# Some Existence Results for a System of Nonlinear Sequential Fractional Differential Equations with Coupled Nonseparated Boundary Conditions

Authors: Muath Awadalla
Journal: Complexity
 (2022)

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## Abstract

This article concerns with the existence and uniqueness theory of solutions for sequential fractional differential system involving Caputo fractional derivatives of order 1<alpha, beta<2 with coupled nonseparated boundary conditions. The standard tools of the fixed point theory were used to establish the main results. Application is introduced to show the validity of our results.


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## Body

## 1. Introduction

This article is devoted to the study of the following nonlinear sequential fractional differential equation.(1)Dcα−1D+λut=ft,ut,vt,Dcα−1ut,Dcβ−1vt,Dcβ−1D+μvt=gt,ut,vt,Dcα−1ut,Dcβ−1vt,uη=ε1vT,Dcα−1uη=ε2Dcα−1vT,vη=δ1uT,Dcβ−1vη=δ2Dcβ−1uT,where Dcα−1 is the fractional derivative of order in Caputo sense, and 1<α,β≤2,0≤t≤T0<η<T,λ,μ∈ℝ+, and ε1,ε2,δ1,δ2∈ℝ.The studied equation, equation (1), is subject to boundary conditions which are coupled but nonseparated. The main goal of the article is to prove the existence of solutions to the problem defined by equation (1).The theory of fractional differential equation subject to boundary conditions has long been of the interest of researchers. Two-point boundary conditions are common in bibliography; however, special attention is given to more complex boundary conditions such as integral and nonlocal multiple points.Differential equations with nonlocal boundary conditions are of high importance in applied sciences. They are useful in modeling some chemical processes and in general to model a process which is located inside a defined region. On the other hand, differential equations with integral boundary conditions are appropriate for modeling problems that involve flowing, such as blood flow problem (see [1, 2]). Other application examples are biology, finances, etc. (see [18–20] for illustration). Some interesting general results on boundary value problems are found in [3, 10–17] and [21–23]. In references [4–9, 24–26], the authors have investigated sequential fractional differential equations.Although many articles have focused on the study of fractional differential equations with multiple boundary conditions, there is no or there are few works in the literature that studied problems with coupled nonseparated fractional boundary conditions. In this regard, we study the problem, equation (1), subject to nonseparated fractional boundary conditions.To prove an existence and uniqueness of a solution to problem defined by equation (1), two main theorems are proposed in the work. Banach’s fixed point theorem is used to establish the first theorem in which uniqueness of a solution to problem (1) is proved. The second theorem establishes existence of the solution to problem (1), using Leray–Schauder’s alternative criterion. After this introductory section of this work, there are four other sections, which are organized following a hierarchical structure. The second section provides a brief review of fractional calculus definitions. The third section is dedicated to the main results of the work. The fourth section provides a numerical example to back up the theoretical results. Finally, conclusion and possible future works are discussed in the fifth section.Notations:(2)ω1t=ε2ρ2q2Δ1ρ12q1+ε2ρ1e−λt+δ1ρ2q2ε1ε2Δ1Δ2ρ12q1+δ1ε12Δ2ρ2e−λT−e−λη+ε1ε2q2Δ1Δ2ρ1q1δ1ε1e−μT−e−μη+ε1ε2q2e−μTΔ1ρ1q1−ρ2q2ε2e−ληΔ1ρ12q1+ε1e−ληρ1ω2t=δ2ρ2Δ1ρ1q1e−λt+ε1δ1δ2ρ2Δ1Δ2ρ1q1e−λT−e−λη+δ2ε1Δ1Δ2q1ε1e−μT−e−μη+δ2Δ1ρ1q1ε1ρ1e−μT−ρ2e−μη,ω3t=ε1δ1ρ2q2Δ1Δ2ρ1q1+ε1δ1Δ2e−λη−e−λT+ε1q2Δ1Δ2q1e−μη−δ1ε1e−μT−ε1q2e−μTΔ1q1−ρ2q2e−ληΔ1ρ1q1+e−λη−ρ2q2Δ1ρ1q1+1e−λt,ω4t=ε1δ1ρ2Δ1Δ2ρ1e−λη−e−λT+ε1Δ1Δ2e−μη−ε1δ1e−μT+1Δ1ρ1ρ2e−λη−ε1ρ1e−μT−ρ2Δ1ρ1e−λt,ω5t=ε1ε2δ1ρ2q2μΔ1Δ2ρ12q1+ε1ε2δ1μΔ2e−λη−e−λT+ε1ε2q2μΔ1Δ2ρ1q1e−μη−δ1ε1e−μT−ε1ε2q2e−μTμΔ1Δ2ρ1q1−ε2ρ2q2e−λημΔ1ρ12q1+ε2e−λημρ1−ε2ρ2q2μΔ1ρ12q1+ε2μρ1e−λt,ω6t=ε1δ1δ2ρ2λΔ1Δ2ρ1q1e−λη−e−λT+ε1δ2λΔ1Δ2q1e−μη−ε1δ1e−μT+δ2λΔ1Δ2ρ1ρ2e−λη−ε1ρ1e−μT−ρ2δ2λΔ1Δ2ρ1e−λt,ω7t=ρ2q2λΔ1ρ1q1+λe−λt+ε1δ1ρ2q2λΔ1Δ2ρ1q1+ε1δ1λΔ2e−λT−e−λη+ε1q2λΔ1Δ2q1δ1ε1e−μT−e−μη+ε1q2e−μTλΔ1q1−ρ2q2e−ληλΔ1ρ1q1+λe−λη,ω8t=μρ2Δ1ρ1e−λt+ε1δ1ρ2Δ1Δ2ρ1e−λT−e−λη+ε1μΔ1Δ2ε1δ1e−μT−e−μη+μΔ1ρ1ε1ρ1e−μT−ρ2e−λη,(3)θ1t=q2ε2Δ1ρ1q1e−μt+δ1ρ2q2ε2Δ1Δ2ρ12q1+δ1ε1Δ2ρ1e−λT−e−λη+q2ε2Δ1Δ2ρ1q1δ1ε1e−μT−e−μη,θ2t=δ2Δ1q1e−μt+δ2Δ1Δ2q1δ1ε1e−μT−e−μη+δ1δ2ρ2Δ1Δ2ρ1q1δ1ε1e−λT−e−λη,θ3t=δ1ρ2q2Δ1Δ2ρ1q1+δ1Δ2e−λη−e−λT+q2Δ1Δ2q1e−μη−δ1ε1e−μT−q2Δ1q1e−μt,θ4t=δ1ρ2Δ1Δ2ρ1e−λη−e−λT+1Δ1Δ2e−μη−δ1ε1e−μT−1Δ1e−μt,θ5t=μδ1ρ2q2ε2Δ1Δ2ρ12q1+δ1ε1Δ2ρ1e−λη−e−λT+q2ε2Δ1Δ2ρ1q1e−μη−μδ1ε1e−μT−q2ε2μΔ1ρ1q1e−μt,θ6t=λδ1δ2ρ2Δ1Δ2ρ1q1e−λη−e−λT+λδ2Δ1Δ2q1e−μη−δ1ρ2e−μT−λδ2Δ1q1e−μt,θ7t=λq2Δ1q1e−μt+λδ1ρ2q2Δ1Δ2ρ1q1+λδ1Δ2e−λT−e−λη+λq2Δ1Δ2q1δ1ε1e−μT−e−μη,θ8t=μΔ1e−μt+μδ1ρ2Δ1Δ2ρ1e−λT−e−λη+μΔ1Δ2δ1ε1e−μT−e−μη,where(4)ρ1=−λΓ2−α∫0ηη−s1−αe−λsds,q1=−μΓ2−α∫0ηη−s1−βe−μsds,ρ2=−με2Γ2−α∫0TT−s1−αe−μsds,q2=−λδ2Γ2−α∫0TT−s1−βe−λsds,Δ1=1−ρ2q2ρ1q1,andΔ2=1−ε1δ1,M1=δ1ε12Δ2e−λT+1+ε1, andM2=−δ1ε1Δ2e−λT+1−1,N1=δ1ε1Δ2,andN2=−δ1Δ2,with(5)K1t,τ=1Γα−1∫τte−λt−rr−sα−2dr,∫0tK1t,τhτdτ=∫0te−λt−rIα−1hrdr,G1t,τ=1Γα−1∫τte−μt−rr−τβ−2dr,∫0tG1t,τhτdτ=∫0te−μt−rIβ−1hrdr,K2t,r=1Γ2−α∫rtt−τ1−αK1τ,rdτ,G2t,r=1Γ2−α∫rtt−τ1−βG1τ,rdτ.Then,(6)1Γ2−α∫0ηη−τ1−α∫0τe−λτ−rIα−1φrdrdτ=∫0ηK2η,rφrdr.In a like manner,(7)1Γ2−α∫0ηη−τ1−β∫0τe−μτ−rIβ−1ψrdrdτ=∫0ηG2η,rψrdr.Lemma 1.
Letϕ,ψ∈C0,T,ℝ, then the solution of the following system:(8)Dcα+λDcα−1ut=ϕt,Dcβ+μDcβ−1vt=ψt,uη=ε1vT,Dcα−1uη=ε2Dcα−1vT,vη=δ1uT,Dcβ−1vη=δ2Dcβ−1uT,is given by(9)ut=∫0tK1t,xφxdx+ω1t∫0Tψxdx+ω2t∫0Tφxdx+ω3t∫0ηφxdx+ω4t∫0ηψxdx+ω5t∫0TK2T,xψxdx+ω6t∫0TG2T,xφxdx+ω7t∫0ηK2η,xφxdx+ω8t∫0ηG2η,xψxdx+M1∫0TG1T,xψxdx+M2∫0ηK1η,xφxdx,(10)vt=∫0tG1t,xψxdx+θ1t∫0Tψxdx+θ2t∫0Tφxdx+θ3t∫0ηφxdx+θ4t∫0ηψxdx+θ5t∫0TK2T,xψxdx+θ6t∫0TG2T,xφxdx+θ7t∫0ηK2η,xφxdx+θ8t∫0ηG2η,xψxdx+N1∫0TG1T,xψxdx+N2∫0ηK1η,xφxdx.Proof.
Solving the sequential linear equations(11)Dcα−1D+λut=ϕt,Dcβ−1D+μvt=ψt,we get(12)ut=c0e−λt+c1+∫0te−λt−sIα−1φsds,vt=d0e−μt+d1+∫0te−μt−sIβ−1ψsds.
Now, we need to find the constantsc0,c1,d0, and d1. Before finding these constants, we find,(13)Dcα−1ut=c0−λ/Γ2−α∫0tt−s1−αe−λsds+1Γ2−α∫0tt−s1−αIα−1φs−λ∫0se−λs−τIα−1φτdτds,Dcβ−1ut=c0−λΓ2−β∫0tt−s1−βe−λsds+1Γ2−β∫0tt−s1−βIβ−1φs−λ∫0se−λs−τIβ−1φτdτds,Dcα−1vt=d0−μΓ2−α∫0tt−s1−αe−μsds+1Γ2−α∫0tt−s1−αIα−1ψs−μ∫0se−μs−τIα−1ψτdτds,Dcβ−1ut=d0−μΓ2−β∫0tt−s1−βe−μsds+1Γ2−β∫0tt−s1−βIβ−1ψs−μ∫0se−μs−τIβ−1ψτdτds.
The first boundary conditionsuη=ε1vT,Dcα−1uη=ε2Dcα−1vT give,(14)c0e−λη+c1+∫0ηK1η,xφxdx=d0ε1e−μT+d1ε1+ε1∫0TG1T,xψxdx,(15)c0ρ1+∫0ηφxdx−λ∫0ηK2η,xφxdx=d0ρ2+ε2∫0Tψxdx−με2∫0TK2T,xψxdx.
From the second boundary conditionsvη=δ1uT,Dcβ−1vη=δ2Dcβ−1uT, we get,(16)d0e−μη+d1+∫0ηG1η,xφxdx=c0δ1e−λT+c1δ1+δ1∫0TG1T,xψxdx,(17)d0q1+∫0ηψxdx−μ∫0ηG2η,xψxdx=c0q2+δ2∫0Tφxdx−λδ2∫0TG2T,xφxdx.
Asimultaneous solution of (15) and (17) implies,(18)d0=q2ε2Δ1ρ1q1∫0Tψxdx+δ2Δ1q1∫0Tφxdx−q2Δ1q1∫0ηφxdx−1Δ1∫0ηψxdx−μq2ε2Δ1ρ1q1∫0TK2T,xψxdx−λδ2Δ1q1∫0TG2T,xφxdx+λq2Δ1q1∫0ηK2η,xφxdx+μΔ1∫0ηG2η,xψxdx,Δ1≠0,(19)c0=ρ2q2ε2Δ1ρ12q1+ε2ρ1∫0Tψxdx+ρ2δ2Δ1ρ1q1∫0Tφxdx−ρ2q2Δ1ρ1q1+1∫0ηφxdx−ρ2Δ1ρ1∫0ηψxdx−μρ2q2ε2Δ1ρ12q1+με2ρ1∫0TK2T,xψxdx−λρ2δ2Δ1ρ1q1∫0TG2T,xφxdx+λρ2q2Δ1ρ1q1+λ∫0ηK2η,xφxdx+μρ2Δ1ρ1∫0ηG2η,xψxdx.
Next, we substitute (18) and (19) into (14) and (16) and solve the result system of the equations; this leads to,(20)d1=δ1ρ2q2ε2Δ1Δ2ρ12q1+δ1ε1Δ2ρ1e−λT−e−λη+q2ε2Δ1Δ2ρ1q1δ1ε1e−μT−e−μη∫0Tψxdx+δ2Δ1Δ2q1δ1ε1e−μT−e−μη+δ1δ2ρ2Δ1Δ2ρ1q1δ1ε1e−λT−e−λη∫0Tφxdx+δ1ρ2q2Δ1Δ2ρ1q1+δ1Δ2e−λη−e−λT+q2Δ1Δ2q1e−μη−δ1ε1e−μT∫0ηφxdx+δ1ρ2Δ1Δ2ρ1e−λη−e−λT+1Δ1Δ2e−μη−δ1ε1e−μT∫0ηψxdx+μδ1ρ2q2ε2Δ1Δ2ρ12q1+δ1ε1Δ2ρ1e−λη−e−λT+q2ε2Δ1Δ2ρ1q1e−μη−μδ1ε1e−μT∫0TK2T,xψxdx+λδ1δ2ρ2Δ1Δ2ρ1q1e−λη−e−λT+λδ2Δ1Δ2q1e−μη−δ1ρ2e−μT∫0TG2T,xφxdx+λδ1ρ2q2Δ1Δ2ρ1q1+λδ1Δ2e−λT−e−λη+λq2Δ1Δ2q1δ1ε1e−μT−e−μη∫0ηK2η,xφxdx+μδ1ρ2Δ1Δ2ρ1e−λT−e−λη+μΔ1Δ2δ1ε1e−μT−e−μη∫0ηG2η,xψxdx+δ1ε1Δ2∫0TG1T,xψxdx−δ1Δ2∫0ηK1η,xφxdx.(21)c1=δ1ρ2q2ε1ε2Δ1Δ2ρ12q1+δ1ε12Δ2ρ2e−λT−e−λη+ε1ε2q2Δ1Δ2ρ1q1δ1ε1e−μT−e−μη+ε1ε2q2e−μTΔ1ρ1q1−ρ2q2ε2e−ληΔ1ρ12q1+ε1e−ληρ1∫0Tψxdx+ε1δ1δ2ρ2Δ1Δ2ρ1q1e−λT−e−λη+δ2ε1Δ1Δ2q1ε1e−μT−e−μη+δ2Δ1ρ1q1ε1ρ1e−μT−ρ2e−μη∫0Tφxdx+ε1δ1ρ2q2Δ1Δ2ρ1q1+ε1δ1Δ2e−λη−e−λT+ε1q2Δ1Δ2q1e−μη−δ1ε1e−μT−ε1q2e−μTΔ1q1−ρ2q2e−ληΔ1ρ1q1+e−λη∫0ηφxdx+ε1δ1ρ2Δ1Δ2ρ1e−λη−e−λT+ε1Δ1Δ2e−μη−ε1δ1e−μT+1Δ1ρ1ρ2e−λη−ε1ρ1e−μT∫0ηψxdx+ε1ε2δ1ρ2q2μΔ1Δ2ρ12q1+ε1ε2δ1μΔ2e−λη−e−λT+ε1ε2q2μΔ1Δ2ρ1q1e−μη−δ1ε1e−μT−ε1ε2q2e−μTμΔ1Δ2ρ1q1−ε2ρ2q2e−λημΔ1ρ12q1+ε2e−λημρ1∫0TK2T,xψxdx+ε1δ1δ2ρ2λΔ1Δ2ρ1q1e−λη−e−λT+ε1δ2λΔ1Δ2q1e−μη−ε1δ1e−μT+δ2λΔ1Δ2ρ1ρ2e−λη−ε1ρ1e−μT∫0TG2T,xφxdx+ε1δ1ρ2q2λΔ1Δ2ρ1q1+ε1δ1λΔ2e−λT−e−λη+ε1q2λΔ1Δ2q1δ1ε1e−μT−e−μη+ε1q2e−μTλΔ1q1−ρ2q2e−ληλΔ1ρ1q1+λe−λη∫0ηK2η,xφxdx+ε1δ1ρ2Δ1Δ2ρ1e−λT−e−λη+ε1μΔ1Δ2ε1δ1e−μT−e−μη+μΔ1ρ1ε1ρ1e−μT−ρ2e−λη∫0ηG2η,xψxdx+δ1ε12Δ2e−λT+1+ε1∫0TG1T,xψxdx+−δ1ε1Δ2e−λT+1−1∫0ηK1η,xφxdx.
By finding the constantsc0,c1,d0, and d1 and substituting them in (12), we get the solution, proof completed.Lemma 2 (see [21]).
For anyϕ,ω∈C0,T;ℝ, we have ∫0tK1t,τϕτdτ−∫0tK1t,τωτdτ≤tα−1/λΓα1−e−λtϕ−ω,∫0tG1t,τϕτdτ−∫0tG1t,τpτdτ≤tβ−1/μΓβ1−e−μtϕ−ω,, ∫0tK2t,τϕτdτ−∫0tK2t,τωτdτ≤t/λ1−e−λtϕ−ω,∫0tG2t,τϕτdτ−∫0tG2t,τωτdτ≤t/μ1−e−μtϕ−ω.


## 2. Preliminaries

Some definitions of fractional are introduced in this section as they are required in sequel of this study.Definition 1 (see [15]).
Consider a real numberγ>0; the Mittag–Leffler function with one parameter is computed as Eγx=∑k=0∞xk/Γγk+1, where x∈ℝ and Γ is the gamma function Γr=∫0∞e−tsr−1ds,∀r>0.Definition 2 (see [15]).
Givenα∈ℝ+ demonstrating the order of the derivative, the Caputo fractional derivative of order α of a function φ:0,+∞⟶ℝ is given by (22)DC0+αφτ=∫0ττ−rp−α−1hprΓn−αdr,p−1<α<p,p∈ℕ∗,hpτ,α∈ℕ.


## 3. Main Results

GivenUα−1=u|u∈C0,T,ℝandDcα−1u∈C0,T,ℝ, and we set uα−1=uα−1+Dcα−1uα−1=sup0≤t≤Tut+sup0≤t≤TDcα−1ut, it is known that Uα−1,.α−1 is a Banach space, see [22].Also, letVβ−1=v|v∈C0,T,ℝandDcβ−1v∈C0,T,ℝ, and the norm vβ−1=vβ−1+Dcα−1vβ−1=sup0≤t≤Tvt+sup0≤t≤TDcβ−1vt, again Vβ−1,.β−1 is a Banach space.It is well known that the product spaceΩ=Uα−1×Vβ−1,.α−1,β−1 is a Banach space with the norm,(23)u,vα−1,β−1=uα−1+vβ−1.Now, substitutingφ,ψ by f⋅,u⋅,v⋅,Dcα−1u⋅,Dcβ−1v⋅,g⋅,u⋅,v⋅,Dcα−1u⋅,Dcβ−1v⋅ in Lemma 1 respectively. Then, we define ℱ:Ω⟶Ω allied to problem (1) as,(24)ℱ≔ℱ1u,vℱ2u,v,where(25)ℱ1u,vt=∫0tK1t,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω1t∫0Tgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω2t∫0Tfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω3t∫0ηfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω4t∫0ηgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω5t∫0TK2T,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω6t∫0TG2T,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω7t∫0ηK2η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω8t∫0ηG2η,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ++M1∫0TG1T,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+M2∫0ηK1η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ.ℱ2u,vt=∫0tG1t,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+θ1t∫0Tgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+θ2t∫0Tfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+θ3t∫0ηfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+θ4t∫0ηgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+θ5t∫0TK2T,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+θ6t∫0TG2T,τfx,ux,vx,Dcα−1ux,Dcβ−1vxdx+θ7t∫0ηK2η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτdx+θ8t∫0ηG2η,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+N1∫0TG1T,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+N2∫0ηK1η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ.Observe that,(26)Dcα−1ℱ1u,vt=∫0tfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ−λ∫0tK2t,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω1t∫0Tgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω2t∫0Tfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω3t∫0ηfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω4t∫0ηgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω5t∫0TK2T,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω6t∫0TG2T,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω7t∫0ηK2η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcα−1ω8t∫0ηG2η,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ.Dcβ−1ℱ2u,vt=∫0tgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ−μ∫0tG2t,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ1t∫0Tgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ2t∫0Tfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ3t∫0ηfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ4t∫0ηgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ5t∫0TK2T,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ6t∫0TG2T,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ7t∫0ηK2η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+Dcβ−1θ8t∫0ηG2η,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ.For computational convenience, we set,(27)Q=Q1+Q2,Q∗=Q1∗+Q2∗,ℛ=ℛ1+ℛ2,ℛ∗=ℛ1∗+ℛ2∗,where(28)Q1=sup0≤t≤T∫0tK1t,xdx+Tω2+ηω3+ω6∫0TG2T,xdx+ω7∫0ηK2η,xdx+M2∫0ηK1η,xdx,Q1∗=Tω1+ηω4+ω5∫0TK2T,xdx+ω8∫0ηG2η,xdx+M1∫0TG1T,xdx,Q2=sup0≤t≤T∫0tdx+λ∫0tK2t,xdx+TDcα−1ω2+ηDcα−1ω3+Dcα−1ω6∫0TG2T,xdx+Dcα−1ω7∫0ηK2η,xdx,Q2∗=TDcα−1ω1+ηDcα−1ω4+Dcα−1ω5∫0TK2T,xdx+Dcα−1ω8∫0ηG2η,xdx,ℛ1=θ2∫0Tdx+θ3∫0ηdx+θ6∫0TG2T,xdx++θ7∫0ηK2η,xdx+N2∫0ηK1η,xdx,ℛ1∗=sup0≤t≤T∫0tG1t,xdx+θ1∫0Tdx+θ4∫0ηdx+θ5∫0TK2T,xdx+θ8∫0ηG2η,xdx+N1∫0TG1T,xdx,ℛ2=Dcβ−1θ2∫0Tdx+Dcβ−1θ3∫0ηdx++Dcβ−1θ6∫0TG2T,xdx+Dcβ−1θ6∫0ηK2η,xdx,ℛ2∗=sup0≤t≤T∫0tdx+μsup0≤t≤T∫0tG2t,xdx+Dcβ−1θ1∫0Tdx+Dcβ−1θ4∫0ηdx+Dcβ−1θ5∫0TK2T,xdx+Dcβ−1θ8∫0ηG2η,xdx.To state the existence result for problem (1), we set the following assumptions:(A1)
The functionsf,g:0,T×ℝ4⟶ℝ are jointly continuous.(A2)
∃Lf,Lg>0 such that(29)ft,u1,v1,u1¯,v1¯−ft,u2,v2,u2¯,v2¯≤Lfu1−u2+u1¯−u2¯+v1−v2+v1¯−v2¯,gt,u1,v1,u1¯,v1¯−gt,u2,v2,u2¯,v2¯≤Lgu1−u2+u1¯−u2¯+v1−v2+v1¯−v2¯.∀t∈0,T,u1,u2,v1,v2,u1¯,u2¯,v1¯,v2¯∈ℝ.(A3)
There exist real numbersk1,k2,τ1,τ2,ς1,ς2,ζ1,ζ2,ξ1,ξ2≥0, such that ft,u,v,u¯,v¯≤k1+τ1u+ς1v+ζ1u¯+ξ1v¯,(30)gt,u,v,u¯,v¯≤k2+τ2u+ς2v+ζ2u¯+ξ1v¯,∀t,u,v,u¯,v¯∈0,T×ℝ4.(A4)
∃, Wα,Wβ,Wα,β∈ℝ+, such that(31)Wα=τ1+ζ1Q+ℛ+τ2+ζ2Q∗+ℛ∗,Wβ=ς1+ξ1Q+ℛ+ς2+ξ2Q∗+ℛ∗,with Wα,β=min1−τ1+ζ1Q+ℛ+τ2+ζ2Q∗+ℛ∗,1−ς1+ξ1Q+ℛ+ς2+ξ2Q∗+ℛ∗Theorem 1.
If both assumptions (A1) and (A2) are satisfied, andLfℛ+LfQ+Lgℛ∗+LgQ∗<1, then the problem (1) has a unique solution on 0,T.Proof.
Define a closed ballBr=u,v∈Uα−1×Vβ−1:u,vα−1,β−1≤r with r≥Nfℛ+Q+Ngℛ∗+Q∗/1−Lfℛ+Q+Lgℛ∗+Q∗, where Nf=sup0≤t≤Tft,0,0,0,0,Ng=sup0≤t≤Tgt,0,0,0,0.
First, we prove thatℱBr⊂Br, ∀u,v∈Br,∀t∈0,T; we have,(32)ℱ1u,vt=∫0tK1t,τfτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω1t∫0Tgτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω2t∫0Tfτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω3t∫0ηfτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω4t∫0ηgτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω5t∫0TK2T,τgτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω6t∫0TG2T,τfτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω7t∫0ηK2η,τfτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+ω8t∫0ηG2η,τgτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+M1∫0TG1T,τgτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ+M2∫0ηK1η,τfτ,uτ,vx,Dcα−1uτ,Dcβ−1vτdτ.
But,(33)ft,ut,vt,Dcα−1ut,Dcβ−1vt=ft,ut,vt,Dcα−1ut,Dcβ−1vt−ft,0,0,0,0+ft,0,0,0,0≤ft,ut,vt,Dcα−1ut,Dcβ−1vt−ft,0,0,0,0+ft,0,0,0,0≤Lfu,vα−1,β−1+sup0≤t≤Tft,0,0,0,0≤Lfr+Nf.Similarly,gt,ut,vt,Dcα−1ut,Dcβ−1vt≤Lgr+Ng.
Then,(34)ℱ1u,v=Lfr+Nf×sup0≤t≤T∫0tK1t,τdτ+Tω2+ηω3+ω6∫0TG2T,τdτ+ω7∫0ηK2η,τdτ+M2∫0ηK1η,τdτ+Lgr+Ng×Tω1+ηω4+ω5∫0TK2T,τdτ+ω8∫0ηG2η,τdτ+M1∫0TG1T,τdτ≤Lfr+NfQ1+Lgr+NgQ1∗,(35)Dcα−1ℱ1u,v=Lfr+Nf×sup0≤t≤T∫0tdτ+λ∫0tK2t,τdτ+TDcα−1ω2+ηDcα−1ω3+Dcα−1ω6∫0TG2T,τdτ+Dcα−1ω7∫0ηK2η,τdτ+Lgr+Ng×TDcα−1ω1+ηDcα−1ω4+Dcα−1ω5∫0TK2T,τdτ+Dcα−1ω8∫0ηG2η,τdτ,≤Lfr+NfQ2+Lgr+NgQ2∗.
(34) and (35) imply(36)ℱ1u,vα−1≤Lfr+NfQ1+Q2+Lgr+NgQ1∗+Q2∗.
Similarly,(37)ℱ2u,vβ−1≤Lfr+Nfℛ1+ℛ2+Lgr+Ngℛ1∗+ℛ2∗.
Combining (36) and (37) leads to(38)ℱu,vα−1,β−1≤Lfr+Nfℛ+Q+Lgr+Ngℛ∗+Q∗=Lfℛ+LfQ+Lgℛ∗+LgQ∗r+Nfℛ+NfQ+Ngℛ∗+NgQ∗≤r.
That is,ℱBr⊂Br.
Now, we show that the operatorℱ is a contraction. ∀u1,v1,u2,v2∈Ω. We have(39)ℱ1u1,v1t−ℱ1u2,v2t≤∫0tK1t,τfτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−fτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω1∫0Tgτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−gτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω2∫0Tfτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−fτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω3∫0ηfτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−fτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω4∫0ηgτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−gτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω5∫0TK2T,τgτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−gτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω6∫0TG2T,τfτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−fτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω7∫0ηK2η,τfτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−fτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+ω8∫0ηG2η,τgτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−gτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+M1∫0TG1T,τgτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−gτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ+M2∫0ηK1η,τfτ,u1τ,v1τ,Dcα−1u1τ,Dcβ−1v1τ−fτ,u2τ,v2τ,Dcα−1u2τ,Dcβ−1v2τdτ,(40)≤LfQ1u1−u2α−1,β−1+v1−v2α−1,β−1+LgQ1∗u1−u2α−1,β−1+v1−v2α−1,β−1.
One can easily show(41)Dcα−1ℱ1u1,v1−Dcα−1ℱ1u2,v2≤LfQ2u1−u2α−1,β−1+v1−v2α−1,β−1+LgQ2∗u1−u2α−1,β−1+v1−v2α−1,β−1.
Combining (40) and (41) leads to(42)ℱ1u1,v1−ℱ1u2,v2α−1≤LfQ+LgQ∗u1−u2α−1,β−1+v1−v2α−1,β−1.
Similarly,(43)ℱ2u1,v1−ℱ2u2,v2β−1≤Lfℛ+Lgℛ∗u1−u2α−1,β−1+v1−v2α−1,β−1.
By (42) and (43), we have(44)ℱu1,v1−ℱu2,v2α−1,β−1≤LfQ+LgQ∗+Lfℛ+Lgℛ∗u1−u2α−1,β−1+v1−v2α−1,β−1,which shows that the operator is a contraction and the proof is completed.Theorem 2.
Suppose the assumptions (A1) and (A3) are satisfied and assume that from (A4) thatWα<1andWβ<1. Then, problem (1) has a solution on 0,T.Proof.
First, we show that the operatorℱ:Ω⟶Ω is completely continuous. ℱ is continuous as a result of the continuity of fandg.
DefineS to be a bounded set in Ω. Then, there exist positive constants l1,l2 such that f⋅,u⋅,v⋅,u¯⋅,v¯⋅≤l1,g⋅,u⋅,v⋅,u¯⋅,v¯⋅≤l2,∀u,v,u¯,v¯∈S.
Then, for anyu,v,u¯,v¯∈S, it follows that(45)ℱ1u,v≤l1Q1+l2Q1∗,(46)Dcα−1ℱ1u,v≤l1ℛ1+l2ℛ1∗.
Combining (45) and (46), we obtain(47)ℱ1u,vα−1≤l1Q1+ℛ1+l2Q1∗+ℛ1∗.
Similarly,(48)ℱ2u,vβ−1≤l1Q2+ℛ2+l2Q2∗+ℛ2∗.
(47) and (48) imply(49)ℱu,vα−1,β−1≤l1Q+ℛ+l2Q∗+ℛ∗.
By equation (49), we showed the operator ℱ is uniformly bounded.
Next, we show thatℱ is equicontinuous; let t1,t2∈0,T with t1<t2, then we have,(50)ℱ1u,vt2−ℱ1u,vt1≤∫0t2K1t2,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ−∫0t1K1t1,τfτ,uτ,vτ,Dcβ−1uτ,Dcβ−1vτdτ+ω1t2−ω1t1∫0Tgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω2t2−ω2t1∫0Tfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω3t2−ω3t1∫0ηfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω4t2−ω4t1∫0ηgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω5t2−ω5t1∫0TK2T,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω6t2−ω6t1∫0TG2T,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω7t2−ω7t1∫0ηK2η,τfτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ+ω8t2−ω8t1∫0ηG2η,τgτ,uτ,vτ,Dcα−1uτ,Dcβ−1vτdτ.
Then, the R.H.S. of (50) approaches zero as t1⟶t2 and implies ℱ1u,vt2−ℱ1u,vt1⟶0 as t1⟶t2.
In a like manner,(51)Dcα−1ℱ1u,vt2−Dcα−1ℱ1u,vt1⟶0,ast1⟶t2.
Also, one can easily show that,ℱ2u,vt2−ℱ2u,vt1⟶0 and Dcβ−1ℱ2u,vt2−Dcβ−1ℱ2u,vt1⟶0,ast1⟶t2.
Accordingly,ℱu,v is equicontinuous, and so ℱ is completely continuous.
Finally, we need to prove that the setX=u,v∈Ω:u,v=θℱu,v,θ∈0,1 is bounded,(52)∀t∈0,T,∀u,v∈X,we have.ut=θℱ1u,v, and vt=θℱ2u,v, then(53)ut≤k1+τ1u+ς1v+ζ1u¯+ξ1v¯Q1+k2+τ2u+ς2v+ζ2u¯+ξ2v¯Q1∗=k1Q1+k2Q1∗+τ1Q1+τ2Q1∗u+ς1Q1+ς2Q1∗v+ζ1Q1+ζ2Q1∗u¯+ξ1Q1+ξ2Q1∗v¯,Dcα−1ut≤k1+τ1u+ς1v+ζ1u¯+ξ1v¯ℛ1+k2+τ2u+ς2v+ζ2u¯+ξ2v¯ℛ1∗=k1ℛ1+k2ℛ1∗+τ1ℛ1+τ2ℛ1∗u+ς1ℛ1+ς2ℛ1∗v+ζ1ℛ1+ζ2ℛ1∗u¯+ξ1ℛ1+ξ2ℛ1∗v¯.
Hence, we have,(54)uα−1≤k1Q1+ℛ1+k2Q1∗+ℛ1∗+τ1Q1+ℛ1+τ2Q1∗+ℛ1∗u+ς1Q1+ℛ1+ς2Q1∗+ℛ1∗v+ζ1Q1+ℛ1+ζ2Q1∗+ℛ1∗u¯+ξ1Q1+ℛ1+ξ2Q1∗+ℛ1∗v¯,vβ−1≤k1Q2+ℛ2+k2Q2∗+ℛ2∗+τ1Q2+ℛ2+τ2Q2∗+ℛ2∗u+ς1Q2+ℛ2+ς2Q2∗+ℛ2∗v+ζ1Q2+ℛ2+ζ2Q2∗+ℛ2∗u¯+ξ1Q2+ℛ2+ξ2Q2∗+ℛ2∗v¯,which imply that(55)uα−1+vβ−1≤k1Q+ℛ+k2Q∗+ℛ∗+τ1Q+ℛ+τ2Q∗+ℛ∗u+ς1Q+ℛ+ς2Q∗+ℛ∗v+ζ1Q+ℛ+ζ2Q∗+ℛ∗u¯+ξ1Q+ℛ+ξ2Q∗+ℛ∗v¯.
But,(56)uα−1=u+u¯,vβ−1=v+v¯.
Then, (55) becomes,(57)uα−1+vβ−1≤k1Q+ℛ+k2Q∗+ℛ∗+τ1+ζ1Q+ℛ+τ2+ζ2Q∗+ℛ∗uα−1+ς1+ξ1Q+ℛ+ς2+ξ2Q∗+ℛ∗vβ−1.Consequently,u,vα−1,β−1≤k1Q+ℛ+k2Q∗+ℛ∗Wα,β,Wα,β≠0,which shows the boundedness of X is bounded. So, by Leray–Schauder’s alternative, ℱ has a fixed point, implying the existence of a solution for the BVP given by (1).


## 4. Example

Given the problem,(58)Dc1/2D+1ut=1936+t2ut+sinvt+Dc1/2ut+Dc1/4vt,0≤t≤3,Dc1/4D−1vt=1764−t4sinut+vt+Dc1/4vt1+Dc1/4vt+tan−1Dc1/2ut,u1=v3,Dc1/2u1=+Dc1/2v3,v1=u3,Dc1/4v1=+Dc1/4u3.Here,(59)ft,ut,vt,Dcα−1ut,Dcβ−1vt=1936+t2ut+sinvt+Dc1/2ut+Dc1/4vt,gt,ut,vt,Dcα−1ut,Dcβ−1vt=1764−t4sinut+vt+Dc1/4vt1+Dc1/4vt+tan−1Dc1/2ut,λ=1,μ=−1,α=32,=β=54,η=1,T=3,ε1=ε2=1,δ1=2,andδ2=1,with(60)ρ1≤1.13,ρ2≤39.31,q1≤2.95,q2≤−12.84,Δ1=152.41,Δ2=−1,M1=−1.1,M2=−1.013,N1=−2,andN2=2.We have,(61)θ1=0.44,θ2=0.004,θ3=0.44,θ4=0.55,θ5=1.056,θ6=3.4,θ7=3.1,θ8=0.05,andω1=0.17,ω2=0.03,ω3=0.19,ω4=0.44,ω5=69,ω6=0.5,ω7=0.17,ω8=0.58.Clearly, the functionsf,g satisfy the hypotheses (A1) and (A2), from the inequalities,(62)ft,u1t,v1t,Dc1/2u1t,Dc1/4v1t−ft,u2t,v2t,Dc1/2u2t,Dc1/4v2t≤1324u1t−u2t+v1t−v2t+Dc1/2u1t−Dc1/2u2t+Dc1/4v1t−Dc1/4v2t.(63)gt,u1t,v1t,Dc1/2u1t,Dc1/4v1t−gt,u2t,v2t,Dc1/2u2t,Dc1/4v2t≤1448u1t−u2t+v1t−v2t+Dc1/2u1t−Dc1/2u2t+Dc1/4v1t−Dc1/4v2t,with(64)Q=Q1+Q2,Q∗=Q1∗+Q2∗,ℛ=ℛ1+ℛ2,ℛ∗=ℛ1∗+ℛ2∗,Q1≤12.54,Q1∗≤23.93,Q2≤10.087,Q2∗≤6.2079,ℛ1≤194.82,ℛ1∗≤89.114,ℛ2≤0.25,ℛ2∗≤24.07.Observe that,Lfℛ+LfQ+Lgℛ∗+LgQ∗=1/324195.07+22.627+1/448113.184+30.1379=0.991814<1; Theorem 1 implies that our problem has unique solution.


## 5. Conclusion and Future Work

In this article, we investigate the existence result of the system of fractional differential equations given in problem (1). For the future work, the researcher may generalize our system by taking an n×1 system of sequential type and may apply another type of fractional derivatives such as Hadamard and Psi-Caputo fractional derivatives.



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*Source: 8992894-2022-02-08.xml*