# Compact and Noncompact Solutions to Generalized Sturm–Liouville and Langevin Equation with Caputo–Hadamard Fractional Derivative

Authors: Ahmed Salem; Noorah Mshary; Moustafa El-Shahed; Faris Alzahrani
Journal: Mathematical Problems in Engineering
 (2021)

---

## Abstract

In this work, through using the Caputo–Hadamard fractional derivative operator with three nonlocal Hadamard fractional integral boundary conditions, a new type of the fractional-order Sturm–Liouville and Langevin problem is introduced. The existence of solutions for this nonlinear boundary value problem is theoretically investigated based on the Krasnoselskii in the compact case and Darbo fixed point theorems in the noncompact case with aiding the Kuratowski’s measure of noncompactness. To demonstrate the applicability and validity of the main gained findings, some numerical examples are included.


---

## Body

## 1. Introduction

Over the past few years, the fractional calculus (which is basically an expansion of the traditional calculus) has provided its remarkable contribution in addressing lots of physical and biological phenomena as well as describing many engineering dilemmas [1]. It has been confirmed in several literature studies that many real-world problems associated with many applied science fields can be described more convenient using the fractional-order differential equations (FoDEs) rather than that of using the ordinary differential equations (ODEs) [2]. In particular, the FoDEs played and still play a significant role in developing lots of models that outline several engineering problems and physical applications such as electromagnetics, porous media, control, and viscoelasticity [3]. For further facts on the basic principles of fractional calculus and the FoDEs, the reader may return to the references [4–7]. The fractional calculus, new interesting research field, is attracting the interest of mathematicians and researchers. With fractional operators, many real-world problems are being investigated, in particular the presenting of mathematical model for the transmission of COVID-19 and mathematical models of HIV/AIDS and drug addiction in prisons [8].The so-called fractional-order Langevin equation (FoLE) (which was first established in [9]) is considered as a practical tool employed in describing some physical phenomena’s evolution [10]. For instance, some operations related to the porous media and the reaction-diffusion can be outlined using this type of equations (see [11–13]). From these motivations, we find that exploring the existence of solutions for such equation will further enrich our understanding and knowledge about this subject. Actually, lots of results related to the solution’s existence and uniqueness of the FoLE (in view of some well-known operators such as the Riemann–Liouville and Caputo operators) were extensively deduced through several research papers (see, e.g., [14–17]). For the purpose of studying such kind of problems, several strategies and schemes could be implemented such as the coincidence theory [18], the upper and lower solution method [19, 20], the fixed point theorems [21, 22], and many others. For example, some convenient findings associated with the solution’s existence of the FoLE have been pointed out in [23]. Besides, this aspect together with the solution’s uniqueness aspect of the same equation have been well explored in [24]. However, for further recent outcomes, we refer the reader to papers [25, 26] and references therein.The Sturm–Liouville operator has several applications in variety fields of science such as engineering and mathematics [27]. The authors in [28] presented an approach to the fractional version of the Sturm–Liouville problem. Kiataramkul et al. [10] combined Sturm–Liouville and Langevin fractional differential equations and they called generalized Sturm–Liouville and Langevin equation (GSLLE). Furthermore, they studied the existence and uniqueness of solutions for GSLLE with antiperiodic boundary conditions.In 1892, another fractional derivative, which is called later as the Hadamard fractional derivative, was proposed by Hadamard. This derivative is distinguished from the others by its construction that includes a logarithmic function of arbitrary exponent. Latest achievements about this derivative and its corresponding Hadamard FoDEs can be found in [29]. In [30], Jarad et al. proposed the so-called Caputo–Hadamard fractional-order derivative, which is a new modification of the Hadamard fractional derivative operator. This new version has proved its suitability in dealing with some physical interpretable initial conditions [31]. In order to get a full overview about this operator and its properties, we refer the reader to the same reference [30]. From the perspective that asserts, there are limited results in literature that addresses the boundary value problems (BVPs) in view of the Caputo–Hadamard derivative [31].This paper attempts to present a theoretical study about the existence of solutions for the nonlinear GSLLE,(1)DCHμαtDCHγ+βtxt=ht,xt,DCHℓxt,subject to the following three point integral boundary conditions:(2)x1=0,xζ=a1JHρ1xζ,T=a2JHρ2xT,where DCHμ is the Caputo–Hadamard derivative operator of order μ, JHγ is the Hadamard integral operator of order γ, and 1≤t≤T, 0<ℓ<γ, 0<γ<1,ρ1,ρ2>0, and 1<μ<2. The functions α:1,T⟶ℝ+, β:1,T⟶ℝ and h:1,T×ℝ×ℝ⟶ℝ are continuous. It is remarkable to point out that when βt=0 for all t∈1,T, then our problem reduces to the Sturm–Liouville fractional BVP and when αt and βt are real constants functions for all t∈1,T, then our problem reduces to the Langevin fractional BVP.Here, we investigate the existence of solution in two cases: in the compact case by utilizing the well-known fixed point theorem due to Krasnoselskii. In the noncompact case, we apply the Kuratowski’s measure of noncompactness technique through using the Darbo’s fixed point theorems. The underlying idea is there are two measures of noncompactness commonly used; these associate numbers to sets in such a way that compact sets all get the measure 0, and other sets get measures that are bigger according to “how far” they are removed from compactness. There are many contributors who used measure of noncompactness to investigate the existence of solution to FoDE [32–34] and references given therein.The remaining of this paper is ordered as follows: Section2 presents briefly some fundamental concepts related to the fractional calculus and fixed point theorems. In Section 3, some mathematical preparations and principles are established to achieve our goals through defining some essential terms and deducing some auxiliary lemmas. Section 4 employs these preparations to attain the main results of this work, which concern with exploring the existence of solution for the GSLLE. Section 5 illustrates two numerical examples followed by Section 6 that outlines the major points and concluded remarks.


## 2. Preliminaries

This part aims to present some fundamental fixed point theorems and essential preliminaries related to the fractional calculus, especially in regard to the Hadamard fractional-order operators.Definition 1 (see [29]).
The Hadamard integral operator of the fractional-order valueμ∈ℝ+ for a function f∈Lpa,b is outlined as(3)JHaμft=1Γμ∫atlogtsμ−1fsdss,where 0≤a≤t≤b<∞.Definition 2 (see [30]).
The Caputo-Hadamard derivative operator of fractional-orderμ for at least n-times differentiable function f:a,∞⟶ℝ is outlined as(4)DCHaμft=1Γn−μ∫atlogtsn−μ−1δnfsdss,where n−1<μ<n, n=μ+1, δ=td/dt, and log⋅=loge⋅.Lemma 1 (see [30]).
Supposef∈Ca,b and μ>0 such that n=μ+1. Then,(5)DCHaμJHaμft=ft.Lemma 2 (see [30]).
Letf∈ACδna,b or Cδna,b and μ>0 such that n=μ+1. Then,(6)JHaμ.DCHaμft=ft−∑k=0n−1cklogtak,where Cδna,b=f:a,b⟶ℂ:δn−1ft∈Ca,b.Lemma 3 (see [30, 29]).
Letμ,ν>0, then we have(i)
JHaμJHaνft=JHaμ+νft,(ii)
JHaμlogν−1t/a=Γν/Γν+μlogμ+ν−1t/a,(iii)
DCHaμJHaμft=JHaμ−νft,μ>ν,(iv)
DCHaμlogν−1t/a=Γν/Γν−νlogν−μ−1t/a,ν≠1,2,…,n−1,n=μ+1,(v)
DCHaμlogν−1t/a=0,ν=1,2,…,n−1,n=μ+1,where f∈Lpa,b, and 1≤p<∞.For simplicity, we will considera=1 in all aforementioned preliminaries, and consequently JHaμ and DHaμ will be denoted by Jμ and Dμ, respectively. However, some essential fixed point theorems are reported below for completeness.Definition 3.
(see [3]). Suppose that ℬ is a Banach space and Υℬ is the collection of subsets of ℬ. The Kuratowski’s measure of noncompactness σ:Υℬ⟶0,∞ is a map outlined as(7)σX=infκ>0:X⊆∪i=1nXiand diamXi≤κ,where(8)diamXi=supx1−x2:x1,x2∈Xi.Lemma 4 (see [3]).
Suppose thatℬ is a Banach space, X, X1, and X2 are bounded sets of ℬ, and σ is the Kuratowski’s measure of noncompactness defined above. Then, we have the following assertions:(i)
X is compact ⟺σX=0.(ii)
σX1+X2≤σX1+σX2, where X1+X2=x1+x2:x1∈X1,x2∈X2.(iii)
σϱX=ϱσX, where ϱX=ϱx:x∈X,ϱ∈ℝ.(iv)
σX+x0=σX, for any x0∈ℬ.(v)
If the mapF:Ca,b,ℝ⟶ℬ is bounded and equicontinuous, it follows(9)σ∫atFsds≤∫atσFsds.Lemma 5 (Ascoli–Arzela theorem [3]).
The familyℱ=ft in C0,T,ℬ has a uniformly-convergent subsequence fntn=1∞, if it is uniformly bounded and equicontinuous maps on 0,T, and ft0 is relatively compact in the Banach space ℬ, for any t0∈0,T.Theorem 1 (Krasnoselskii’s fixed point theorem [3, 26]).
Suppose thatCε is a closed convex bounded set of a Banach space ℬ, and S1,S2:ℬ⟶ℬ are two operators satisfying(1)
S1x1+S2x2∈Cε, for each x1,x2∈Cε.(2)
S2 is a contraction mapping.(3)
S1 is continuous and S1Cε is a compact subset of ℬ.
Then,S=S1x+S2x has a fixed point.Theorem 2 (Darbo’s fixed point theorem [3]).
Suppose thatCε is a closed convex bounded set of a Banach space ℬ, and S:Cε⟶Cε is a continuous mapping so that(10)σSW≤kσW,for all closed subsets W∈Cε, where 0≤k<1. Then, S has a fixed point in Cε.


## 3. Auxiliary Results

This section presents some auxiliary results that will provide a basis for introducing some results associated with the existence of the solutions for the GSLLE.Lemma 6.
The linear BVP given in the following(11)DCHμαtDCHγ+βtxt=gt,subjects to the conditions in (2) which is equivalent to the following fractional-order integral equation:(12)xt=JHγJHμgt−βtxtαt+χt,Tχζ,Ta1JHρ1xζ−JHγJHμgζ−βζxζαζ+χt,ζχT,ζa2JHρ2xT−JHγJHμgT−βTxTαT,where g:1,T⟶ℝ is a continuous function and the other concepts are defined as in the Introduction section, χζ,T≠0,(13)χt,s=b1tb2s−b2tb1s=b1tb2tb1sb2s,(14)bit=1Γγ∫1tlogtsγ−1logi−1sdssαs,i∈1,2.Proof.
Applying the Hadamard fractional-order integralJHμ to both sides of (11) yields(15)αtDCHγxt+βtxt−c0+c1logt=JHμght.
That is,(16)DCHγxt=JHμgt−βtxt+c0+c1logtαt.
Again, takingJHγ to both sides of the equation above yields(17)xt=JHγJHμgt−βtxt+c0+c1logtαt+c2.
Now, using the first condition (x1=0 given in (2)) yields c2=0, and then we have(18)xt=JHγJHμgt−βtxtαt+b1tc0+b2tc1.
On the other hand, using the other two boundary conditions given in (2) leads us to the following two assertions:(19)a1JHρ1xζ=JHγJHμgζ−βζxζαζ+b1ζc0+b2ζc1,a2JHρ2xT=JHγJHμgT−βTxTαT+b1Tc0+b2Tc1.
Solving these two equations gives, respectively, the following two expressions ofc0 and c1:(20)c0=1χζ,Ta1b2TJHρ1xζ−a2b2ζJHρ2xT−b2TJHγJHμgζ−βζxζαζ+b2ζJHγJHμgT−βTxTαT,(21)c1=−1χζ,Ta1b1TJHρ1xζ−a2b1ζJHρ2xT−b1TJHγJHμgζ−βζxζαζ+b1ζJHγJHμgT−βTxTαT.
Finally, substitutingc0 and c1 in (18) yields the desired result.Lemma 7.
Lett∈1,T and s∈1,T. Then, for the function χt,s defined in (13), we can deduce the following assertion:(22)χ0s=maxt∈1,Tχt,s=α2b1sγb2sα11+γγγΓγ+2,where α1=mint∈1,Tαt and α2=maxt∈1,Tαt.Proof.
Since the functionα:1,T⟶ℝ+ is continuous, then it attains its minimum and maximum which leads to(23)1α2JHγlogi−1t≤bit≤1α1JHγlogi−1t,i=1,2,where bi,i=1,2 are defined in (14). In view of Lemma 3, we get(24)1α2Γγ+ilogγ+i−1t≤bit≤1α1Γγ+ilogγ+i−1t,i=1,2.
Whence,(25)χt,s≤b1sftα2Γγ+2,where(26)ft=α2b2s1+γα1b1slogγt−logγ+1t.
Differentiation gives(27)f′t=γ+1tlogγ−1tα2b2sγα1b1s−logt,which means that the function f is increasing on 1,t0 and decreasing on t0,T where t0=eα2b2sγ/α1b1s. Thus, ft≤ft0 for all t∈1,T which implies the desired result.In this work, the Banach space of all continuous real-valued functionsℬ=C1,T,ℝ will be considered in which it is equipped with the usual maximum norm:(28)x=max1≤t≤Txt,x∈ℬ.Define the space(29)Y=xt:xt∈ℬ,DCHℓxt∈ℬ,t∈1,T,as a subset of ℬ, then one can easily check that Y represents a Banach subspace equipped with the norm:(30)xY=maxx,DCHℓx=maxmax1≤t≤Txt,max1≤t≤TDCHℓxt.In light of Lemma6, the B.V.P. given in (1) and (2) can be converted into a fixed point problem. This means that the function xt must hold the two assertions: Sxt=xt and DCHℓSxt=DCHℓxt for all t∈1,T, where S:ℬ⟶ℬ and DCHℓS:ℬ⟶ℬ are the two operators defined on ℬ. For achieving this objective, we define the operator S as follows:(31)Sxt=S1xt+S2xt,(32)DCHℓSxt=DCHℓS1xt+DCHℓS2xt,where(33)S1xt=JHγJHμht,xt,DCHℓxtαt−χt,Tχζ,TJHγJHμhζ,xζ,DCHℓxζαζ−χt,ζχT,ζJHγJHμhT,xT,DCHℓxTαT,S2xt=−JHγβtxtαt+χt,Tχζ,Ta1JHρ1xζ+JHγβζxζαζ+χt,ζχT,ζa2JHρ2xT+JHγβTxTαT,(34)CHDℓS1xt=JHγ−ℓJHμh,xt,CHDℓxtαt−CHDℓχt,Tχζ,TJHγJHμhζ,xζ,DCHℓxζαζ−CHDℓχt,ζχT,ζJHγJHμhT,xT,DCHℓxTαT,DCHℓS2xt=−JHγ−ℓβtxtαt+DCHℓχt,Tχζ,Ta1JHρ1xζ+JHγβζxζαζ+DCHℓχt,ζχT,ζa2JHρ2xT+JHγβTxTαT.In a similar manner in Lemma7, we can deduce as follows.Lemma 8.
Lett∈1,T and s∈1,T. Then, for the function χt,s defined in (13), we can deduce the following assertion:(35)χℓs=maxt∈1,TDCHℓχt,s=α2b1sγ−ℓb2sα11+γ−ℓγ−ℓγ−ℓΓγ−ℓ+2,where α1=mint∈1,Tαt and α2=maxt∈1,Tαt.Lemma 9.
Letχt,τ be defined as in (13), t1,t2∈1,T, and 1<τ<T. Then, for all ε>0, there exists δ=δε>0 such that χt2,τ−χt1,τ<ε whenever t2−t1<δ.Proof.
Letε>0, t2>t1 and t2−t1<δ where(36)δ=minexpα1εΓ1+γb2τ1/γ−1,expα1εΓ1+γb1τlogT1/γ−1.
Then, from (14) and the fact that 0<γ<1, we have(37)b1t2−b1t1=1Γγ∫1t2logt2sγ−1dssαs−∫1t1logt1sγ−1dssαs≤1α1Γγ∫1t1logt1sγ−1−logt2sγ−1dss+∫t1t2logt2sγ−1dss=1α1Γγ+12logt2t1γ+logγt1−logγt2≤2α1Γγ+1logt2t1γ.
Sincet2−t1<δ, then t2/t1<1+δ/t1<1+δ. Hence,(38)b1t2−b1t1≤2α1Γγ+1logγ1+δ<εb2τ.
Similarly, we obtain(39)b2t2−b2t1≤2logTα1Γγ+1logγ1+δ<εb1τ.
Therefore,(40)χt2,τ−χt1,τ=b2τb1t2−b1t1+b1τb2t2−b2t1<ε,which ends the proof.The previous discussion together with the aforesaid proposed results represent a preparation for the next section that presents the main outcomes of the problem under consideration.


## 4. The Existence of Solution for the GSLLE

This part intends to introduce some theoretical outcomes associated with the existence of solution for the GSLLE given in (1) and (2). From now on, we will utilize the following appropriate conditions:H1: the function h:1,T×ℝ×ℝ⟶ℝ is continuous.H2: there exists a positive function Q:1,T⟶ℝ+ such that(41)ht,xt,Dℓxt≤Qt,∀t∈1,T.H3: there exists a constant L>0 such that(42)ht,x1,y1−ht,x2,y2≤Lmaxx1−x2,y1−y2,for all t∈1,T and x1,x2,y1,y2∈ℬ.For convenience, we define(43)λit=ailogρitΓρi+1,i=1,2,(44)ψℓγ=logγTα1Γγ−ℓ+1+χℓTlogγζ+χℓζlogγTα1χζ,TΓγ+1,(45)Ψℓ=β∗ψℓγ+χℓTλ1ζ+χℓζλ2Tχζ,T,where β∗=maxt∈1,Tβt.


### 4.1. Compactness Case

In this subsection, we assume the compactness of solution and investigate the existence of it through employing the Krasnoselskii’s fixed point theorem to deduce the first result about the existence of solution of the GSLLE.Theorem 3.
SupposeH1 and H2 hold. If(46)maxΨ0,Ψℓ<1,where Ψℓ is defined as in (45), then there is at least one solution to the B.V.P (1) and (2) in 1,T.Proof.
For the purpose of proving this theorem, the closed ballCϵ is defined in the following manner:(47)Cε=xt∈Y:Dℓxt∈Y,xY≤ε,where ε is the radius and assumed so that(48)ε≥Qmaxψ0γ+μ,ψℓγ+μ−maxΨ0,Ψℓ,where ψℓ⋅ is defined as in (44).
One can confirm thatCε is a closed convex bounded set. Actually, this paves the way to implement the Krasnoselskii’s fixed point theorem. To achieve this goal, we intend to prove that S1x1+S2x2∈Cε, ∀x1t,x2t∈Cε. So, to perform this task, we start with the following assertion:(49)S1x1t≤1αtJHγ+μht,x1t,DCHℓx1t+χt,Tχζ,TJHγ+μhζ,x1ζ,DCHℓx1ζ+χt,ζχT,ζJHγ+μhT,x1T,DCHℓx1T≤Qα1Γγ+μ∫1tlogtsγ+μ−1dss+χ0Tχζ,T∫1ζlogζsγ+μ−1dss+χ0ζχT,ζ∫1TlogTsγ+μ−1dss=Qα1Γγ+μ+1logγ+μt+χ0Tlogγ+μζ+χ0ζlogγ+μTχζ,T.
It is known thatlogt is positive and increasing on 1,T which implies that logγ+μt≤logγ+μT. Consequently, we have(50)S1x1≤Qψ0γ+μ.
On the other hand, we have(51)S2x2≤JHγβtx2tαt+χt,Tχζ,Ta1Jρ1x2ζ+JHγβζx2ζαζ+χt,ζχT,ζa2JHρ2x2T+JHγβTx2TαT≤xβ∗logγTα1Γγ+1+χTχζ,Ta1logρ1ζΓρ1+1+β∗logγζα1Γγ+1+χζχT,ζa2logρ2TΓρ2+1+β∗logγTα1Γγ+1=xβ∗ψγ+χTλ1ζ+χζλ2Tχζ,T≤εΨ0.
Now, from (50) and (51), we obtain(52)S1x1+S2x2≤Qψ0γ+μ+εΨ0.
Similarly, we obtain(53)DCHℓS1x1≤Qψℓγ+μ,DCHℓS2x2≤εΨℓ.
Accordingly, the two inequalities above imply(54)DCHℓS1x1+DCHℓS2x2≤Qψℓγ+μ+εΨℓ.
Based on (52) and (54), we can obtain the following assertion:(55)S1x1+S2x2Y≤Qmaxψ0γ+μ,ψℓγ+μ+εmaxΨ0,Ψℓ.
That is,(56)S1x1+S2x2Y≤ε,∀x1,x2∈Cε,∀t∈1,T.
Hence, the first condition given in Theorem1 is satisfied. For the purpose of showing the second condition of the same theorem which requires to prove that S2 is a contraction mapping, we intend to take into account the term S2x1−S2x2 in accordance with the following manner:(57)S2x1−S2x2≤JHγβtx1t−x2tαt+χt,Tχζ,Ta1JHρ1x1ζ−x2ζ+JHγβζx1ζ−x2ζαζ+χt,ζχT,ζa2JHρ2x1T−x2T+JHγβTx1T−x2TαT≤β∗logγTα1Γγ+1+χ0Tχζ,Ta1logρ1ζΓρ1+1+β∗logγζα1Γγ+1+χ0ζχζ,Ta2logρ2TΓρ2+1+β∗logγTα1Γγ+1x1−x2.
In other words,(58)S2x1−S2x2≤Ψ0x1−x2.
Similarly, we can attain the following inequality:(59)DCHℓS2x1−DCHℓS2x2≤Ψℓx1−x2.
Again, applying (58) and (59) into the norm ⋅Y defined above yields(60)S1x1−S2x2Y=maxΨ0,Ψℓx1−x2Y.
Due tomaxΨ0,Ψℓ<1 as we previously assumed, then S2 is indeed contraction mapping.
In the meantime, to address the last condition of Theorem1, we should first note that S1 is a continuous operator on Cε due to that the function h so does. Therefore, the rest claim that should be proved is to show that such operator is compact. In accordance with Lemma 5, it is sufficient to prove that S1Cε is equicontinuous and uniformly bounded. In regard to the boundedness issue, its enough to track the following assertions:(61)S1xY=maxmax1≤t≤TS1xt,max1≤t≤TDCHℓS1xt≤Qmaxψ0γ+μ,ψℓγ+μ.
Thus, it remains to prove thatS1 is an equicontinuous operator. For this purpose, we intend to estimate the term S1xt2−S1xt1, where t1,t2∈1,T with t2>t1, as follows:(62)S1xt2−S1xt1≤1Γγ+μ∫1t1logt2sγ+μ−1−logt1sγ+μ−1hs,x1s,DCHℓx1sαsdss+1Γγ+μ∫t1t2logt2sγ+μ−1hs,xs,DCHℓxsαsdss+χt2,T−χt1,Tχζ,TΓγ+μ∫1ζlogζsγ+μ−1hs,xs,DCHℓxsαsdss+χt2,ζ−χt1,ζχT,ζΓγ+μ∫1TlogTsγ+μ−1hs,xs,DCHℓxsαsdss≤Qα1Γγ+μ+1logγ+μt2−logγ+μt1+Qα1Γγ+μ+1logt2t1γ+μ+χt2,T−χt1,TQlogζγ+μα1χζ,TΓγ+μ+1+χt2,ζ−χt1,ζQlogTγ+μα1χT,ζΓγ+μ+1=0.
Note that with aiding the results in Lemma9, the right-hand side goes uniformly to zero as t2⟶t1. By the same method, we can deduce that(63)DCHℓS1xt2−DCHℓS1xt1,goes uniformly to zero as t2⟶t1. Thus, based on Lemma 5, one can conclude that S1 is an equicontinuous operator, proving the third condition of Krasnoselskii’s Theorem 1, and hence there is at least one solution of the main problems (1) and (2) on 1,T, as required.


### 4.2. Noncompactness Case

The existence of solution for the FoLE in the case of noncompactness of solution can be further explored by utilizing Kuratowski’s measure of noncompactness through applying Darbo’s fixed point Theorem2. This matter can be addressed by considering the next existence result.Theorem 4.
Assume that the two assumptionsH1 and H3 are satisfied. If(64)maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓ<1,where ψℓ⋅ and Ψℓ are defined in (44) and (45), respectively, then the main problems (1) and (2) have at least one solution on 1,T.Proof.
Due to the matter of finding a solution of the main problem is just considered as a fixed point problem, we define operatorS:ℬ⟶ℬ and its fractional derivative DℓS:ℬ⟶ℬ as previously mentioned in (31) and (31), respectively. To begin the proof of this result, we first need to show that the operator S:Cε⟶Cε is continuous where Cε is defined in (47), but with radius(65)ε≥Nmaxψ0γ+μ,ψℓγ+μ−maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓ,and N=supt∈1,Tht,0,0<∞. Obviously, the set Cε is closed, bounded, and convex nonempty subset of the Banach space Y. Now, we need to prove that S(Cε) ⊂Cε and S is continuous on Cε.
To show thatS:Cε⟶Cε is continuous, we let xnn∈ℕ to be a sequence of a Banach space Y such that xn⟶x as n⟶∞. If t∈1,T, then(66)S1xnt−S1xt≤1α1JHγ+μht,xnt,DCHℓxnt−ht,xt,DCHℓxt+1α1JHγ+μhζ,xnζ,DCHℓxnζ−hζ,xζ,DCHℓxζ+1α1JHγ+μhT,xnT,DCHℓxnT−hT,xt,DCHℓxT.
Consequently, from the assumptionH3, we have(67)S1xnt−S1xt≤Lxn−xYα1Γγ+μ+1logγ+μT+χTlogγ+μζ+χζlogγ+μTχζ,T≤Lψ0γ+μxn−xY.
Also,(68)S2xnt−S2xt≤β∗α1JHγxnt−xt+χt,TχT,ζa1JHρ1xnζ−xζ+β∗α1JHγxnζ−xζ+χt,ζχT,ζa2JHρ2xnT−xT+β∗α1JHγxnT−xT≤β∗logγTα1Γγ+1xn−xY+χTχζ,Ta1logρ1ζΓρ1+1+β∗logγζα1Γγ+1xn−xY+χζχζ,Ta2logρ2TΓρ2+1+β∗logγTα1Γγ+1xn−xY=Ψ0xn−xY.
Hence, we deduce that(69)S1xn−S1x≤Lψ0γ+μ+Ψ0xn−xY.
Similarly, we can attain the following assertion:(70)DCHℓxn−DCHℓx≤Lψℓγ+μ+Ψℓxn−xY.
In conclusion, we get(71)Sxn−SxY≤maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓxn−xY.
Asn⟶∞, then Sxn−SxY⟶0 which implies that the operator S is continuous on the space Y.
In order to prove that S(Cε)⊂Cε, take x∈Cε. In view of (61), we get(72)S1xY≤Lε+Nmaxψ0γ+μ,ψℓγ+μ.
Indeed, fort∈1,T, we have(73)S2xt≤β∗εlogγ+μTα1Γγ+1+χTχζ,Ta1εlogρ1ζΓρ1+1+β∗εlogγζα1Γγ+1+χζχζ,Ta2εlogρ2TΓρ2+1+β∗εlogγTα1Γγ+1εΨ0.
Similarly, we can deduce thatDCHℓS2xt≤εΨℓ which with the inequality above lead to(74)S2xY≤εmaxΨ0,Ψℓ.
Based on (72) and (74), we observe that SxY≤ε which concludes that S(Cε)⊂Cε.
The final step of this proof is to show that the operatorS:Cε⟶Cε satisfies the inequality of the Kuratowski measure of noncompactness in Theorem 2. To do this, let W be a closed subset of Cε such that there exists Wi,i=1,2,…,n and W⊆∪i=1nWi. Then,(75)σYSW=infκ:supSx1t−Sx2tY≤κ,x1,x2∈Wi=infκ:supmaxSx1t−Sx2t,DCHℓSx1t−DCHℓSx2t,x1,x2∈Wi≤κ=maxinfκ:supSx1t−Sx2t,x1,x2∈Wi≤κinfκ:supDCHℓSx1t−DCHℓSx2t,x1,x2∈Wi≤κ=maxσSW,σDCHℓSW.
Also, let there existU⊆∪i=1nUi and V⊆∪i=1nVi such that(76)Ui=JHμht,xit,DCHℓxitαt:xi∈Wi,t∈1,T,Vi=ht,xit,DCHℓxit:xi∈Wi,t∈1,T.
In view of continuity of the functionsh and α, the sets U and V/α are bounded and equicontinuous and so with the last item of Lemma 4, we obtain(77)σU≤JHμσVαt≤1α1JHμσV,and with Definition 3, we get(78)σV=infκ:supht,xi1t,DCHℓxi1t−ht,xi2t,DCHℓxi2t≤κ,xi1,xi2∈Wi≤infκ:supLmaxxi1−xi2,DCHℓxi1−DCHℓxi2≤κ,xi1,xi2∈Wi=Linfκ/L:supxi1−xi2Y≤κL,xi1,xi2∈Wi=LσW,which implies that(79)σU≤Llogμtα1Γμ+1σW.
From the second and third items of Lemma4, we get(80)σS1W≤σJHγHJμht,xt,DCHℓxtαt+χTχζ,TσJHγJHμhζ,xζ,DCHℓxζαt+χζχζ,TσHJγHJμhT,xT,CHDℓxTαt,x∈W≤HJγσUt+HJγγσUT+HJγσUζ≤Lα1Γγ+μ+1logγ+μT+χTlogγ+μζ+χzetalogγ+μTχζ,TσW=Lψ0γ+μσW.
In the similar way, we have(81)σS2W≤σJHγβtxtαt+χt,TχT,ζa1JHρ1xζ+JHγβζxζαζ+χt,ζχζ,Ta2JHρ2xT+JHγβTxTαT,x∈W≤Ψ0σW.
Hence,(82)σSW≤σS1W+σS2W≤Lψ0γ+μ+Ψ0σW.
Similarly, we can follow the same procedure in regard toDCHℓSW. In other word, we can obtain the following assertion:(83)σDCHℓSW≤σDCHℓS1W+σDCHℓS2W≤ψℓγ+μ+ΨℓσW.
Substituting (82) and (83) into (75) concludes that(84)σYSW≤maxLψ0γ+μ+Ψ0,ψℓγ+μ+ΨℓσW.
In view of the Darbo fixed point theorem, we confirm that the operatorS has a fixed point which represents a solution of the main problems (1) and (2).


## 4.1. Compactness Case

In this subsection, we assume the compactness of solution and investigate the existence of it through employing the Krasnoselskii’s fixed point theorem to deduce the first result about the existence of solution of the GSLLE.Theorem 3.
SupposeH1 and H2 hold. If(46)maxΨ0,Ψℓ<1,where Ψℓ is defined as in (45), then there is at least one solution to the B.V.P (1) and (2) in 1,T.Proof.
For the purpose of proving this theorem, the closed ballCϵ is defined in the following manner:(47)Cε=xt∈Y:Dℓxt∈Y,xY≤ε,where ε is the radius and assumed so that(48)ε≥Qmaxψ0γ+μ,ψℓγ+μ−maxΨ0,Ψℓ,where ψℓ⋅ is defined as in (44).
One can confirm thatCε is a closed convex bounded set. Actually, this paves the way to implement the Krasnoselskii’s fixed point theorem. To achieve this goal, we intend to prove that S1x1+S2x2∈Cε, ∀x1t,x2t∈Cε. So, to perform this task, we start with the following assertion:(49)S1x1t≤1αtJHγ+μht,x1t,DCHℓx1t+χt,Tχζ,TJHγ+μhζ,x1ζ,DCHℓx1ζ+χt,ζχT,ζJHγ+μhT,x1T,DCHℓx1T≤Qα1Γγ+μ∫1tlogtsγ+μ−1dss+χ0Tχζ,T∫1ζlogζsγ+μ−1dss+χ0ζχT,ζ∫1TlogTsγ+μ−1dss=Qα1Γγ+μ+1logγ+μt+χ0Tlogγ+μζ+χ0ζlogγ+μTχζ,T.
It is known thatlogt is positive and increasing on 1,T which implies that logγ+μt≤logγ+μT. Consequently, we have(50)S1x1≤Qψ0γ+μ.
On the other hand, we have(51)S2x2≤JHγβtx2tαt+χt,Tχζ,Ta1Jρ1x2ζ+JHγβζx2ζαζ+χt,ζχT,ζa2JHρ2x2T+JHγβTx2TαT≤xβ∗logγTα1Γγ+1+χTχζ,Ta1logρ1ζΓρ1+1+β∗logγζα1Γγ+1+χζχT,ζa2logρ2TΓρ2+1+β∗logγTα1Γγ+1=xβ∗ψγ+χTλ1ζ+χζλ2Tχζ,T≤εΨ0.
Now, from (50) and (51), we obtain(52)S1x1+S2x2≤Qψ0γ+μ+εΨ0.
Similarly, we obtain(53)DCHℓS1x1≤Qψℓγ+μ,DCHℓS2x2≤εΨℓ.
Accordingly, the two inequalities above imply(54)DCHℓS1x1+DCHℓS2x2≤Qψℓγ+μ+εΨℓ.
Based on (52) and (54), we can obtain the following assertion:(55)S1x1+S2x2Y≤Qmaxψ0γ+μ,ψℓγ+μ+εmaxΨ0,Ψℓ.
That is,(56)S1x1+S2x2Y≤ε,∀x1,x2∈Cε,∀t∈1,T.
Hence, the first condition given in Theorem1 is satisfied. For the purpose of showing the second condition of the same theorem which requires to prove that S2 is a contraction mapping, we intend to take into account the term S2x1−S2x2 in accordance with the following manner:(57)S2x1−S2x2≤JHγβtx1t−x2tαt+χt,Tχζ,Ta1JHρ1x1ζ−x2ζ+JHγβζx1ζ−x2ζαζ+χt,ζχT,ζa2JHρ2x1T−x2T+JHγβTx1T−x2TαT≤β∗logγTα1Γγ+1+χ0Tχζ,Ta1logρ1ζΓρ1+1+β∗logγζα1Γγ+1+χ0ζχζ,Ta2logρ2TΓρ2+1+β∗logγTα1Γγ+1x1−x2.
In other words,(58)S2x1−S2x2≤Ψ0x1−x2.
Similarly, we can attain the following inequality:(59)DCHℓS2x1−DCHℓS2x2≤Ψℓx1−x2.
Again, applying (58) and (59) into the norm ⋅Y defined above yields(60)S1x1−S2x2Y=maxΨ0,Ψℓx1−x2Y.
Due tomaxΨ0,Ψℓ<1 as we previously assumed, then S2 is indeed contraction mapping.
In the meantime, to address the last condition of Theorem1, we should first note that S1 is a continuous operator on Cε due to that the function h so does. Therefore, the rest claim that should be proved is to show that such operator is compact. In accordance with Lemma 5, it is sufficient to prove that S1Cε is equicontinuous and uniformly bounded. In regard to the boundedness issue, its enough to track the following assertions:(61)S1xY=maxmax1≤t≤TS1xt,max1≤t≤TDCHℓS1xt≤Qmaxψ0γ+μ,ψℓγ+μ.
Thus, it remains to prove thatS1 is an equicontinuous operator. For this purpose, we intend to estimate the term S1xt2−S1xt1, where t1,t2∈1,T with t2>t1, as follows:(62)S1xt2−S1xt1≤1Γγ+μ∫1t1logt2sγ+μ−1−logt1sγ+μ−1hs,x1s,DCHℓx1sαsdss+1Γγ+μ∫t1t2logt2sγ+μ−1hs,xs,DCHℓxsαsdss+χt2,T−χt1,Tχζ,TΓγ+μ∫1ζlogζsγ+μ−1hs,xs,DCHℓxsαsdss+χt2,ζ−χt1,ζχT,ζΓγ+μ∫1TlogTsγ+μ−1hs,xs,DCHℓxsαsdss≤Qα1Γγ+μ+1logγ+μt2−logγ+μt1+Qα1Γγ+μ+1logt2t1γ+μ+χt2,T−χt1,TQlogζγ+μα1χζ,TΓγ+μ+1+χt2,ζ−χt1,ζQlogTγ+μα1χT,ζΓγ+μ+1=0.
Note that with aiding the results in Lemma9, the right-hand side goes uniformly to zero as t2⟶t1. By the same method, we can deduce that(63)DCHℓS1xt2−DCHℓS1xt1,goes uniformly to zero as t2⟶t1. Thus, based on Lemma 5, one can conclude that S1 is an equicontinuous operator, proving the third condition of Krasnoselskii’s Theorem 1, and hence there is at least one solution of the main problems (1) and (2) on 1,T, as required.


## 4.2. Noncompactness Case

The existence of solution for the FoLE in the case of noncompactness of solution can be further explored by utilizing Kuratowski’s measure of noncompactness through applying Darbo’s fixed point Theorem2. This matter can be addressed by considering the next existence result.Theorem 4.
Assume that the two assumptionsH1 and H3 are satisfied. If(64)maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓ<1,where ψℓ⋅ and Ψℓ are defined in (44) and (45), respectively, then the main problems (1) and (2) have at least one solution on 1,T.Proof.
Due to the matter of finding a solution of the main problem is just considered as a fixed point problem, we define operatorS:ℬ⟶ℬ and its fractional derivative DℓS:ℬ⟶ℬ as previously mentioned in (31) and (31), respectively. To begin the proof of this result, we first need to show that the operator S:Cε⟶Cε is continuous where Cε is defined in (47), but with radius(65)ε≥Nmaxψ0γ+μ,ψℓγ+μ−maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓ,and N=supt∈1,Tht,0,0<∞. Obviously, the set Cε is closed, bounded, and convex nonempty subset of the Banach space Y. Now, we need to prove that S(Cε) ⊂Cε and S is continuous on Cε.
To show thatS:Cε⟶Cε is continuous, we let xnn∈ℕ to be a sequence of a Banach space Y such that xn⟶x as n⟶∞. If t∈1,T, then(66)S1xnt−S1xt≤1α1JHγ+μht,xnt,DCHℓxnt−ht,xt,DCHℓxt+1α1JHγ+μhζ,xnζ,DCHℓxnζ−hζ,xζ,DCHℓxζ+1α1JHγ+μhT,xnT,DCHℓxnT−hT,xt,DCHℓxT.
Consequently, from the assumptionH3, we have(67)S1xnt−S1xt≤Lxn−xYα1Γγ+μ+1logγ+μT+χTlogγ+μζ+χζlogγ+μTχζ,T≤Lψ0γ+μxn−xY.
Also,(68)S2xnt−S2xt≤β∗α1JHγxnt−xt+χt,TχT,ζa1JHρ1xnζ−xζ+β∗α1JHγxnζ−xζ+χt,ζχT,ζa2JHρ2xnT−xT+β∗α1JHγxnT−xT≤β∗logγTα1Γγ+1xn−xY+χTχζ,Ta1logρ1ζΓρ1+1+β∗logγζα1Γγ+1xn−xY+χζχζ,Ta2logρ2TΓρ2+1+β∗logγTα1Γγ+1xn−xY=Ψ0xn−xY.
Hence, we deduce that(69)S1xn−S1x≤Lψ0γ+μ+Ψ0xn−xY.
Similarly, we can attain the following assertion:(70)DCHℓxn−DCHℓx≤Lψℓγ+μ+Ψℓxn−xY.
In conclusion, we get(71)Sxn−SxY≤maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓxn−xY.
Asn⟶∞, then Sxn−SxY⟶0 which implies that the operator S is continuous on the space Y.
In order to prove that S(Cε)⊂Cε, take x∈Cε. In view of (61), we get(72)S1xY≤Lε+Nmaxψ0γ+μ,ψℓγ+μ.
Indeed, fort∈1,T, we have(73)S2xt≤β∗εlogγ+μTα1Γγ+1+χTχζ,Ta1εlogρ1ζΓρ1+1+β∗εlogγζα1Γγ+1+χζχζ,Ta2εlogρ2TΓρ2+1+β∗εlogγTα1Γγ+1εΨ0.
Similarly, we can deduce thatDCHℓS2xt≤εΨℓ which with the inequality above lead to(74)S2xY≤εmaxΨ0,Ψℓ.
Based on (72) and (74), we observe that SxY≤ε which concludes that S(Cε)⊂Cε.
The final step of this proof is to show that the operatorS:Cε⟶Cε satisfies the inequality of the Kuratowski measure of noncompactness in Theorem 2. To do this, let W be a closed subset of Cε such that there exists Wi,i=1,2,…,n and W⊆∪i=1nWi. Then,(75)σYSW=infκ:supSx1t−Sx2tY≤κ,x1,x2∈Wi=infκ:supmaxSx1t−Sx2t,DCHℓSx1t−DCHℓSx2t,x1,x2∈Wi≤κ=maxinfκ:supSx1t−Sx2t,x1,x2∈Wi≤κinfκ:supDCHℓSx1t−DCHℓSx2t,x1,x2∈Wi≤κ=maxσSW,σDCHℓSW.
Also, let there existU⊆∪i=1nUi and V⊆∪i=1nVi such that(76)Ui=JHμht,xit,DCHℓxitαt:xi∈Wi,t∈1,T,Vi=ht,xit,DCHℓxit:xi∈Wi,t∈1,T.
In view of continuity of the functionsh and α, the sets U and V/α are bounded and equicontinuous and so with the last item of Lemma 4, we obtain(77)σU≤JHμσVαt≤1α1JHμσV,and with Definition 3, we get(78)σV=infκ:supht,xi1t,DCHℓxi1t−ht,xi2t,DCHℓxi2t≤κ,xi1,xi2∈Wi≤infκ:supLmaxxi1−xi2,DCHℓxi1−DCHℓxi2≤κ,xi1,xi2∈Wi=Linfκ/L:supxi1−xi2Y≤κL,xi1,xi2∈Wi=LσW,which implies that(79)σU≤Llogμtα1Γμ+1σW.
From the second and third items of Lemma4, we get(80)σS1W≤σJHγHJμht,xt,DCHℓxtαt+χTχζ,TσJHγJHμhζ,xζ,DCHℓxζαt+χζχζ,TσHJγHJμhT,xT,CHDℓxTαt,x∈W≤HJγσUt+HJγγσUT+HJγσUζ≤Lα1Γγ+μ+1logγ+μT+χTlogγ+μζ+χzetalogγ+μTχζ,TσW=Lψ0γ+μσW.
In the similar way, we have(81)σS2W≤σJHγβtxtαt+χt,TχT,ζa1JHρ1xζ+JHγβζxζαζ+χt,ζχζ,Ta2JHρ2xT+JHγβTxTαT,x∈W≤Ψ0σW.
Hence,(82)σSW≤σS1W+σS2W≤Lψ0γ+μ+Ψ0σW.
Similarly, we can follow the same procedure in regard toDCHℓSW. In other word, we can obtain the following assertion:(83)σDCHℓSW≤σDCHℓS1W+σDCHℓS2W≤ψℓγ+μ+ΨℓσW.
Substituting (82) and (83) into (75) concludes that(84)σYSW≤maxLψ0γ+μ+Ψ0,ψℓγ+μ+ΨℓσW.
In view of the Darbo fixed point theorem, we confirm that the operatorS has a fixed point which represents a solution of the main problems (1) and (2).


## 5. Numerical Examples

In this part, the main B.V.P. reported in (1) and (2) is addressed via two numerical examples. These examples are handled for demonstrating the two results formulated in the previous section.


### 5.1. Example 1

Consider the B.V.P. as follows:(85)Dc7/4e−tDc2/3+130t1/2+1100xt=e−t5+etxtDc1/4xt1+xtDc1/4xt,t∈1,2,subject to the conditions:(86)x1=0,x32=71000J2/3x32,x2=13500J2/3x2.Comparing this problem with the main problem yields to outline thatμ=7/4, γ=2/3, ℓ=1/4, ζ=3/2, T=2, a1=7/1000, and a2=13/500,ρ1=2/5,ρ2=4/7. On the other hand, we have αt=t3/2 and βt=1/30t1/2+1/100, which give α1=e−2,α2=e−1 and β∗=0.0571. Moreover, we have(87)ht,xt,Dc1/2xt=e−t5+etxtDc1/2xt1+xtDc1/2xt,which is continuous and satisfies(88)ht,xt,Dc1/2xt≤Qt=1et5+et.This implies that the second conditionH2 is satisfied with Q=0.0476634. By carrying out Mathematica software, it is easy to obtain χT,ζ=2.03966,(89)χ0T=0.602723,χ0ζ=0.403831,χℓT=1.21289,χℓζ=0.689803,ψ0γ=9.00499,ψℓγ=11.3628,Ψ0=0.566214,Ψℓ=0.741573,which implies that maxΨ0,Ψℓ=0.741573<1. Therefore, based on Theorem 1, we conclude that the B.V.P. (85) and (86) have at least one solution on 1,2.


### 5.2. Example 2

Consider the following B.V.P.:(90)Dc5/3e−tDc1/2+t1/2+1xt=xtxt+1sin4t10t+92+Dc1/3xtDc1/3xt+1cos2tt+12+t25,t∈1,π,subject to the conditions:(91)x1=0,x43=9100J1/2x43,xπ=5171J1/2xπ.Comparing (90) and (91) with (1) and (2) leads us to obtain μ=5/3, γ=1/2, ℓ=1/3, ζ=4/3, T=π, a1=9/100, a2=5/171, ρ1=3/7, and ρ2=3/8. In the meantime, we obtain αt=e−t and βt=t1/2+1 which implies α1=e−π, α2=e−1, and β∗=π+1. Furthermore, we have(92)ht,xt,Dc1/3xt=xtxt+1sin4t10t+92+Dc1/3xtDc1/3xt+1cos2tt+12+t25,which is a continuous function. In order to see that it satisfies the third condition H3, we have(93)ht,x1t,Dc1/3x1t−ht,x2t,Dc1/3x2t≤110t+92x1t−x2t+1tt+12Dc1/3x1t−Dc1/3x2t≤1192x1t−x2t+122Dc1/3x1t−Dc1/3x2t≤1192+122maxx1t−x2t,Dc1/3x1t−Dc1/3x2t,which implies that the function h satisfies the third condition H3 with L=365/1444. By carrying out Mathematica software, it is easy to obtain χT,ζ=22.0176,(94)χ0T=1.51194,χ0ζ=0.694439,χℓT=6.60417,χℓζ=1.58684,ψ0γ+μ=29.7801,ψℓγ+μ=32.902,Ψ0=0.181856,Ψℓ=0.582983,which implies that(95)maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓ=max0.510466,0.946042=0.946042<1.Hence, according to Theorem4, we deduce that there is at least one solution to the B.V.P. (90) and (91) on the interval 1,π.


## 5.1. Example 1

Consider the B.V.P. as follows:(85)Dc7/4e−tDc2/3+130t1/2+1100xt=e−t5+etxtDc1/4xt1+xtDc1/4xt,t∈1,2,subject to the conditions:(86)x1=0,x32=71000J2/3x32,x2=13500J2/3x2.Comparing this problem with the main problem yields to outline thatμ=7/4, γ=2/3, ℓ=1/4, ζ=3/2, T=2, a1=7/1000, and a2=13/500,ρ1=2/5,ρ2=4/7. On the other hand, we have αt=t3/2 and βt=1/30t1/2+1/100, which give α1=e−2,α2=e−1 and β∗=0.0571. Moreover, we have(87)ht,xt,Dc1/2xt=e−t5+etxtDc1/2xt1+xtDc1/2xt,which is continuous and satisfies(88)ht,xt,Dc1/2xt≤Qt=1et5+et.This implies that the second conditionH2 is satisfied with Q=0.0476634. By carrying out Mathematica software, it is easy to obtain χT,ζ=2.03966,(89)χ0T=0.602723,χ0ζ=0.403831,χℓT=1.21289,χℓζ=0.689803,ψ0γ=9.00499,ψℓγ=11.3628,Ψ0=0.566214,Ψℓ=0.741573,which implies that maxΨ0,Ψℓ=0.741573<1. Therefore, based on Theorem 1, we conclude that the B.V.P. (85) and (86) have at least one solution on 1,2.


## 5.2. Example 2

Consider the following B.V.P.:(90)Dc5/3e−tDc1/2+t1/2+1xt=xtxt+1sin4t10t+92+Dc1/3xtDc1/3xt+1cos2tt+12+t25,t∈1,π,subject to the conditions:(91)x1=0,x43=9100J1/2x43,xπ=5171J1/2xπ.Comparing (90) and (91) with (1) and (2) leads us to obtain μ=5/3, γ=1/2, ℓ=1/3, ζ=4/3, T=π, a1=9/100, a2=5/171, ρ1=3/7, and ρ2=3/8. In the meantime, we obtain αt=e−t and βt=t1/2+1 which implies α1=e−π, α2=e−1, and β∗=π+1. Furthermore, we have(92)ht,xt,Dc1/3xt=xtxt+1sin4t10t+92+Dc1/3xtDc1/3xt+1cos2tt+12+t25,which is a continuous function. In order to see that it satisfies the third condition H3, we have(93)ht,x1t,Dc1/3x1t−ht,x2t,Dc1/3x2t≤110t+92x1t−x2t+1tt+12Dc1/3x1t−Dc1/3x2t≤1192x1t−x2t+122Dc1/3x1t−Dc1/3x2t≤1192+122maxx1t−x2t,Dc1/3x1t−Dc1/3x2t,which implies that the function h satisfies the third condition H3 with L=365/1444. By carrying out Mathematica software, it is easy to obtain χT,ζ=22.0176,(94)χ0T=1.51194,χ0ζ=0.694439,χℓT=6.60417,χℓζ=1.58684,ψ0γ+μ=29.7801,ψℓγ+μ=32.902,Ψ0=0.181856,Ψℓ=0.582983,which implies that(95)maxLψ0γ+μ+Ψ0,Lψℓγ+μ+Ψℓ=max0.510466,0.946042=0.946042<1.Hence, according to Theorem4, we deduce that there is at least one solution to the B.V.P. (90) and (91) on the interval 1,π.


## 6. Conclusion

This paper has introduced two novel results associated with the existence of solutions for the fractional-order Langevin problem that has been formulated in the sense of Caputo–Hadamard fractional-order derivative operator. These two results have been shown in light of two well-known fixed point theorems, namely, the Krasnoselskii’s theorem in the case of compactness and Darbo’s theorem in the case of noncompactness. Two numerical examples have been demonstrated for completeness and illustrate the implementations of our idea on scientific problems. Here, we applied this idea on fractional differential equations; we think that applying this idea on the fractional differential inclusion will be useful in the future study.



---

*Source: 9995969-2021-11-23.xml*