# Monotone Iterative Technique for Conformable Fractional Differential Equations with Deviating Arguments

Authors: Huiling Chen; Shuman Meng; Yujun Cui
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

This paper is concerned with the existence of extremal solutions for periodic boundary value problems for conformable fractional differential equations with deviating arguments. We first build two comparison principles for the corresponding linear equation with deviating arguments. With the help of new comparison principles, some sufficient conditions for the existence of extremal solutions are established by combining the method of lower and upper solutions and the monotone iterative technique. As an application, an example is presented to enrich the main results of this article.


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## Body

## 1. Introduction

In recent years, people have been paying attention to the progress of the fractional differential equations. In fact, it is the generalization of the ordinary differential equations to a noninteger order. Significantly, fractional differential equations appear more frequently in different fields of science and engineering, such as viscoelasticity, circuit, and neuron modeling [1–3]. Gradually, fractional differential equations are increasingly regarded as effective assistants. We have observed that many papers are exploring the existence of solutions of boundary value problems for fractional differential equations by using nonlinear functional analysis methods such as fixed point theorems, fixed point index on cone, variational methods and critical point theory, the theory of Mawhin coincidence degree, and the upper and lower solution method; see the monographs of Kilbas et al. [1], Podlubny [2], Diethem [3], the papers [4–26], and the references therein. Among them, the monotone iterative technique is an ingenious and effective method that offers theoretical, as well constructive existence results for nonlinear problems via linear iterates [9–15, 17, 23, 26]. It yields monotone sequences that converge to the extremal solutions in a sector generated by the upper and lower solutions. For example, the authors of [22] adopted the method of monotone iteration combined with the method of upper and lower solutions to consider the following system of nonlinear fractional differential equations:(1)Dαvt=ft,vt,wt,t∈0,T,Dαwt=gt,wt,vt,t∈0,T,t1−αvt|t=0=x0,t1−αwt|t=0=y0,where 0<T<∞, f,g∈C0,T×ℝ×ℝ,ℝ, x0,y0∈ℝ, and x0≤y0. In addition, [15, 24] used these methods to study the initial value problems for nonlinear fractional differential equations with no deviating arguments. On the basis of [22], Jian et al. [13] successfully investigated the following nonlinear fractional order differential systems with deviating arguments:(2)Dαvt=ft,vt,vθt,wt,wθt,t∈0,1,Dαwt=gt,wt,wθt,vt,vθt,t∈0,1,t1−αvt|t=0=x0,t1−αwt|t=0=y0,where θ∈C0,1,0,1. They introduce two well-defined monotone sequences that converge to the solution of the system and, then, establish the existence and uniqueness of the solution of the system. Finally, a numerical iterative scheme is introduced to obtain an accurate approximate solution for the systems.Motivated by the abovementioned papers, in this paper, we devote ourselves to the existence of solutions to the following boundary value problems with deviation arguments:(3)Dδϕt=ft,ϕt,ϕθt,t∈0,T,ϕ0=ϕT,where δ∈0,1, θ∈C0,1,0,1, and f∈C0,1×ℝ×ℝ,ℝ, and Dδϕ is the conformable fractional derivative of order δ. The conformable fractional calculus which was introduced in the work of Khalil et al. [27], then developed by Abdeljawad [28], have been receiving a lot of attention due to the wide application in physics and engineering [29, 30]. The reader is referred to [14, 16, 17, 27–33] and references therein for some recent advances in conformable fractional calculus and its applications.In this paper, by establishing two comparison results and using the monotone iterative technique combined with the method of upper and lower solutions, some sufficient conditions are presented for the existence of extremal solutions for periodic boundary value problem (3).


## 2. Preliminaries

Definition 1 (See [27]).
Letf:0,+∞⟶ℝ and t>0, and the conformable fractional derivative of order 0<α≤1 is defined by(4)Dαft=limρ⟶0ft+ρt1−α−ftρ,for t>0, and the conformable fractional derivative at 0 is defined as Dαf0=limt⟶0+Dαft. If f is differentiable, then Dαft=t1−αf′t.Definition 2 (See [27]).
Letα∈0,1. The conformable fractional integral of a function f:0,+∞⟶ℝ of order α is denoted as(5)Iαft=∫0tsα−1fsds.Lemma 1 (See [32]).
LetT>0. Assume that f∈C0,T and Dαf∈C0,T∩L0,T with 0<α≤1. Then, we have(6)IαDαft=ft−f0.Lemma 2 (See [27]).
Letα∈0,1, l1,l2,q,K∈ℝ, and the functions f, h be α-differentiable on 0,+∞. Then,(a)
DαK=0 for all constant functions ft=K(b)
Dαl1f+l2f=l1Dαft+l2Dαht(c)
Dαtq=qtq−α(d)
Dαfh=ftDαht+htDαft(e)
Dαf/h=hDαf−fDαh/h2 when ht≠0Lemma 3 (See [34]).
LetA:X⟶X linear operator, rA be the spectral radius of A, and A=maxϕ=1Aϕ. Then,(1)
rA≤A(2)
ifrA<1, then ℐ−A−1 exists and ℐ−A−1=∑n=0∞An, where ℐ stands for the identity operator
It is given thatT>0. Let E=C0,T; then, E is a Banach space with the norm x=maxt∈0,Txt.Let us introduce the following values and functions which will be used in the rest paper.(7)K1=Kδ,l=e−K1Tδ,M=eK1TδeK1Tδ−1,Ψ1t≡Ml,Ψ2t≡M,t∈0,T,M¯=Ml+δN2M2l2TδKδ2−N2M2l2T2δ−δ2NMKδ2−N2M2T2δ,M˜=Ml−M2NTδδ+δN2M2l2TδKδ2−N2M2l2T2δ−N3M3T2δKδ2−N2M2T2δ.For the forthcoming analysis, we first consider the following two boundary value problems for a linear differential fractional equations:(8)Dδϕt+Kϕt=ht,t∈0,T,ϕ0=ϕT+a,(9)Dδϕt+Kϕt+Nϕθt=ht,t∈0,T,ϕ0=ϕT+a.Lemma 4.
LetK>0, a∈ℝ, and h∈E. Then, problem (8) has the unique solution:(10)ϕt=∫0TGt,shsds+aΨt,where Ψt=1/1−e−K1Tδe−K1tδ and(11)Gt,s=eK1TδeK1Tδ−1e−K1tδ−sδsδ−1,0<s≤t≤T,1eK1Tδ−1e−K1tδ−sδsδ−1,0≤t<s≤T.Proof.
Multiply both sides of the first equation of (8) by eK1tδ, namely,(12)eK1tδDδϕt+KeK1tδϕt=eK1tδht.
By using Lemma2 (d), equation (12) is equivalent to(13)DδeK1tδϕt=eK1tδht.
In view of Lemma1 and Definition 2, we get(14)eK1tδϕt−ϕ0=∫0tsδ−1eK1sδhsds,so(15)ϕt=e−K1tδϕ0+∫0tsδ−1eK1sδhsds.
The boundary conditionϕ0=ϕT+a leads to(16)ϕ0=ϕT+a=e−K1Tδϕ0+∫0Tsδ−1eK1sδhsds+a.
Clearly,(17)ϕ0=1eK1Tδ−1∫0Tsδ−1eK1sδhsds+a1−e−K1Tδ.
Substituting (17) into (15), it follows that linear problem (8) has the following integral representation of the solution:(18)ϕt=e−K1tδ1eK1Tδ−1∫0Tsδ−1eK1sδhsds+∫0tsδ−1eK1sδhsds+a1−e−K1Tδe−K1tδ,=eK1TδeK1Tδ−1∫0tsδ−1e−K1tδ−sδhsds+1eK1Tδ−1∫tTsδ−1e−K1tδ−sδhsds+a1−e−K1Tδe−K1tδ,=∫0TGt,shsds+aΨt.
This completes the proof.
For all0<δ≤1, Green’s function G admits the following properties:(19)1eK1Tδ−1sδ−1≤Gt,s≤eK1TδeK1Tδ−1sδ−1,t∈0,T,s∈0,T.
Namely,(20)Mlsδ−1≤Gt,s≤Msδ−1,t∈0,T,s∈0,T.
In addition, forΨ given in Lemma 4, we can get(21)Ψ1t=e−K1Tδ1−e−K1Tδ≤Ψt≤11−e−K1Tδ=Ψ2t.
We define the operatorA on E by(22)Aht=∫0TGt,shsds,h∈E.
It is easy to see thatA:E⟶E is a positive linear continuous operator.Lemma 5.
A=1/K.Proof.
By direct computation, one has(23)∫0TGt,sds=eK1TδeK1Tδ−1∫0tsδ−1e−K1tδ−sδds+1eK1Tδ−1∫tTsδ−1e−K1tδ−sδds,=eK1TδδK1eK1Tδ−11−e−K1tδ+1δK1eK1Tδ−1eK1Tδ−K1tδ−1,=1δK1=1K.
Then, for anyh∈E, we have(24)Ah=maxt∈0,TAht≤maxt∈0,T∫0TGt,sds⋅h=1Kh,which implies that A≤1/K. On the other hand, take h0t≡1, then h0∈E, h0=1, and(25)Ah0=maxt∈0,TAht=∫0TGt,sds=1Kh0.
This yieldsA≥1/K. Therefore, A=1/K. This completes the proof.
We recall thatl=e−K1Tδ. Then, l∈0,1. For ∀h∈C0,T,0,+∞, it follows from (20) that(26)Aht=∫0TGt,shsds≤M∫0Tsδ−1hsds,t∈0,T,Aht=∫0TGt,shsds≥Ml∫0Tsδ−1hsds,t∈0,T.
The abovementioned two inequalities show that(27)Aht≥lAhs,∀t,s∈0,T,∀h∈C0,T,0,+∞.
Based on the above analysis, we have the following result on (9).Lemma 6.
LetK>0, 0≤N<K, a∈ℝ, θ∈C0,T,0,T, and h∈E. Then, problem (9) has a unique solution.Proof.
From Lemma4, it follows that ϕ∈E is a solution of (9) if and only if(28)ϕt=∫01Gt,s−Nϕθs+hsds+aΨt.
Now, we introduce an operatorB:E⟶E as follows:(29)Bϕt=Nϕθt,t∈0,T.
It is easy to see thatB is a positive linear operator with B=N. Thus, (28) reduces to(30)I+ABϕt=Aht+aΨt.
Note from Lemma5 that AB≤A⋅B=N/K<1. Thus, it follows from Lemma 3 that I+AB−1 exists and(31)I+AB−1=∑i=0∞−1iABi=I−AB+AB2+⋯+−1nABn+⋯.
Therefore, the unique solution of (9) is given by(32)ϕt=∑i=0∞−1iABiAht+a∑i=0∞−1iABiΨt.
The proof is complete.
Now, we present two comparison results.Lemma 7.
LetK>0, 0≤N≤Kl2, a∈ℝ, and θ∈C0,T,0,T. Assume that ϕ∈E satisfies Dδϕ∈E and(33)Dδϕt≤−Kϕt−Nϕθt,t∈0,T,ϕ0≤ϕT.
Then,ϕt≤0 for all t∈0,T.Proof.
Takeht=Dδϕt+Kϕt+Nϕθt, a=ϕ0−ϕT. Then,(34)ht≤0,a≤0.
Applying Lemma6, (32) holds, and (32) can be expressed by(35)ϕt=∑i=0∞AB2iI−ABAht+a∑i=0∞AB2iI−ABΨt.
Sinceh≤0, it implies that h0t≡−Ah0≥0. Thus, from (27), we obtain(36)−Ah≥lh0,−Ah≤1lh0.
With the help of positivity of operatorAB, the definition of operator B, and (23), we have(37)−ABAh≤1lABh0=NlKh0.
Consequently, we conclude that(38)I−ABAh≤−lh0+NlKh0=−l−NlKh0≤0.
On the other hand, by (21), we infer that(39)I−ABΨt=e−K1tδ1−e−K1Tδ−N∫0TGt,se−K1θsδ1−e−K1Tδds≥e−K1Tδ1−e−K1Tδ−N1−e−K1Tδ∫0TGt,sds,=11−e−K1Tδl−NK≥0.
Hence,ϕt≤0 holds for all t∈0,T that follow from a≤0 and (35). This completes the proof.Lemma 8.
LetK>0, 0≤NMTδ<δ, 0≤N<K, M˜>0, M¯>0, and θ∈C0,T,0,T. Assume that ϕ∈E satisfies Dδϕ∈E and (33). Then, ϕt≤0 for all t∈0,T.Proof.
Take againht=Dδϕt+Kϕt+Nϕθt, a=ϕ0−ϕT. Then,(40)ht≤0,a≤0.
Applying Lemma6, (32) holds, and (32) can be expressed by(41)ϕt=∑i=0∞AB2iAht−∑i=0∞AB2i+1Aht+a∑i=0∞ABiΨt−∑i=0∞AB2i+1Ψt.
Taking notice of the fact thatht≤0, by (20), we have(42)Aht=∫0TGt,shsds≤Ml∫0Tsδ−1hsds,and for n≥1,(43)AB2nAht=N2n∫0TGt,s∫0TGθs,τ2n−1∫0TGθτ2n−1,τ2n−2⋯∫0TGθτ2,τ1∫0TGθτ1,τ0hτ0dτ0dτ1,…,dτ2n−1ds≤N2n∫0TGt,sdsMl∫0Tsδ−1ds2n−1∫0TMlτ0δ−1hτ0dτ0,=N2nM2nl2nKTδδ2n−1∫0Tτ0δ−1hτ0dτ0,and for n≥1,(44)AB2n+1Aht=N2n+1∫0TGt,s∫0TGθs,τ2n∫0TGθτ2n,τ2n−1…∫0TGθτ2,τ1∫0TGθτ1,τ0hτ0dτ0dτ1,…,dτ2nds≤N2n+1∫0TGt,sdsMl∫0Tsδ−1ds2n∫0TMlτ0δ−1hτ0dτ0,=N2n+1M2n+1KTδδ2n∫0Tτ0δ−1hτ0dτ0.
These lead us to(45)∑i=0∞AB2iAht−∑i=0∞AB2i+1Aht≤Ml+∑i=1∞N2iM2il2iKTδδ2i−1−∑i=0∞N2i+1M2i+1KTδδ2i∫0Tsδ−1hsds=M¯∫0Tsδ−1hsds≤0.
By (20)–(23) and the positivity of operator AB, we have(46)Ψt≥e−K1Tδ1−e−K1Tδ=Ml,ABΨt≤ABΨ2t=N∫0TGt,sΨ2θsds,≤M2N∫0Tsδ−1ds=M2NTδδ,and for n≥1,(47)AB2nΨt≥AB2nΨ1t=N2n∫0TGt,s∫0TGθs,τ2n−1∫0TGθτ2n−1,τ2n−2…∫0TGθτ2,τ1Ψ1θτ1dτ1dτ2,…,dτ2n−1ds≥N2nΨ1t∫0TGt,sds∫0TMlsδ−1ds2n−1,=N2nKMlMlTδδ2n−1=N2nM2nl2nT2n−1δKδ2n−1,and for n≥1,(48)AB2n+1Ψt≤AB2n+1Ψ2t=N2n+1∫0TGt,s∫0TGθs,τ2n∫0TGθτ2n,τ2n−1…∫0TGθτ2,τ1Ψ2θτ1dτ1dτ2,…,dτ2nds≤N2n+1Ψ2t∫0TGt,sds∫0TMlsδ−1ds2,=N2n+1KMlMTδδ2n=N2n+1M2n+1Kδ2nT2nδ.
These, together with the fact thata≤0, ensure that(49)a∑i=0∞ABiΨt−∑i=0∞AB2i+1Ψt=aΨt−ABΨt+∑i=1∞ABiΨt−∑i=1∞AB2i+1Ψt≤aMl−M2NTδδ+∑i=1∞N2iM2il2iT2i−1δKδ2i−1−∑i=1∞N2i+1M2i+1Kδ2iT2iδ,=aM˜≤0.
Thus, by (41), (45), and (49), we have that ϕt≤0 for all t∈0,T, and the lemma is proved.


## 3. Main Results

Now, we are in the position to prove the existence of extremal solutions of (3) by using the monotone iterative method of lower and upper solutions. To this end, we define the lower and upper solutions of (3).Definition 3.
A functionu0∈E satisfying Dδu0∈E is called a lower solution of problem (3) if it satisfies(50)Dδu0t≤ft,u0t,u0θt,t∈0,T,u00≤u0T.
Analogously, a functionw0∈E satisfying Dδw0∈E is called an upper solution of (3) if the inequalities(51)Dδw0t≥ft,w0t,w0θt,t∈0,T,w00≥w0T,hold.Theorem 1.
Assume that the following conditions hold:H1θ∈C0,T,0,TH2: the functions u0 and w0 are lower and upper solutions of problem (3), respectively, such that u0t≤w0t on 0,TH3f∈C0,T×ℝ2,ℝ and there exist K>0, N≥0 such that(52)ft,x,z−ft,x¯,z¯≥−Kx−x¯−Nz−z¯,for allt∈0,T, u0t≤x¯≤x≤w0t, u0t≤z¯≤z≤w0tH4: the inequality N≤Kl2 holds or the inequalities NMTδ<δ, N<K, M˜>0, M¯>0 hold
Then, (3) has minimal and maximal solution u,w in the sector u0,w0, which can be obtained by monotone iterative sequences starting from u0 and w0, where u0,w0=z∈E:u0t≤zt≤w0t,t∈0,T.Proof.
Fork=1,2,…, let us define(53)Dδukt+Kukt+Nukθt=ft,uk−1t,uk−1θt+Kuk−1t+Nuk−1θt,t∈0,T,uk0=ukT,(54)Dδwkt+Kwkt+Nwkθt=ft,wk−1t,wk−1θt+Kwk−1t+Nwk−1θt,t∈0,T,wk0=wkT.
By Lemma6, for any k=1,2,…, we know that linear problems (53) and (54) have a unique solution ukt, wkt, respectively, which implies that the sequences ukt, wkt are well defined. Furthermore, ukt, wkt can be expressed as(55)ukt=I+AB−1Aℱuk−1t,wkt=I+AB−1Aℱwk−1t,where ℱ:E⟶E is a bounded operator defined by(56)ℱut=ft,ut,uθt+Kut+Nut,u∈E.
By the integral expression of operatorA, it is easy to see that A is completely continuous. Hence, I+AB−1Aℱ is completely continuous.
Firstly, let us prove that(57)u0≤u1≤w1≤w0.
To do this, letvt=u0t−u1t. By the definition of the lower solution, we get(58)Dδvt=Dδu0t−Dδu1t≤ft,u0t,u0θt−ft,u0t,u0θt+Ku1t−u0t+Nu1θt−u0θt,=−Kvt−Nvθt,t∈0,T,v0=u00−u10≤u0T−u1T=vT.
This shows, by Lemma7 or Lemma 8, that vt≤0 on 0,T, and hence, u0≤u1. Similarly, we can deduce that w1≤w0.
Now, letvt=u1t−w1t; by H2 and H3, we obtain(59)Dδvt=Dδu1t−Dδw1t,=ft,u0t,u0θt−Ku1t−u0t−Nu1θt−u0θt−ft,w0t,w0θt+Kw1t−w0t+Nw1θt−w0θt≤−Ku0t−w0t−Nu0θt−w0θt−Ku1t−u0t−w1t+w0t+N−u1θt+u0+w1θt−w0θt,=−Kvt−Nvθt,t∈0,T,v0=u10−w10=u1T−w1T=vT.
Then, from Lemma7 or Lemma 8, we get vt≤0, which yields u1≤w1.
Secondly, we need to show thatu1 and w1 are the lower and upper solutions of problem (3), respectively. In fact, it follows from H2 and H3 that(60)Dδu1t=ft,u0t,u0θt−Ku1t−u0t−Nu1θt−u0θt−ft,u1t,u1θt+ft,u1t,u1θt≤−Ku0t−u1t−Nu0θt−u1θt−Ku1t−u0t−Nu1θt−u0θt+ft,u1t,u1θt,=ft,u1t,u1θt,u10=u1T,which show that v1 is a lower solution of problem (3). Similarly, we can conclude that w1 is an upper solution of problem (3).
Repeating the foregoing arguments, we can prove that the sequencesukt, wkt are lower and upper solutions of problem (3), respectively, and satisfy the following inequality:(61)u0≤u1⋯≤uk≤⋯≤wk≤⋯≤w1≤w0.
Obviously, the sequencesukt, wkt are uniformly bounded in E and by (55) and the complete continuity of operator I+AB−1Aℱ, and it follows that ukt, wkt are relatively compact. This, together with the monotonicity of the sequences ukt, wkt, guarantees that the sequences ukt, wkt converge uniformly to u,w, respectively, and that u,w∈v0,w0 are solutions of (3).
Finally, we prove the minimal and maximal property ofu and w on v0,w0. We assume that z∈v0,w0 is any solution of (3) and there exists a positive integer k such that ukt≤zt≤wkt for t∈0,T.
Letvt=ukt−zt, then(62)Dδvt=Dδukt−Dδzt,=ft,uk−1t,uk−1θt−Kukt−uk−1t−Nukθt−uk−1θt−ft,zt,zθt≤−Kuk−1t−zt−Nuk−1θt−zθt−Kukt−uk−1t−Nukθt−uk−1θt,=−Kvt−Nvθt,v0=uk0−z0=ukT−zT=vT,undoubtedly, at≤0, namely, ukt≤zt. By a similar method, we can show that zt≤wkt. Thus, uk≤z≤wk, k=1,2,…. It is easy to find that ut≤z≤wt when k⟶∞. That is u, w are minimal and maximal solutions of (1) in the sector u0,w0. The proof is completed.
Then, by applying Lemma7 or Lemma 8, we get vt≤0, that is ukt≤zt on 0,T. Similarly, we can show that zt≤wkt on 0,T. Notice that u0t≤zt≤w0t on 0,T. So, ukt≤zt≤wkt hold for every k from mathematical induction. Hence, by taking k⟶+∞, we have ut≤zt≤wt on 0,T. The proof is complete.Example 1.
We consider the following BVP:(63)D1/2ϕt=−131+2π21+ϕt3+260πcosϕ2t24,t∈0,1,ϕ0=ϕ1.
Obviously,δ=1/2, T=1, θt=t2, and(64)ft,u,v=−131+2π21+u3+260πcosv24.
Letu0t=−1,w0t=2π; then,(65)D1/2u0t=0<260πcos14=ft,u0t,u0θt,t∈0,1,u00=u01=−1,D1/2w0t=0>−1+2π3=ft,w0t,w0θt,t∈0,1,w00=w01=2π.
This shows thatu0,w0 are lower and upper solutions of (63). On the other hand, it is easy to verify that H3 holds for K=1 and N=1/60. Furthermore, we have(66)K1=2,l=e−2∈0,1,M=11−e−2≈1.1565,N=160<1=K,NMTδ≈0.0193<12=δ,M¯≈0.1388>0,M˜≈0.1119>0.
Hence, all conditions of Theorem1 hold. Therefore, equation (63) has the extremal solution in v0,w0.



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