# On the Existence and Uniqueness of Solution to Fractional-Order Langevin Equation

Authors: Ahmed Salem; Noorah Mshary
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

In this work, we give sufficient conditions to investigate the existence and uniqueness of solution to fractional-order Langevin equation involving two distinct fractional orders with unprecedented conditions (three-point boundary conditions including two nonlocal integrals). The problem is introduced to keep track of the progress made on exploring the existence and uniqueness of solution to the fractional-order Langevin equation. As a result of employing the so-called Krasnoselskii and Leray-Schauder alternative fixed point theorems and Banach contraction mapping principle, some novel results are presented in regarding to our main concern. These results are illustrated through providing three examples for completeness.


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## Body

## 1. Introduction

Unquestionably, the fractional calculus, which requires the order of the functional operator of calculus to be turned into the fractional case, is a forceful mathematical argument for providing much and more flexibility in dealing with many real-world applications [1, 2]. More particularly, the fractional-order differential equations (FODEs), which are considered a cornerstone of this field, have been extensively employed in modeling several phenomena and numerous scientific applications such as diffusion modeling [3], robot manipulators [4], economics [5], and many more [6–9]. In view of its outstanding importance, lots of mathematicians have paid their attentions to deeply explore the initial and/or boundary value problems (BVPs) established based on such equations (see [10–13]). Actually, these problems, which may have different boundary conditions, describe several physical real-world issues like thermoelasticity, underground water flow, and heat conduction. One of the most significant of those problems is known as the Langevin equation which was formulated for the first time in 1908 by Langevin, who still much earlier described the progress of some physical matters in fluctuating settings [14, 16]. It came out, later on, that such equation had assisted physicians successfully to identify some techniques related to the abnormal diffusion. It also, afterward, succeeded in describing some other techniques in economic field that contain the cost index operating [17]. Besides, its generalization played, and still plays, a significant duty in describing the noise causes in the field of critical dynamics [18]. In 1996, Mainardi and Pironi fractionalized such equation and hence turned it into the fractional-order case to be named later the fractional-order Langevin equation (FoLE) [19]. Several articles then made this equation up at the top of their works (see, e.g., [20–22]). It has been recently shown that such equation, for example, but not limited to, plays notable role in outlining the reaction-diffusion models [23, 24] and describes some processes associated with porous media [25, 26].In the same framework, many efforts have been underway to explore the two most main aspects associated with the solutions of the FoLEs: the existence and uniqueness [27]. In view of studying such two aspects, several recent articles have been published. In particular, through using some special boundary conditions, some applicable results related to the existence of solutions for the FoLE have been reported in [28]. Both aspects have been investigated when the coupled incommensurate fractional-order systems were handled by Salem and Alnegga and Salem et al. in [15, 29]. However, in more recent times, numerous outcomes associated with those aspects have, moreover, been informed widely (see [1, 14, 21, 30–34]). In more concrete terms, some useful theoretical tools, like the so-called Banach contraction principle and Krasnoselskii’s fixed point theorem, have been employed to explore the existence and uniqueness of solutions for several nonlinear FODEs such as the implicit FODEs [35], the fractional-order Cauchy problems [36].As per the best of our knowledge, there are few published works that address the FoLE with nonlocal integral boundary conditions as informed in [14]. In view of this motivation, this work examines the existence and uniqueness of solution for the nonlinear FoLE with three-point boundary conditions involving two nonlocal integrals. Actually, this boundary value problem, with two different values of fractional order, has the following form:(1)cDrcDs+ℓzt=ft,zt,cDρzt,t∈0,T,where cDr is the Caputo derivative operator of order r in which k−1<r≤k and k∈ℕ. In this work, we assume that the function f:0,T×ℝ×ℝ→ℝ is a continuous function, and 1<r<2, 0<ρ≤s≤1, and ℓ∈ℝ are the dissipative parameters. We take here the function f, in general form, constitutes from position (displacement) zt and fractional velocity Dcρztof the particle at time t∈0,T,T>0. This function perhaps contains external force field, position-dependent phenomenological fluid friction coefficient, intensity of the stochastic force, or zero-mean Gaussian white noise term.The equation (1) subjects to the following boundary conditions:(2)z0=0,zT=α∫0Tzudu,zζ=β∫0ζzudu,where α,β∈ℝ and 0<ζ<T. It is remarkable to see that the first condition indicates the particle begins its motion from the origin point. Also, all conditions can be interpreted informally as “the displacement of the particle at a certain point proportionates the signed area of the region in the tz-plane that is bounded by the graph of z and the t-axis from the origin to this certain point”.One more observation should be recorded; all hypotheses and conditions in this work are assumed on the basis of the various views inspired by the aforementioned cited articles. Anyhow, here is how the rest of this paper is arranged. Section2 introduces some primary preliminaries associated with fractional calculus and some fixed point theorems, while Section 3 deduces some auxiliary lemmas that pave the way for establishing novel theorems related to the existence and uniqueness of the solution for our main problem. Such theorems are illustrated and shown in Section 4, while Section 5 involves one illustrative example to verify all findings, followed by the last section that summarizes the achievements of the whole work.


## 2. Preliminaries

In the whole of this work, the Caputo derivative operator and its inverse operator, the Riemann-Liouville (R-L) integral, are adopted since they suite most physical systems. These two operators as well as some basic concepts and essential preliminaries about fractional calculus are briefly presented here together with some theoretical results associated with the fixed point theorem.Definition 1. [37].
The R-L integral operator of orderrfor a continuous functionz:0,∞→ℝis outlined as(3)Irzt=1Γr∫0tt−vr−1zvdv,where r>0.Definition 2. [37].
The R-L derivative operator of orderrfor ak-times absolutely continuous functionz:0,∞→ℝis outlined as(4)Drzt=1Γk−rddtk∫0tt−vk−r−1zvdv=ddtkIk−rzt,where k−1≤r<k and k∈ℕ.Definition 3. [37].
The Caputo derivative operator of orderr for a k-times absolutely continuous function z:0,∞→ℝ is outlined as(5)cDrzt=1Γk−r∫0tt−vk−r−1zkvdv,where k−1<r≤k and k∈ℕ.The following remark states some important relations chosen from the famous books [37]. Actually, such relations will be used during the whole of this work.Remark 4.
Suppose thatz:0,∞→ℝis a continuous function, ris a positive real. Then,
(i)IrIszt=Ir+szt,s>0
(ii)cDrIszt=Is−rzt, for r∈0,s, while cDrIrzt=zt
(iii)Irtρ=Γρ+1Γρ+r+1tρ+r,ρ>−1
(iv)cDrtρ=Γρ+1Γρ−r+1tρ−r,ρ>−1,ρ≠0,1,⋯,r
(v)cDrtρ=0,ρ=0,1,⋯,rLemma 5.
SupposecDrzis a continuous function, then(6)IrDcrzt=zt+c0+c1t+⋯+ck−1tk−1for k∈ℕ, where k−1<r≤k.Next, some primary keys and essential theorems related to the fixed point theorem are stated for the purpose of getting novel results.Definition 6. [38, 39].
A contraction of a normed spaceXis a mappingS:X→Xthat meets(7)Sz1−Sz2≤δz1−z2,∀z1,z2∈Xfor some δ<1.Theorem 7. (Banach fixed point theorem, Granas and Dugundji [38, 39]).
There is a unique fixed point for every contraction mapping on a complete metric space.Theorem 8. (Krasnoselskii fixed point theorem [38]).
For a convex setBε of a Banach space Xand a mappingSz=S1z+S2zsuch that(1)
S1z1+S2z2∈Bε, for each z1,z2∈Bε(2)
S2 represents a contraction mapping(3)
S1 is continuous and compactThen,S has a fixed point.Theorem 9. [39].
LetO be an open set of the Banach space X=C0,T,ℝ+ such that O≠ϕ. If the operator S:O¯→X is relatively compact, then either(1)
S has a fixed point z∗∈O¯, or(2)
There existsγ∈0,1 and z∗∈∂O such that γSz=z, where ∂O is the boundary of O itself


## 3. Some Deduced Auxiliary Lemmas

In this part, some auxiliary lemmas are introduced in order to pave the way for exploring the existence and uniqueness of the solution for our main problem. In other words, such lemmas will play a major role in transforming the main issue to a fixed point problem. For this purpose, let us introduce the following two results, with noting that the functionft,zt,cDρzt given in (1) will be indicated by ft for simplicity.Lemma 10.
Letf∈C0,T,ℝ. Then, the unique root of the problem given in (1)-(2) can be expressed as(8)zt=Is+rft−ℓIszt+χ1tΛαT+χ2tΛβζ,where(9)χ1t=tst−ζT−ζTs,(10)χ2t=tsT−tT−ζζs,(11)Λαt=αIzt−Is+rft+ℓIszt.Proof.
Applying the integral operatorIr to both sides of (1) and then using Lemma 5. yield(12)cDszt=−ℓIzt+Irft+c2t+c1.
Again, applyingIr to both sides of (12) gives(13)zt=Is+rft−ℓIszt+c2ts+1Γs+2+c1tsΓs+1+c0,where ci’s are arbitrary constants, and i=1,2,3. Applying the first condition of (2) yields c0=0, and hence(14)zt=Is+rft−ℓIszt+c2ts+1Γs+2+c1tsΓs+1.
By using the two other conditions given in (2), we obtain the following:(15)c1TsΓs+1+c2Ts+1Γs+2=ΛαT,(16)c1ζsΓs+1+c2ζs+1Γs+2=Λβζ.
Solving the two equations (15) and (16) gives the following:(17)c1Γs+1=TT−ζζsΛβζ−ζT−ζTsΛαT,(18)c2Γs+2=1T−ζTsΛαT−1T−ζζsΛβζ.Substituting (17) and (18) into (14) yields the result. Conversely, it is easy by aiding with Remark 4. and Lemma 5. to see that the solution (8) satisfies the fractional differential equation (1) and the boundary conditions in (2).Lemma 11.
Letχ1andχ2be defined as given in (9) and (10), respectively. Then,(19)χ1=maxt∈0,Tχ1t=maxζs+1sss+1s+1T−ζTs,1,(20)χ2=maxt∈0,Tχ2t=Ts+1sss+1s+1T−ζζs.Proof.
Lett,ζ∈0,T. Define the function ut as follows:(21)ut=ts∣t−ζ∣=tsζ−t,0≤t≤ζ,tst−ζ,ζ≤t≤T.
One observes that when0≤t≤ζ, we have u′t=ts−1sζ−s+1t, which means that ut is increasing on 0,t0 and it is decreasing on t0,ζ, where t0=sζ/s+1. Hence, we can deduce that ut has a maximum value at t=t0. On the other hand, when ζ<t≤T, we obtain u′t=ts−1s+1t−sζ, which implies that ut is increasing on ζ,T, and so ut has a maximum value at t=T. Therefore, the result holds due to the following:(22)χ1=maxut0T−ζTs,uTT−ζTs.
In similar manner, one can deduce the second statement of this lemma.To follow up on the progress made in exploring the existence and uniqueness of solutions for the FoLEs, some further results are established below, which will surely serve us in the next sections. In this regard, letC=C0,T,ℝ be Banach space of continuous real-valued functions equipped with the usual supremum norm(23)z=sup0≤t≤Tzt,z∈C.Define the following set(24)P=zt:zt∈C,cDρzt∈C,t∈0,T.One can easily verify thatP is a Banach subspace equipped with the following norm:(25)zP=maxz,Dρzc=maxsup0≤t≤T∣zt,sup0≤t≤TcDρzt∣.In view of Lemma10., the main problem given in (1)-(2) will be transformed into its equivalent fixed point problem. In other words, the function zt should satisfy the two statements; Szt=zt and cDρSzt=cDρzt, for all t∈0,T, where S:C→C is an operator defined on the Banach space C. In order to achieve this goal, consider the operator S as defined a little while ago. That is,(26)Szt=Is+rft−ℓIszt+χ1tΛαT+χ2tΛβζ,(27)cDρSzt=Is+r−ρft−ℓIs−ρzt+χ3tΛαT+χ4tΛβζ,where Λα· is defined in (11) and(28)χ3t=cDρχ1t=Γs+2ts−ρΓs−ρ+2T−ζTst−ζs+1−ρs+1,(29)χ4t=cDρχ2t=−Γs+2ts−ρΓs−ρ+2T−ζζst−Ts+1−ρs+1.Next, we introduce the following auxiliary result based on all provided aforementioned principles.Lemma 12.
Letχ3tandχ4tare defined as in (28) and (29), respectively. Then,(30)χ3=χ3=Γs+1Γs−ρ+2T−ζmaxζs−ρ+1Tss−ρs+1s−ρ,s+1T−s−ρ+1ζTρ,(31)χ4=χ4=Γs+1Ts−ρ+1Γs−ρ+2T−ζζsmaxs−ρs+1s−ρ,ρ.Proof.
Lett0=ζs+1−ρ/s+1. Define the function(32)ut=ts−ρt−t0,t∈0,T.
If0≤t≤t0, then ut=ts−ρt0−t and(33)u′t=ts−ρ−1s−ρt0−s−ρ+1twhich leads to ut is increasing on 0,t1 and is decreasing on t1,t0 where(34)t1=s−ρs−ρ+1t0=ζs−ρs+1<t0.
This concludes thatut1 is a maximum value.
Ift0<t≤T, then ut=ts−ρt−t0 and(35)u′t=ts−ρ−1s−ρ+1t−s−ρt0>ts−ρ−1t0>0which leads to ut is increasing on t0,T. This concludes that uT is a maximum value. In similar manner, one can deduce the second statement of this lemma.


## 4. Existence and Uniqueness Results

This section proposes some theoretical results related to the existence and uniqueness of the solution for the main target problem, the FoLE given in (1)-(2). First of all, we establish some mathematical assumptions that will be used from now on. Such assumptions are given below.(36)Δ=ℓΔs+αχ1T+βχ2ζ,(37)Δ′=∣ℓ∣Δ′s+∣α∣χ3T+∣β∣χ4ζ.where(38)Δs=1+χ1Ts+χ2ζsΓs+1,(39)Δ′s=Ts−ρΓs−ρ+1+χ3Ts+χ4ζsΓs+1,and χi, i=1,2,3,4 are given in (19), (20), (30), and (31), respectively.In the same vein, we also introduce some further conditions below to establish our proposed results. In fact, these conditions have been recently introduced in [40] for the purpose of investigating the same issue of ours, but they were adopted for handling another problem.H1 The function ft,zt,cDρzt is continuous on 0,TH2 There exist a positive function ψ:0,T→ℝ+ such that(40)∣ft,zt,cDρzt∣qtH3 There exist two constants M1,M2>0 such that(41)∣ft,z1t,cDρz1−ft,z2,cDρz2∣≤M1∣z1−z2∣+M2cDρz1−cDρz2,t∈0,TH4 There exist a positive function h:0,T→ℝ+ and other nondecreasing function Ψ:ℝ+→ℝ+ such that(42)∣ft,zt,cDρzt∣≤htΨzP,t∈0,TOn the other side, by going back to (26) and (27), one can rewrite each of Szt and cDρSzt, as a sum of two operators S1 and S2. In other words, Szt=S1z+S2zt and cDρSzt=cDρS1z+cDρS2zt, where(43)S1zt=Is+rft−χ1tIs+rfT−χ2tIs+rfζ,S2zt=−ℓIszt+χ1tαIzT+ℓIszT+χ2tβIzζ+ℓIszζ,cDρS1zt=Is+r−ρft−χ3tIs+rfT−χ4tIs+rfζ,cDρS2zt=−ℓIs−ρzt+χ3tαIzT+ℓIszT+χ4tβIzζ+ℓIszζ.After this preparation, we find ourselves ready to present the results that are at the heart of the matter. Here is the first one that deduced on the basis of the Krasnoselskii theorem.Theorem 13.
SupposeH1 and H2 hold. If(44)maxΔ,Δ′<1,where Δ and Δ′ are defined in (36) and (37), respectively, then the problem given in (1)-(2) has at least one solution on 0,T.Proof.
To prove this theorem, define the closed ball,Bε, as follows:(45)Bε=zt∈P:cDρzt∈P,ztP≤ε,where P and zP are defined in Section 3, with radius(46)ε≥qmaxΔs+r,Δ′s+r1−maxΔ,Δ′,where Δ· and Δ′· are defined in (38) and (39), respectively.
One can easily verify thatBε is a convex closed set. However, for the purpose of implementing on the Krasnoselskii theorem given in 2.2, we need firstly to show that S1z1+S2z2∈Bε, for each z1t,z2t∈Bε. To carry out this step, we begin with(47)∣S1z1∣≤Is+rft,z1t,Dρz1t+∣χ1tIs+rfT,z1T,Dρz1T∣+χ2tIs+rfζ,z1ζ,Dρz1ζ≤qtΓs+r∫0tt−vs+r−1dv+∣χ1t∣∫0TT−vs+r−1dv+∣χ2t∣∫0ζζ−vs+r−1dv=ts+r+χ1tTs+r+χ2tζs+rΓs+r+1qt≤1+χ1tTs+r+χ2tζs+rΓs+r+1qt,∣S2z2∣≤∣ℓ∣Is∣z2t∣+∣χ1t∣∣α∣I∣z2T∣+∣ℓ∣Is∣z2T∣+∣χ2t∣∣β∣I∣z2ζ∣+∣ℓ∣Is∣z2ζ∣≤1+χ1tTs+χ2tζsΓs+1ℓ+αχ1tT+βχ2tζz2.
Due toz2t∈Bε, then ∣z2t∣≤z2t≤z2tP≤ε. This matter implies to(48)S1z1+S2z2≤1+χ1∣Ts+r+χ2ζs+rΓs+r+1q+1+χ1Ts+χ2ζsΓs+1∣ℓ∣+∣α∣χ1T+∣β∣χ2ζε=qΔs+r+∣ℓ∣Δs+∣α∣χ1T+∣β∣χ2ζε=qΔs+r+Δε.
Thus, the last inequality impliesS1z1+S2z2≤ε. In a similar manner, we can show that(49)∣cDρS1z1∣≤Ts+r−ρΓs+r−ρ+1+χ3Ts+r+χ4ζs+rΓs+r+1q,∣cDρS2z2∣≤Ts−ρΓs−ρ+1+χ3Ts+χ4ζsΓs+1ℓ+αχ3T+βχ4ζε,which implies that(50)cDρS1z1+cDρS2z2≤qΔ′s+r+∣ℓ∣Δ′s+∣α∣χ3T+∣β∣χ4ζε=qΔ′s+r+Δ′ε.
These conclude that(51)S1z1+S2z2P=maxsupt∈0,T∣S1z1+S2z2,supt∈0,TcDρS1z1+cDρS2z2∣=maxqΔs+r+Δε,qΔ′s+r+Δ′ε=qmaxΔs+r,Δ′s+r+max+Δ,Δ′ε≤ε,which means that S1z1+S2z2∈Bε for each z1t,z2t∈Bε. Thus, the first part of Krasnoselskii Theorem 8 has been shown. The next step is to show that the operator S2 is contraction. For this purpose, we estimate ∣S2z1−S2z2∣ as follows:(52)S2z1−S2z2≤ℓIsz1t−z2t+χ1tαIz1T−z2T+ℓIsz1T−z2T+χ2tβIz1ζ−z2ζ+ℓIsz1ζ−z2ζ≤1+χ1Ts+χ2ζsΓs+1ℓ+αχ1T+βχ2ζz1−z2≤ℓΔs+αχ1T+βχ2ζz1−z2P=Δz1−z2P.
Similarly, we can prove that(53)∣cDρS2z1−cDρS2z2∣≤∣ℓ∣Δ′s+∣α∣χ3T+∣β∣χ4ζz1−z2P=Δ′z1−z2P.These lead to(54)S2z1−S2z2P≤maxΔ,Δ′z1−z2P,which implies that S2 is contraction if maxΔ,Δ′<1. At the same time, as the function f is assumed to be continuous according to the assumption H1, the operator S1 is also continuous on Bε. To complete the proof of this theorem, it remains to show that S1 is compact. In regard to Arzela-Ascoli theorem, it is enough to show that S1Bε is uniformly bounded and equicontinuous. For the boundedness, one can easily verify that S1z satisfies the following perspective.(55)S1zP≤qmaxΔs+r,Δ′s+r.
In the meantime, to show thatS1 is an equicontinuous operator, let t1,t2∈0,T such that t1<t2. Then, we have(56)∣S1zt2−S1zt1∣≤1Γs+r∫0t1t2−τs+r−1−t1−τs+r−1×∣fτ,zτ,Dρzτ∣dτ+1Γs+r∫t1t2t2−τs+r−1∣fτ,zτ,Dρzτ∣dτ+∣χ1t1−χ1t2∣Γs+r∫0TT−τs+r−1∣fτ,zτ,Dρzτ∣dτ+∣χ2t1−χ2t2∣Γs+r∫0ζζ−τs+r−1∣fτ,zτ,Dρzτ∣dτ.
Again, using conditionH2 to turn the inequality above to be in the form(57)∣S1zt2−S1zt1∣≤qΓs+r∫0t1t2−τs+r−1−t1−τs+r−1dτ+∫t1t2t2−τs+r−1dτ+∣t2st2−ζ−t1st1−ζ∣TsT−ζ∫0TT−τs+r−1dτ+∣t2sT−t2−t1sT−t1∣ζsT−ζ∫0ζζ−τs+r−1dτ.
Consequently, the above inequality leads us to another one of the form(58)∣S1zt2−S1zt1∣≤qΓs+r+1t2s+r−t1s+r+t2s+1−t1s+1Tr+ζr+ζTt2−t1Tr−1+ζr−1T−ζ.
Observe that ast2→t1, the right-hand side goes to zero uniformly. This means that such side does not depend never on z. Therefore, as per Arzela-Ascoli theorem, the operator S1 is equicontinuous. Hence, in view of Krasnoselskii theorem, we conclude that (1)-(2) has at least one solution.In the next argument, we will investigate the uniqueness of solution for the problem under consideration. Of course, Banach fixed point theorem, reported in Theorem7, will be utilized in this investigation. However, the following result is established for achieving that purpose.Theorem 14.
SupposeH1 and H3 hold. If(59)Ψ=M1+M2maxΔs+r,Δ′s+r+maxΔ,Δ′<1.Then, the problem given in (1)-(2) has a unique solution on 0,T.Proof.
Consider the setBε as given in (45) with(60)ε≥WmaxΔs+r,Δ′s+r1−Ψ,where W=sup0tT∣ft,0,0∣. For each t∈0,T and z∈Bε, one might estimate the term ∣ft,zt,cDρzt∣ as follows:(61)∣ft,zt,cDρzt∣=∣ft,zt,cDρzt−ft,0,0+ft,0,0∣≤∣ft,zt,cDρzt−ft,0,0∣+∣ft,0,0∣≤M1∣zt+M2cDρzt∣+W≤M1+M2zP+W≤M1+M2ε+W.
The next step is now to show thatSBε⊆Bε, ∀z∈Bε. Based on the arguments given in the previous theorem, one can obtain the following assertions.(62)∣S1zt∣≤M1+M2ε+WΔs+r,∣cDρS1zt∣≤M1+M2ε+WΔ′s+r,∣S2zt∣≤Δε,∣cDρS2zt∣≤Δ′ε.
Applying the norm overt gives(63)S1zP≤M1+M2ε+WmaxΔs+r,Δ′s+r,S2zP≤εmaxΔ,Δ′.
Therefore,(64)SzP≤S1zP+S2zP≤M1+M2ε+WmaxΔs+r,Δ′s+r+εmaxΔ,Δ′=Ψε+WmaxΔs+r,Δ′s+r≤ε.
Now, in accordance to Theorem7, we need to show that the mapping S is contraction. In order to achieve this goal, we should notice that the operator S2 is contraction as shown in the previous theorem. Hence, it remains to verify that S1 is also contraction. For this purpose, let z1,z2∈Bε, t∈0,T and consider ∣S1z1t−S1z2t∣ as follows:(65)∣S1z1t−S1z2t∣≤Is+r∣ft,z1t,cDρz1t−ft,z2t,cDρz2t∣+∣χ1t∣Is+r∣fT,z1T,cDρz1T−fT,z2T,cDρz2T+χ2tIs+rfζ,z1ζ,cDρz1ζ−fζ,z2ζ,cDρz2ζ.
Using conditionH3 yields(66)∣S1z1t−S1z2t∣≤1Γs+r∫0tt−τs+r−1M1∣z1−z2+M2cDρz1−cDρz2∣dτ+∣χ1t∣Γs+r∫0TT−τs+r−1M1∣z1−z2+M2cDρz1−cDρz2∣dτ+∣χ2t∣Γs+r∫0ζζ−τs+r−1M1∣z1−z2+M2cDρz1−cDρz2∣dτ,≤M1∣z1t−z2t+M2cDρz1t−cDρz2t∣Δs+r.
Similarly, we can obtain the following inequality:(67)∣cDρS1z1t−cDρS1z2t∣≤M1∣z1t−z2t+M2cDρz1t−cDρz2t∣Δ′s+r.
Taking under the norm·P implies(68)S1z1−S1z2P≤M1+M2maxΔs+r,Δ′s+rz1−z2P.
Using the two deduced assertions in (54) and (68) implies(69)Sz1−Sz2P≤S1z1−S1z2P+S2z1−S2z2P≤M1+M2maxΔs+r,Δ′s+r+maxΔ,Δ′z1−z2P≤Ψz1−z2P,which shows that S is a contraction mapping, and this finishes the proof.In the same context, we introduce the following further result which also explores the existence matter of our main problem by taking into account, this time, Theorem9.Theorem 15.
SupposeH1andH4hold. If(70)λmaxhΨzPΔs+r+Δ,hΨzPΔ′s+r+Δ′>1.
Then, the problem given in (1)-(2) has at least one solution on 0,T, for some λ,ε>0, where Δ,Δ,Δ· and Δ′· are defined, respectively, in (36), (37), (38) and (39).Proof.
SupposeO=z∈C:cDρz∈C,zP<λ is an open subset of the Banach space in C=C0,T,ℝ+. Assume that S:O¯→∈C is an operator defined as given in (26) with its fractional derivatives in (27), which are also previously defined as a sum of two operators S1 and S2. Observe that S is continuous since S1 and S2 are so. From this point, we find that, it is the right time, to apply on Theorem 9. For that purpose and according to Arzela-Ascoli theorem, we need to show that S is compact. In other words, we need to show that such operator is bounded and also equicontinuous. For the boundedness, let t∈0,T and consider the term ∣Szt∣ as follows:(71)∣Szt∣≤Is+r∣ft,zt,Dρzt∣+∣ℓ∣Is∣zt∣+∣χ1t∣∣α∣I∣zT∣+Is+r∣fT,zT,DρzT∣+∣ℓ∣Is∣zT∣+∣χ2t∣∣β∣I∣zζ∣+Is+r∣fζ,zζ,Dρzζ∣+∣ℓ∣Is∣zζ∣.
Using conditionH4 yields(72)∣Szt∣≤hΨzPΓs+r∫0tt−τs+r−1dτ+∣ℓ∣zΓs+1ts+∣χ1t∣∣α∣TzP+hΨzPΓs+r∫0TT−τs+r−1dτ+∣ℓ∣zΓs+1Ts+∣χ2t∣∣β∣ζzP+hΨzPΓs+r∫0ζζ−τs+r−1dτ+∣ℓ∣zΓs+1ζs.
Consequently, the above inequality leads to the following assertion.(73)Szt=sup0tT∣Szt∣≤hΨzPΔs+r+zPΔ.
Similarly, we can easily obtain the following other assertion.(74)DcρSz=sup0tTcDρSzt∣≤hΨzPΔ′s+r+zPΔ′.
By the last two inequalities, one can deduce that the operatorS is bounded. On the other hand, to prove that this operator is also equicontinuous, we should recall that it has been previously formulated as a sum of two operators S1 and S2, with noting that the first one, S1, has appeared as an equicontinuous operator through the proof of Theorem 13.. Therefore, to complete the proof of this theorem, it remains to show that S2 is also equicontinuous. In order to achieve this objective, let t1,t2∈0,T such that t1<t2. Then, we have(75)∣S2zt2−S2zt1∣≤∣ℓ∣Γs×∫0t1t1−τs−1−t2−τs−1∣zτ∣dτ+∫t1t2t2−τs−1∣zτ∣dτ+∣χ1t2−χ1t1∣∣α∣∫0T∣zτ∣dτ+∣ℓ∣Γs∫0TT−τs−1∣zτ∣dτ+∣χ2t2−χ2t1∣∣β∣∫0ζ∣zτ∣dτ+∣ℓ∣Γs∫0ζζ−τs−1∣zτ∣dτ.
Actually, the above inequality leads to the following other one(76)∣S2zt2−S2zt1∣≤z∣ℓ∣Γs+12t2−t1s+t1s−t2s+∣t2st2−ζ−t1st1−ζ∣TsT−ζ∣α∣T+∣ℓ∣TsΓs+1+∣t2sT−t2−t1sT−t1∣ζsT−ζ∣β∣ζ+∣ℓ∣ζsΓs+1≤z2∣ℓ∣Γs+1t2−t1s+t2s+1−t1s+1+ζt2s−t1sTsT−ζ∣α∣T+∣ℓ∣TsΓs+1+t2s+1−t1s+1+Tt2s−t1sζsT−ζ∣β∣ζ+∣ℓ∣ζsΓs+1.
One observes that the right-hand side of inequality above goes to zero uniformly ast2→t1. This, however, implies that S2 is definitely equicontinuous via Arzela-Ascoli theorem. The last step in this proof that should be considered is showing the boundedness of the solution of the equation z=γSz, where γ∈0,1. In this regard, let t∈0,T and z be such solution, then by considering (4.12, 4.12), we have(77)zP=γSzP≤SzP=maxSz,DρSzc≤hΨzPmaxΔs+r,Δ′s+r+zPmaxΔ,Δ′.
Using the previous inequality allows one to reveal the following inequality.(78)ztPhΨzPmaxΔs+r,Δ′s+r+zPmaxΔ,Δ′≤1.
Obviously, due toztP<λ as previously defined in O, then the last assertion contradicts assumption (70). Hence, by Theorem 9, one can conclude that S has a fixed point z∈O¯, which finishes the proof.


## 5. Illustrative Examples

This section introduces three examples to demonstrate the proposed results related to the main issue in this work. For simplicity in calculations, these three examples are unified under the same values of our problem’s parameters (1)–(2). In other words, we take r=8/5, s=2/3, ρ=1/2, ζ=1/2, T=1, ℓ=1/75, α=1/150 and β=3/200. Setting these values in the terms (19), (20), (30) and (31) simplifies their expressions to be in the following forms χ1=1, χ2=1.03413, χ3=1.80715 and χ4=1.80456. Accordingly, the following terms are also estimated via (36)-(39) as Δs=2/3=2.937, Δs+r=34/15=0.854, Δ=0.054, Δ′s=2/3=4.339, Δ′s+r=34/15=1.455 and Δ′=0.083.Example 1.
Consider the following problem.(79)D8/5D2/3+175zt=e−t7+et∣ztD1/2zt∣1+∣ztD1/2zt∣,t∈0,1,subject to the following conditions(80)z0=0,z1=1150∫01zudu,z12=3200∫01/2zudu.Comparing (79)-(80) with (1)-(2) yields the following continuous function on 0,1(81)ft,zt,D1/2zt=e−t7+et∣ztD1/2zt∣1+∣ztD1/2zt∣.Note thatH1 is satisfied, and, on the other hand, we have(82)∣ft,zt,D1/2zt∣≤e−t7+et∣ztD1/2zt∣1+∣ztD1/2zt∣,which implies(83)∣ft,zt,D1/2zt∣≤qt=e−t7+et.Therefore, conditionH2 is also satisfied with q=1/8. Now, based on the above estimations, one can verify the following statement:(84)maxΔ,Δ′=0.083<1.This means that the problem given in (79)-(80) has a unique solution on 0,1, as per Theorem 13..Example 2.
Consider the following problem:(85)D8/5D2/3+175zt=lnt2t.∣zt∣2+∣zt∣+∣D1/2zt∣101+t2∣D1/2zt∣+1,subject to the same conditions given in (80). One can notice that the function f has the form(86)ft,zt,D1/2zt=lnt2t.∣zt∣2+∣zt∣+∣D1/2zt∣101+t2∣D1/2zt∣+1,which satisfies H1 and implies(87)∣ft,z1t,D1/2z1t−ft,z2t,D1/2z2t∣≤14∣z1−z2∣+110D1/2z1−D1/2z2.
This definitely means that conditionH3 is hold, with M1=1/4 and M2=1/10. Besides, one can have the following assertion.(88)Ψ=M1+M2maxΔs+r,Δ′s+r+maxΔ,Δ′=0.5926<1,which implies that the problem (85) has a unique solution on 0,1 according to Theorem 14.Example 3.
Consider the following problem.(89)D8/5D2/3+175zt=e−t4maxsup0≤t≤1∣zt,sup0≤t≤1cD1/2zt∣51+t2,subject to the same conditions given in (80). In this example, we have the function f is of the form(90)ft,zt,D1/2zt=e−t4maxsup0≤t≤1∣zt,sup0≤t≤1cD1/2zt∣51+t2.
In fact, the above function satisfiesH1 and implies(91)∣ft,zt,D1/2zt∣≤htΨzP,where ht=e−t4/51+t2 and ΨzP=zP<λ, in which λ is assumed as given in (70). In other words, condition H4 is satisfied with h=1/5, while the assumption given in (70) will be hold if(92)max0.1708zP+0.054,0.291zP+0.083<λ.
This, consequently, impliesλ>0.117. It means that the problem (89), in view of Theorem 15., has at least one solution on 0,1, for λ>0.117.


## 6. Conclusion

The existence and uniqueness of solution for the fractional-order Langevin equation have been studied and explored in this article. This problem has been addressed subject to novel three-point boundary conditions involving two nonlocal integrals. In view of this study, some new applicable conditions and results have been established associated with our main target. These results have been illustrated through three numerical examples.



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