# Fractional Langevin Equations with Nonseparated Integral Boundary Conditions

Authors: Khalid Hilal; Lahcen Ibnelazyz; Karim Guida; Said Melliani
Journal: Advances in Mathematical Physics
 (2020)

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## Abstract

In this paper, we discuss the existence of solutions for nonlinear fractional Langevin equations with nonseparated type integral boundary conditions. The Banach fixed point theorem and Krasnoselskii fixed point theorem are applied to establish the results. Some examples are provided for the illustration of the main work.


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## Body

## 1. Introduction

Fractional derivatives give an excellent description of memory and hereditary properties of different processes. Properties of the fractional derivatives make the fractional-order models more useful and practical than the classical integral-order models.Several researchers in the recent years have employed the fractional calculus as a way of describing natural phenomena in different fields such as physics, biology, finance, economics, and bioengineering (for more details see [1–9] and many other references).With the recent outstanding development in fractional differential equations, the Langevin equation has been considered a part of fractional calculus, and thus, important results have been elaborated (see [10–14]).The Langevin equation was first introduced by Langevin in 1908; it is a fundamental theory of the Brownian motion to describe the evolution of physical phenomena in fluctuating environments [15, 16]. The fractional model of the Langevin equation as a generalization of the classical one gives a fractional Gaussian process parametrized by two indices, and this fractional model is more flexible for modeling fractal processes [17, 18].The fractional Langevin equation is extensively studied in the literature from both the theoretical and numerical point of views (for more details see [19–25]). In [26], the authors studied a nonlinear Langevin equation involving two fractional orders in different intervals. In [27], the authors discussed the existence theory for a nonlinear Langevin equations with nonlocal multipoint and multistrip boundary conditions. In [28], fractional Langevin equations with nonlocal integral boundary conditions have been investigated by Salem et al. In [14], an antiperiodic boundary value problem for the Langevin equation involving two fractional orders has been studied.Recently, in [29], the authors discussed the nonlinear fractional differential equations with nonseparated type integral boundary conditions; however, the fractional Langevin equations involving nonseparated integral boundary conditions have not been investigated yet; that is why, in this work and motivated by all the works cited above, we study the existence and uniqueness of the fractional Langevin equations with nonseparated integral boundary conditions as follows:
(1)DcβDcα+λxt=ft,xt,Ipxt,t∈0,1,x0+μx1=σ1∫01gs,xsds,Dcαx0+μDcαx1=σ2∫01hs,xsds,Dc2αx0+μDc2αx1=σ3∫01ks,xsds,where 0<α<1, 1<β≤2, p>0, λ,μ,σ1,σ2,σ3∈ℝ∗ with μ≠−1, and cDβ,cDα are the Caputo fractional derivatives and f:0,1×ℝ×ℝ⟶ℝ and g,h,k:0,1×ℝ⟶ℝ are given continuous functions.This paper is divided into four sections, in which the second provides some notations and basic known results, in the third section, we study the existence and uniqueness of solutions to problem (1), and in the fourth section, we give two examples to illustrate our results.


## 2. Preliminaries and Notations

In this section, we give some notation, definitions, and lemma which are needed throughout this paper.Definition 1 (see [5]).
The fractional integral of orderα>0 with the lower limit zero for a function f can be defined as
(2)Iαft=1Γα∫0tt−sα−1fsds.Definition 2 (see [5]).
The Caputo derivative of orderα>0 with the lower limit zero for a function f can be defined as
(3)Dcαf=1Γn−α∫0tt−sn−α−1fnsds,
wheren∈ℕ, 0≤n−1<α<n, and t>0.Theorem 3 (see [30]).
LetM be a bounded, closed, convex, and nonempty subset of a Banach space X. Let A and B be operators such that
(I)
Ax+By∈M whenever x,y∈M(II)
A is compact and continuous(III)
B is a contraction mapping
Then, there existsz∈M such that z=Az+Bz.Lemma 4 (see [5]).
Letα,β≥0; then, the following relations hold:
(4)Iαtβ=Γβ+1Γα+β+1tα+β,Dcαtβ=Γβ+1Γβ−α+1tβ−α.Lemma 5 (see [5]).
Letn∈ℕ and n−1<α<n. If f is a continuous function, then, we have
(5)IαDcαft=ft+α0+a1t+a2t2+⋯+an−1tn−1.Lemma 6.
Lety∈C0,1,ℝ. Then, a unique solution of the bondary value problem
(6)DcβDcα+λxt=yt,t∈0,1,x0+μx1=σ1∫01gs;xsds,Dcαx0+μDcαx1=σ2∫01hs;xsds,Dc2αx0+μDc2αx1=σ3∫01ks;xsdsis given by
(7)xt=1Γα+β∫0tt−sα+β−1ysds−λΓα∫0tt−sα−1xsds+A1t∫01hs;xsds+A2t∫01gs;xsds+A3t∫01ks;xsds+A4tΓβ−α∫011−sβ−α−1ysds+A5tΓβ∫011−sβ−1ysds+μλ1+μΓα∫011−sα−1xsds−μ1+μΓα+β∫011−sα+β−1ysds,where
(8)A1t=tασ21−λΓ2−σΓα+11+μ+tα+1λσ2Γ2−αΓ2+α+μ1+μ2Γα+1−11+μΓα+2Γ2−αλσ2−μσ21+μ2Γα+1,A2t=tαλσ1Γα+11+μ−σ11+μ+μλσ11+μ2Γα+1,A3t=tαΓ2−ασ3Γα+11+μ+tα+1Γ2−ασ3Γα+2μ+μΓ2−ασ31+μ2Γα+1−Γ2−ασ31+μΓα+2,A4t=tαΓ2−αμΓα+11+μ−tα+1Γ2−αΓα+2+Γ2−αμ1+μΓα+2−μ21+μ2Γα+1,A5t=tαμΓα+11+μ−μ21+μ2Γα+1.Proof.
By applying Lemma5, we have
(9)cDα+λxt=Iβyt+a0+a1t,Dcaxt=Iβyt+a0+a1t−λxt,xt=Iα+βyt+Iαa0+Iαa1t−Iαλxt+a2,where a0,a1,a2∈ℝ.
Using the conditionDc2αx0+μDc2αx1=σ3∫01ks;xsds, we obtain
(10)a1=Γ2−ασ3μ∫01ks;xsds+λσ2μ∫01hs;xsds−1Γβ−α∫011−sβ−αysds.
By conditionsDcαx0+μDcαx1=σ2∫01ks;xsds and x0+μx1=σ1∫01gs;xsds, we have
(11)a0=−Γ2−ασ31+μ∫01ks,xsds+1−λΓ2−ασ21+μ∫01hs,xsds+dλσ11+μ∫01gs,xsds+Γ2−αμ1+μΓβ−α∫011−sβ−α−1ysds−μ1+μΓβ∫011−sβ−1ysds,a2=μλ1+μΓα∫011−sα−1xsds+σ11+μ−μλσ1Γα+11+μ2∫01gs,xsds−μ1+μΓα+β∫011−sα+β−1ysds+μ21+μ2ΓβΓα+1∫011−sβ−1ysds+Γ2−αΓβ−αμ1+μΓα+2−μ21+μ2Γα+1∫011−sβ−α−1ysds+μ21+μ2Γα+1−μ1+μΓα+2Γ2−αλσ2μ−μσ21+μ2Γα+1∫01hs,xsds.Substituting the value ofa0, a1, and a2, we obtain the desired results. And by direct computation, one can obtain the converse of the lemma.


## 3. Main Results

Denote byX the Banach space of all continuous functions from 0,1⟶ℝ endowed with norm x=supxt:t∈0,1.By Lemma6, we transform problem (1) into a fixed point problem as x=Px, where P:X⟶X is given by
(12)Pxt=1Γα+β∫0tt−sα+β−1fs,xs,Ipxsds−λΓα∫0tt−sα−1xsds+A1t∫01hs;xsds+A2t∫01gs;xsds+A3t∫01ks;xsds+A4tΓβ−α∫011−sβ−α−1fs,xs,Ipxsds+A5tΓβ∫011−sβ−1fs,xs,Ipxsds+μλ1+μΓα∫011−sα−1xsds−μ1+μΓα+β∫011−sα+β−1fs,xs,Ipxsds.Theorem 2.
Suppose thatf:0,1×ℝ2⟶ℝ and h,g,k:0,1×ℝ⟶ℝ are continuous functions satisfying
H1—there exist positive constants q1,q2 such that
(13)ft,x1,y1−ft,x2,y2≤q1x1−x2+q2y1−y2,for all x1,x2,y1,y2∈ℝ, t∈0,1.
H2—there exist positive constants q3,q4,q5 such that
(14)gt,x−gt,y≤q3x−y,ht,x−ht,y≤q4x−y,kt,x−kt,y≤q5x−y,∀x,y∈ℝ.
Then there exist a unique solution for boundary value problem (1) provided that r1<1, where
(15)r1=q1+q2Γp+11Γα+β+1+A4Γβ−α+1+A5Γβ+1+μ1+μΓα+β+1+∣λ∣Γα+1+μλ1+μΓα+1+A1q4+A2q3+A3q5and Ai=maxt∈0,1∣Ait∣ for i=1,2,..,5.Proof.
We setsup0≤t≤1ft,0,0=M0, sup0≤t≤1gt,0=M1, sup0≤t≤1ht,0=M2, sup0≤t≤1kt,0=M3.
LetBr=x∈X:x≤r the ball with radius r, where r≥r2/1−r1, with
(16)r2=M0Γα+β+1+M0A4Γβ−α+1+A1M1+A2M2+A3M3+A5M0Γβ+1+μM01+μΓα+β+1.Then,Br is a closed, convex, and nonempty subset of the Banach space X.Our aim is to prove that the operatorP has a unique fixed point on Br.We show that PBr⊆Br.Forx∈Br, t∈0,1, we have
(17)Pxt≤1Γα+β∫0tt−sα+β−1∣fs,xs,Ipxs∣ds+λΓα∫0tt−sα−1∣xs∣ds+A1t∫01hs;xsds+A2t∫01gs;xsds+A3t∫01ks;xsds+A4tΓβ−α∫011−sβ−α−1∣fs,xs,Ipxs∣ds+A5tΓβ∫011−sβ−1∣fs,xs,Ipxs∣ds+μλ1+μΓα∫011−sα−1∣xs∣ds+μ1+μΓα+β∫011−sα+β−1∣fs,xs,Ipxs∣ds≤1Γα+β∫0tt−sα+β−1∣fs,xs,Ipxs−fs,0,0∣+∣fs,0,0∣ds+λΓα∫0tt−sα−1∣xs∣ds+∣A1t∣∫01∣hs;xs−hs;0∣+∣hs;0∣ds+∣A2t∣∫01∣gs;xs−gs;0∣+∣gs;0∣ds+∣A3t∣∫01∣ks;xs−ks;0∣+∣ks;0∣ds+∣A4t∣Γβ−α∫011−sβ−α−1∣fs,xs,Ipxs−fs,0,0∣+fs,0,0ds+A5tΓβ∫011−sβ−1fs,xs,Ipxs−fs,0,0+fs,0,0ds+μλ1+μΓα∫011−sα−1∣xs∣ds+μ1+μΓα+β∫011−sα+β−1×fs,xs,Ipxs−fs,0,0+fs,0,0ds,which implies that
(18)Px≤1Γα+β+1q1+q2Γp+1x+M0+λΓα+1x+A1q4x+M2+A2q3x+M1+A3q5x+M3+A4Γβ−α+1q1+q2Γp+1x+M0+A5Γβ+1q1+q2Γp+1x+M0+μλ1+μΓα+1x+μ1+μΓα+β+1q1+q2Γp+1x+M0≤q1+q2Γp+11Γα+β+1+A4Γβ−α+1+A5Γβ+1+μ1+μΓα+β+1+λΓα+1+μλ1+μΓα+1+A1q4+A2q3+A3q5x+M0Γα+β+1+M0A4Γβ−α+1+A1M2+A2M1+A3M3+A5M0Γβ+1+μM01+μΓα+β+1≤r1r+r2≤r.Now, forx,y∈Br and for t∈0;1,
(19)Pxt−Pyt≤1Γα+β∫0tt−sα+β−1fs,xs,Ipxs−fs,ys,Ipysds+λΓα∫0tt−sα−1xs−ysds+A1t∫01hs;xs−hs;ysds+A2t∫01gs;xs−gs;ysds+A3t∫01ks;xs−ks;ysds+A4tΓβ−α∫011−sα−β−1fs,xs,Ipxs−fs,ys,Ipysds+A5tΓβ∫011−sβ−1fs,xs,Ipxs−fs,ys,Ipysds+μλ1+μΓα∫011−sα−1xs−ysds+μ1+μΓα+β∫011−sα+β−1fs,xs,Ipxs−fs,ys,Ipysds.Thus,(20)Px−Py≤r1x−y.Sincer1<1, then operator P is a contraction mapping. Therefore, boundary value problem (1) has a unique solution.Theorem 7.
Assume thatH1 and H2 hold, f:0;1×ℝ2⟶ℝ is a continuous function. Further, we suppose
H3—ft,x,y≤mt;ht,x≤ρt;kt,x≤ψt;∣gt,x∣≤φt,∀t,x,y∈0,1×ℝ2 with m,ρ,φ,ψ∈C0,1;ℝ+.Then, boundary value problem (1) has at least one solution on 0,1 if Q<1 and r1′<1, where
(21)Q=A1q4+A2q3+A3q5+A4Γβ−α+1q1+q2Γp+1+A5Γβ+1q1+q2Γp+1+μλ1+μΓα+1+μ1+μΓα+β+1q1+q2Γp+1,r′1=λΓα+1+μλ1+μΓα+1.Proof.
Consider the closed ballBr′=x∈X:x≤r′ with the radius r′≥r′2/1−r′1, where
(22)r′2=mΓα+β+1+A1ρ+A2ϕ+A3ψ+A4mΓβ−α+1+A5mΓβ+1+μm1+μΓα+β+1.
We introduce the decompositionP=P1+P2, where
(23)P1xt=1Γα+β∫0tt−sα+β−1fs,xs,Ipxsds−λΓα∫0tt−sα−1xsds,P2xt=A1t∫01hs;xsds+A2t∫01gs;xsds+A3t∫01ks;xsds+A4tΓβ−α∫011−sβ−α−1fs,xs,Ipxsds+A5tΓβ∫011−sβ−1fs,xs,Ipxsds+μλ1+μΓα∫011−sα−1xsds−μ1+μΓα+β∫011−sα+β−1fs,xs,Ipxsds.
Forx,y∈Br, we have
(24)P1xt+P2yt≤1Γα+β∫0tt−sα+β−1∣fs,xs,Ipxs∣ds+λΓα∫0tt−sα−1∣xs∣ds+∣A1t∣∫01∣hs;ys∣ds+A2t∫01∣gs;ys∣ds+∣A3t∣∫01∣ks;ys∣ds+A4tΓβ−α∫011−sβ−α−1∣fs,ys,Ipys∣ds+∣A5t∣Γβ∫011−sβ−1∣fs,ys,Ipys∣ds+μλ1+μΓα∫011−sα−1∣ys∣ds+μ1+μΓα+β∫011−sα+β−1fs,ys,Ipysds≤1Γα+β∫0tt−sα+β−1msds+λΓα∫0tt−sα−1xsds+A1∫01ρsds+A2∫01ϕsds+A3∫01ψsds+A4Γβ−α∫011−sβ−α−1msds+A5Γβ∫011−sβ−1msds+μλ1+μΓα∫011−sα−1ysds+μ1+μΓα+β∫011−sα+β−1msds,which implies that
(25)P1x+P2y≤mΓα+β+1+λΓα+1r′+A1ρ+A2ϕ+A3ψ+A4mΓβ−α+1+A5mΓβ+1+μλ1+μΓα+1r′+μm1+μΓα+β+1≤r′.Thus,P1x+P2y⊂Br′.Forx,y∈Br′, we have
(26)P2xt−P2yt≤A1t∫01hs;xs−hs;ysds+A2t∫01gs;xs−gs;ysds+A3t∫01ks;xs−ks;ysds+A4tΓβ−α∫011−sα−β−1fs,xs,Ipxs−fs,ys,Ipysds+A5tΓβ∫011−sβ−1fs,xs,Ipxs−fs,ys,Ipysds+∣μλ∣∣1+μ∣Γα∫011−sα−1xs−ysds+μ1+μΓα+β∫011−sα+β−1fs,xs,Ipxs−fs,ys,Ipysds≤A1∫01q4xs−ysds+A2∫01q3xs−ysds+A3∫01q5xs−ysds+A4Γβ−α∫011−sβ−α−1q1xs−ys+q2Ipxs−Ipysds+A5Γβ∫011−sβ−1q1xs−ys+q2Ipxs−Ipysds+∣μλ∣∣1+μ∣Γα∫011−sα−1xs−ysds+∣μ∣1+μΓα+β∫011−sα+β−1q1xs−ys+q2Ipxs−Ipysds,which implies that
(27)P2x−P2y≤A1q4x−y+A2q3x−y+A3q5x−y+A4Γβ−α+1q1+q2Γp+1x−y+A5Γβ+1q1+q2Γp+1x−y+∣μλ∣∣1+μ∣Γα+1x−y+∣μ∣∣1+μ∣Γα+β+1q1+q2Γp+1x−y≤A1q4+A2q3+A3q5+A4Γβ−α+1q1+q2Γp+1+A5Γβ+1q1+q2Γp+1+∣μλ∣∣1+μ∣Γα+1+∣μ∣∣1+μ∣Γα+β+1q1+q2Γp+1x−y≤Qx−y,since Q<1, then P2 is a contraction.Next, we show thatP1 is compact and continuous. Continuity of f implies that the operator P1 is continuous.Since,P1x≤m/Γα+β+1+λ/Γα+1r′, therefore, P1 is uniformly bounded on Br′. Suppose that 0≤t1<t2≤1. We have
(28)P1xt2−P1xt1≤1Γα+β∫0t1t2−sα+β−1−t1−sα+β−1fs,xs,Ipxsds+∫t1t2t2−sα+β−1fs,xs,Ipxsds≤+λΓα∫0t1t2−sα−1−t1−sα−1xsds+∫t1t2t2−sα−1xsdsmΓα+β+12t2−t1α+β+t1α+β−t2α+β+λrΓα+12t2−t1α+t1α−t2α.Ast1→t2, the above expression tends to be zero independently from x∈Br′. This implies that P1 is relatively compact on Br′. Then, by the Arzela-Ascoli theorem, operator P1 is compact on Br′.Therefore, according to the Krasnoselskii fixed point theorem, problem (1) has at least one solution on Br′.


## 4. Example

Example 8.
Consider the following boundary value problem:(29)cD3/4cD1/3+1300xt=1500+t2sinx+1Γ3/4∫0tt−s−1/4xds,t∈0,1,x0+x1=1200∫01xs300+xsds,cD1/3x0+cD1/3x1=1200∫011s+23xs30+xsds,cD2/3x0+cD2/3x1=1200∫011s+42xs30+xsds.
We chooseβ=4/3, α=1/3, λ=1/300, μ=1, σ1=1/200, σ2=1/200, σ3=1/200, p=1/4, and the continuous functions
(30)ft,x,I1/4x=1500+t2sinx+1Γ3/4∫0tt−s−1/4xds,gt,x=xt300+xt,ht,x=1t+23xt30+xt,kt,x=1t+42xt30+xt.
Clearly,q1=q2=1/500, q3=1/300, q4=1/240, and q5=1/480,
Then, we haver1≈0.07<1.
Thus, all the assumptions of Theorem2 hold. Then, problem (29) has a unique solution.Example 9.
Consider the following boundary value problem:(31)cD3/4cD1/3+1300xt=1t+10215+x2+∫0tt−s−1/415+x2ds,t∈0,1,x0+x1=1200∫011s+10130+xsds,cD1/3x0+cD1/3x1=1200∫011s+23xs30+xsds,cD2/3x0+cD2/3x1=1200∫011s+42xs30+xsds.
Here,β=3/4, α=1/3, λ=1/30, μ=1, σ1=1/200, σ2=1/200, σ3=1/200, p=1/4, and the continuous functions
(32)ft,x,I1/4x=1t+10215+x2+∫0tt−s−1/415+x2ds,gt,x=1t+10xt30+xt,ht,x=1t+23xt30+xt,kt,x=1t+42xt30+xt.
Clearly,q1=q2=1/500, q3=1/300, q4=1/240, and q5=1/480; thus, we have Q≈0.057<1 and r′1≈0,0065<1.
Then, problem (31) has a least one solution.


## 5. Conclusion

In this paper, we studied the existence and uniqueness of fractional Langevin equations with nonseparated integral boundary conditions. First of all, we transformed the problem into an equivalent fixed point problem; second of all, we utilized the Banach contraction principle and the Krasnoselskii fixed point theorem to prove the existence and uniqueness of solutions.



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