International Journal of Analysis and Applications Volume 19, Number 3 (2021), 465-476 URL: https://doi.org/10.28924/2291-8639 DOI: 10.28924/2291-8639-19-2021-465 EXISTENCE AND LOCATION OF A UNIQUE SOLUTION OF CAPUTO-LIOUVILLE TYPE LANGEVIN EQUATION WITH FINITELY MANY NONLINEARITIES AND NONLOCAL BOUNDARY CONDITIONS BASHIR AHMAD1,∗, AHMED ALSAEDI1, HANAN AL-JOHANY1, SOTIRIS K. NTOUYAS2 1Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia 2Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece ∗Corresponding author: bashirahmad−qau@yahoo.com Abstract. In this paper, we discuss the existence of a unique solution of Caputo-Liouville type Langevin equation involving two fractional orders and finitely many nonlinearities, equipped with nonlocal boundary conditions via Banach contraction mapping principle. The location of the unique solution of the given problem is also presented. In addition, we discuss the existence of solutions for the problem at hand by means of Krasnosel’skĭi’s fixed point theorem. Examples are constructed for the illustration of the obtained results. The paper concludes with some interesting remarks. 1. Introduction Fractional order differential equations received overwhelming attention of many researchers as these equa- tions extensively appear in the mathematical modeling of several scientific and technical phenomena. Ex- amples include physics, biology, chemistry, control theory, electrical circuits, wave propagation, blood flow phenomena, signal and image processing, etc. For further details, see [1]- [5]. Received February 28th, 2021; accepted April 9th, 2021; published April 28th, 2021. 2010 Mathematics Subject Classification. 26A33, 34A08, 34B15. Key words and phrases. Langevin equation; nonlinearities; nonlocal boundary condition; uniqueness; location. ©2021 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 465 https://doi.org/10.28924/2291-8639 https://doi.org/10.28924/2291-8639-19-2021-465 Int. J. Anal. Appl. 19 (3) (2021) 466 Langevin equation, formulated in terms of integer-order derivatives by Langevin [6] in 1908, is a well- known equation of mathematical physics, which is used to describe the evolution of physical phenomena, such as Brownian motion in fluctuating environments. Langevin equation is also known as a stochastic differential equation as it is related to the fast motion of microscopic variables of the dynamical systems. However, the failure of classical Langevin equation to describe the complex systems led to its several generalizations, which successfully modeled the physical phenomena in disordered regions [7], anomalous diffusion processes in complex and viscoelastic environment [8, 9], etc. Among these generalizations includes the one obtained by replacing the ordinary derivative by fractional order derivative in it; the resulting form is known as fractional Langevin equation and can take care of the fractal and memory properties of the phenomena under investigation. Applications of fractional Langevin equation include motor control system [10], single-file diffusion [11], transformation of the Fokker- Planck equation into the Wiener process [12], association of Kramers-Fokker-Planck equation with Langevin equation [13], etc. In order to obtain a more flexible model for fractal processes, Lim et al. [14] introduced a new form of Langevin equation involving two different fractional orders. For some recent results on fractional Langevin equation, for instance, see [15]- [24] and references therein. Modern tools of functional analysis have played a key role in developing the theory (existence and unique- ness of solutions) for fractional order initial and boundary value problems, for example, see [25]- [30]. In this paper, motivated by the recent development on Langevin equation, we study the following nonlocal boundary value problem involving Langevin equation with finitely many nonlinearities: (1.1) cDα(cDβ + µ)y(t) = m∑ i=1 aifi(t,y(t)), 0 < α ≤ 1, 1 < β ≤ 2, (1.2) y(0) = 0, y(ξ1) = 0, y(1) = ω y(ξ2), 0 < ξ1 < ξ2 < 1, where cD denotes the Caputo-Liouville fractional derivative operator, fi : [0, 1] × R → R are continuous functions, and ω ∈ R. By using Banach contraction mapping principle we prove the existence of a unique solution of boundary value problem (1.1)-(1.2) and moreover we study the location of the unique solution. An existence result is also obtained via Krasnosel’skĭi’s fixed point theorem. The paper is organized as follows. In Section 2 we recall some basic definitions and properties from fractional calculus and solve a linear variant of the boundary value problem (1.1)-(1.2). The main results are presented in Section 3. Examples illustrating the obtained results are also constructed. 2. Preliminaries Let us recall some basic definitions on fractional calculus. Int. J. Anal. Appl. 19 (3) (2021) 467 Definition 2.1. ( [2, 3]). The Riemann–Liouville fractional integral Iαa y of order α > 0 for a function y ∈ L1[a,b],−∞ < a < b < +∞, existing almost everywhere on [a,b], is defined by Iαa y (t) = 1 Γ (α) t∫ a (t−s)α−1 y (s)ds, where Γ denotes the Euler gamma function. Definition 2.2. [2, 3]. Let y,y(m) ∈ L1[a,b]. Then the Riemann–Liouville fractional derivative Dαay of order α ∈ (m− 1,m],m ∈ N, existing almost everywhere on [a,b], is defined as Dαay (t) = dm dtm Im−αa y (t) = 1 Γ (m−α) dm dtm t∫ a (t−s)m−1−α y (s)ds. The Caputo fractional derivative cDαay of order α ∈ (m− 1,m],m ∈ N is defined as cDαay (t) = D α a [ y (t) −y (a) −y′ (a) (t−a) 1! − . . .−y(m−1) (a) (t−a)m−1 (m− 1)! ] . Remark 2.1. [2]. If y ∈ ACm[a,b], then the Caputo fractional derivative cDαay of order α ∈ (m−1,m],m ∈ N, existing almost everywhere on [a,b], is defined as cDαay(t) = I m−α a y (m) (t) = 1 Γ (m−α) t∫ a (t−s)m−1−α y(m) (s)ds. Proposition 2.1. ( [2]) For κ > 0 and α > 0 with n− 1 < α ≤ n, and y ∈ L1[a,b], we have the following properties: (i) Iαa I κ ay(t) = I κ aI α a y(t) = I α+κ a y(t); (ii) Iαa (t−a) η = Γ (η + 1) Γ (α + η + 1) (t−a)α+η, η > −1; (iii) cDαa [I α a y (t)] = y(t); (iv) Iαa [ cDαay (t)] = y (t) − n−1∑ p=0 y(p) (a) (t−a)p p! , y ∈ Cn[a,b]. In the sequel, we write Iσ and cDσ instead of Iσ0 and cDσ0 respectively. To study the nonlinear problem (1.1)-(1.2), we first solve its linear variant in the following lemma. Lemma 2.1. For a given ρ ∈ C([0, 1],R), the unique solution of the boundary value problem (2.1) cDα(cDβ + µ)y(t) = ρ(t) 0 < α ≤ 1, 1 < β ≤ 2, (2.2) y(0) = 0, y(ξ1) = 0, y(1) = ωy(ξ2), Int. J. Anal. Appl. 19 (3) (2021) 468 is given by y(t) = 1 Γ(β + α) ∫ t 0 (t−s)β+α−1ρ(s)ds− µ Γ(β) ∫ t 0 (t−s)β−1y(s)ds + ρ1(t) [∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ρ(s)ds−µ ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) y(s)ds ] + ρ2(t) [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ρ(s)ds−µ ∫ 1 0 (1 −s)β−1 Γ(β) y(s)ds −ω ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ρ(s)ds + µω ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) y(s)ds ] , (2.3) where ρ1(t) = 1 ∆Γ(β + 1) ( t(1 −ωξβ2 ) − (1 −ωξ2)t β ) , ρ2(t) = 1 ∆Γ(β + 1) ( tβξ1 − tξ β 1 ) ,(2.4) and it is assumed that (2.5) ∆ = [ξ β 1 (1 −ωξ2) − ξ1(1 −ωξ β 2 )] Γ(β + 1) 6= 0. Proof. Applying the Riemann-Liouville integral operator Iα to both sides of (2.1) and using Proposition 2.1 (iv), we obtain (2.6) (cDβ + µ)y(t) + c0 = I αρ(t), where c0 ∈ R is an unknown constant. Next, operating Iβ on both sides of (2.6), we obtain y(t) = ∫ t 0 (t−s)β+α−1 Γ(β + α) ρ(s)ds− µ Γ(β) ∫ t 0 (t−s)β−1y(s)ds − c0t β Γ(1 + β) − c1 − c2t. (2.7) Using the conditions y(0) = 0 and y(ξ1) = 0 in (2.7) yields c1 = 0 and (2.8) c0ξ β 1 Γ(β + 1) + c2ξ1 = A, where A = ∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ρ(s)ds−µ ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) y(s)ds. Now using y(1) = ωy(ξ2) in (2.7), we obtain (2.9) (1 −ωξβ2 ) Γ(β + 1) c0 + (1 −ωξ2)c2 = B, where B = ∫ 1 0 (1 −s)β+α−1 Γ(β + α) ρ(s)ds−µ ∫ 1 0 (1 −s)β−1 Γ(β) y(s)ds −ω ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ρ(s)ds + µω ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) y(s)ds. Int. J. Anal. Appl. 19 (3) (2021) 469 Solving (2.8) and (2.9) for c0 and c2, we find that (2.10) c0 = 1 ∆ (σ3A− ξ1B), c2 = 1 ∆ (σ1B −σ2A), where (2.11) ∆ = σ3σ1 − ξ1σ2, σ1 = ξ β 1 Γ(β + 1) , σ2 = (1 −ωξβ2 ) Γ(β + 1) , σ3 = (1 −ωξ2). Inserting c1 = 0 and the values of c0 and c2 from (2.10) into (2.7), together with (2.11), leads to the solution (2.3). The converse follows by direct computation. This completes the proof. � Definition 2.3. A function y ∈ C([0, 1] ,R) is a solution of the boundary value problem (1.1)-(1.2) if and only if it satisfies the integral equation: y(t) = ∫ t 0 (t−s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds−µ ∫ t 0 (t−s)β−1 Γ(β) y(s)ds + ρ1(t) [∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds−µ ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) y(s)ds ] + ρ2(t) [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds−µ ∫ 1 0 (1 −s)β−1 Γ(β) y(s)ds −ω ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds + µω ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) y(s)ds ] . (2.12) 3. Uniqueness and location results In the following theorem, we prove the existence of a unique solution for the problem (1.1)-(1.2) by applying Banach contraction mapping principle. Theorem 3.1. Let fi(t,y) : [0, 1] × R → R, i = 1, . . . ,m be continuous functions satisfying the Lipschitz condition: (A1) |fi(t,x) −fi(t,y)| ≤ Li |x−y| , ∀t ∈ [0, 1] , x,y ∈ R, Li > 0, i = 1, 2, . . . ,m. Then, the boundary value problem (1.1)-(1.2) has a unique solution on [0, 1] if (3.1) Ω1 < 1, where Ω1 = ( ∑m i=1 |ai|Li) Γ(β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )] + |µ| Γ(β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2 ( 1 + |ω|ξβ2 )] , (3.2) and ρ̄1 = max t∈[0,1] |ρ1(t)|, ρ̄2 = max t∈[0,1] |ρ2(t)|. Int. J. Anal. Appl. 19 (3) (2021) 470 Proof. To transform the problem (1.1) and (1.2) into a fixed-point problem, we introduce an operator N : C([0, 1] ,R) → C([0, 1] ,R) as (Ny)(t) = ∫ t 0 (t−s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds−µ ∫ t 0 (t−s)β−1 Γ(β) y(s)ds + ρ1(t) [∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds−µ ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) y(s)ds ] + ρ2(t) [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds−µ ∫ 1 0 (1 −s)β−1 Γ(β) y(s)ds −ω ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds + µω ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) y(s)ds ] , (3.3) where C([0, 1],R) is the Banach space of all continuous functions from [0, 1] into R equipped with the norm ‖y‖ = supt∈[0,1] |y(t)| . Observe that the fixed points of the operator N are solutions of the problem (1.1) and (1.2) by Definition 2.3. Further, it is an immediate consequence of the dominated convergence theorem that Ny ∈ C([0, 1],R) for every y ∈ C([0, 1],R). The proof will be complete by means of Banach contraction mapping principle once it is shown that that the operator N defined by (3.3) is a contraction. For x,y ∈ R and ∀t ∈ [0, 1], we have ‖(Nx) − (Ny)‖ = sup t∈[0,1] ∣∣∣∣∣ ∫ t 0 (t−s)β+α−1 Γ(β + α) ( m∑ i=1 ai[fi(s,x(s)) −fi(s,y(s))] ) ds −µ ∫ t 0 (t−s)β−1 Γ(β) [x(s) −y(s)]ds +ρ1(t) [∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ( m∑ i=1 ai[fi(s,x(s)) −fi(s,y(s))] ) ds −µ ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) [x(s) −y(s)]ds ] +ρ2(t) [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ( m∑ i=1 ai[fi(s,x(s)) −fi(s,y(s))] ) ds −µ ∫ 1 0 (1 −s)β−1 Γ(β) [x(s) −y(s)]ds −ω ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ( m∑ i=1 ai[fi(s,x(s)) −fi(s,y(s))] ) ds +µω ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) [x(s) −y(s)]ds ]∣∣∣∣∣ ≤ ( ∑m i=1 |ai|Li) Γ(β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )] ‖x−y‖ Int. J. Anal. Appl. 19 (3) (2021) 471 + |µ| Γ(β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2 ( 1 + |ω|ξβ2 )] ‖x−y‖, which leads to ‖(Nx) − (Ny)‖≤ Ω1 ‖x−y‖ . Evidently, we deduce by (3.1) that N : C([0, 1] ,R) → C([0, 1] ,R) is a contraction. Hence, by Banach contraction mapping principle, the operator N has a unique fixed point, which corresponds to a unique solution of the boundary value problem (1.1)-(1.2) on [0, 1]. This completes the proof. � Example 3.1. Consider the following boundary value problem: cD 1 2 ( cD 3 2 + 1 5 ) y(t) = 3∑ i=1 aifi(t,y(t)), t ∈ [0, 1], y(0) = 0, y (1 3 ) = 0, y(1) = y (2 3 ) . (3.4) Here α = 1/2,β = 3/2,µ = 1/5,ω = 1,ξ1 = 1/3,ξ2 = 2/3,m = 3,a1 = 1/2,a2 = 1,a3 = 3/4 and f1 (t,y) = 1 √ t2 + 100 |y| |y| + 1 + e−t, f2 (t,y) = 1 t2 + 20 tan−1 y + 2, f3 (t,y) = 1 15 ( e−t t2 + 1 ) sin y + t2 1 + t2 . It is easy to find that |fi (t,x) −fi (t,y)| ≤ Li |x−y| , i = 1, 2, 3, with L1 = 1/10,L2 = 1/20,L3 = 1/15 and 3∑ i=1 aiLi = 3/20. Furthermore, we have |∆| = |[ξ β 1 (1−ωξ2)−ξ1(1−ωξ β 2 )]| Γ(β+1) ≈ 0.000499, ρ̄1 = maxt∈[0,1] |ρ1(t)| = ρ1(t)|t=0.830537 ≈ 1.437777, ρ̄2 = maxt∈[0,1] |ρ2(t)| = ρ2(t)|t=1 ≈ 1.605694 and Ω1 = 0.826088 < 1. Clearly all the assumptions of Theorem 3.1 are satisfied. Therefore the problem (3.4) has a unique solution on [0, 1]. In the following result, we present location of the unique solution of the boundary value problem (1.1)- (1.2). Theorem 3.2. Let the hypotheses of Theorem 3.1 hold. Then the unique solution y of problem (1.1)-(1.2) satisfies (3.5) ‖y‖≤ Ω2 1 − Ω1 , where Ω1 is given by (3.2) and (3.6) Ω2 = ( m∑ i=1 Mi ){ 1 Γ(β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )]} , with Mi = supt∈[0,1] |fi (t, 0)| . Int. J. Anal. Appl. 19 (3) (2021) 472 Proof. By Theorem 3.1, the solution (2.12) of the boundary value problem (1.1)-(1.2) is unique. In view of the inequality: ∣∣∣ m∑ i=1 aifi(s,y(s)) ∣∣∣ ≤ m∑ i=1 |ai|(Li‖y‖ + Mi), where Mi = supt∈[0,1] |fi (t, 0)| , it follows from (2.12) that ‖y‖≤ m∑ i=1 |ai|(Li‖y‖ + Mi) sup t∈[0,1] {∫ t 0 (t−s)β+α−1 Γ(β + α) ds + |ρ1(t)| ∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ds + |ρ2(t)| [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ds + |ω| ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ds ]} + |µ|‖y‖ sup t∈[0,1] {∫ t 0 (t−s)β−1 Γ(β) ds + |ρ1(t)| ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) ds + |ρ2(t)| [∫ 1 0 (1 −s)β−1 Γ(β) ds + |ω| ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) ds ]} ≤‖y‖ { ( ∑m i=1 |ai|Li) Γ(β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )] + |µ| Γ(β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2 ( 1 + |ω|ξβ2 )]} + ( m∑ i=1 Mi ){ 1 Γ(β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )]} = ‖y‖Ω1 + Ω2. Solving the above inequality for ‖y‖ yields ‖y‖≤ Ω2 1 − Ω1 . This completes the proof. � Example 3.2. In relation to Example 3.1, it is found that ‖y‖≤ δ, where δ ≈ 35.008543. 4. An existence result In this section, we establish an existence result for the boundary value problem (1.1)-(1.2) via Kras- nosel’skĭi’s fixed point theorem [31]. Theorem 4.1. Let fi : [0, 1] × R → R be continuous functions satisfying the condition (A2): |fi(t,y)| 6 h(t), ∀ i = 1, 2, . . . ,m, (t,y) ∈ [0, 1] × R, h ∈ C([0, 1],R+). Then the boundary value problem (1.1)-(1.2) has at least one solution on [0, 1], provided that (4.1) Q1 := |µ| Γ(β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2(1 + |ω|ξ β 2 ) ] < 1. Int. J. Anal. Appl. 19 (3) (2021) 473 Proof. Consider Br = {y ∈ C([0, 1],R) : ‖y‖≤ r}, with r > Q2‖h‖ 1 −Q1 , where (4.2) Q2 := m∑ i=1 |ai| { 1 Γ (β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )]} and ‖h‖ = sup t∈[0,1] |h(t)|. Then we define the operators P and Q on Br to C([0, 1],R) as P(t) = ∫ t 0 (t−s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds +ρ1(t) ∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds +ρ2(t) [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds −ω ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds ] ,(4.3) Q(t) = −µ ∫ t 0 (t−s)β−1 Γ(β) y(s)ds−µρ1(t) ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) y(s)ds −µρ2(t) [∫ 1 0 (1 −s)β−1 Γ(β) y(s)ds−ω ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) y(s)ds ] .(4.4) We show that Px + Qy ∈ Br. For x,y ∈ Br, we find that ‖(Px) + (Qy)‖ ≤ sup t∈[0,1] {∫ t 0 (t−s)β+α−1 Γ(β + α) ( m∑ i=1 |ai||fi(s,y(s))| ) ds + |µ| ∫ t 0 (t−s)β−1 Γ(β) |y(s)|ds +|ρ1(t)| [∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ( m∑ i=1 |ai||fi(s,y(s))| ) ds +|µ| ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) |y(s)|ds ] + |ρ2(t)| [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ( m∑ i=1 |ai||fi(s,y(s))| ) ds +|µ| ∫ 1 0 (1 −s)β−1 Γ(β) |y(s)|ds + |ω| ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ( m∑ i=1 |ai||fi(s,y(s))| ) ds +|µ||ω| ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) |y(s)|ds ]} ≤ m∑ i=1 |ai|‖h‖ { 1 Γ (β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )]} + |µ|r Γ (β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2 ( 1 + |ω|ξβ2 )] ≤ Q2 ‖h‖ + Q1r ≤ r Int. J. Anal. Appl. 19 (3) (2021) 474 Thus, Px + Qy ∈ Br. Next we show that Q is a contraction mapping. For x,y ∈ C([0, 1],R) and each t ∈ [0, 1], we obtain ‖(Qx) − (Qy)‖ ≤ |µ| ∫ t 0 (t−s)β−1 Γ(β) |x(s) −y(s)|ds + |µ||ρ1(t)| ∫ ξ1 0 (ξ1 −s)β−1 Γ(β) (x(s) −y(s)|ds +|µ||ρ2(t)| [∫ 1 0 (1 −s)β−1 Γ(β) |x(s) −y(s)|ds +|ω| ∫ ξ2 0 (ξ2 −s)β−1 Γ(β) |x(s) −y(s)|ds ] ≤ |µ| Γ(β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2(1 + |ω|ξ β 2 ) ] ‖x−y‖, which is a contraction in view of the condition (4.1). We show that P is compact and continuous. The continuity of fi implies that the operator P is continuous. P is uniformly bounded on Br as ‖Px‖≤ Q2 ‖h‖ , where Q2 is given by (5.1). We shall prove now that P is equicontinuous. For t1, t2 ∈ [0, 1] with t1 > t2, we have |Py(t1) −Py(t2)| ≤ ∣∣∣∫ t1 0 (t1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) − ∫ t2 0 (t2 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) )∣∣∣ +|ρ1(t1) −ρ1(t2)| ∫ ξ1 0 (ξ1 −s)β+α−1 Γ(β + α) ( m∑ i=1 aifi(s,y(s)) ) ds +|ρ2(t1) −ρ2(t2)| [∫ 1 0 (1 −s)β+α−1 Γ(β + α) ( m∑ i=1 |ai||fi(s,y(s))| ) ds +|ω| ∫ ξ2 0 (ξ2 −s)β+α−1 Γ(β + α) ( m∑ i=1 |ai||fi(s,y(s))| ) ds ] ≤ ∑m i=1 |ai|‖h‖ Γ (β + α + 1) { 2 (t1 − t2) β+α + |tβ+α1 − t β+α 2 | + |ρ1(t1) −ρ1(t2)|ξ β+α 1 +|ρ2(t1) −ρ2(t2)|(1 + |ω|ξ β+α 2 ) } , which is independent of y and tends to zero as t1 −t2 → 0. So P is equicontinuous. Hence, by Arzelá-Ascoli Theorem, P is compact on Br. Thus all the assumptions of Krasnosel’skĭi’s fixed point theorem [31] are verified and hence its conclusion implies that the boundary value problem (1.1)-(1.2) has at least one solution on [0, 1]. The proof is completed. � 5. Concluding remarks We have discussed the solvability of Caputo-Liouville type Langevin equation involving two fractional orders and finitely many nonlinearities subject to nonlocal boundary conditions by means of standard fixed Int. J. Anal. Appl. 19 (3) (2021) 475 point theorems. It is imperative to notice that the right-hand side of (1.1) provides a leverage to consider a variety of nonlinearities, for instance, some of the terms in the given sum may be of the type fi(t,y) =∫ t 0 ki(t,s)y(s)ds or ∫ t 0 gi(s,y(s))ds or I pgi(t,y(t)) or some of the functions may be non-Lipschitz type. Here are two examples. (a): Choosing the right-hand side of (1.1) as Riemann-Liouville type integral nonlinearities of the form:∑m i=1 aiI qifi(t,y(t)), qi > 0 instead of ∑m i=1 aifi(t,y(t)), the results for the problem (1.1) and (1.2) obtained in the previous sections become the ones for the problem with Riemann-Liouville type integral nonlinearities by replacing Ω1 and Q2 with Ω̂1 and Q̂2 respectively, where Ω̂1 = 1 Γ(β + α + 1) m∑ i=1 (|ai|Li) Γ(qi + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )] + |µ| Γ(β + 1) [ 1 + ρ̄1ξ β 1 + ρ̄2 ( 1 + |ω|ξβ2 )] , Q̂2 := m∑ i=1 { |ai| Γ(qi + 1)Γ (β + α + 1) [ 1 + ρ̄1ξ β+α 1 + ρ̄2 ( 1 + |ω|ξβ+α2 )]} . (b): By a simple manipulation, we can obtain the results for the problem (1.1) and (1.2) with the right-hand side of (1.1) of the form: m∑ i=1 aifi(t,y(t)) + κ∑ j=1 ajI qjgj(t,y(t)), qj > 0, where fi and gj are given continuous functions. Conflicts of Interest: The author(s) declare that there are no conflicts of interest regarding the publication of this paper. 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