# Antiperiodic Boundary Value Problems for Impulsive Fractional Functional Differential Equations via Conformable Derivative

Authors: Jingfeng Wang; Chuanzhi Bai
Journal: Journal of Function Spaces
 (2018)

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## Abstract

In this paper, by using the lower and upper solution method and the monotone iterative technique, we investigate the existence of solutions to antiperiodic boundary value problems for impulsive fractional functional equations via a recent novel concept of conformable fractional derivative. An example is given to illustrate our theoretical results.


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## Body

## 1. Introduction

In the past decades, fractional differential equations play important roles in describing many phenomena and processes occurring in engineering and scientific disciplines; for instance, see [1–7]. Besides the research of fractional differential equations, the impulsive differential equation is found to be an effective tool to study some problems of medicine, engineering, biology, and physics [8–10]. In various fields, such as physics, engineering, and chemistry, many models come down to antiperiodic boundary value problems, so there have been many papers focused on the subject of fractional differential equations with impulsive antiperiodic boundary value conditions; one can refer to [11–14].Recently, Khalil et al. [15] introduced the conformable fractional derivative, which was a new well-behaved definition, depending just on the basic limit definition of the derivative. For details and applications of this concept, see [16–18]. Fractional differential equations via conformable fractional derivatives have recently received considerable attention; see [19–22] and the references therein. In [23], the authors studied the following periodic boundary value problem for impulsive conformable fractional integrodifferential equations:(1)tkDαxt=ft,xt,Fxt,Sxt,t∈J=0,T,t≠tk,Δxt=tk=Ikxtk,k=1,2,…,m,x0=xT,where aDα denotes the conformable fractional derivative of order 0<α≤1 starting from a∈{t0,…,tm}, 0=t0<t1<⋯<tm<tm+1=T, f∈C(J×R3,R):(2)Fxt=∫0tlt,sxsds,Sxt=∫0Tht,sxsds,l∈CD,R+,D=t,s∈J2:t≥s,h∈CJ2,R+,Ik∈CR,R,Δxtk=xtk+-xtk-.They obtained the existence of solutions for (1) by using the monotone iterative method.Motivated by the above-mentioned work and a recent paper [24], in this article, we discuss the existence of solutions to antiperiodic boundary value problems for impulsive conformable fractional functional differential equations:(3)tkDαxt=ft,xt,xwt,t∈J=0,T,t≠tk,Δxt=tk=Ikxtk,k=1,2,…,m,x0=-xT,xt=x0,t∈-r,0,where f∈C(J×R2,R), aDα denotes the conformable fractional derivative of order 0<α≤1 starting from a∈{t0,…,tm}, 0=t0<t1<⋯<tm<tm+1=T, w∈C(J,J+), t-r≤w(t), t∈J and tk<w(t)≤t, t∈(tk,tk+1], J+=[-r,T], r>0, Ik∈C(R,R), Δx(tk)=x(tk+)-x(tk-).To our knowledge, the work on the antiperiodic boundary value problems for impulsive fractional conformable functional differential equations is not to be initiated yet. By applying the method of lower and upper solutions coupled with the monotone iterative technique, we obtain the existence of extreme solutions for problem (3).The rest of this paper is arranged as follows. Section2 contains some preliminary notations, definitions, and basic results about conformable fractional calculus. In Section 3, we establish comparison principle and define the upper and lower solutions, and then we obtain the existence of extreme solutions for problem (3) by means of the monotone iterative technique. Finally, in Section 4, an example is given to show the effectiveness of the results obtained.Remark 1.
Ifα=1, then BVP (3) is reduced to BVP (1) in [24].


## 2. Preliminaries

LetJ-=J+∖{t1,t2,…,tm}, J0=[t0,t1], Jk=(tk,tk+1] for k=1,2,…,m be subinterval of J:(4)PCJ+,R=x:J+→R:xtis  continuous  everywhere  except  for  sometkat  whichxtk-andxtk+exist  andxtk-=xtk,k=1,2,…,m.And let E={x∈PC(J+,R):x(t)=x(0),t∈[-r,0]} with norm(5)xE=supt∈J+⁡xt,and then E is a Banach space. A function x∈E is called a solution of problem (3) if it satisfies (3).Fora,b∈J with a≤b, new notations are introduced in [23] as follows:(6)eM/αa,b=eM/αb-aα,a,b∈ti,ti+1,i=0,1,…,m,eM/αb-tiα·eM/αti-aα,a∈ti-1,ti,b∈ti,ti+1,i=1,…,m,eM/αb-tqα·∏a<ti-1<ti<beM/αti-ti-1α·eM/αtp-aα,a<ti-1<ti<b,i=2,…,m,where tq=max⁡tk;a<tk<b, tp=min⁡tk;a<tk<b for some k∈{1,2,…,m} and ∏0(·)=1.Leth:J→R be a function given by(7)ht=h0t,t∈J0,h1t,t∈J1,⋮hmt,t∈Jm.The impulsive integral notation is defined as(8)∫abhsd^s=∫atphp-1sds+∫tptp+1hpsds+⋯+∫tqbhqsds,a,b∈J,where a≤tp<⋯<tq≤b.Property 2 (see [17]).
Leta≤c≤b≤d be nonnegative real numbers. The following relations hold:(i)
eM/α(a,c)eM/α(c,b)=eM/α(a,b),(ii)
eM/α(a,b)eM/α(c,d)=eM/α(a,d)eM/α(c,b).Definition 3 (see [17]).
The conformable fractional derivative starting from a pointa of a function f:[a,∞)→R of order 0<α≤1 is defined by(9)aDαft=limε→0⁡ft+εt-a1-α-ftε,provided that the limit exists.Definition 4 (see [17]).
Letα∈(0,1]. The conformable fractional integral starting from a point a of a function f:[a,∞)→R of order α is defined as(10)aIαft=∫ats-aα-1fsds.Remark 5.
Iff is differentiable, then aDαf(t)=(t-a)1-αf′(t). In addition, if aDαf exists on [a,∞), then we say that f is α-differentiable on [a,∞).Lemma 6 (see [17]).
Letα∈(0,1], k1,k2,p,λ∈R, and the functions f,g be α-differentiable on [a,∞). Then(i)
aDα(k1f+k2g)=k1aDα(f)+k2aDα(g)(ii)
aDα(t-a)p=p(t-a)p-α(iii)
aDαλ=0 for all constant functions f(t)=λ(iv)
aDα(fg)=faDαg+gaDαf(v)
aDαf/g=gaDαf-faDαg/g2 for all functions g(t)≠0Theorem 7 (see [23]).
Let an interval[c,d]⊂[a,∞) and let f:[a,∞)→R be given function satisfying the following:(i)
f is continuous on [c,d](ii)
f is α-differentiable for some α∈(0,1)
Then there exists a constante∈(c,d), such that aDαf(e)=f(d)-f(c)/1/α(d-a)α-1/α(c-a)α.


## 3. Main Results

We now consider the following antiperiodic boundary value problem:(11)tkDαxt+Mxt=σt-Nxwt,t≠tk,t∈J,Δxt=tk=-Lkxtk+Ikηtk+Lkηtk,k=1,2,…,m,x0=-xT,xt=x0,t∈-r,0,where M>0, N≥0, 0≤Lk<1 are constants and σ∈PC(J,R).Lemma 8.
x ∈ P C ( J , R ) is a solution of BVP (11) if and only if x∈E is a solution of the impulsive integral equation:(12)xt=∫0TG1t,sσs-Nxwsd^s+∑k=1mG2t,tk-Lkxtk+Ikηtk+Lkηtk,t∈J,∫0TG10,sσs-Nxwsd^s+∑k=1mG20,tk-Lkxtk+Ikηtk+Lkηtk,t∈-r,0,where(13)G1t,s=s-thα-1e-M/αth,teM/αth,s1+e-M/α0,T,0≤s<t≤T,-s-thα-1e-M/α0,te-M/αth,TeM/αth,s1+e-M/α0,T,0≤t≤s≤T,and(14)G2t,s=e-M/αs,t1+e-M/α0,T,0≤s<t≤T,-e-M/α0,te-M/αs,T1+e-M/α0,T,0≤t≤s≤T,with th=max⁡tk;k=0,1,…,mandtk≤s.Proof.
For convenience, let(15)Bt=σt-Nxwt,τtk=-Lkxtk+Ikηtk+Lkηtk.If x∈PC(J,R) is a solution of BVP (11), then for t∈J0=[t0,t1], we have by Lemma 6 that(16)t0DαeM/αt-t0αxt=eM/αt-t0αt0Dαxt+eM/αt-t0αMxt.Thus, multiplying by eM/α(t0,t) both sides of the first equation of (11), we obtain(17)t0DαeM/αt0,txt=eM/αt0,tBt.The conformable fractional integral of order α from t0 to t (t∈J0) of (17) yields(18)xt=xt0e-M/αt0,t+∫t0ts-t0α-1e-M/αt0,teM/αt0,sBsds,t∈J0.For t∈J1=(t1,t2], multiplying by eM/α(t1,t) both sides of the first equation of (11), we get(19)t1DαeM/αt1,txt=eM/αt1,tBt.Applying the conformable fractional integral of order α to both sides of (19) for t∈J1, we obtain(20)xt=xt1+e-M/αt1,t+∫t1ts-t1α-1e-M/αt1,teM/αt1,sBsds.By using Property 2, x(t1+)=x(t1)+τ(t1), and(21)xt1=xt0e-M/αt0,t1+∫t0t1s-t0α-1e-M/αt0,t1eM/αt0,sBsds, (20) implies that(22)xt=xt0e-M/αt0,t+∫t0t1s-t0α-1e-M/αt0,teM/αt0,sBsds+τt1e-M/αt1,t+∫t1ts-t1α-1e-M/αt1,teM/αt1,sBsds.In the same way, for t∈Jk, we derive(23)xt=xt0e-M/αt0,t+∑t0<tk<t∫tk-1tks-tk-1α-1e-M/αtk-1,teM/αtk-1,sBsds+∑t0<tk<tτtke-M/αtk,t+∫tlts-tlα-1e-M/αtl,teM/αtl,sBsds,where tl=max⁡tk;k=0,1,…,mandtk<t.
Note thatt0=0, and we impose t=T in (23). From antiperiodic boundary value condition x(0)=-x(T), we deduce that(24)x0=-11+e-M/α0,T∑k=1m∫tk-1tks-tk-1α-1e-M/αtk-1,TeM/αtk-1,sBsds+∫tmTs-tmα-1e-M/αtm,TeM/αtm,sBsds+∑k=1mτtke-M/αtk,T.Substituting (24) into (23), we have(25)xt=-e-M/α0,t1+e-M/α0,T∑k=1m∫tk-1tks-tk-1α-1e-M/αtk-1,TeM/αtk-1,sBsds+∫tmTs-tmα-1e-M/αtm,TeM/αtm,sBsds+∑k=1mτtke-M/αtk,T+∑0<tk<t∫tk-1tks-tk-1α-1e-M/αtk-1,teM/αtk-1,sBsds+∑0<tk<tτtke-M/αtk,t+∫tlts-tlα-1e-M/αtl,teM/αtl,sBsds=11+e-M/α0,T-∑k=1m∫tk-1tks-tk-1α-1e-M/α0,te-M/αtk-1,TeM/αtk-1,sBsds-∫tmTs-tmα-1e-M/α0,te-M/αtm,TeM/αtm,sBsds-∑k=1mτtke-M/α0,te-M/αtk,T+∑0<tk<t∫tk-1tks-tk-1α-1e-M/α0,Te-M/αtk-1,teM/αtk-1,sBsds+∑0<tk<tτtke-M/α0,Te-M/αtk,t+∫tlts-tlα-1e-M/αtl,teM/αtl,sBsds+∑0<tk<t∫tk-1tks-tk-1α-1e-M/αtk-1,teM/αtk-1,sBsds+∑0<tk<tτtke-M/αtk,t+∫tlts-tlα-1e-M/αtl,teM/αtl,sBsdsBy using Property 2 (ii), we get(26)xt=11+e-M/α0,T-∫tTs-thα-1e-M/α0,te-M/αth,TeM/αth,sBsd^s-∑t<tk<Tτtke-M/α0,te-M/αtk,T+∑0<tk<tτtke-M/αtk,t+∫0ts-thα-1e-M/αth,teM/αth,sBsd^s,t∈0,T,which implies that (12) holds.
Conversely, assumex(t) is a solution of (12); then by direct calculus, we can easily obtain that x(t) satisfies fractional impulsive antiperiodic boundary value problem (11). The proof is completed.Denotea=max{t1-t0,t2,-t1,…,tm+1-tm}. Now we establish the comparsion result.Lemma 9.
Assume thatα∈(0,1], M>0, N,N1≥0, 0≤Lk<1, k=1,2,…,m. If(27)eM/α0,Tα1+e-M/α0,Tm+1aαN+11+e-M/α0,T∑k=1mLk<1,then (11) has a unique solution.Proof.
For eachx∈E, we define an operator A by(28)Axt=∫0TG1t,sσs-Nxwsd^s+∑k=1mG2t,tk-Lkxtk+Ikηtk+Lkηtk,t∈J,∫0TG10,sσs-Nxwsd^s+∑k=1mG20,tk-Lkxtk+Ikηtk+Lkηtk,t∈-r,0,where G1,G2 are given by Lemma 8; then Ax∈E.
It is easy to check that(29)eM/αc,d≤eM/α0,T,for0≤c≤d≤T,and(30)e-M/αc,d≤1,for0≤c≤d≤T.For t∈(tj,tj+1], j∈{0,1,…,m}, and by (29)-(30), we obtain(31)
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.Moreover, we get(32)supt,s∈J2⁡G2t,s≤11+e-M/α0,T.Thus, ∀x,y∈E, we obtain(33)Ax-AyE=supt∈J⁡∫0TG1t,sσs-Nxwsd^s+∑k=1mG2t,tk-Lkxtk+Ikηtk+Lkηtk-∫0TG2t,sσs-Nywsd^s-∑k=1mG2t,tk-Lkytk+Ikηtk+Lkηtk=supt∈J⁡∫0TG1t,sNyws-xwsd^s+∑k=1mLkG2t,tkyxk-xtk≤supt∈J⁡N∫0TG1t,sd^s+∑k=1mG2t,tkLkx-yE≤eM/α0,Tα1+e-M/α0,Tm+1aαN+11+e-M/α0,T∑k=1mLkx-yE.By (27) and Banach fixed point theorem, A has a unique fixed point which is the unique solution of problem (11). The proof is completed.Lemma 10.
Let0<α≤1. Suppose that x∈E satisfies(34)tkDαxt+Mxt+Nxwt≤0,t∈J-,Δxtk≤-Lkxtk,k=1,2,…,m,x0≤0,xt=x0,t∈-r,0,where constants M>0, N,N1≥0, 0<Lk≤1. Assume in addition that(35)aααm+1M+N+∑k=1mLk≤1.Then x(t)≤0 for all t∈J+.Proof.
Suppose, to the contrary, that there must bet∗∈(0,T] such that x(t∗)>0. Let t∗∈[0,t∗) with t∗∈Ji, i=0,1,…,m, such that x(t∗)=inf⁡xt:t∈0,t∗=-b<0. It is easy to show that(36)tkDαxt≤-Mxt+Nxwt≤bM+N.Suppose t∗∈(tj,tj+1] for j∈{0,1,…,m}. It is easy to know that t∗<t∗; then i≤j. By Theorem 7, we have(37)xt∗-xtj≤xt∗-xtj+-Ljxtj=1αt∗-tjαtjDαxrj-Ljxtj≤baααM+N+Lj,rj∈tj,t∗,xtj-xtj-1≤xtj-xtj-1+-Ljxtj-1=1αtj-tj-1αtj-1Dαxrj-1-Lj-1xtj-1≤baααM+N+Lj-1,rj-1∈tj-1,tj,⋯xti+1-xt∗=1αti+1-tiα-1αt∗-tiαtiDαxr∗≤baααM+N,r∗∈t∗,ti+1.Summing up the above inequalities, we obtain(38)xt∗-xt∗≤baααm+1M+N+∑k=1mLk,which implies that(39)0<xt∗≤xt∗+baααm+1M+N+∑k=1mLk=-b+baααm+1M+N+∑k=1mLk.Thus(40)aααm+1M+N+∑k=1mLk>1,which contradicts (35). The proof is completed.Definition 11.
The functionsμ0,ν0∈E are said to be related lower and upper solutions for BVP (3) if(41)μ0t≤ν0t,t∈J+,tkDαμ0t≤ft,μ0,μ0wt,t∈J-,Δμ0tk≤Ikμ0tk,k=1,2,…,m,μ00≤-ν0T,μ0t≤μ00,t∈-r,0.And(42)tkDαν0t≥ft,ν0,ν0wt,t∈J-,Δν0tk≥Ikν0tk,k=1,2,…,m,ν00≥-μ0T,ν0t≥ν00,t∈-r,0.
Forμ0,ν0∈E, we write μ0≤ν0 if μ0(t)≤ν0(t) for all t∈J+, and(43)μ0,ν0=x∈E:μ0t≤xt≤ν0t,t∈J+.Theorem 12.
Letμ0,ν0∈E be coupled lower and upper solutions of BVP (3). Suppose that the following conditions hold:
(H1) The functionf∈PC(J×R2,R) satisfies(44)ft,x,y-ft,x¯,y¯≥-Mx-x¯-Ny-y¯,for μ0(t)≤x¯(t)≤x(t)≤ν0(t), μ0(w(t))≤y¯(t)≤y(t)≤ν0(w(t)), t∈J-
(H2) All functionsIk∈C(R,R) satisfy(45)Ikxtk-Ikytk≥-Lkxtk-ytk,where μ0(tk)≤y(tk)≤x(tk)≤ν0(tk), k=1,2,…,m
(H3) ConstantsM>0, N≥0, 0≤Lk<1 in (H1) and (H2) satisfy (27) and (35).
Then there exists monotone sequence{μn},{νn}⊂E which converges uniformly to the minimal and maximal solutions x∗,x∗ of BVP (3), respectively, such that(46)μ0≤μ1≤⋯≤μn≤x∗≤x≤x∗≤νn≤⋯≤ν1≤ν0,on J, where x is a solution of BVP (3) such that μ0(t)≤x(t)≤ν0(t) on J+.Proof.
We construct two sequences{μn}, {νn} which are the solutions of(47)tkDαμnt+Mμnt+Nμnwt=ft,μn-1t,μn-1wt+Mμn-1t+Nμn-1wt,t∈J-,Δμntk=-Lkμntk+Ikμn-1tk+Lkμn-1tk,k=1,2,…,m,μn0=-νn-1T,μnt=μn0,t∈-r,0,and(48)tkDανnt+Mνnt+Nνnwt=ft,νn-1t,νn-1wt+Mνn-1t+Nνn-1wt,t∈J-,Δνntk=-Lkνntk+Ikνn-1tk+Lkνn-1tk,k=1,2,…,m,νn0=-μn-1T,νnt=νn0,t∈-r,0.It follows from (H3) and Lemma 9 that problem (47) has a unique solution. Similarly, we also conclude that problem (48) has a unique solution too.
We prove that these sequences satisfy the following properties:(i)
μn-1≤μn,νn≤νn-1,n≥1(ii)
ifμn-1≤νn-1, then μn≤νn,n≥1
To prove (i), putp(t)=μ0(t)-μ1(t); then we get by Definition 11 and (47) that(49)tkDαpt=tkDαμ0t-tkDαμ1t≤ft,μ0,μ0wt--Mμ1t-Nμ1wt+ft,μ0,μ0t+Mμ0t+Nμ0wt=-Mpt-Npwt,t∈J-,and(50)Δptk=Δμ0tk-Δμ1tk≤Ikμ0tk--Lkμ1tk+Ikμ0tk+Lkμ0tk=-Lkptk,k=1,2,…,m,p0=μ00-μ10=μ00+ν0T≤0,and(51)pt=p0,t∈-r,0.By using (H3) and Lemma 10, we deduce that p(t)≤0, which implies μ0(t)≤μ1(t) for all t∈J+, i.e., μ0≤μ1. Analogously, ν1≤ν0. By mathematics induction, we can obtain that {μn} is a nondecreasing sequence and {νn} is a nonincreasing sequence.
To prove (ii), we first show thatμ1≤ν1, if μ0≤ν0. Let q=μ1-ν1, using (47), (48), and (H1), we have(52)q0=μ10-ν10=-ν0T+μ0T≤0,qt=q0,t∈-r,0,tkDαqt+Mqt+Nqwt=tkDαμ1t+Mμ1t+Nμ1wt-tkDαν1t+Mν1t+Nν1wt=ft,μ0t,μ0wt+Mμ0t+Nμ0wt-ft,ν0t,ν0wt+Mν0t+Nν0wt=ft,μ0t,μ0wt-ft,ν0t,ν0wt-Mν0t-μ0t-Nν0wt-μ0wt≤0,t∈J-,and by (H2)(53)Δqtk=-Lkμ1tk+Ikμ0tk+Lkμ0tk--Lkν1tk+Ikν0tk+Lkν0tk≤-Lkqtk,k=1,2,…,m.Thus, from (H3) and Lemma 10, we obtain q(t)≤0, which yields μ1≤ν1. Still by mathematical induction, we have μn≤νn, n≥1.
Following (i) and (ii) above, we have(54)μ0≤μ1≤⋯≤μn≤⋯≤νn≤⋯ν1≤ν0,for  alln∈N.Obviously, {μn} and {νn} satisfy (47) and (48), respectively. Thus, there exist x∗ and x∗ on J, such that limn→∞μn=x∗ and limn→∞νn=x∗ uniformly on J. It is easy to see that x∗ and x∗ are solutions of BVP (3) in [μ0,ν0].
Finally, we provex∗ and x∗ are extreme solutions of BVP (3) in [μ0,ν0]. Assume that x(t) is any solution of problem (3), which satisfies μ0(t)≤x(t)≤ν0(t), t∈J+. In the following, we will prove that if μn(t)≤x(t)≤νn(t) for some positive integral n, then there holds μn+1(t)≤x(t)≤νn+1(t) on J+. Let p=μn+1-x; then for t∈J+ we have(55)p0=μn+10-x0=-νnT+xT≤0,pt=μn+1t-xt=μn+10-x0=p0,t∈-r,0,tkDαpt=tkDαμn+1t-tkDαxt=-Mμn+1t-Nμn+1wt+ft,μnt,μnwt+Mμnt+Nμnwt-ft,xt,xwt=-Mμn+1t-xt-Nμn+1wt-xwt-ft,xt,xwt-ft,μnt,μnwt+Mxt-μnt+Nxwt-μnwt≤-Mpt-Npwt,t∈J-,and(56)Δptk=Δμn+1tk-Δxtk=-Lkμn+1tk+Ikμntk+Lkμntk-Ikxtk=-Lkμn+1tk-xtk-Ikxtk-Ikμntk+Lkxtk-μntk≤-Lkptk,k=1,2,…,m.By Lemma 10, we have p(t)≤0 for all t∈J+, which leads to μn+1≤x on t∈J+. Similarly, we can show that x≤νn+1 on J+. By mathematical induction, we obtain that μn≤x≤νn for each n. Taking n→∞, we have x∗(t)≤x(t)≤x∗(t) for t∈J+. The proof is finished.Remark 13.
Letα=1; then Theorem 12 is generalization of the result of Theorem 3.1 in [24].


## 4. An Example

Example 1.
Consider the problem(57)tkD2/3xt=-16sin⁡xt-12-35xwt,t∈0,12∖14,Δx14=120x14,x0=-x12,xt=x0,t∈-1,0,where(58)wt=56t,t∈0,14,32t-16,t∈14,13,t,t∈13,12.Thus(59)ft,xt,xwt-ft,x¯t,x¯wt=-16sin⁡xt-12-16sin⁡x¯t-12-35xwt+35x¯wt≥-16xt-x¯t-35xwt-x¯wtfor all t∈J, x(t),x(w(t)),x¯(t),x¯(w(t)))∈R, x(t)≥x¯(t), x(w(t))≥x¯(w(t))).LetM=1/6, N=3/5; then condition (H1) holds. Since I1(x)=1/20x, condition (H2) also holds with L1=1/20. We easily know that α=2/3, T=1/2, m=1, t1=1/4, a=1/4, r=1, w∈C(J,J+), t-r≤w(t), t∈0,1/2, and t1<w(t)≤t for t∈t1,1/2. Moreover, we have(60)eM/α0,12=eM/α1/2-1/4α+eM/α1/4-0α=2e1/4·1/42/3=2.2086,eM/α0,1/2α1+e-M/α0,1/2m+1aαN+11+e-M/α0,1/2∑k=1mLk=0.5612+0.0178=0.5790<1,and(61)aααm+1M+N+∑k=1mLk=0.9128+0.05=0.9628<1,which implies that hypothesis (H3) is satisfied.Setting(62)μ0t=-58,t∈-1,14,-23,t∈14,12,ν0t=-14t+23,t∈-1,14,-14t+34,t∈14,12.Obviously,(63)μ0t≤ν0tfort∈J+=-1,12,μ0t=μ00,ν0t≥ν00,t∈-1,0,μ00=-58=-ν012,ν00=23=-μ012,Δμ014=-23--58=-124≤-132=120μ014,and(64)Δν014=34-23=112>29960=120ν014.We see that(65)ft,μ0t,μ0wt=-16sin⁡μ0t-12-35μ0wt=-16sin⁡-98+38,t∈0,14,-16sin⁡-76+25,t∈14,12>0=tkD2/3μ0t,t∈0,12∖14.Moreover, we have(66)ft,ν0t,ν0wt=-16sin⁡ν0t-12-35ν0wt=-16sin⁡-14t+16-35-524t+59,t∈0,14,-16sin⁡-14t+14-35-1432t-16+34,t∈14,13,-16sin⁡-14t+14-35-14t+34,t∈13,12≤-16sin⁡548+132-13,t∈0,14,-16sin⁡16+340-1940,t∈14,13,-16sin⁡18+340-920,t∈13,12=-0.3194,t∈0,14,-0.4276,t∈14,13,-0.3958,t∈13,12,0D2/3ν0t=0D2/3-14t+23=-14t1/3≥-14141/3=-0.1575,t∈0,14,and(67)1/4D2/3ν0t=1/4D2/3-14t+34=-14t-141/3≥-14141/3=-0.1575,t∈14,12,which implies that(68)ft,ν0t,ν0wt≤tkD2/3ν0t,t∈0,12∖14.Hence, μ0 and ν0 are coupled lower and upper solutions of (57). Thus, applying Theorem 12, we have the existence of monotone sequences that approximate the extremal solutions of (57) in the interval -1,1/2.



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*Source: 7643123-2018-12-31.xml*