On first and second order linear Stieltjes differential equations

J. Math. Anal. Appl. 511 (2022) 126010 Contents lists available at ScienceDirect Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa On first and second order linear Stieltjes differential equations Francisco J. Fernández ∗, Ignacio Marquéz Albés, F. Adrián F. Tojo CITMAga, 15782, Santiago de Compostela, Spain Departamento de Estatística, Análise Matemática e Optimización, Universidade de Santiago de Compostela, 15782, Facultade de Matemáticas, Campus Vida, Santiago, Spain a r t i c l e i n f o a b s t r a c t Article history: Received 21 September 2021 Available online 13 January 2022 Submitted by S. Hencl Keywords: Stieltjes derivative Second order Uniqueness Existence Green’s function This work deals with the obtaining of solutions of first and second order Stieltjes differential equations. We define the notion of Stieltjes derivative on the whole domain of the functions involved, provide a notion of n-times continuously Stieltjesdifferentiable functions and prove existence and uniqueness results of Stieltjes differential equations in the space of such functions. We also present the Green’s functions associated to the different problems and an application to the Stieltjes harmonic oscillator. © 2022 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). 1. Introduction There has been a recent surge in the study of Stieltjes differential equations focused on obtaining applicable results comparable to those available for classical derivatives [2–14,16,17,19]. These works center their attention in the procuring of solutions of first order differential equations and systems. The theory developed starts with the obtaining of simple solutions, like the solution of the first order linear problem [3,4], which is identified with the exponential, in order to, later, prove existence and uniqueness results in more general settings [7,13,16]. Some of these works also provide interesting practical applications [5,9]and others generalize the framework in several ways, such as allowing for sign changing derivators [4], considering several different derivators [16]or generalizing the concept of Stieltjes derivative [15]. In any case, all of the aforementioned works restrict themselves to the first order case. The reason behind this is that, in order to study higher order problems, the notion of higher order Stieltjes derivative has to be correctly defined, which is not obvious. In fact, the first difficulty lies on the mere definition of the Stieltjes derivative, which, to the best of our knowledge, is nowhere defined in the literature on the whole domain of definition of the function, something which impedes taking a second derivative. *Corresponding author. E-mail addresses: fja[email protected] (F.J. Fernández), [email protected] (I. Marquéz Albés), [email protected] (F.A.F. Tojo). https://doi.org/10.1016/j.jmaa.2022.126010 0022-247X/© 2022 The Author(s). Published by Elsevier Inc. This is an open access article under the CC BY-NC-ND license (http://creativecommons.org/licenses/by-nc-nd/4.0/). 2F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 In this work we provide this definition, which enables us to study second order problems. First, we consider the Stieltjes derivative in the whole of the domain of the function, which allows us to talk about the space of continuously Stieltjes-differentiable functions in the same way we speak of the space of continuously differentiable functions with the usual derivative. We can then explore the first order problem in this space, obtaining existence and uniqueness results that mirror those of the previous works. In fact, we profit from the opportunity of revisiting the solution of the first order linear problem to provide a constructive way of obtaining its solution. All of these steps are also taken with a further generalization: our functions are allowed to take real or complex values. Furthermore, we obtain the explicit expression of the Green’s function of the first order linear problem with initial conditions and we construct the Stieltjes versions of the sine and cosine functions using the complex version of the Stieltjes exponential. Once we have studied the first order problem with various degrees of regularity (something the subsequent spaces of n-times continuously Stieltjes-differentiable functions allow), we move on to study second order problems. First, we present existence and uniqueness results for the homogeneous second order problem with constant coefficients and then we study the non homogeneous case with varying degrees of regularity. Here we also obtain the explicit expression of the Green’s function of the second order linear problem with initial conditions. All this work is then illustrated with an application to the Stieltjes harmonic oscillator for which we also analyze the resonance effect. Finally, in order to validate the explicit solutions obtained, we compare them with the numerical approximation of the corresponding first order linear system using the numerical scheme introduced in [2]. The structure of this work is as follows: In Section 2we present some preliminary concepts and we prove several results related to Lebesgue-Stieltjes integral. In Section 3we introduce the space of bounded Stieltjes differentiable functions and analyze some of its properties. We study the first order linear Stieltjes differential equation in Section 4, including in the complex case. In this section we also define the complex Stieltjes exponential and the Stieltjes version of the sine and cosine functions. In Section 5we study the homogeneous Stieltjes second order problem with constant coefficients, the non homogeneous case and we also obtain an explicit solution for both situations. Finally, in Section 6we present an application to the Stieltjes harmonic oscillator. We obtain the explicit solution of the overdamped, critically damped and underdamped cases, and provide an example in which the resonance effect appears. In order to validate the explicit solution obtained, we compare it with the numerical solution of the corresponding first order linear system. 2. Preliminaries Let [a, b] ⊂Rbe an interval, Fthe field Ror Cand g:R →Ra left-continuous non-decreasing function. We will refer to such functions as derivators. For these functions, we define the set Dg={dn}n∈Λ (where Λ ⊂N) as the set of all discontinuity points of g, namely, Dg={t ∈R :Δ +g(t) >0}where Δ+g(t) := g(t+) −g(t), t ∈R, and g(t+) denotes the right hand side limit of gat t. We also define Cg:= {t∈R:gis constant on (t−ε, t +ε)forsomeε>0}. Observe that Cgis open in the usual topology of R, so we can write Cg= n∈ Λ{(an,b n)}(2.1) where  Λ⊂Nand (ak, bk) ∩(aj, bj) =∅for k=j. With this notation, we denote N− g:= {an}n∈ Λ\Dg, N+ g:= {bn}n∈ Λ\Dgand Ng:= N− g∪N+ g. F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 3 Remark 2.1. For the aims of this paper, we will assume without loss of generality that g(a) =0. Furthermore, we will also assume that gis continuous at x =a. As pointed out in [3, p. 21] and [12, Proposition 4.28], the continuity assumption has no impact in the study of differential equations, which is our final goal. Finally, in order to properly define the Stieltjes derivative in the whole [a, b], we will also ask that [a, b] \Cg=∅. We define gB:R →Ras: gB(t)=⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩  s∈[a,t)∩Dg Δ+g(s),t>a, − s∈[t,a)∩Dg Δ+g(s),t≤a. It is clear that gBis a left-continuous and non-decreasing function. Moreover, the map gC:R →Rgiven by gC(t):=g(t)−gB(t), is also non-decreasing and continuous. We say gCthat is the continuous part of gand gBis the jump part of g. Observe that both gBand gCare continuous at x =aand gC(a) =gB(a) =0. Throughout this work we consider the Lebesgue–Stieltjes measure space (R, Mg, μg), where Mgand μg are the σ-algebra and measure constructed in an analogous fashion to the classical Lebesgue measure, where the length of [c, d)is given by μg([c, d)) =g(d) −g(c). The interested reader may refer to [11]for details concerning this measure space. We must emphasize that, in the case of considering g(t) =t, we recover the classic Lebesgue measure space that we will denote by (R, L, μ) ≡(R, MId, μId) where Id is the identity function. Furthermore, we can define the measure space associated with the continuous part, (R, MgC, μgC), the jump part, (R, MgB, μgB), and the one associated with the derivator itself, (R, Mg, μg). If we denote by B(τu)the Borel σ-algebra associated to τu, the usual topology of R, we have that B(τu) ⊂M gand also B(τu) ⊂M gM, with M=C, B. We must mention that if E⊂R, μ∗ gM(E) ≤μ∗ g(E), for M=C, B, being μ∗ gMand μ∗ gthe outer measures associated to gMand grespectively, M=C, B. We also have that if E⊂R is a bounded set, then μ∗ g(E) <∞. We have the following lemma that, in particular, provides us with a relationship between the σ-algebras Mg, MgCand MgB. Lemma 2.2. The following properties hold for the maps g, gCand gB: 1. Given an element E∈M gthere exists H∈Gδ(that is, His a countable intersection of open sets) and N∈M gsuch that E⊂H, N⊂H, μg(N) =0and E=H\N. 2. Given an element E∈M gthere exists F∈Fσ(that is, Fis a countable union of closed sets) and N∈M gsuch that μg(N) =0, F∩N=∅and E=F∪N. 3. Mg⊂M gC. 4. MgB=P(R). Proof. Since B(τu) ⊂M gand gis left-continuous, we have that μg(E)=inf n∈N μg([an,b n)) : E⊂ n∈N [an,b n)=inf n∈N μg((an,b n)) : E⊂ n∈N (an,b n). Indeed on the one hand given {(an, bn)}n∈Nsuch that E⊂n∈N(an, bn), we have that 4F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 μ∗ g(E)≤μ∗ g n∈N (an,b n)≤ n∈N μ∗ g((an,b n)), therefore, μg(E) ≤inf n∈Nμg((an,b n)) : E⊂n∈N(an,b n). On the other hand, given ε >0and {[an, bn)}n∈Nsuch that E⊂n∈N[an, bn), we have, thanks to the left-continuity of g, that there exists {(˜an, bn)}n∈Nsuch that [an, bn) ⊂(˜an, bn)and μ∗ g((˜an, bn)) ≤μ∗ g([an, bn)) +ε/2n, n ∈N. Thus, n∈Nμ∗ g(˜an, bn) ≤n∈Nμ∗ g([an, bn)) +εand we conclude, taking the infimum in both sides of inequality, that inf n∈Nμg((an,b n)) : E⊂n∈N(an,b n)≤μg(E). Now, we can proceed as in [1, Corollaries 15.5 and 15.8] to obtain 1 and 2, respectively. Now, for 3, given an element E∈M g, there exists F∈Fσand N∈M gsuch that μg(N) =0, F∩N=∅ and E=F∪N. Now, F∈B(τu) ⊂M cand μ∗ gC(N) ≤μ∗ g(N) =0so we have that N∈M gCsince (R, MgC, μgC)is a complete measure space. Therefore, E⊂M gC. Finally, for E∈P(R), we have that E=(E\DgB) ∪(E∩DgB). Now, (E\DgB) ⊂CgBand then μ∗ gB(E\DgB) =0, so E\DgB∈M gB. Finally E∩DgB∈B(τu) ⊂M gB. Therefore E∈M gB, which finishes the proof of 4.  We denote by L1 g([a, b); F)the set of functions f:[a, b) →Fsuch that their real and imaginary parts, that is, Re(f)and Im(f) respectively, are measurable and [a,b)|f| dμg<∞. For this class of functions we define  [a,b) fdμg= [a,b) Re(f)dμg+i [a,b) Im(f)dμg. Lemma 2.3. Given a function f∈L 1 g([a, b); F),  [a,t) fdμg= [a,t) fdμgC+ s∈[a,t)∩Dg f(s)Δ+g(s),∀t∈[a, b]. Proof. Given a function f∈L 1 g([a, b); F), thanks to Lemma 2.2 and the fact that μgC(E) ≤μg(E)for all E∈M g, we have that f∈L 1 gC([a, b); F). Now thanks to [18, Theorems 6.3.13, 6.12.3 and 6.12.7], and separating the real and imaginary part if necessary, we have the desired result.  Corollary 2.4. Given E∈M gand taking f=χE(the characteristic function associated to E) in Lemma 2.3 we have that μg(E)=μgC(E)+  s∈E∩Dg Δ+g(s). We now introduce a tool that will allow us to transform Lebesgue-Stieltjes integrals with respect to gC into the usual Lebesgue ones. In particular, in light of Lemma 2.3, this means that we will have a way of transforming any Lebesgue-Stieltjes integral into a Lebesgue one. Definition 2.5 (Pseudo-inverse of gC). Given an interval [a, b]and a derivator g:R →R, we define the pseudo-inverse of the continuous part gCin the interval [0, gC(b)] by: γ:x∈[0,gC(b)] →γ(x)=mint∈[a, b]:gC(t)=x∈[a, b].(2.2) In [7, Proposition 5.1] we can find some of the properties of the pseudo-inverse of a continuous derivator mapping the real line onto the real line. For our context, by extending linearly the map goutside of the interval [a, b]we can obtain the required property, which leads to the following result. F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 5 Proposition 2.6. We have the following properties for the pseudo-inverse of the continuous part gCin the interval [0, gC(b)]: •For all x ∈[0, gC(b)], gC(γ(x)) =x. •For all t ∈[a, b], γ(gC(t)) ≤t. •For all t ∈[a, b], t /∈CgC∪N+ gC, γ(gC(t)) =t. •The map γis strictly increasing. •The map γis left-continuous everywhere and continuous at every x ∈[0, gC(b)], x /∈gC(Cg). Now we are ready to prove the following result. Proposition 2.7. Given an interval [a, b]and a derivator g:R →R: 1. The continuous part gC:([a, b],MgC)→([0,gC(b)],L) is a measurable morphism.1 2. The pseudo-inverse of the continuous part gC γ:([0,gC(b)],L)→([a, b],MgC) is a measurable morphism. Proof. Let us prove the two statements separately. 1. Let us consider a subset E⊂[0, gC(b)] such that E∈L. We have that there exists F∈Fσand N∈L with μ(N) =0such that F∩N=∅and E=F∪N. It is clear that (gC)−1(F) ∈B(τu)so, if we prove that μ∗ gC((gC)−1(N)) =0, where μ∗ gCis the outer Lebesgue-Stieltjes measure, we will have finished. Now, since μ(N) =0, given ε >0, there exists a countable disjoint family {[cn,  dn)}n∈Nsuch that N⊂n∈N[cn,  dn)and n∈N( dn−cn) <ε. We have that (gC)−1([ck,  dk)) =[γ(ck), γ( dk)), for all k∈N, thus (gC)−1(N) ⊂n∈N[γ(cn), γ( dn)). Finally, μ∗ gC((gC)−1N)≤ n∈N μ∗ gC[γ(cn),γ( dn)) =  n∈NgC(γ( dn)) −gC(γ(cn))= n∈N ( dn−cn)<ε. Since ε >0was arbitrarily chosen, we have that μ∗ gC((gC)−1N) =0, which finishes the proof of 1. 2. Let us consider a subset E⊂[a, b]such that E∈M gC. We have that there exists F∈Fσand N∈M gC such that μg(N) =0, F∩N=∅and E=F∪N. Thus, we conclude that γ−1(E) =γ−1(F) ∪γ−1(N). Now, since γis strictly increasing, it is a Borel map, so we have that γ−1(F) ∈B(τu) ⊂L. Hence, if we prove that μ∗(γ−1(N)) =0, where μ∗is the outer Lebesgue measure, we are done. The proof in this case is analogous to the previous one, the only difference lies in that, given an interval [c, d), we have that γ−1([c, d)) ⊂[gC(c), gC(d)], thus μ∗(γ−1([c, d))) ≤gC(d) −gC(c) =μ∗ gC([c, d)).  The following Corollary is in the line of [2, Lemma 1]. 1Given two measurable spaces (X, ΣX)and (Y, ΣY), we say that a function f:X→Yis a measurable morphism if f−1(F) ∈ΣX, for all F∈ΣY. 6F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 Corollary 2.8. Given a function f∈L 1 g([a, b); F), for every t ∈[a, b],  [a,t) fdμg= [a,t) fdμgC+ s∈[a,t)∩Dg f(s)Δ+g(s)=  [0,gC(t))  fdμ+ s∈[a,t)∩Dg f(s)Δ+g(s), where  f=f◦γ, γ:t ∈[0, gC(b)] →γ(t)is given by (2.2)and μdenotes the Lebesgue measure. Proof. We write (X, ΣX) =([a, t), MgC)and (Y, ΣY) =([0, gC(t)), L). We have, thanks to Proposition 2.7, that  f:Y→Fis a measurable function and gC:(X, ΣX) →(Y, ΣY)is a measurable morphism, which (cf. [20, Exercise 1.4.38]) ensures that  Y fdgC ∗μgC= X ( f◦gC)dμgC, where gC ∗μgC:E∈ΣY→gC ∗μgC(E)=μgC((gC)−1(E)) is the pushforward measure in (Y, ΣY). However, given an element (c, d) ⊂[0, gC(t)), it is clear that gC ∗μgC(c, d) =μgC((gC)−1(c, d)) =d −c. In particular, (cf. [1, Theorem 13.8]) gC ∗μgC=μ. Therefore,  Y fdgC ∗μgC= [0,gC(t))  fdμ. Finally, since μgC(CgC∪N+ gC) =0and γ(gC(s)) =sfor all s ∈[a, b]\(CgC∪N+ gC), we have that  [a,t) ( f◦gC)dμgC= [a,t) fdμgC. Finally, we recall a concept of continuity introduced in [3]as well as some of its properties. To that end we define the g-topology, τg, as the family of those sets U⊂Rsuch that for every x ∈Uthere exists δ>0 such that if y∈Rsatisfies |g(y) −g(x)| <δthen y∈U. Then, the following definition can be understood as the continuity of a function f:(I, τg) →(F, τu), see [15, Lemma 6]. Definition 2.9 (g-continuous function). A function f:[a, b] →Fis g-continuous at a point t ∈[a, b], or continuous with respect to gat t, if for every ε >0, there exists δ>0such that |f(t) −f(s)| <ε, for every s ∈[a, b]with |g(t) −g(s)| <δ. If fis g-continuous at every point t ∈[a, b], we say that fis g-continuous on [a, b]. Proposition 2.10 ([3, Proposition 3.2]). If f:[a, b] →Ris g-continuous on [a, b], then 1. fis continuous from the left at every t0∈(a, b]; 2. if gis continuous at t0∈[a, b), then so is f; 3. if gis constant on some [α, β] ⊂[a, b], then so is f. In particular, g-continuous functions on [a, b]are continuous on [a, b]when gis continuous on [a, b). F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 7 3. The space of bounded g-differentiable functions In the literature –see, for instance, [3,4,11,15]– authors use the following definition of Stieltjes derivative. Definition 3.1. We define the Stieltjes derivative, or g–derivative, of function f:[a, b] →Rat a point t ∈[a, b]\Cgas f g(t)=⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ lim s→t f(s)−f(t) g(s)−g(t),t/∈Dg, lim s→t+ f(s)−f(t) g(s)−g(t),t∈Dg, provided the corresponding limits exist and, in that case, we say that fis g–differentiable at t. In particular, for t ∈N+ g∪N− g, the g-derivative at tmust be understood in the following sense: f g(t)=⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ lim s→t+ f(s)−f(t) g(s)−g(t),t∈N+ g, lim s→t− f(s)−f(t) g(s)−g(t),t∈N− g. (3.1) Remark 3.2. Observe that the points of Cgare excluded from the definition of g–derivative. This is because the corresponding limit cannot be considered at those points since they are in a neighborhood where the corresponding function is not defined. Observe also that the previous definition is also valid for functions with values in C. Remark 3.3. Taking into account Definition 3.1 and given a function f:[a, b] →R, the following conditions will be necessary for the existence of the g-derivative in all of the points of [a, b]\Cg: •If a ∈[a, b] \Cg, then a /∈N− g. Indeed if a ∈N− g, to calculate the g-derivative at awe need to know the values of fto the left of a, which are not defined. Observe that Cg∩Ng=∅therefore the previous condition is equivalent to a /∈N− g. •If b ∈[a, b] \Cg, then b /∈N+ g∪Dg. Indeed if b ∈N+ g∪Dg, to calculate the g-derivative at bwe need to know the values of fto the right of b, which are not defined. Observe that Cg∩Ng=Cg∩Dg=∅ therefore the previous condition is equivalent to b /∈N+ g∪Dg. • There exists f(t+)for every t ∈(a, b) ∩Dg(which is also a sufficient condition for the existence of the g-derivative at that point). •Given t ∈(a, b) ∩N− gand ε >0, there exists δ>0such that, if s <twith g(t) −g(s) <δthen, |f(s) −f(t)| <ε. We say, in that case, that fis g-continuous from the left at t. To check this fact it is enough to observe that gis left continuous (in the usual sense) at t. The function fmight not be g-continuous at t. Indeed, take for instance g:t∈R→g(t)=⎧ ⎪ ⎨ ⎪ ⎩ t, t ≤1, 1,1≤t≤2, t−1,t≥2. (3.2) Then, f:t∈[0,3] →f(t)=t, 0≤t≤1, t+1,1<t≤3, 8F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 is g-differentiable at t =1since lim s→1− f(s)−f(1) g(s)−g(1) = lim s→1− s−1 s−1=1. Observe that gis continuous at t =1, but fis not, so fcannot be g-continuous at that point. •Given t ∈(a, b) ∩N+ g, and ε >0, there exists δ>0such that, if s >twith g(s) −g(t) <δthen, |f(s) −f(t)| <ε. We say, in that case, that fis g-continuous from the right at t. To check this fact it is enough to observe that gis right continuous (in the usual sense) at t. Observe that, once again, the fmight not be g-continuous at such points. Indeed, take for instance gas in (3.2)and f:t∈[0,3] →f(t)=t, 0≤t<2, t+1,2≤t≤3. In this case, fis g-differentiable at t = 2 but fis not g-continuous at such point. •Given t ∈(a, b)\(Cg∪Dg∪Ng), fis g-continuous at t. In particular, fis continuous at tsince gis continuous at those points. We conclude that, interestingly enough, the g-differentiability of a function at a point of Ngdoes not imply the g-continuity of the function at the point. The g-differentiability of a function only guarantees the g-continuity at the points of (a, b)\(Cg∪Dg∪Ng). Definition 3.4 (C1 g([a, b]; F)space). Let g:R →Rbe such that a /∈N− gand b /∈N+ g∪Dg. We say that f:[a, b] →Fbelongs to C1 g([a, b]; F)if the following conditions are met: 1. f∈C g([a, b]; F), 2. ∃f g(x), for every x ∈[a, b]\Cg, 3. ∃h ∈C g([a, b]; F)such that h(x) =f g(x), for every x ∈[a, b]\Cg. Unless necessary, we will write C1 g([a, b]) instead of C1 g([a, b]; F)for brevity. Let us show now that if we assume that b /∈Cg(observe that, in that case, the hypothesis [a, b] \Cg=∅ is trivially satisfied) the previous definition is consistent insofar as the function given by 3, if it exists, it is unique. Proposition 3.5. Let [a, b] ⊂Rbe a closed interval, g:R →Ra derivator such that a /∈N− gand b /∈ Cg∪N+ g∪Dgand f∈C g([a, b]; F)be g-differentiable at every x ∈[a, b]\Cg. If h1, h2∈C g([a, b]) are such that h1(x) =h2(x) =f g(x), for every x ∈[a, b]\Cg, then h1=h2. Proof. Let us show that h1(x) =h2(x)for every x ∈Cg. Given x∈Cg, there exists a unique connected component of Cg, (an, bn), such that x∈(an, bn). Let us see that h1(x) =h2(x) =f g(bn). Indeed, since h1 is g-continuous, we have, by Proposition 2.10, that h1is constant on (an, bn)and left-continuous, therefore, h1(x) =h1(bn) =f g(bn)for every x ∈(an, bn). The case of h2is proven analogously.  Remark 3.6. Observe that if b ∈Cg, given a function f∈C 1 g([a, b]) the function h ∈C g([a, b]) such that f g(x) =h(x)for every x ∈[a, b] \Cgis not uniquely defined in a neighborhood of point bsince we can not compute the g-derivative at x =bn, with b ∈(an, bn) ⊂Cg. F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 9 A possible way of defining the g-derivative at the points of Cg, which is coherent with the definition of the space C1 g, follows from the previous proof. Indeed, we can generalize Definition 3.1 in the following terms. Definition 3.7. Let [a, b] ⊂Rbe a closed interval and g:R →Ra derivator such that a /∈N− gand b /∈Cg∪N+ g∪Dg. We define the Stieltjes derivative, or g–derivative, of a function f:[a, b] →Fat a point t ∈[a, b]as f g(t)= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ lim s→t f(s)−f(t) g(s)−g(t),t/∈Dg∪Cg, lim s→t+ f(s)−f(t) g(s)−g(t),t∈Dg, lim s→b+ n f(s)−f(bn) g(s)−g(bn),t∈(an,b n)⊂Cg, (3.3) with an, bnas in (2.1); provided the corresponding limits exist. In that case, we say that fis g–differentiable at t. The g-derivative in the points Ngmust be understood as in (3.1). Remark 3.8. It follows from the Definition 3.7 that, for t ∈Dg, f g(t)exists if and only if f(t+)exists and, in that case, f g(t)=f(t+)−f(t) Δ+g(t). Similarly, for any t ∈(an, bn) ⊂Cg, we have that f g(t)exists if and only if f g(bn)exists and, in that case, f g(t) =f g(bn). The following result which includes some basic properties of the Stieltjes derivative is a generalization of [12, Proposition 3.13]. Proposition 3.9. Let [a, b] ⊂Rbe a closed interval and g:R →Ra derivator such that a /∈N− gand b /∈Cg∪N+ g∪Dg. Given an element t ∈[a, b]we denote by: t∗=t, t /∈Cg, bn,t∈(an,b n)⊂Cg, with an, bnas in (2.1). If f1, f2are two g-differentiable functions at t, then: •The function λ1f1+λ2f2is g-differentiable at tfor any λ1, λ2∈Rand (λ1f1+λ2f2) g(t)=λ1(f1) g(t)+λ2(f2) g(t). •The product f1f2is g-differentiable at tand (f1f2) g(t)=(f1) g(t)f2(t∗)+(f2) g(t)f1(t∗)+(f1) g(t)(f2) g(t)Δ+g(t∗).(3.4) •If f2(t∗)(f2(t∗)+(f2) g(t) Δ+g(t∗)) =0, the quotient f1/f2is g-differentiable at tand f1 f2 g (t)= (f1) g(t)f2(t∗)−(f2) g(t)f1(t∗) f2(t∗)(f2(t∗)+(f2) g(t)Δ +g(t∗)) (3.5) 16 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 v(t)=vB(t)vC(t), where vB∈ACgB([0, T]; F)is the unique solution of the problem v gB(t)−β(t)v(t)=0,g B−a.e. t ∈[0,T), v(0) = 1,(4.5) given by vB(t)= s∈[0,t)∩Dg!1+β(s)Δ+g(s)"; (4.6) and vC∈ACgC([0, T]; F)is the unique solution of v gC(t)−β(t)v(t)=0,g C−a.e. t ∈[0,T), v(0) = v0,(4.7) given by vC(t)=u(gC(t)),(4.8) where u ∈AC([0, T]; F)is the unique solution of u(t)= β(t)u(t),a.e.t∈[0,gC(T)), u(0) = v0,(4.9) where  β=β◦γand γis provided by Definition 2.5. Furthermore, vcan be written as v(t)=v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠,(4.10) with  β(t)=⎧ ⎪ ⎨ ⎪ ⎩ β(t),t∈[0,T)\Dg, ln (1 + β(t)Δ+g(t)) Δ+g(t),t∈[0,T)∩Dg. Proof. Existence and uniqueness: If vsolves (4.1), then (x, y) where x := Re vand y:= Im vsolves the real system ⎧ ⎪ ⎨ ⎪ ⎩ x g(t)−Re β(t)x(t)+Imβ(t)y(t)=0,g−a.e. t ∈[0,T), y g(t)−Im β(t)x(t)−Re β(t)y(t)=0,g−a.e. t ∈[0,T), x(0) = Re v0,y(0) = Im v0, (4.11) and vice-versa, that is, the function x +iy, where (x, y)is a solution of (4.11), solves (4.1). Now, it is easy to see that (4.11)satisfies the conditions of [6, Theorem 4.3] with L =| Re(β)| +| Im(β)|, so it has a unique solution on [0, T]. Hence, (4.1)has a unique solution there as well. F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 17 Expression of the solution: Given the nature of problem (4.1) where the g-derivative has to be a multiple of itself it is only natural to use an ansatz of the form v(t)=v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠, with  β∈L 1 g([0, T); F). Given that Mg⊂M gCand Mg⊂M gB, it is clear that if  β∈L 1 g([0, T); F), then  β∈L 1 gC([0, T); F)and  β∈L 1 gB([0, T); F). Furthermore, v(t)=v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμgB+ [0,t) β(s)dμgC⎞ ⎟ ⎠ =v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμgB⎞ ⎟ ⎠exp ⎛ ⎜ ⎝ [0,t) β(s)dμgC⎞ ⎟ ⎠=vB(t)vC(t), where vB(t) := exp [0,t) β(s)dμB gand vC(t) := v0exp [0,t) β(s)dμC g. From the definition we deduce that vB∈AC gB([0, T]; F)and vC∈AC gC([0, T]; F). Hence, given the gB-continuity of vB, we have that (vB) g(t) =0for every t ∈[0, T)\(Dg∪Cg) and, thanks to the gC-continuity of vC, it holds that (vC) g(t) =0 for every t ∈[0, T) ∩Dg. Thus, by Proposition 3.9, v g(t)=(vB) g(t)vC(t),t∈[0,T)∩Dg, vB(t)(vC) g(t),g−a.e. t ∈[0,T)\(Dg∪Cg). This implies that we will have a different equation for each of the components of the solution: (vB) g(t)=β(t)vB(t),t∈[0,T)∩Dg,(4.12) (vC) g(t)=β(t)vC(t),g−a.e. t ∈[0,T)\(Dg∪Cg).(4.13) We will start studying equation (4.12). For t ∈[0, T) ∩Dgwe have that (vB) g(t)=vB(t+)−vB(t) Δ+g(t)=(vB) gB(t). Now, if we develop equation (4.12): vB(t+)−vB(t) Δ+g(t)=β(t)vB(t) and we get that vB(t+)=vB(t)(1 + β(t)Δ +g(t)),t∈[0,T)∩Dg.(4.14) In order to get a solution candidate for equation (4.12), define h(t) =ln(1+β(t)Δ+g(t))/Δ+g(t)if t ∈Dg, h(t) =0if t /∈Dg. Then, taking into account that μgB(t) =0for every t /∈Dg, we define 18 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 H(t):=exp⎛ ⎜ ⎝ [0,t) h(t)dμgB⎞ ⎟ ⎠=exp⎛ ⎜ ⎝ [0,t) ln(1 + β(s)Δ+g(s)) Δ+g(s)dμgB⎞ ⎟ ⎠ =exp⎛ ⎝ s∈[0,t)∩Dg ln(1 + β(s)Δ+g(s))⎞ ⎠ =exp⎛ ⎝ s∈[0,t)∩Dg!ln 1+β(s)Δ+g(s)+iArg(1 + β(s)Δ+g(s))"⎞ ⎠. To show that His well defined, let us check that the series  s∈[0,t)∩Dg ln 1+β(s)Δ+g(s)and  s∈[0,t)∩Dg Arg(1 + β(s)Δ+g(s)) are absolutely convergent. We have that  s∈[0,T )∩Dgln 1+β(s)Δ+g(s)= s∈Aln 1+β(s)Δ+g(s)+ s∈Bln 1+β(s)Δ+g(s), where A=s∈[0,T)∩Dg:1+β(s)Δ+g(s)≥1=s∈[0,T)∩Dg:ln 1+β(s)Δ+g(s)≥0, B=s∈[0,T)∩Dg:1+β(s)Δ+g(s)<1=s∈[0,T)∩Dg:ln 1+β(s)Δ+g(s)<0. In order to bound the sum on Ait is enough to take into account that 0 ≤ln(1 +x) ≤xfor every x ∈[0, ∞):  s∈Aln 1+β(s)Δ+g(s)= s∈A ln 1+β(s)Δ+g(s)≤ s∈A ln !1+|β(s)|Δ+g(s)|"≤ s∈A|β(s)|Δ+g(s)<∞, because β∈L 1 gB([0, T), F). Now, let us focus on the sum on B. For any s ∈B, taking into account that 1 +β(s)Δ+g(s) =0, we have that 0<1+β(s)Δ +g(s)2=[1+Re(β(s)) Δ+g(s)]2+[Im(β(s)) Δ+g(s)]2 =1+2Re(β(s)) Δ+g(s)+|β(s)Δ +g(s)|2<1. In particular, 2 Re(β(s)) Δ+g(s) +|β(s) Δ+g(s)|2<0which yields Re(β(s)) <0. Now, we can consider the following sets: B1=#s∈B:0<1+2Re(β(s))Δ+g(s)+|β(s)Δ+g(s)|2<1 2$, B2=#s∈B:1 2≤1+2Re(β(s))Δ+g(s)+|β(s)Δ+g(s)|2<1$. Observe that B=B1∪B2. The definition of B1implies that 1>2|Re(β(s))|Δ+g(s)−|β(s)Δ+g(s)|2>1 2,∀s∈B1. Therefore, F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 19 |Re(β(s))|Δ+g(s)>1 4,∀s∈B1. Hence, we have that B1is finite since, otherwise, we would have that β/∈L 1 gB([0, T), F), which is a contradiction. For the elements in the set B2we have that: 1 2≥2|Re(β(s))|Δ+g(s)−|β(s)Δ+g(s)|2>0,∀s∈B2. Thus, if we take into account that ln(1/(1 −x)) ≤2x, for every x ∈[0, 1/2], ln 1+β(s)Δ+g(s)=1 2ln !1+2Re(β(s))Δ+g(s)+|β(s)Δ+g(s)|2" =1 2ln !1/!1+2Re(β(s))Δ+g(s)+|β(s)Δ+g(s)|2"" =1 2ln !1/!1−!2|Re(β(s))|Δ+g(s)−|β(s)Δ+g(s)|2""" ≤2!2|Re(β(s))|Δ+g(s)−|β(s)Δ+g(s)|2"≤4|Re(β(s))|Δ+g(s). Hence,  s∈Bln 1+β(s)Δ+g(s)<∞. Let us now bound the term associated with the argument. Taking into account that |atan(x)| ≤|x|for every x ∈R, we have that  s∈[0,T )∩DgArg(1 + β(s)Δ+g(s))≤ s∈[0,T )∩Dg |Im(β(s))Δ+g(s)| |1+Re(β(s))Δ+g(s)|. Let us divide the set [0, T) ∩Dginto the subsets  B1=s∈[0,T)∩Dg:|Re(β(s))|Δ+g(s)>1/2,  B2=([0,T)∩Dg)\ B1. Observe that  B1must be of finite cardinality. On the other hand, given t ∈ B2, 1+Re(β(s))Δ+g(s)≥1 2. Thus,  s∈B2 |Im(β(s))Δ+g(s)| |1+Re(β(s))Δ+g(s)|≤2 s∈B2Im(β(s))Δ+g(s)<∞. Hence, we conclude that His well defined. In order to prove that His a solution of (4.12), we observe that, given t ∈[0, T) ∩Dg, H(t+) = lim s→t+exp ⎛ ⎜ ⎝ [0,s) ln(1 + β(s)Δ+g(s)) Δ+g(s)dμgB⎞ ⎟ ⎠ 20 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 = lim s→t+exp ⎛ ⎜ ⎝ [0,t) ln(1 + β(s)Δ+g(s)) Δ+g(s)dμgB+ln(1+β(t)Δ+g(t)) +  (t,s) ln(1 + β(s)Δ+g(s)) Δ+g(s)dμgB⎞ ⎟ ⎠ =(1 + β(t)Δ+g(t)) exp ⎛ ⎜ ⎝ [0,t) ln(1 + β(s)Δ+g(s)) Δ+g(s)dμgB⎞ ⎟ ⎠=(1+β(t)Δ+g(t))H(t), so equation (4.14)holds and vB:= His a solution of (4.12). Observe that, given any set A ⊂[0, T)\Dg, we have A =(A\Dg) ∪(A ∩(Dg\Dg)) thus μ∗ gB(A) ≤μ∗ gB(A\Dg) +μ∗ gB(A ∩(Dg\Dg)) ≤μ∗ gB(A\Dg) + μ∗ g(Dg\Dg) =0. Therefore vBsatisfies (4.5) and, moreover, vB(t)=exp⎛ ⎝ s∈[0,t)∩Dg ln(1 + β(s)Δ+g(s))⎞ ⎠= s∈[0,t)∩Dg!1+β(s)Δ+g(s)". Let us now study equation (4.13). First, observe that, given an element t ∈[0, T)\(Dg∪Cg), there exists δ>0such that gis continuous on (t −δ, t +δ). In the case t ∈N− gwe further know that gis strictly increasing on the interval (t −δ, t], and constant on (t, t +δ). In the case t ∈N+ g, gwould be constant on (t −δ, t)and strictly increasing on [t, t +δ). In any case (observe that, if t ∈N− gwe have to take the limit from the left and in the case t ∈N+ gthe limit from the right, respectively): (vC) g(t) = lim s→t vC(s)−vC(t) g(s)−g(t)= lim s→t vC(s)−vC(t) gC(s)−gC(t)=(vC) gC(t). Hence, taking into account that μ∗ g(A) =0 ⇔μ∗ gC(A) =0for any A ⊂[0, T)\Dg, together with the fact that Cg=CgC, we see that equation (4.13)is equivalent to (vC) gC(t)=β(t)vC(t),g C−a.e. t ∈[0,T)\(Dg∪CgC).(4.15) Let us observe that μgC(Dg∪CgC) ≤μgC(Dg\Dg) +μgC(Dg) +μgC(CgC) =0, since μgC(Dg\Dg) ≤ μg(Dg\Dg) =0by hypothesis. Therefore, (4.15)is equivalent to: (vC) gC(t)=β(t)vC(t),g C−a.e. t ∈[0,T).(4.16) Now we will see that vC(t) := u(gC(t)), with u ∈AC([0, gC(T)]; F)the solution of (4.9)satisfies equation (4.16). On the one hand, we have that  β=β◦γ∈L 1([0, gC(T)]; F). Indeed, the measurability is a consequence of Proposition 2.7. Now, using a similar argument as the one in the proof of Corollary 2.8:  [0,gC(T)) | β|dμ= [0,T ) |β|dμgC≤ [0,T ) |β|dμg<∞. Thus, (4.9)admits a unique solution u(t)=v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμ⎞ ⎟ ⎠∈AC([0,gC(T)]; F). In particular, vC(t) =u(gC(t)) is such that (vC) g(t) =0for every t ∈Dg. Indeed, F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 21 (vC) g(t) = lim s→t+ vC(s)−vC(t) g(s)−g(t)= lim s→t+ u(gC(s)) −u(gC(t)) g(s)−g(t)=0, thanks to the continuity of the composition u ◦gC. On the other hand, since u ∈AC([0, gC(T)]; F)is the solution of (4.9), there exists a Lebesgue-null set N⊂[0, gC(T)] such that u(t)= β(t)u(t),∀t∈[0,gC(T)]\N. In particular, u(gC(t)) =  β(gC(t)) u(gC(t)),∀t∈[0,T]\(gC)−1(N), whence, by Proposition 4.1, (vC) gC(t)=u(gC(t)) =  β(gC(t)) u(gC(t)),∀t∈[0,T]\(gC)−1(N). Taking into account that γ(gC(t)) =tfor every t ∈[0, T]\(CgC∪N+ gC), that μgC(CgC∪N+ gC) =0and that μgC((gC)−1(N)) =0(see the proof of Proposition 2.7), we deduce that (vC) gC(t)=β(t)vC(t),g C-a.e. t∈[0,T). Last, in regard to vC, using a reasoning similar to the one used in the proof of the Corollary 2.8, we have that vC(t)=u(gC(t)) = v0exp ⎛ ⎜ ⎝ [0,gC(t))  β(s)dμ⎞ ⎟ ⎠=v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμgC⎞ ⎟ ⎠. Finally, let us check that v:= vCvBis in the space ACg([0, T]; F). To show this, let us define  β(t)=⎧ ⎪ ⎨ ⎪ ⎩ β(t),t∈[0,T)\Dg, ln (1 + β(t)Δ+g(t)) Δ+g(t),t∈[0,T)∩Dg, and check that v(t)=v0exp ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠=v0exp ⎛ ⎜ ⎝ [0,t)\Dg β(s)dμg+ s∈[0,t)∩Dg β(s)Δ+g(s)⎞ ⎟ ⎠. Indeed, on the one hand, v(t)=v0⎡ ⎣ s∈[0,t)∩Dg!1+β(s)Δ+g(s)"⎤ ⎦exp ⎛ ⎜ ⎝ [0,gC(t))  β(s)dμ⎞ ⎟ ⎠ =v0exp ⎛ ⎜ ⎝ [0,gC(t))  β(s)dμ+ s∈[0,t)∩Dg ln (1 + β(s)Δ+g(s)) Δ+g(s)Δ+g(s)⎞ ⎟ ⎠. Now, thanks to the fact that μ(Dg) =0as it is a countable set, we see that 22 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010  [0,gC(t)) (β◦γ)(s)dμ= [0,gC(t)) ( β◦γ)(s)dμ. Thus, by Corollary 2.8,  [0,t) β(s)dμg= [0,gC(t))  β(s)dμ+ s∈[0,t)∩Dg ln (1 + β(s)Δ+g(s)) Δ+g(s)Δ+g(s). Finally, it is clear that  β∈L 1 g([0, T); F), therefore v∈ACg([0, T); F).  Remark 4.3. We must take into account the following remarks: 1. If gCis constant, then the solution of (4.1)is reduced to v0vBand the hypothesis μg(Dg\Dg) =0is not necessary. 2. The hypothesis μg(Dg\Dg) =0that appears in the statement of Theorem 4.2 has been used to express the solution of (4.1)as the product of the solutions of the problems (4.5)and (4.7). This hypothesis is not essential to guarantee the existence of a solution of problem (4.1). Even in the case μg(Dg\Dg) =0, we will have (4.10)is well defined and a valid solution of problem (4.1). Indeed, since  β∈L 1 g([0, T); F), we have that ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠  g (t)= β(t),g-a.e. t∈[0,T). Therefore, (4.2) ensures that ⎛ ⎜ ⎝exp ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠⎞ ⎟ ⎠  g (t)=β(t)exp ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠,g-a.e. t∈[0,T)\Dg.(4.17) Now, given t ∈[0, T) ∩Dg, lim s→t+exp ⎛ ⎜ ⎝ [0,s) β(s)dμg⎞ ⎟ ⎠= lim s→t+exp ⎛ ⎜ ⎝ [0,t) β(s)dμg+ln(1+β(t)Δ+g(t)) +  (t,s) β(s)dμg⎞ ⎟ ⎠ =(1 + β(t)Δ+g(t)) exp ⎛ ⎜ ⎝ [0,t) β(s)dμg⎞ ⎟ ⎠, so equation (4.17)is also satisfied for the points of Dg. Remark 4.4. The previous result is a generalization of the results in [3, Section 6] for several reasons. 1. The solution obtained is valid in the complex case, whereas in [3]it is only applied to the real case. The generalization to the complex case is immediate considering the complex exponential and the principal branch of the complex logarithm. 2. We have proven that the hypothesis  s∈[0,T )∩Dgln 1+β(s)Δ+g(s)<∞ F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 23 occurring in [3, Definition 6.1 and Lemma 6.5] is not necessary, it being a direct consequence of β∈ L1 g([0, T); F)and g(T) <∞. This was also proven in [13, Lemma 3.1] for the real case. 3. The solution obtained generalizes that in [3, Lemma 6.5]. Indeed, in the particular case β∈L 1 g([0, T); R) and given that 1 +β(t)Δ+g(t) =0for every t ∈[0, T) ∩Dg, we have that, for every t ∈[0, T) ∩Dg, Arg(1 + β(t)Δ+g(t)) = π, 1+β(t)Δ+g(t)<0, 0,1+β(t)Δ+g(t)>0. Hence, if we write T− β:= {t ∈[0, T) ∩Dg:1 +β(t)Δ+g(t) <0}and T+ β={t ∈[0, T) ∩Dg:1 +β(t)Δ+g(t) > 0}(observe that T− βis of finite cardinality), we have that ln !1+β(t)Δ+g(t)"=ln 1+β(t)Δ+g(t),t∈T+ β, ln 1+β(t)Δ+g(t)+iπ, t ∈T− β. Taking into account the previous observations, v(t)=exp ⎛ ⎜ ⎝ [0,t)\Dg β(s)dμg+ t∈[0,t)∩Dg ln 1+β(t)Δ+g(t)+i s∈[0,t)∩T− β π⎞ ⎟ ⎠ =cos ⎛ ⎜ ⎝ s∈[0,t)∩T− β π⎞ ⎟ ⎠exp ⎛ ⎜ ⎝ [0,t)\Dg β(s)dμg+ t∈[0,t)∩Dg ln 1+β(t)Δ+g(t)⎞ ⎟ ⎠. Hence, if T− β={t1, ..., tk}and tk+1 := T, we get v(t)= ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ exp ⎛ ⎜ ⎝ [0,t)\Dg β(s)dμg+ t∈[0,t)∩Dg ln 1+β(t)Δ+g(t)⎞ ⎟ ⎠,t∈[0,t 1], cos(jπ)exp⎛ ⎜ ⎝ [0,t)\Dg β(s)dμg+ t∈[0,t)∩Dg ln 1+β(t)Δ+g(t)⎞ ⎟ ⎠,t∈(tj,t j+1], j=1,...,k, which is precisely the solution in [3, Lemma 6.5]. 4. In the case that there exists some element t ∈[0, T) ∩Dgsuch that 1 +β(t)Δ+g(t) =0, the set T0 β:= {t∈[0,T)∩Dg:1+β(t)Δ+g(t)=0} is of finite cardinality and, therefore, if we denote by t0 β:= min T0 βif T0 β=∅, t0 β:= Totherwise, we have that v(t)=⎧ ⎨ ⎩ u(gC(t)) s∈[0,t)∩Dg!1+β(s)Δ+g(s)",t∈[0,t 0 β], 0,t∈(t0 β,T]. Taking into account that we are assuming that gis continuous at t =0, we have that t0 β=minT0 β>0. Thus, v(t) =0for every t ∈[0, t0 β]. 24 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 Definition 4.5. Given an element β∈L 1 g([0, T); F)and v0=1, we denote the solution of problem (4.1) constructed in Theorem 4.2 by expg(β; 0, t) ∈AC g([0, T]; F)and call it the complex g-exponential map or just g-exponential map. In the following result we present some important properties of the complex g-exponential function. Proposition 4.6. Let β, β1, β2∈L 1 g([0, T); F). The following properties hold: 1. If a =Reβand b =Imβthen expg(β;0,t)= u∈[0,t)∩Dg!1+a(u)Δ+g(u)+ib(u)Δ+g(u)"exp ⎛ ⎜ ⎝ [0,gC(t)) (a◦γ)dμ⎞ ⎟ ⎠ ·⎡ ⎢ ⎣cos ⎛ ⎜ ⎝ [0,gC(t)) (b◦γ)dμ⎞ ⎟ ⎠+isin ⎛ ⎜ ⎝ [0,gC(t)) (b◦γ)dμ⎞ ⎟ ⎠⎤ ⎥ ⎦. (4.18) 2. expg(β;0,t)=exp g(β; 0, t), for every t ∈[0, T]. 3. Given n ∈N, expg(β; 0, t)n=exp g(pn(β); 0, t) ∈ACg([0, T]; F), where pn(β)(t)=nβ(t)+ n  k=2 n kβ(t)kΔ+g(t)k−1,n∈N. 4. Given n ∈N, expg(β; 0, t)−n=exp g(qn(β); 0, t) ∈ACg([0, t0 β]; F), where qn(β)(t)=−pn(β)(t) 1+pn(β)(t)Δ +g(t),n∈N. Observe that expg(β; 0, t)−nis not well defined in (t0 β, T]since expg(β; 0, ·) =0in that set. 5. For all t ∈[0, T), expg(β1;0,t)exp g(β2;0,t)=exp g(β1+β2+β1β2Δ+g;0,t).(4.19) Proof. 1. Indeed, expg(a+bi;0,t) =exp⎛ ⎜ ⎝ [0,t)\Dg a(s)dμg+i [0,t)\Dg b(s)dμg⎞ ⎟ ⎠exp ⎛ ⎝ u∈[0,t)∩Dg ln(1 + (a(u)+ib(u))Δ+g(u))⎞ ⎠ =exp⎛ ⎜ ⎝ [0,t)\Dg a(s)dμg⎞ ⎟ ⎠ u∈[0,t)∩Dg!1+a(u)Δ+g(u)+ib(u)Δ+g(u)" ·⎡ ⎢ ⎣cos ⎛ ⎜ ⎝ [0,t)\Dg b(s)dμg⎞ ⎟ ⎠+isin ⎛ ⎜ ⎝ [0,t)\Dg b(s)dμg⎞ ⎟ ⎠⎤ ⎥ ⎦. Now the formula is obtained reasoning as in Corollary 2.8. F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 25 2. This property is clear from the definition of the complex conjugate. 3. Observe that pn(β) ∈L 1 g([0, T); F)since pn(β)L1 g([0,T );F)≤nβL1 g([0,T )+ n  k=2 n k t∈[0,T )∩Dg!|β(t)|Δ+g(t)"k<∞. Thus the solution of problem (4.1) where we consider pn(β)instead of βis given by v=vBvCwhere vB(t)= s∈[0,t)∩Dg1+nβ(s)+ n  k=2 n kβ(s)kΔ+g(s)k−1Δ+g(s), = s∈[0,t)∩Dg1+nβ(s)Δ+g(s)+ n  k=2 n kβ(s)kΔ+g(s)k, = s∈[0,t)∩Dgn  k=0 n kβ(s)kΔ+g(s)k= s∈[0,t)∩Dg!1+β(s)Δ+g(s)"n =⎛ ⎝ s∈[0,t)∩Dg!1+β(s)Δ+g(s)"⎞ ⎠ n , vC(t)=exp⎛ ⎜ ⎝ [0,t)nβ(s)+ n  k=2 n kβ(s)kΔ+g(s)k−1dμgC⎞ ⎟ ⎠=exp⎛ ⎜ ⎝n [0,t) β(s)dμgC⎞ ⎟ ⎠ =⎡ ⎢ ⎣exp ⎛ ⎜ ⎝ [0,t) β(s)dμgC⎞ ⎟ ⎠⎤ ⎥ ⎦ n . Hence, expg(β; 0, t)n=exp g(pn(β); 0, t). 4. Observe that qn(β) ∈L 1 g([0, T); F)since qnL1 g([0,t0 βT);F)≤nβL1 g([0,T )+ t∈[0,T )∩Dg |pn(β)(t)Δ +g(t)| |1+pn(β)(t)Δ +g(t)|<∞, because pn(β) ∈L 1 g([0, T); F). Therefore, the solution to problem (4.1), where we consider qn(β)instead of β, is given by v=vBvCwhere vB(t)= s∈[0,t)∩Dg1−nβ(s)+n k=2 !n k"β(s)kΔ+g(s)k−1 1+!nβ(s)+n k=2 !n k"β(s)kΔ+g(s)k−1"Δ+g(s)Δ+g(s), = s∈[0,t)∩Dg1 1+!nβ(s)+n k=2 !n k"β(s)kΔ+g(s)k−1"Δ+g(s) =⎛ ⎝ s∈[0,t)∩Dg!1+β(s)Δ+g(s)"⎞ ⎠ −n , vC(t)=exp⎛ ⎜ ⎝ [0,t) −nβ(s)+n k=2 !n k"β(s)kΔ+g(s)k−1 1+!nβ(s)+n k=2 !n k"β(s)kΔ+g(s)k−1"Δ+g(s)dμgC⎞ ⎟ ⎠ 32 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 expg(x;0,s)−1expg(z;0,s)=exp g−x 1+xΔ+g;0,s expg(z;0,s) =exp g−x 1+xΔ+g+z−xzΔ+g 1+xΔ+g;0,s =exp gz−x 1+xΔ+g;0,s . Therefore,  [0,t) expg(x;0,s)−1expg(z;0,s) 1+xΔ+g(s)dμg(s)= [0,t) 1 1+xΔ+g(s)expgz−x 1+xΔ+g;0,s dμg(s) =(z−x)−1 [0,t)expgz−x 1+xΔ+g;0,· g (s)dμg(s) =(z−x)−1+expgz−x 1+xΔ+g;0,t −expgz−x 1+xΔ+g;0,0, =(z−x)−1-expg(x;0,t)−1expg(z;0,t)−1.. Finally, v(t)=exp g(x;0,t)+exp g(x;0,t)(z−x)−1-expg(x;0,t)−1expg(z;0,t)−1. =exp g(x;0,t)+(z−x)−1-expg(z;0,t)−expg(x;0,t).. Observe that, differentiating v gagain, we obtain that v g−(x +z) v g+x zv=0, so, for any values P, Q ∈C, taking x =(−P+/P2−4Q)/2, z=P−x, vsolves the equation v g+Pv  g+Q v=0. This fact illustrates how we can obtain a solution of a second order problem from a first order problem. In the next section we study this type of problems. 5. Linear g-differential problems of second order with constant coefficients In this section we consider g-differential problems of second order with constant coefficients. Since we will assume that the coefficients are constant, we will look for solutions in the space BC2 g([0, T]; F). Once again, we assume that 0 /∈N− gand T/∈N+ g∪Dg∪Cg. 5.1. The homogeneous case Let us consider the second order homogeneous linear Cauchy problem ⎧ ⎪ ⎨ ⎪ ⎩ v g(t)+Pv  g(t)+Qv(t)=0,∀t∈[0,T], v(0) = x0, v g(0) = v0, (5.1) where P, Q, x0, v0∈F. We start by defining what we understand as a solution of problem (5.1). Definition 5.1. We say v∈BC2 g([0, T]; F)is a solution of (5.1)if it satisfies the equation v g(t)+Pv  g(t)+Qv(t)=0,∀t∈[0,T] and the initial conditions v(0) =x0and v g(0) =v0. We have the following lemma, whose proof is straightforward from the linearity of the g-derivative. F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 33 Lemma 5.2. Let x0, v0∈Fand v1, v2∈BC2 g([0, T]; F)be such that (vk) g(t)+P(vk) g(t)+Qv k(t)=0,∀t∈[0,T],k=1,2.(5.2) If v1(0) (v2) g(0) −v2(0) (v1) g(0) =0, then v=c1v1+c2v2is a solution of (5.1), where c1=(v2) g(0) x0−v0v2(0) v1(0) (v2) g(0) −v2(0) (v1) g(0), c2=v0v1(0) −(v1) g(0) x0 v1(0) (v2) g(0) −v2(0) (v1) g(0). Theorem 5.3. For (5.1), the following hold: •If P2−4 Q =0, then, defining λ1=(−P+/P2−4Q)/2and λ2=(−P−/P2−4Q)/2, we have that v(t)=v0−λ2x0 λ1−λ2expg(λ1;0,t)−v0−λ1x0 λ1−λ2expg(λ2;0,t) is a solution of (5.1). Furthermore, v∈BC∞ g([0, T]; F)and it is the unique solution in that space. •If P2−4 Q =0, then, taking λ =−P/2, v(t)=x0expg(λ;0,t)+(v0−λx 0)exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s) is a solution of (5.1). Furthermore, v∈BC∞ g([0, T]; F)and it is the unique solution in that space. Proof. We consider the characteristic equation of problem (5.1), λ2+Pλ+Q=0. If P2−4 Q =0, let v1=exp g(λ1; 0, t)and v2(t) =exp g(λ2; 0, t). By Corollary 4.19 we have that v1, v2∈BC ∞ g([0, T]; F) ⊂BC 2 g([0, T]; F). Furthermore, it can be checked that both functions satisfy (5.2). On the other hand, v1(0) (v2) g(0) −v2(0) (v1) g(0) = λ2−λ1=0. Hence, by Lemma 5.2, there exists a solution of problem (5.1)given by v(t)=v0−λ2x0 λ1−λ2expg(λ1;0,t)−v0−λ1x0 λ1−λ2expg(λ2;0,t). If P2−4 Q =0we get the double root λ =−P/2of the characteristic equation. Observe that the left hand side of the equation occurring in (5.1)can be written as (∂g+P/2)2vwhere ∂gdenotes the g-derivative operator. Hence, we define v1(t) := expg(λ; 0, t), which is a solution of (∂g+P/2)v=0and consider the unique solution of (v2) g(t)=λv 2(t)+v1(t),g−a.e. t ∈[0,T), v2(0) = 0. 34 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 Since (∂g+P/2)v1=0, it is clear that (∂g+P/2)2v2=0. Furthermore, v2(0) =0and (v2) g(0) =1, so v2is the solution we are looking for. By Corollary 4.19, v1∈BC ∞ g([0, T]; F) and, applying Corollary 4.21, v2∈BC∞ g([0, T]; F)as well. Now, thanks to Proposition 4.12, we have that: v2(t)=exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg. Since v2(0) =0, (v2) g(0) =1we have that: v1(0) (v2) g(0) −v2(0) (v1) g(0) = 1 =0. Thus, by Lemma 5.2, there exists a solution of problem (5.1)given by v(t)=x0expg(λ;0,t)+(v0−λx 0)exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s). Finally, if we define u(t) =v g(t) ∈BC∞ g([0, T]) we have that the pair of functions (u, v) ∈[ACg([0, T]; F)]2 satisfies the following system of differential equations: ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩v u g (t)=01 −Q−Pv(t) u(t), v(0) =x0,u(0) = v0. Thanks to [3, Theorem 7.3] we have that the previous system has a unique solution in [ACg([0, T]; F)]2, therefore vis the unique solution of (5.2)in the space BC∞ g([0, T]; F).  5.2. The non homogeneous case In this section we focus on the non homogeneous version of the second order linear problem, namely, ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ v g(t)+Pv  g(t)+Qv(t)=f(t),∀t∈[0,T], v(0) = x0, v g(0) = v0, (5.3) where P, Q, x0, v0∈Fare constant values and f∈ACg([0, T]; F). Since the coefficients are constant, it will be the regularity of the term fthat determines the additional regularity of the solution. As in the previous sections, we will see that it is possible to prove the uniqueness of solution when we consider the solution in the space BC2 g([0, T]; F). Theorem 5.4. Let f∈BCn g([0, T]; F)and assume 1 +λ Δ+g(t) =0, for all t ∈[0, T) ∩Dgand λ ∈Fsuch that λ2+Pλ +Q =0. Then, problem (5.3)has a unique solution v∈BCn+2 g([0, T], F)given by v(t)=x0expg(λ2;0,t)+(v0−λ2x0)exp g(λ2;0,t)· [0,t) expg(λ2;0,s)−1 1+λ2Δ+g(s)expg(λ1;0,s)dμg(s) +exp g(λ2;0,t) [0,t) expg(λ2;0,s)−1 1+λ2Δ+g(s)expg(λ1;0,s)·⎛ ⎜ ⎝ [0,s) expg(λ1;0,r)−1 1+λ1Δ+g(r)f(r)dμg(r)⎞ ⎟ ⎠dμg(s), (5.4) F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 35 where (x −λ1)(x −λ2) =x2+Px +Q. Proof. Let be λ1, λ2the two complex eigenvalues of the characteristic polynomial x2+Px +Q =0. Assume v2∈C n+2 g([0, T], F)is a solution of problem (5.3). Observe that, if we define v1=(v2) g−λ2v2, it is clear that v1∈Cn+1 g([0, T], F)and (v1) g−λ1v1=f, v1(0) =(v2) g(0) −λ2v2(0) =v0−λ2x0, so v1has to solve the problem (v1) g(t)=λ1v1(t)+f(t),g−a.e. t ∈[0,T) v(0) = v0−λ2x0.(5.5) By Corollary 4.21, problem (5.5)has a unique solution in Cn+1 g([0, T], F), so v1is that unique solution. Furthermore, by definition, v1=(v2) g−λ2v2and v2(0) =x0, so v2solves the problem (v2) g(t)=λ2v2(t)+v1(t),g−a.e. t ∈[0,T) v(0) = x0.(5.6) By Corollary 4.21, problem (5.6)has a unique solution in Cn+2 g([0, T], F), so v2is that unique solution. This implies that, if a solution in Cn+2 g([0, T], F)of problem (5.3) exists, it has to be unique. In order to obtain that unique solution it is enough to retrace the steps we have taken to prove the uniqueness. Let v1be the unique solution of problem (5.5)in Cn+1 g([0, T], F)and let v2be the unique solution of problem (5.6)in Cn+2 g([0, T], F). Clearly v2is a solution of problem (5.3). In order to obtain the explicit expression of the solution, observe that, by Proposition 4.12, v1is of the form v1(t)=(v0−λ2x0)exp g(λ1;0,t)+exp g(λ1;0,t) [0,t) expg(λ1;0,s)−1f(s) 1+λ1Δ+g(s)dμg(s), and v2of the form v2(t)=x0expg(λ2;0,t)+exp g(λ2;0,t) [0,t) expg(λ2;0,s)−1v1(s) 1+λ2Δ+g(s)dμg(s) =x0expg(λ2;0,t)+(v0−λ2x0)exp g(λ2;0,t) [0,t) expg(λ2;0,s)−1expg(λ1;0,s) 1+λ2Δ+g(s)dμg(s) +exp g(λ2;0,t) [0,t) expg(λ2;0,s)−1expg(λ1;0,s) 1+λ2Δ+g(s) · [0,s) expg(λ1;0,r)−1f(r) 1+λ1Δ+g(r)dμg(r)dμg(s). Now, thanks to Proposition 4.6, expg(λ2;0,t)−1expg(λ1;0,t)=exp g(−λ2/(1 + λ2Δ+g(t)); 0,t)exp g(λ1;0,t) =exp g((λ1−λ2)/(1 + λ2Δ+g(t)); 0,t). Thus, 36 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 v2(t)=x0expg(λ2;0,t)+(v0−λ2x0)exp g(λ2;0,t) [0,t) 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s dμg(s) +exp g(λ2;0,t) [0,t) 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s  · [0,s) expg(λ1;0,r)−1f(r) 1+λ1Δ+g(r)dμg(r)dμg(s). Remark 5.5. Note that, for λ ∈Fsuch that λ2+Pλ +Q =0, the condition 1 +λ Δ+g(t) =0can only happen for a finite number of t ∈[0, T) ∩Dg. Remark 5.6. From the previous expression we can derive the expression of Green’s function of problem (5.3) just by equating  R G(t, r)f(r)dμg(r)=exp g(λ2;0,t) [0,t) 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s  · [0,s) expg(λ1;0,r)−1f(r) 1+λ1Δ+g(r)dμg(r)dμg(s). (5.7) Now, if we consider the product measure space ([0, T], Mg·Mg, μg·μg), we have, by Fubini’s Theorem [1, Theorem 10.10],  R G(t, r)f(r)dμg(r)=exp g(λ2;0,t) [0,T ]·[0,T ] 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s  ·expg(λ1;0,r)−1f(r) 1+λ1Δ+g(r)χ[0,s)(r)χ[0,t)(s)dμg·dμg =exp g(λ2;0,t) [0,T ] [0,T ] expg(λ1;0,r)−1expgλ1−λ2 1+λ2Δ+g;0,s  ·(1 + λ1Δ+g(r))−1(1 + λ2Δ+g(s))−1f(r)χ(r,t)(s)χ[0,t)(r)dμg(s)dμg(r) =exp g(λ2;0,t) [0,t) expg(λ1;0,r)−1f(r) 1+λ1Δ+g(r) ·⎛ ⎜ ⎝ (r,t) 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s dμg(s)⎞ ⎟ ⎠dμg(r). Therefore, for t, r∈[0, T], F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 37 G(t, r)=exp g(λ2;0,t)exp g(λ1;0,r)−1(1 + λ1Δ+g(r))−1χ[0,t)(r) · (r,t) 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s dμg(s) =exp g(λ2;0,t)exp g(λ1;0,r)−1(1 + λ1Δ+g(r))−1χ[0,t)(r) · [r,t) 1 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s dμg(s) −expg(λ2;0,t)exp g(λ2;0,r)−1(1 + λ1Δ+g(r))−1(1 + λ2Δ+g(r))−1Δ+g(r)χ[0,t)(r). Observe that: •If λ1=λ2, v(s)=exp gλ1−λ2 1+λ2Δ+g;0,s ∈ACg([r, T ]; F) is the solution of ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ v g(s)= λ1−λ2 1+λ2Δ+g(s)v(s),g−a.e. s ∈[r, T ), v(r)=exp gλ1−λ2 1+λ2Δ+g;0,r. Therefore,  [r,t) λ1−λ2 1+λ2Δ+g(s)expgλ1−λ2 1+λ2Δ+g;0,s dμg(s) =exp gλ1−λ2 1+λ2Δ+g;0,t −expgλ1−λ2 1+λ2Δ+g;0,r =exp g(λ1;0,t)exp g(λ2;0,t)−1−expg(λ1;0,r)exp g(λ2;0,r)−1. Thus, the Green’s function in the case λ1=λ2has the following expression: G(t, r)=exp g(λ2;0,t)exp g(λ1;0,r)−1(1 + λ1Δ+g(r))−1χ[0,t)(r) ·(λ1−λ2)−1!expg(λ1;0,t)exp g(λ2;0,t)−1−expg(λ1;0,r)exp g(λ2;0,r)−1" −expg(λ2;0,t)exp g(λ2;0,r)−1(1 + λ1Δ+g(r))−1(1 + λ2Δ+g(r))−1Δ+g(r)χ[0,t)(r) =+(λ1−λ2)−1expg(λ1;0,t)exp g(λ1;0,r)−1(1 + λ1Δ+g(r))−1χ[0,t)(r) −(λ1−λ2)−1expg(λ2;0,t)exp g(λ2;0,r)−1(1 + λ1Δ+g(r))−1χ[0,t)(r) −expg(λ2;0,t)exp g(λ2;0,r)−1(1 + λ1Δ+g(r))−1(1 + λ2Δ+g(r))−1Δ+g(r)χ[0,t)(r) =+(λ1−λ2)−1expg(λ1;0,t)exp g(λ1;0,r)−1(1 + λ1Δ+g(r))−1χ[0,t)(r) −(λ1−λ2)−1expg(λ2;0,t)exp g(λ2;0,r)−1(1 + λ2Δ+g(r))−1χ[0,t)(r). (5.8) •If λ1=λ2, we have the following expression for the Green’s function: G(r, t)=exp g(λ;0,t)exp g(λ;0,r)−1(1 + λΔ+g(r))−1χ[0,t)(r)· (r,t) 1 1+λΔ+g(s)dμg(s) 38 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 =exp g(λ;0,t)exp g(λ;0,r)−1(1 + λΔ+g(r))−1χ[0,t)(r) ·⎛ ⎜ ⎝ [0,t) 1 1+λΔ+g(s)dμg(s)− [0,r) 1 1+λΔ+g(s)dμg(s)−Δ+g(r) 1+λΔ+g(r)⎞ ⎟ ⎠(5.9) =+exp g(λ;0,t)⎛ ⎜ ⎝ [0,t) 1 1+λΔ+g(s)dμg(s)⎞ ⎟ ⎠expg(λ;0,r)−1(1 + λΔ+g(r))−1χ[0,t)(r) −expg(λ;0,t)⎛ ⎜ ⎝ [0,r) 1 1+λΔ+g(s)dμg(s)⎞ ⎟ ⎠expg(λ;0,r)−1(1 + λΔ+g(r))−1χ[0,t)(r) −expg(λ;0,t)exp g(λ;0,r)−1(1 + λΔ+g(r))−2Δ+g(r)χ[0,t)(r). Remark 5.7. Observe that we can arrive to expressions (5.8)and (5.9)using an integration by parts argument in formula (5.7). Indeed, given two elements h1, h2∈ACg([0, T]; F)we have that h1h2∈ACg([0, T]; F)and (h1h2) g(t)=(h1) g(t)h2(t)+h1(t)(h2) g(t)+(h1) g(t)(h2)(t)Δ +g(t),g−a.e. t ∈[0,T]. Observe that we are explicitly excluding the points of Cgin the above formula. In particular, for t ∈[0, T], h1(t)h2(t)−h1(0) h2(0) =  [0,t) (h1) g(s)h2(s)dμg(s)+  [0,t) h1(s)(h2) g(s)dμg(s) + [0,t) (h1) g(s)(h2)(s)Δ +g(s)dμg(s). (5.10) Now we study two cases: •Case λ1=λ2. Let us consider h1(t)=(λ1−λ2)−1expg(λ1;0,t)exp g(λ2;0,t)−1, h2(t)=  [0,t) expg(λ1;0,r)−1 1+λ1Δ+g(r)f(r)dμg(r). We have that h1, h2∈ACg([0, T]; F), so  [0,t) expg(λ2;0,s)−1 1+λ2Δ+g(s)expg(λ1;0,s)⎛ ⎜ ⎝ [0,s) expg(λ1;0,r)−1 1+λ1Δ+g(r)f(r)dμg(r)⎞ ⎟ ⎠dμg(s) =(λ1−λ2)−1expg(λ1;0,t)exp g(λ2;0,t)−1 [0,t) expg(λ1;0,r)−1 1+λ1Δ+g(r)f(r)dμg(r) − [0,t) (λ1−λ2)−1expg(λ1;0,s)exp g(λ2;0,s)−1expg(λ1;0,s)−1 1+λ1Δ+g(s)f(s)dμg(s) − [0,t) expg(λ2;0,s)−1 1+λ2Δ+g(s)expg(λ1;0,s)expg(λ1;0,s)−1 1+λ1Δ+g(s)f(s)Δ +g(s)dμg(s), (5.11) F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 39 and we recover the expression of Green’s function in (5.8). Observe that by substituting (5.11)in (5.4) we obtain v2(t)=v0−λ2x0 λ1−λ2expg(λ1;0,t)−v0−λ1x0 λ1−λ2expg(λ2;0,t) +(λ1−λ2)−1expg(λ1;0,t) [0,t) expg(λ1;0,s)−1 1+λ1Δ+g(s)f(s)dμg −(λ1−λ2)−1expg(λ2;0,t) [0,t) expg(λ2;0,s)−1 1+λ2Δ+g(s)f(s)dμg. (5.12) We have that vh(t)=v0−λ2x0 λ1−λ2expg(λ1;0,t)−v0−λ1x0 λ1−λ2expg(λ2;0,t)∈BC∞ g([0,T]; F) is the solution of the homogeneous equation (5.1)and vp(t)=+(λ1−λ2)−1expg(λ1;0,t) [0,t) expg(λ1;0,s)−1 1+λ1Δ+g(s)f(s)dμg −(λ1−λ2)−1expg(λ2;0,t) [0,t) expg(λ2;0,s)−1 1+λ2Δ+g(s)f(s)dμg is a particular solution in the space BCn+2 g([0, T], F)of the non homogeneous equation (5.3)that satisfies vp(0) =(vp) g(0) =0. •Case λ1=λ2. We have that expression (5.4) reduces to v2(t)=x0expg(λ;0,t)+(v0−λx0)exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s) +exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)⎛ ⎜ ⎝ [0,s) expg(λ;0,r)−1 1+λΔ+g(r)f(r)dμg(r)⎞ ⎟ ⎠dμg(s). (5.13) Define h1(t)=  [0,t) 1 1+λΔ+g(s)dμg(s), h2(t)=  [0,t) expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s). We have that h1, h2∈ACg([0, T]; F). Hence, by formula (5.10),  [0,t) 1 1+λΔ+g(s)⎛ ⎜ ⎝ [0,s) expg(λ;0,r)−1 1+λΔ+g(r)f(r)dμg(r)⎞ ⎟ ⎠dμg(s) 40 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 = [0,t) 1 1+λΔ+g(s)dμg(s) [0,t) expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s) (5.14) − [0,t) ⎛ ⎜ ⎝ [0,s) 1 1+λΔ+g(r)dμg(r)⎞ ⎟ ⎠expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s) − [0,t) expg(λ;0,s)−1 (1 + λΔ+g(s))2f(s)Δ +g(s)dμg(s). Substituting expression (5.14)in(5.13)we obtain v2(t)=x0expg(λ;0,t)+(v0−λx0)exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s) +exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s) [0,t) expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s) −expg(λ;0,t) [0,t) ⎛ ⎜ ⎝ [0,s) 1 1+λΔ+g(r)dμg(r)⎞ ⎟ ⎠expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s) −expg(λ;0,t) [0,t) expg(λ;0,s)−1 (1 + λΔ+g(s))2f(s)Δ +g(s)dμg(s). Observe that vh(t)=x0expg(λ;0,t)+(v0−λx0)exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s)∈BC∞ g([0,T]; F) is the solution of the homogeneous equation (5.3)and vp(t)=exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg(s) [0,t) expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s) −expg(λ;0,t) [0,t) ⎛ ⎜ ⎝ [0,s) 1 1+λΔ+g(r)dμg(r)⎞ ⎟ ⎠expg(λ;0,s)−1 1+λΔ+g(s)f(s)dμg(s) −expg(λ;0,t) [0,t) expg(λ;0,s)−1 (1 + λΔ+g(s))2f(s)Δ +g(s)dμg(s) is a particular solution of the non homogeneous equation (5.3)in the space BCn+2 g([0, T], F)that satisfies vp(0) =(vp) g(0) =0. 6. The Stieltjes harmonic oscillator In this section we present an application related to the real solution of the Stieltjes harmonic oscillator (g-harmonic oscillator). Let g:R →Rbe a derivator such that 0 /∈N− gand T/∈N+ g∪Dg∪Cgand denote by gCits continuous part. We consider the following equation: F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 41 ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ v g(t)+2ζω 0v g(t)+ω2 0v(t)=0,g−a.e. t ∈[0,T), v(0) = x0, v g(0) = v0, (6.1) where x0and v0are real numbers and: •ω0is the undamped angular frequency of the oscillator, ω0=0k m, where m >0is the mass of the oscillator and k>0is a measure of the stiffness of the spring; •ζis the damping ratio, ζ=c 2√mk , with c >0, the viscous damping coefficient (resistance of the medium). If ζ>1we have an overdamped oscillator, if ζ=1the oscillator is critically damped and, if ζ<1, the oscillator is underdamped. Observe that the solutions of the characteristic equation are given by: λ=1 2−2ζω 0±14ζ2ω2 0−4ω2 0=−ζω 0±ω0/ζ2−1. Assume that 1 +λ Δ+g(t) =0for all t ∈[0, T) ∩Dgand for all λsolution of the characteristic equation. We have the following real solution of the g-harmonic oscillator in terms of the damping ratio: •If ζ>1, we have two real solutions of the characteristic equation, λ1=−ζω 0−ω0/ζ2−1and λ2=−ζω 0+ω0/ζ2−1. Thus, the solution of (6.1)is given by v(t)=v0−λ2x0 λ1−λ2expg(λ1;0,t)−v0−λ1x0 λ1−λ2expg(λ2;0,t), =v0−λ2x0 λ1−λ2exp(λ1gC(t)) s∈[0,t)∩Dg!1+λ1Δ+g(s)" −v0−λ1x0 λ1−λ2expg(λ2gC(t)) s∈[0,t)∩Dg!1+λ2Δ+g(s)". •If ζ=1, we have one real solution of the characteristic equation, λ =−ζω 0. Thus, the solution of (6.1) is given by v(t)=x0expg(λ;0,t)+(v0−λx 0)exp g(λ;0,t) [0,t) 1 1+λΔ+g(s)dμg, =exp(λg C(t)) s∈[0,t)∩Dg!1+λΔ+g(s)"·⎡ ⎣x0+(v0−λx 0)⎛ ⎝gC(t)+  s∈[0,t)∩Dg Δ+g(s) 1+λΔ+g(s)⎞ ⎠⎤ ⎦. •If ζ<1, we have a pair of conjugate complex solutions, λ1=−ζω 0+i ω0/1−ζ2and λ2=−ζω 0− i ω0/1−ζ2. If we denote by a =−ζω 0and b =ω0/1−ζ2, we have that the solution is given by 48 F.J. Fernández et al. / J. Math. Anal. Appl. 511 (2022) 126010 Fig. 6.6. Comparison between the exact solution and the numerical approximation (vertical lines have to be understood as jumps and not as a multivalued function). Finally, in Fig. 6.6, we can see the comparison between the exact solution and the numerical approximation for h =1.e −1and h =1.e −2. Acknowledgments The authors would like to thank the anonymous referee for their comments, suggestions and corrections, as they have greatly contributed to improve the quality of the manuscript. The authors were partially supported by Xunta de Galicia, project ED431C 2019/02, and by the Agencia Estatal de Investigación (AEI) of Spain under grant MTM2016-75140-P, co-financed by the European Community fund FEDER. Ignacio Márquez Albés was partially supported by Xunta de Galicia under grant ED481B-2021-074. References [1] R.G. Bartle, The Elements of Integration and Lebesgue Measure, Wiley Classics Library, John Wiley & Sons, Inc., New York, 1995. [2] F.J. Fernández, F.A.F. Tojo, Numerical solution of Stieltjes differential equations, Mathematics 8(9) (2020). [3] M. Frigon, R. 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