encontramos dos apéndices. El primero profundiza en un tema que se mencionó en el Capítulo 5, la obtención de una versión hiperbólica de la fórmula para la suma de fasores. La obtención de dicha fórmula da lugar a un capítulo muy didáctico –publicado en [166]– en el cual se desgrana, desde el punto de vista matemático, el formalismo de fasores tan comúnmente utilizado en el ámbito de la física y la ingeniería eléctrica. El segundo apéndice contiene el código del programa de Mathematica desarrollado en el Capítulo 7 y una referencia a la biblioteca electrónica Wolfram Library Archive desde el cual se puede descargar. 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