Existence of extremal solutions for discontinuous Stieltjes differential equations

López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 https://doi.org/10.1186/s13660-020-02316-w RESEARCH Open Access Existence of extremal solutions for discontinuous Stieltjes differential equations Rodrigo López Pouso1and Ignacio Márquez Albés1* *Correspondence: ignacio[email protected] 1Departamento de Estatística, Análise Matemática e Optimización, Faculty of Mathematics, Universidade de Santiago de Compostela, Santiago de Compostela, Spain Abstract Stieltjes differential equations, which contain equations with impulses and equations on time scales as particular cases, simply consist in replacing usual derivatives by derivatives with respect to a nondecreasing function. In this paper we prove new results for the existence of extremal solutions for discontinuous Stieltjes differential equations. In particular, we prove that the pointwise infimum of upper solutions of a Stieltjes differential equation is the minimal solution under certain hypotheses. These results can be adapted to the context of both difference equations and impulsive differential equations. MSC: 34A36; 34K05 Keywords: Upper solution; Ordinary differential equations; Impulsive differential equations; Dynamic equations; Time scales 1 Introduction Let us consider the initial value problem xg(t)=ft,x(t),t∈I=[0,1],x(0)=0, (1.1) wherexg(t)denotestheStieltjesderivativeoftheunknownwithrespecttoanondecreasing and left-continuous function g:R−→Rasintroducedin[7]. Theaimofthispaperistoreplicatetheresultsobtainedin[4]forODEsinthemoregeneral context of Stieltjes differential equations. That is, to solve as satisfactorily as possible the following problem: to find the weakest sufficient conditions over the right-hand side f∈L1 g(I)sothattheminimalsolutionsolutionistheleastuppersolutionandthemaximal oneisthegreatestlowersolution.In[5]wecanfindsomeresultsregardingtheexistenceof extremal solutions of this type of equation in the presence of a pair of well-ordered lower anduppersolutions.In[6]theauthorsfollowedthislineofresearchworkinginthecontext ofmeasuredifferentialequations,andthenadaptedtheresultsobtainedtotheframework ofStieltjesdifferentialequations.Therefore,thispapercomplements,inasense,thestudy initiated in these papers. We have organised the paper as follows. In Sect. 2, we present the basic definitions and results required for this paper. In Sect. 3, we are looking for some necessary conditions ©The Author(s) 2020. 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López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 2 of 21 for fthat assure that the infimum of all upper solutions of (1.1) is a solution. Then, in Sect.4weobtainanewexistenceresultfromthoseprovedintheprevioussection.Finally, in Sect. 5, we present a result that guarantees the existence of the extremal solutions for Stieltjesdifferentialequations. We thenadapttheresults obtained todifferenceequations and impulsive differential equations. As a final comment, note that in this paper we work on the interval I= [0,1] and the initial condition x(0) = 0 for simplicity, but the results are true for any other interval ˜ I= [a,b] and any other initial condition x(a)=xa,xa∈R, by doing the obvious changes. 2Preliminaries Let g:R→Rbe a nondecreasingandleft-continuous function. Inordertorecall the definition oftheStieltjes derivative ofafunction with respect to g(or simply the g-derivative of a function) as presented in [7], we need to define the sets Cg=s∈R:gis constant on (s–ε,s+ε)forsomeε>0, and Dg={t∈R:g(t)>0},whereg(t)=g(t+)–g(t)andg(t+) denotes the limit of gat t from the right. Now the g-derivative of a function x:I−→Rat a point t∈I\Cgis xg(t)=⎧ ⎨ ⎩ lims→tx(s)–x(t) g(s)–g(t)if t/∈Dg, lims→t+x(t)–x(t) g(s+)–g(t)if t∈Dgand t<1, provided that the corresponding limit exists. Note that, for a point t∈Dg,xg(t)existsif and only if x(t+)exists. Noticethatwedonotdefineg-derivativesatpointst∈Cg,noritisnecessarybecauseCg is a null-measure set for μg(the Lebesgue–Stieltjes measure induced by g), see [7,Proposition 2.5]. Therefore, thedifferential equation in (1.1) is not really defined for t∈I∩Cg. The following result, the fundamental theorem of calculus for the Lebesgue–Stieltjes integral [7, Theorem 5.4], establishes the relation between Stieltjes derivatives and the Lebesgue–Stieltjes integral for a particularly interesting set of functions. Theorem2.1(FundamentalTheoremofCalculusfortheLebesgue–Stieltjesintegral) Let a,b∈R,a<b,and F :[a,b]−→R.The following conditions are equivalent. (1) The function Fis absolutely continuous on [a,b]with respect to g(also expressed as g-absolutely continuous on [a,b]or F∈ACg([a,b]))according to the following definition:to each ε>0,there is some δ>0such that,for any family {(an,bn)}m n=1 of pairwise disjoint open subintervals of [a,b],the inequality m  n=1g(bn)–g(an)<δ implies m  n=1F(bn)–F(an)<ε. López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 3 of 21 (2) The function Ffulfills the following properties: (a) There exists F g(t)for g-almost all t∈[a,b)(i.e., for all texcept on a set of μg measure zero); (b) F g∈L1 g([a,b)),the set of Lebesgue–Stieltjes integrable functions with respect to μg;and (c) For each t∈[a,b],we have F(t)=F(a)+[a,t)F g(s)dμg. (2.1) In this paper we consider integration in the Lebesgue–Stieltjes sense mainly, and we shall call “g-measurable” any function (or set) which is measurable with respect to the Lebesgue–Stieltjesσ-algebrageneratedbyg.Moreover,integralssuchasthatin(2.1)shall be denoted also as [a,t)F g(s)dg(s). For properties of g-absolutely continuous functions, we refer the readers to [2,7]. One of themain properties is that every g-absolutelycontinuousfunction is also g-continuous in the sense of the following definition. Definition 2.2 ([7, Definition 3.1]) A function F:[a,b]⊂R→Ris g-continuous at s ∈ [a,b]if,foreveryε>0,thereexistsδ>0suchthat t∈A,g(t)–g(s)<δ⇒ f(t)–f(s)<ε. We say that fis g-continuous on A if it is g-continuous at every point t0∈[a,b]. We shall denote by BCg([a,b]) the set of all g-continuous functions that are also bounded. It is shown in [7, Definition 5.5] that ACg([a,b]) is a subset of this set. Hence, the next result gives, indirectly, some properties of g-absolutely continuous functions. Proposition 2.3 ([7, Definition 3.2]) If F :[a,b]→Ris a g-continuous function on [a,b], then (1) Fis continuous from the left at every s∈(a,b]; (2) if gis continuous at s∈[a,b),then so is F; (3) if gis constant on some [c,d]⊂[a,b],then so is F. Further properties about the behaviour of g-absolutely continuous functions can be found in another result from the same paper. Proposition2.4([7,Proposition 5.2]) If F is g-absolutely continuous on [a,b],then it has bounded variation. For the purpose of this paper, we shall also recall the following result. López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 4 of 21 Proposition 2.5 ([2, Proposition 5.6]) Let S⊂ACg(I)be such that {F(t0):F∈S}is bounded.Assume that there exists h ∈L1 g([t0,t1)) such that F g(t)≤h(t)for g-almost all t ∈[t0,t1), and for all F ∈S. Then Sis relatively compact in BCg(I). As a final note for this section, we establish the definition of a solution of (1.1), as well as other relevant definitions such as lower and upper solutions. Definition2.6 Afunctionx:I−→Ris a solution of (1.1)ifx∈ACg(I), x(0)=0 and xg(t)=ft,x(t),g-a.a. t∈I. We say that xmin istheminimalsolutionif xmin isasolutionandxmin ≤xon Iforanyother solutionx.Themaximalsolutionisdefinedinananalogouswaywiththeobviouschanges. When both the minimal and the maximal solutions exist, we call them the extremal solutions. Definition2.7 Afunctionu:I−→Ris an upper solution of (1.1)ifu∈ACg(I), u(0)≥0 and ug(t)≥ft,u(t),g-a.a. t∈I. Afunctionl:I−→Ris a lower solution of (1.1)ifl∈ACg(I), l(0)≤0and l g(t)≤ft,l(t),g-a.a. t∈I. 3 Properties of the infimum of upper solutions Consider problem (1.1). We will assume that fsatisfies the following hypothesis: (H1) There exists M∈L1 g(I)such that |f(t,x)|≤M(t)for g-a.a. t∈I,allx∈R. Remark 3.1 If fsatisfies a local boundedness condition, such as (H1∗)ForeachR>0,thereexistsMR∈L1 g(I)such that |f(t,x)|≤MR(t)for g-a.a. t∈I, all x∈R,|x|≤R, we can study the existence of local solutions. To do so, we fix R>0,wedefine ˜ f(t,x)=ft,max–R,min{x,R}, and westudy (1.1)withfreplaced by ˜ f,whichsatisfies(H1).Observe thatsolutionsofthe new problem are local solutions of the former one. Inthefollowing,weshalldenotethesetofadmissibleuppersolutionsfor(1.1)asfollows: U=u∈ACg(I):u(0)≥0, ug(t)≥ft,u(t)g-a.e. on I,ug(t)≤M(t)g-a.e. on I, López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 5 of 21 and define uinf(t):=inf{u(t):u∈U},t∈I.Notethatuinf(0) =0 as the function ugiven by u(t)=[0,t)M(s)dg(s)belongstoUand, trivially, u(0)=0. Sincetheaimofthispaperistofindoutsomeconditionsguaranteeingthatthefunction uinf is the minimal solution of the problem, we first need to obtain conditions that assure that uinf ∈ACg(I)and|(uinf)g|≤M. In order to do so, we need the following lemma, in which the first condition for our goal, due to Antunes Monteiro and Slavík (see condition (C4) in [1]), will appear. Lemma3.2 Consider β1,β2,...,βn∈U.If f verifies (H1) and (H2) For all t∈I∩Dg,the map u∈R→u+f(t,u)(g(t+)–g(t)) is nondecreasing, then the function βmin(t)=min{β1(t),β2(t),...,βn(t)},t∈I,is an element of U. Proof To prove this result, it suffices to show that given β1,β2∈U,βmin(t)=min{β1(t), β2(t)},t∈I,belongstoU. First of all, note that βmin ∈ACg(I) since we can write it as the difference of two g-absolutely continuous functions: βmin(t)=β1(t)–β2(t) 2–|β1(t)–β2(t)| 2. Moreover, βmin(0) ≥0 trivially, and so, all that is left to prove is that for g-a.a. t∈I (βmin)g(t)≥f(t,βmin(t)) and |(βmin)g(t)|≤M(t). Let E={t∈I:∃(β1)g(t),(β2)g(t),(βmin)g(t)},andlett0∈E.Notethatt0/∈Cgsince there exist g-derivatives atthatpoint. We distinguish two possible cases: either β1≥β2ona set S⊂[0,1] such that t0∈[S∩(t0,1)]or β1<β2on (t0,t0+δ)forsomeδ> 0. Assume the first one holds. If β1(t0)≥β2(t0), then (βmin)g(t0)=lim t→t+ 0 βmin(t)–βmin(t0) g(t)–g(t0)=lim t→t+ 0,t∈S∩(t0,1) βmin(t)–βmin(t0) g(t)–g(t0) =lim t→t+ 0 β2(t)–β2(t0) g(t)–g(t0)=(β2)g(t0)≥ft0,β2(t0)=ft0,βmin(t0). Otherwise,β2(t0)>β1(t0),andsot0∈Dg.Notethatβ1(t+ 0)=limt→t+ 0,t∈S∩(t0,1) β1(t)≥β2(t+ 0). Hence, using hypothesis (H2), (βmin)g(t0)=βmin(t+ 0)–βmin(t0) g(t+ 0)–g(t0)=β2(t+ 0)–β1(t0) g(t+ 0)–g(t0)=β2(t0)+g(t0)(β2)g–β1(t0) g(t0) ≥β2(t0)+g(t0)f(t0,β2(t0))–β1(t0) g(t0)≥β1(t0)+g(t0)f(t0,β1(t0))–β1(t0) g(t0) =ft0,β1(t0)=ft0,βmin(t0). Thus (βmin)g(t)≥f(t,βmin(t)) for g-a.a. t∈I.Moreover,|(βmin)g)|≤M. Indeed, if β1(t0)≥ β2(t0), then it is clear. If β1(t0)<β2(t0), we have (βmin)g(t0)≥f(t0,βmin(t0))≥–M(t0)and (βmin)g(t0)=β2(t+ 0)–β1(t0) g(t0)≤β1(t+ 0)–β1(t0) g(t0)=(β1)g(t0)≤M(t0). The case β1<β2on (t0,t0+δ)forsomeδ>0issimilar.  López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 6 of 21 Using the previous lemma, one can show that uinf verifies some of the required properties. Lemma3.3 If f satisfies hypotheses (H1)–(H2), then uinf ∈ACg(I)and (uinf)g(t)≤M(t), g-a.a.t ∈I. Proof Let s,t∈Ibe such that s<t. By definition of uinf,givenε>0,thereexistu1,u2∈U such that 0≤u1(t)–uinf(t)<ε 2,0≤u2(s)–uinf(s)<ε 2. Define u(z)=min{u1(z),u2(z)}for all z∈I. By Lemma 3.2,u∈U.Moreover,0≤u(t)– uinf(t)<ε/2, 0≤u(s)–uinf(s)<ε/2. Hence, uinf(t)–uinf(s)≤uinf(t)–u(t)+u(t)–u(s)+u(s)–uinf(s) <ε 2+[s,t)Mdg+ε 2=[s,t)Mdg+ε. Since ε> 0 is arbitrary, we have that |uinf(t)–uinf(s)|≤[s,t)Mdg.Now,usingthatM∈ L1 g(I), it is easy to check using standard arguments that uinf ∈ACg(I). Moreover, for each s∈Ithat (uinf)g(s)exists,defineΦs(t)=[s,t)Mdg,t∈I,t>s.NotethatΦsis g-absolutely continuous, so, by the fundamental theorem of calculus [7, Theorem 2.4], we have that (uinf)g(t)=lim t→s+|uinf(t)–uinf(s)| g(t)–g(s)≤lim t→s+ Φs(t)–Φs(s) g(t)–g(s)=(Φs)g(s)=M(s). Now, since uinf ∈ACg(I), we have that (uinf)g(s) exists for g-a.a. s∈I,andtheresultfollows.  Furthermore, one can show that uinf can be approximated by a sequence of U. Lemma 3.4 If f verifies (H1)–(H2), then there exists a nonincreasing sequence {un}⊂U that converges uniformly on I to uinf. Proof For each t∈[0,1],defineu0(t)=[0,t)M(τ)dg(τ)∈U.Assumethatu1,u2,...,un–1 ∈ Uhave been defined. For every i∈{0,1,...,n–1},chooseyi∈Usatisfying the following inequalities: uinfi n≤yii n≤uinfi n+1 n. Define un=min{un–1,y0,...,yn–1}.Thenun∈Uby Lemma 3.2; moreover, the sequence {un}∞ n=1 is nonincreasing. Furthermore, {un}∞ n=1 verifies Proposition 2.5 since, for each n∈N,|(un)g(t)|≤M(t)forg-a.a. t∈Iand {un(0) : n∈N}={0}as 0 ≤un(0) ≤u1(0) = 0. Hence, {un}isarelativelycompactset,andthereforethereexistsasubsequence{unk}that converges uniformly in BCg(I)toalimit,sayv.Since{un}is a monotone sequence, it also converges uniformly to v. Therefore, it is enough to show that v=uinf. López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 7 of 21 Sinceun≥uinf foralln∈N,wehavethatv≥uinf.Assumethatv=uinf.Thenthereexists t0∈Isuchthat v(t0)>uinf(t0).Bothfunctionsbelongto BCg(I),soProposition2.3ensures that they are left-continuous. Hence, there exist c>0andδ>0suchthatuinf(t)<v(t)–c for all t∈(t0–δ,t0]. Consider n∈Nsuch that 1/n<min{c,δ}.Thenuinf(t)<v(t)–c≤ un(t)–c≤un(t)–1/nforallt∈(t0–δ,t0],andsouinf(t)+1/n<un(t)forallt∈(t0–1/n,t0]. Now, for some i=0,1,...,n,i/n∈(t0–1/n,t0], and so uinf(i/n)+1/n<un(i/n), which is a contradiction. Therefore, v=uinf. In the last two theorems of this section, we study the behaviour of fover the graph of uinf, from which one can obtain conditions over fso that uinf is a solution. Theorem3.5 Consider (1.1)under hypotheses (H1)–(H2). Then,for g-a.a.t∈I, (uinf)g(t)≥ft,uinf(t)χI1(t)+ liminf y→(uinf(t))+f(t,y)χI2(t), where I1={t∈I:uinf(t)=u(t),ug(t)≥f(t,u(t)) for some u∈U}∪Dgand I2=I\I1. Proof First, note that hypotheses (H1)–(H2) guarantee that uinf ∈ACg(I), and therefore (uinf)gexists g-almost everywhere. Let s∈I1\Dgbe such that (uinf)g(s) exists, and let u∈Ube the corresponding function to the definition of I1.Then(uinf)g(s)=ug(s)≥f(t,u(s)) = f(t,uinf(s)). On the other hand, for s∈Dg, consider a sequence {un}∞ n=1 ⊂Uas in Lemma 3.4.Weknowthat,foralln,it holds that uns+≥un(s)+g(s)fs,un(s)≥uinf(s)+g(s)fs,uinf(s). Hence,since{un}convergesuniformlytouinf,itfollowsfromtheMoore–Osgoodtheorem [3, Chapter VII, Theorem 2] that uinfs+≥uinf(s)+g(s)fs,uinf(s), or equivalently, (uinf)g(s)≥f(s,uinf(s)). Finally, we study (uinf)gon I2. To do so, we consider again a sequence {un}∞ n=1 ⊂Uas in Lemma 3.4.Since|(un)g|is uniformly L1 g-bounded on I,wehavethatliminfn→∞(un)g∈ L1 g(I). Moreover, by Fatou’s lemma, for ˜ t<t, uinf(t)–uinf(˜ t)=liminf n→∞ un(t)–un(˜ t)=liminf n→∞ [˜ t,t)(un)gdg ≥[˜ t,t) liminf n→∞ (un)gdg. Now, since uinf ∈ACg(I), we have that uinf(t)–uinf(˜ t)=[˜ t,t)(uinf)gdg.Hence, (uinf)g(t)≥liminf n→∞ (un)g(t)≥liminf n→∞ ft,un(t)g-a.a. t∈I. Now, if s∈I2and uinf(s)=un(s)forsomen, the definition of I2implies that s/∈Dgand either (un)g(s)doesnotexistor(un)g(s)<f(s,un(s)). The set  n∈Nt∈I\Dg:(un)g(s)∪t∈I\Dg:(un)g(s)<fs,un(s) López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 8 of 21 is a null-measure set with respect to the g-measure.Hence,forg-a.a. t∈I2,wehavethat uinf(t)<un(t)foralln∈N,andso,since{un(t)}isoneoftheinfinitelymanysequencesthat converge to uinf(t)+,wehavethat (uinf)g(t)≥liminf n→∞ ft,un(t)≥liminf y→(uinf(t))+f(t,y), which concludes the proof.  Remark 3.6 It follows from Theorem 3.5 that if the following condition is satisfied liminf y→(uinf(t))+f(t,y)≥ft,uinf(t),forg-a.a. t∈I, then (uinf)g≥f(t,uinf(t)), i.e., uinf is an upper solution. Note,however,thatforallt∈I∩Dg,uinf isa“solution”,i.e.,(uinf(t))g=f(t,uinf(t))aslong ashypotheses(H1)–(H2)aresatisfied.Indeed,wealreadyknowthat(uinf)g(t)≥f(t,uinf(t)) for t∈I∩Dg. Reasoning by contradiction, assume that there exists t0∈I∩Dsuch that (uinf)g(t0)>f(t0,uinf(t0)), or equivalently, uinf(t+ 0)>uinf(t0)+g(t0)f(t0,uinf(t0)) = a.Then one can find z0∈(a,uinf(t+ 0)). Define u(t)=⎧ ⎨ ⎩ uinf(t)ift∈[0,t0], z0+(t0,t)M(τ)dg(τ)ift∈(t0,1]. First, note that ug(t0)=u(t+ 0)–u(t0) g(t0)=z0–uinf(t0) g(t0)>a–uinf(t0) g(t0)=ft0,uinf(t0)=ft0,u(t0). Moreover, |ug|≤Mtrivially and u∈ACg(I)asitisdefinedasapiecewisefunctionof ACg(I)functions.Hence,u∈U, which is a contradiction, as u(t+ 0)=z0<uinf(t+ 0). Therefore,inordertodeterminetheconditionsguaranteeingthatuinf isasolution,there is no need to see what happens at points of Dgas long as (H1)–(H2) hold. We now prove the following lemma that we will need in order to obtain a necessary condition for uinf being an upper solution. Lemma3.7 Let M : [0,1]→[0,∞]be a g-integrable function.If F ⊂[0,1] is a set of positive g-measures,then there exists F1⊂Fsuchthat,for all s∈F1, lim t→s+ g(t)–g(s) μg([s,t)∩F)=1, lim t→s+ 1 μg([s,t)∩F)[s,t)\FM(τ)dg(τ)=0. Proof First, let G:I=[0,1]→Rbe the map given by G(0)=0, G(t)=[0,t)χF(s)dg(s), ∀t∈(0,1], whereχFdenotesthecharacteristicfunctionofthesetF.ClearlyχF∈L1 g((0,1])andtherefore it is trivial that G∈ACg(I). Hence, there exists a set F0⊂Fsuch that μg(F\F0)=0 López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 9 of 21 and there exists Gg(s) for all s∈F0.Moreover,Gg(s)=χF(s)=1 for all s∈F0.Thus, 1=Gg(s)=lim t→s+ G(t)–G(s) g(t)–g(s)=lim t→s+[0,t)χF(τ)dg(τ)–[0,s)χF(τ)dg(τ) g(t)–g(s) =lim t→s+[s,t)∩Fdg(τ) g(t)–g(s)=lim t→s+ μg([s,t)∩F) g(t)–g(s). Consider now the map H:I→Rdefined as H(0)=0, H(t)=[0,t)M(s)·χI\Fdg(s), ∀t∈(0,1]. Once again, since M0=M·χI\F∈L1 g((0,1]), it follows that H∈ACg(I), and therefore there exists F1⊂F0such that μg(F0\F1)=0andH g(s) exists for all s∈F1.Moreover, H g(s)=M(s)·χI\F(s)=0 for all s∈F1.Hence, 0=H g(s)=lim t→s+ H(t)–H(s) g(t)–g(s)=lim t→s+[0,t)M0(τ)dg(τ)–[0,s)M0(τ)dg(τ) g(t)–g(s). Now, since s∈F1⊂F0,wehavethat 0=lim t→s+[s,t)M0(τ)dg(τ) g(t)–g(s)=lim t→s+[s,t)M0(τ)dg(τ) g(t)–g(s)·lim t→s+ g(t)–g(s) μg([s,t)∩F) =lim t→s+[s,t)M0(τ)dg(τ) g(t)–g(s)·g(t)–g(s) μg([s,t)∩F)=lim t→s+[s,t)M0(τ)dg(τ) μg([s,t)∩F), and so, for all s∈F1,wehave lim t→s+ g(t)–g(s) μg([s,t)∩F)=1, lim t→s+ 1 μg([s,t)∩F)[s,t)\FM(τ)dg(τ)=0.  We can now state and prove the following necessary condition for uinf being an upper solution. Theorem 3.8 Consider problem (1.1)under hypotheses (H1)–(H2). Assume (uinf)g(t)≥ f(t,uinf(t)) for g-a.a.t∈I.Then: (a) The set J={t∈I\Dg:(uinf)g(t)>limsupy→(uinf(t))–f(t,y)}is a countable union of sets which contain no positive measure set.Specifically,J=n,m∈NJn,m,where Jn,m=t∈I\Dg:(uinf)g(t)–1 n>supf(t,y):uinf(t)– 1 m<y<uinf(t). (b) (uinf)g(t)≤limsupy→(uinf(t))–f(t,y)for g-a.a.t∈I\Dgprovided that,for all n,m∈N, the set Jn,mis g-measurable. Proof For each t∈J,thereexistsn∈Nsuch that (uinf)g(t)–1 n>limsup y→(uinf(t))–f(t,y)=inf ε>0sup uinf(t)–ε<y<uinf(t)f(t,y). López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 16 of 21 Note that the set Sxis nonempty. Indeed, since {xn}converges uniformly on Ito x,there exists N∈Nsuch that x(t)–ε 3<xN(t)<x(t)+ε 3,∀t∈I. Define s(t)=xN(t)–qfor some q∈(ε/3,2ε/3)∩Q.Itiseasytoseethats∈S. Lemma4.4 Letx∈ACg(I).For all t0∈(0,1),allε>0,ally∈(x(t0)–ε,x(t0))andall δ>0, there exists s ∈Sxsuch that y –δ<s(t0)<y.Analogously,for all t0∈(0,1), all ε>0,all y∈(x(t0)–ε,x(t0)) and all δ>0,there exists s∈Sxsuch that y<s(t0)<y+δ. Proof We shall prove the first part of the statement, as the second part is analogous. Fix t0∈(0,1), ε>0,y∈(x(t0)–ε,x(t0)) and δ>0.Take˜ δ∈(0,δ]suchthatx(t0)–ε<y–˜ δ. Since {xn}→xuniformly on Iand y∈(x(t0)–ε,x(t0)), we can find j,N∈Nbig enough so that x(t0)–j–1 jε<y–˜ δ<y<x(t0)–ε jand x(t)– ε 2j<xN(t)<x(t)+ ε 2j,∀t∈I. The function s(t)=xN(t)–xN(t0)+qfor some q∈(y–˜ δ,y)∩Qverifies the statement of the lemma. Indeed, first s∈Sxsince conditions 2 and 3 are trivially fulfilled and s(t)=xN(t)–xN(t0)+q<x(t)+ ε 2j–x(t0)+ ε 2j+y=x(t)–x(t0)+ε j+y <x(t)–x(t0)+ε j+x(t0)–ε j=x(t); s(t)=xN(t)–xN(t0)+q>x(t)– ε 2j–x(t0)– ε 2j+y–˜ δ=x(t)–x(t0)–ε j+y–˜ δ >x(t)–x(t0)–ε j+x(t0)–j–1 jε=x(t)–ε. Moreover, s(t0)=xN(t0)–xN(t0)+q=q∈(y–˜ δ,y)∩Q⊂(y–δ,y)∩Q. ThefollowingtheoremgivesasufficientconditionforJn,mbeingmeasurable,andtherefore, a useful result to turn uinf into a solution. Theorem4.5 Let N ⊂Ibeag-nullmeasureset,and let f :I×R→Rbe a function such that f(·,q)is g-measurable for each q∈Q.If,for all t ∈I\Nandallx∈R,we have maxliminf y→x–f(t,y),liminf y→x+f(t,y)≥f(t,x), then,for all x ∈ACg(I)and all ε>0,the mapping t∈I→supf(t,y):x(t)–ε<y<x(t) is g-measurable. López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 17 of 21 Proof Fix x∈ACg(I)andε>0.DefineSxas before. Then Sxis a countable family of functions. Indeed, since Dis countable, the set Dmis countable for each m∈N.Foreach ω=(ω1,...,ωm)∈Dm,letusdenotebySωa set of step functions of Sxthat are constant on the intervals whose extreme points are consecutive numbers of ω.Itiseasytoseethat each Sωis countable, and so Sxis countable as it can be written as Sx= m∈N ω∈DmSω. Hence, given that f(·,s(·)) is g-measurable on (0,1) for s∈S, it is enough to show that σ=σ0,where σ(t):= sup y∈(x(t)–ε,x(t))f(t,y), σ0(t):=sup s∈Sft,s(t). It is obvious that σ(t)≥σ0(t) on (0,1). To prove that σ0≥σon (0,1) \N,fixt0∈(0,1) \ Nand take a sequence {yn}n∈N⊂(x(t0)–ε,x(t0)) such that limn→∞ f(t0,yn)=σ(t0). Our assumptionsguaranteethat,foreachn,wehavethateitherliminfy→y– nf(t0,y)≥f(t0,yn)or liminfy→y+ nf(t0,y)≥f(t0,yn).Assumethatthefirstcaseholdsastheotheroneisanalogous. By definition, we have f(t0,yn)≤liminf y→y– nf(t0,y)= lim r→0+inf yn–r<z<ynf(t0,z). Thenthereexistsδ>0suchthatinfyn–δ<z<ynf(t0,z)≥f(t0,yn)–1/n.Hence,foreachn∈N, by Lemma 4.4,thereexistssn∈Sxsuch that yn–δ<sn(t0)<yn,andso ft0,sn(t0)≥inf yn–δ<z<ynf(t0,z)≥f(t0,yn)–1 n. Therefore, σ0:= sups∈Sf(t0,s(t0)) ≥f(t0,sn(t0)) ≥f(t0,yn)–1/n.Sincethisholdsforeach n∈N, σ0(t0)≥lim n→∞f(t0,yn)–1 n=lim n→∞f(t0,yn)=σ(t0), and so σ=σ0on (0,1)\N. 5 Existence of extremal solutions One can verify that analogous arguments work for the set of admissible lower solutions: L=l∈ACg(I):l(0)≤0, l g(t)≤ft,l(t)g-a.e. on I,l g≤Mg-a.e. on I, and lsup(t)=sup{l(t):l∈L}for all t∈I, obtaining analogous results. Hence, combining Theorems 4.2 and 4.5 and their analogues for lsup, one can obtain the following result. Theorem 5.1 Let f : [0,1] ×R→Rbe a mapping satisfying (H1)–(H3). If f(·,q)is gmeasurable for all q ∈Qand for g-a.a.t∈Iandallx∈R,it holds that minlimsup y→x–f(t,y),limsup y→x+f(t,y) López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 18 of 21 ≤f(t,x)≤maxliminf y→x–f(t,y),liminf y→x+f(t,y), then uinf is the maximal solution of (1.1)and lsup is the minimal one. Next we illustrate the applicability of Theorem 5.1 in a family of examples with nonmonotone discontinuities accumulating around the initial condition. Example5.2 Let g:R−→Rbe an arbitrary nondecreasing and left-continuous function and φ: [0,1] −→ Rbe a nondecreasing g-absolutely continuous function on [0,1] such that φ(0) =0 (take, for instance, φ(t)=λ(g(t)–g(0)), λ>0). We shall prove by means of Theorem 5.1 that (1.1) has the minimal and the maximal solutions for f(t,x)=⎧ ⎨ ⎩ 2+sin 1 x+φ(t) if t∈I\Dgand x>0, 2otherwise, where square brackets mean integer part. We remark that fis discontinuous and nonmonotone with respect to xon every neighbourhood of the initial condition. First,observethat f(t,x)∈(1,3)forall(t,x)∈I×R,whichimplies(H1);second,foreach fixed t∈I∩Dg,wehavethatf(t,·) is constantly equal to 2, which implies (H2). Now for (H3). Since φ(t)≥0 for all t∈I, we deduce that discontinuities can only occur at points (t,x)suchthatx=0or 1 x+φ(t)=nfor some n∈N. Therefore, we define γ0(t)= 0 for all t∈Iand, for each n=1,2,..., γn(t)=1 n–φ(t) for all t∈[0,I]. Notice that, for each fixed t∈[0,1], the mapping f(t,·) is continuous on R\∞ n=0{γn(t)} (it might also be continuous at some points x=γn(t)forsomen∈N, but this is not important). Therefore, for each fixed t∈[0,1], the mapping f(t,·)satisfies(4.1)onR\ ∞ n=0{γn(t)}. It remains to show that the curves γn,n=0,1,...,eithersatisfythedifferential equation, or they satisfy (4.2)and(4.3). Given n=0,1,...,γnis nonincreasing, the definition of g-derivative yields (γn)g(t)≤0<1≤minft,γn(t),liminf y→(γn(t))+f(t,y), limsup y→(γn(t))–f(t,y)for g-a.a. t∈[0,1]. Hence, we have that γn,n=0,1,...,satisfies(4.3). Moreover, (γn)g(t)≥limsup y→(γn(t))–f(t,y), n=0,1,... only occurs for t∈Awith μg(A)= 0. Therefore, (H3) is satisfied. López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 19 of 21 Finally, wecheck that f(·,q)isg-measurablefor all q∈Qand that, for g-a.a. t∈Iand all x∈R,wehave minlimsup y→x–f(t,y),limsup y→x+f(t,y)≤f(t,x)≤maxliminf y→x–f(t,y),liminf y→x+f(t,y). The last part follows from the fact that, for each fixed t∈[0,1], the mapping f(t,·)iscontinuousfromtheleftatevery x∈R.Indeed,thisistrivialift∈Dg;otherwise,observethat f(t,x)= 2 for all x≤0, f(t,x)=2forx>γ1(t)andforn=1,2,...,wehave f(t,x)=2+sin(n) for all x∈γn+1(t),γn(t),x>0. To deduce that f(·,q)isg-measurable for each q∈Q,justnotethatf(·,q)assumesa finite number of different values on corresponding Borel-measurable subsets of [0,1], hencef(·,q)isaBorel-measurablefunction,whichimpliesthatf(·,q)isg-measurablesince Lebesgue–Stieltjes measures are Borel measures. 5.1 Applications to difference equations Any difference equation of the form xn+1 –xn=f(n,xn), n=0,1,2,...,N,x0given (5.1) can be expressed as a g-differential equation xg(t)=ft,x(t)g-a.a. t∈I=[0,N],x(0)=x0, (5.2) where g(t)=min{n∈Z:n≥t}. Indeed, given a solution of (5.2)x∈ACg(I), and bearing in mind that Cg=I\Zand Dg=Z, for all n≥1, we have xg(n)=x(n+)–x(n) g(n+)–g(n)=x(n+1)–x(n), (5.3) and so xn+1 –xn=xg(n)=f(n,x(n)) = f(n,xn). Conversely, if x:I∩Z→Ris a solution of (5.1), we define ˜ x(t)=x(g(t)) for all t∈I. First of all, note that ˜ x∈ACg(I)since˜ xgexists g-a.e. Isince for n∈Dg=I\Cg, ˜ xg(n)=x(g(n+))–x(g(n)) g(n+)–g(n)=x(n+1)–x(n)∈R and, moreover, ˜ xg∈L1 g(I)as I˜ xgdg =N  i=0 ˜ xg(i)g(i)= N  i=0 x(i+1)–x(i)<∞. Finally, for t∈Ifixed, t∈[tk,tk+1)forsomek=0,1,2,...,N, ˜ x(0)+[0,t)˜ xgdg =x(0)+  {i∈I∩Z:i<t}˜ xg(i)g(i)=x(0)+  {i∈I∩Z:i<t}x(i+1)–x(i) López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 20 of 21 =x(k)=˜ x(t). Then it follows from (5.3)that˜ xis a solution of (5.2). Recalling Remark 3.6,fsatisfying conditions (H1)–(H2) was enough to guarantee that (uinf)g(t)=f(t,uinf(t))fort∈I∩Dg.Then,ifthereexistsM:I∩Z→Rsuchthat|f(n,x)|≤ M(n) for all n∈I∩Z,allx∈Rand for all n∈I∩Zthe mapping u∈R→u+f(n,u)is nondecreasing, we can assure that uinf is the maximal solution of (5.1). Note that this problem has a unique solution trivially; however, we have proved that such a solution is theinfimumofalltheuppersolutionsoftheproblem.Analogousargumentsworkforlsup. 5.2 Applications to impulsive differential equations It has been shown in [7] that an impulsive problem of the form ⎧ ⎨ ⎩ x(t)=f(t,x(t)) for a.a. t∈I\J, x(t+)=x(t)+It(x(t)) if t∈J,(5.4) where J={tk∈I:k∈N}, can be treated as a Stieltjes differential equation of the form xg(t)=F(t,x(t)), where g(t)=t+ {k∈N:tk<t} 2–k,F(t,x)=⎧ ⎨ ⎩ f(t,x)ift∈I\J, 2kItk(x)ift∈J,t=tk. Then, using Theorem 5.1, one can obtain a result assuring the existence of extremalsolutions for impulsive differential equations. Corollary5.3 Consider (5.4). Suppose that the following conditions are satisfied: 1. fis L1(I)-bounded and,for each k∈N,there exists αk∈Rsuch that |Itk|≤αk; 2. For all k∈N,the map u∈R→u+Itk(u)is nondecreasing; 3. Either limsup y→x–f(t,y)≤f(t,x)≤liminf y→x+f(t,y), a.a. t∈I,∀x∈R, (5.5) or there exists a family of functions γn:[an,bn]⊂I→R,n∈N,with the following properties: (i) γ nexistsfora.a.t∈Iand γ n∈L1(I); (ii) for all k∈N,γn(t+ k)exists and k∈N|γn(t+ k)–γn(tk)|<∞; (iii) for all t∈I, γn(t)=γn(0)+[0,t)γ n(s)ds+ tk∈[0,t) (γnt+ k–γn(tk); (iv) for a.a.t∈Iand all x∈R\{n∈N:an≤t≤bn}{γn(t)},inequality (5.5)holds,while for each n∈Nand a.a.t∈[an,bn],we have either (γn)(t)=f(t,γn(t)) or (γn)(t)≥ft,γn(t)whenever (γn)(t)≥liminf y→(γn(t))+f(t,y), (5.6) López Pouso and Márquez Albés Journal of Inequalities and Applications (2020) 2020:47 Page 21 of 21 (γn)(t)≤ft,γn(t)whenever (γn)(t)≤limsup y→(γn(t))–f(t,y). (5.7) 4. For all q∈Q,the map f(·,q)is Borel-measurable; 5. For almost all t∈I\Jand all x∈R, minlimsup y→x–f(t,y),limsup y→x+f(t,y)≤f(t,x)≤maxliminf y→x–f(t,y),liminf y→x+f(t,y) and for all k∈Nand all x∈R, minlimsup y→x–Itk(y),limsup y→x+Itk(y)≤Itk(x)≤maxliminf y→x–Itk(y),liminf y→x+Itk(y). Then uinf is the maximal solution of (5.4)and lsup is the minimal one. Acknowledgements The authors would like to thank the anonymous referees for their encouraging reports and comments which helped to improve this manuscript. Funding Rodrigo López Pouso was partially supported by Ministerio de Economía y Competitividad, Spain, and FEDER, Project MTM2016-75140-P and Xunta de Galicia under grant ED431C 2019/02. Ignacio Márquez Albés was supported by Ministerio de Economía y Competitividad, Spain, and FEDER, Project MTM2016-75140-P and Xunta de Galicia under grants ED481A-2017/095 and ED431C 2019/02. Availability of data and materials Not applicable. Competing interests The authors declare that they have no competing interests. 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