# Existence and Uniqueness for Coupled Systems of Hilfer Type Sequential Fractional Differential Equations Involving Riemann–Stieltjes Integral Multistrip Boundary Conditions

Authors: Ayub Samadi; Sotiris K. Ntouyas; Suphawat Asawasamrit; Jessada Tariboon
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this paper, we study a coupled system of Hilfer type sequential fractional differential equations supplemented with Riemann–Stieltjes integral multistrip boundary conditions. The standard tools of the fixed point theory are employed to prove the existence and uniqueness results for the considered problem. Examples are constructed for the illustration of the obtained results.


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## Body

## 1. Introduction

Fractional calculus is a generalization of the classical calculus. Fractional differential equations become another necessary tool in solving real-life problems in different research areas such as physics, biology, engineering, and mechanics, see for example the monographs and papers [1–11].Boundary value problems of fractional differential equations represent an important and interesting branch of applied analysis. Usually, the researchers have given attention in studying fractional differential equations involving Caputo or Riemann–Liouville fractional derivative. But, Caputo or Riemann–Liouville derivative was not considered appropriate in studying some new models in engineering for example. To avoid the difficulties, some new type fractional order derivative operators were introduced in the literature such as Hadamard, Erdeyl-Kober, and Katugampola. Hilfer in [12] introduced a new derivative, which generalizes both Riemann–Liouville and Caputo derivatives. For some applications involving Hilfer fractional derivative, the interested reader is referred to [13–16] and references cited therein.In [17], Nuchpong et al. investigated a new class of boundary value problems for fractional differential equations for involving sequential Hilfer type fractional derivative and subject to Riemann–Stieltjes integral multistrip boundary conditions of the form(1)HDα,β+kHDα−1,βuz=fz,uz,z∈c,d,uc=0,ud=λ∫cdusdHs+∑i−1nμi∫ηiμiusds,where HDα,β denotes the Hilfer fractional derivative operator of order α, 1<α<2, and parameter β, 0≤β≤1, f:c,d×ℝ⟶ℝ is a continuous function; ∫cdxsdHs is the Riemann–Stieltjes integral with respect to the function H:c,d⟶ℝ, c≥0, k,μi∈ℝ,c<ηi<ξi≤d,i=1,2,…,n. Existence and uniqueness results are established by using basic tools from fixed point theory.The study of system of Hilfer type was initiated by Wongcharoen et al. [18], by presenting the following system of fractional differential equations:(2)HDα1,β1uz=fz,uz,vz,z∈c,d,HDα2,β2vz=gz,uz,vz,z∈c,d,uc=0ud=∑i=1mθ¯iIϕivξi,vc=0vd=∑j=1nζ¯jIϕ¯juzi,in which HDα1,β1 and HDα2,β2 indicate the Hilfer fractional derivatives of orders α1 and α2, 1<α1,α2<2, and parameters β1,β2, 0≤β1,β2≤1, f,g:c,d×ℝ×ℝ⟶ℝ are continuous functions, c≥0, θ¯i,ζ¯i∈ℝ, and Iϕi,Iϕ¯j are the Riemann–Liouville fractional integrals of order ϕi>0,ϕ¯j>0, i=1,2,…,m, j=1,2,…,n.Inspired by the forenamed studies, this article considers the existence and uniqueness of solutions for the following coupled system of Hilfer type fractional differential equations with Riemann–Stieltjes integral multistrip boundary conditions of the form(3)HDα1,β1+σ1HDα1−1,β1uz=f1z,uz,vzz∈c,d,HDα2,β2+σ2HDα2−1,β2vz=f2z,uz,vzz∈c,d,uc=0,ud=λ1∫cdvsdH1s+∑i=1nμi∫ηiξivsds,vc=0,vd=λ2∫cdusdH2s+∑r=1pνr∫ζrθrusds,in which HDα1,β1 and HDα2,β2 are the Hilfer fractional derivatives of orders 1<α1,α2<2 and parameters β1,β2, 0≤β1,β2≤1, f1,f2:c,d×ℝ×ℝ⟶ℝ are continuous functions, ∫cd⋅dH1s, ∫cd⋅dH2s are the Riemann–Stieltjes integrals with respect to the functions Hi:c,d⟶ℝ, i=1,2, c≥0, μi,νr∈ℝ, ηi,ξi,ζr,θr∈c,d, i=1,2,…,n, r=1,2,…,p, λ1,λ2,σ1,σ2∈ℝ.The remaining of this article has been regulated as follows: In section2 some concepts, lemmas and theorems are recalled which will be applied throughout this paper. In Section 3, an auxiliary lemma has been proved which concerns a linear variant of system (3) and it is used to convert the coupled system (3) into a fixed point problem. The classical fixed point theorems have been applied in order to obtain the results regarding the existence/uniqueness in Section 4. Thus, the classical Banach fixed point theorem is applied to obtain uniqueness result, while Leray-Schauder alternative and Krasnosel’skiĭ’s fixed point theorems are applied to present existence results. Examples are also constructed to illustrate the obtained results.


## 2. Preliminaries

Now, the following items are reminded which will be applied to fulfil the main results in the next steps.Throughout the paper, the Banach space of all continuous mappings fromc,d to ℝ are denoted by Y=Cc,d,ℝ which is equipped with the norm y=supyz;z∈c,d. It is clear that the space Y×Y, equipped with norm defined by x,y=x+y, is a Banach space.Also,ACnc,d,ℝ is the n-times absolutely continuous functions defined as(4)ACnc,d,ℝ=f:c,d⟶ℝ;fn−1∈ACc,d,ℝ.For a real valued functiong:0,∞⟶ℝ, the Riemann–Liouville fractional integral of order η>0 is defined by Iηgt=∫0tt−sη−1/Γηgsds, in which the right-hand side is defined point-wise on 0,∞, see [2]. Besides, for the function g, the Riemann–Liouville fractional derivative of order δ>0 is defined by RLDδgt=1/Γn−δd/dtn∫0tgs/t−ss−n+1ds, in which n=δ+1, where δ denotes the integer part of a real number δ, see [2], while the Caputo fractional derivative is defined by CDsgt=1/Γn−δ∫0t1/t−sδ−n+1d/dsngsds, provided that the right-hand side exists.Also, the Hilfer fractional derivative of orderα and parameter β of a function is defined by(5)HDα,βut=Iβn−αDnI1−βn−αut,where n−1<α<n,0≤β≤1,t>a,D=d/dt, see [12]. Note that if β=0 and β=1, the Hilfer derivative is reduced to the Riemann–Liouville and Caputo fractional derivatives, respectively.The following lemma will be applied to prove a lemma in the next section which presents a pattern of existence of solutions for system (1.3).Lemma 1 (see [13]).
Leth∈Lc,d, n1−1<α¯≤n1,n1∈ℕ, 0≤β¯≤1, In1−α¯1−β¯h∈ACkc,d. Then, we have the following relation:(6)Iα¯HDα¯,β¯hz=hz−∑k=0n1−1z−ck−n1−α¯1−β¯Γk−n1−α¯1−β¯+1limz⟶c+dkdzkI1−βn1−α¯hz.Finally, we collect the fixed point theorems applied to prove the main results in this paper.Lemma 2 (Banach fixed point theorem, [19]).
LetD be a closed set in X and T:D⟶D satisfies(7)Tu−Tv≤λu−v,for someλ∈0,1,for all u,v∈D.Then T admits a unique fixed point inD.Lemma 3 (Leray–Schauder alternative [20]).
Let the setω be closed bounded convex in X and O an open set contained in ω with 0∈O. Then, for the continuous and compact T:U¯⟶ω, either(1)
aT admits a fixed point in U¯, or(2)
aa there exists u∈∂U and μ∈0,1 with u=μTu.Lemma 4 (Krasnosel’skiĭ fixed point theorem, [21]).
LetN indicate a closed, bounded, convex, and nonempty subset of a Banach space Y and C,D be operators such that (i) Cx+Dy∈N where x,y∈N, (ii) C is compact and continuous, and (iii) D is a contraction mapping. Then, there exists z∈N such that z=Cz+Dz.


## 3. An Auxiliary Result

Lemma 5.
Letc≥0, 1<α1,α2<2, 0≤β1,β2≤1, γ1=α1+2β1−α1β1, γ2=α2+2β2−α2β2, h1,h2∈Cc,d,ℝ and Θ≠0. Then, the solution of the system(8)HDα1,β1+σ1HDα1−1,β1uz=h1z,z∈c,d,HDα2,β2+σ2HDα2−1,β2vz=h2z,z∈c,d,uc=0,ud=λ1∫cdvsdH1s+∑i=1nμi∫ηiξivsds,vc=0,vd=λ2∫cdusdH2s+∑r=1pνr∫ζrθrusds,is given by(9)uz=−σ1∫czusds+Iα1h1z+z−cγ1−1ΘΓγ1G3−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2h2sdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2h2sds+σ1∫cdusds−Iα1h1d+G2−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1h1sdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1h1sds+σ2∫cdvsds−Iα2h2d,(10)vz=−σ2∫czvsds+Iα2h2z+z−cγ2−1ΘΓγ2G1−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1h1sdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1h1sds+σ2∫cdvsds−Iα2h2d+G4−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2h2sdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2h2sds+σ1∫cdusds−Iα1h1d,where(11)G1=d−cγ1−1Γγ1,G2=λ1∫cdz−cγ2−1Γγ2dH1z+∑i=1nμiξi−cγ2−ηi−cγ2Γγ2+1,G3=d−cγ2−1Γγ2,G4=λ2∫cdz−cγ1−1Γγ1dH2z+∑r=1pνrθr−cγ1−ζr−cγ1Γγ1+1,(12)Θ=G1G3−G2G4.Proof .
Letu,v be a solution of system (5). By Lemma 2, we have(13)uz=k1z−cγ1−2Γγ1−1+k2z−cγ1−1Γγ1−−σ1∫czusds+Iα1h1z,vz=d1z−cγ2−2Γγ2−1+d2z−cγ2−1Γγ2−σ2∫czvsds+Iα2h2z,where k1,k2,d1,d2 are the arbitrary constants, since 1−β12−α1=γ1 and 1−β12−α2=γ2. Applying uc=0 and vc=0, we deduce that k1=d1=0. Thus, the previous equations become(14)uz=k2z−cγ1−1Γγ1−σ1∫czusds+Iα1h1z,(15)vz=d2z−cγ2−1Γγ2−σ2∫czvsds+Iα2h2z.
Now, applying the boundary conditionsud=λ1∫cdvsdH1s+∑i=1nμi∫ηiξivsds and vd=λ2∫cdusdH2s+∑r=1p∫ζrθrusds, we get(16)k2d−cγ1−1Γγ1−d2λ1∫cdz−cγ2−1Γγ2dH1z+∑i=1nμiξi−cγ2−ηi−cγ2Γγ2+1=−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2h2sdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2h2sds+σ1∫cdusds−Iα1h1d,d2d−cγ2−1Γγ2−k2λ2∫cdz−cγ1−1Γγ1dH2z+∑r=1pνrθr−cγ1−ζr−cγ1Γγ1+1=−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1h1sdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1h1sds+σ2∫cdvsds−Iα2h2d.
Consequently, we have the system(17)k2G1−d2G2=Ω1,d2G3−k2G4=Ω2,where(18)Ω1=−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2h2sdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2h2sds+σ1∫cdusds−Iα1h1d,Ω2=−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1h1sdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1h1sds+σ2∫cdvsds−Iα2h2d.
By solving the above system, we have(19)k2=Ω1G3+Ω2G2Θ,d2=G1Ω2+G4Ω1Θ.Substituting the values of k2 and d2 into (10) and (14), respectively, we obtain the solutions (9) and (10). We can obtain the converse by direct computation. The proof is finished.


## 4. Existence and Uniqueness Result

Due to Lemma5, we define an operator Q:Y×Y⟶Y×Y by(20)Qu,vz≔Q1u,vz,Q2u,vz,where(21)Q1u,vz=−σ1∫czusds+Iα1f1z,uz,vz+z−cγ1−1ΘΓγ1G3−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2f2s,us,vsdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds−Iα1f1d,ud,vd+G2−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1f1s,us,vsdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1f1s,us,vsds+σ2∫cdvsds−Iα2f2d,ud,vd,(22)Q2u,vz=−σ2∫czvsds+Iα2f2z,uz,vz+z−cγ2−1ΘΓγ2G1−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1f1s,us,vsdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1f1s,us,vsds+σ2∫cdvsds−Iα2f2d,ud,vd+G4−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2f2s,us,vsdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds−Iα1f1d,ud,vd.For convenience, we set(23)C1=1Γα1+1+d−cγ1−1ΘΓγ1G31Γα1+1+G2d−cγ1−1ΘΓγ1λ21Γα1+1∫cds−cα1dH2s+1Γα1+2∑r=1pνrθr−cα1+1−ζr−cα1+1,C2=d−cγ1−1ΘΓγ1G31Γα2+2∑i=1nμiξi−cα2+1−ηi−cα2+1+λ1Γα2+1∫cds−cα2dH1s+G2d−cγ1−1ΘΓγ11Γα2+1,C3=σ1d−c+G3d−cγ1−1ΘΓγ1d−cσ1+G2d−cγ1−1ΘΓγ1d−cσ1λ2∫cds−cdH2s+G2d−cγ1−1ΘΓγ1σ1∑r=1pνr12θr−c2−ζr−c2,C4=d−cγ1−1ΘΓγ1G3σ2λ1∫cds−cdH1s+d−cγ1−1ΘΓγ1G3σ2∑i=1nμi12ξi−c2−ηi−c2+G2d−cγ1−1ΘΓγ1d−cσ2,(24)D1=1Γα2+1+d−cγ2−1ΘΓγ2G11Γα2+1+G4d−cγ2−1ΘΓγ2λ11Γα2+1∫cds−cα2dH1s+1Γα2+2∑i=1nμiξi−cα2+1−ηi−cα2+1,D2=d−cγ2−1ΘΓγ2G11Γα1+2∑r=1pνrθr−cα1+1−ζr−cα1+1+λ2Γα1+1∫cds−cα1dH2s+G4d−cγ2−1ΘΓγ21Γα1+1,D3=σ2d−c+G1d−cγ2−1ΘΓγ2d−cσ2+G4d−cγ2−1ΘΓγ2d−cσ2λ1∫cds−cdH1s+G4d−cγ2−1ΘΓγ2σ2∑i=1nμi12ξi−c2−ηi−c2,D4=d−cγ2−1ΘΓγ2G1σ1λ2∫cds−cdH2s+d−cγ2−1ΘΓγ2G1σ1∑r=1pνr12θr−c2−ζr−c2+G4d−cγ2−1ΘΓγ2d−cσ1.Now, Banach’s fixed point theorem is applied to present the following uniqueness result.Theorem 1.
LetΘ≠0 and f1,f2:c,d×ℝ2⟶ℝ be two continuous functions such that for all z∈c,d and u¯i,v¯i∈ℝ,i=1,2, we have(25)f1z,u¯1,v¯1−f1z,u¯2,v¯2≤ℓ1u¯1−u¯2+v¯1−v¯2,f2z,u¯1,v¯1−f2z,u¯2,v¯2≤ℓ2u¯1−u¯2+v¯1−v¯2.where ℓ1,ℓ2 are positive constants and u¯i,v¯i∈ℝ,i=1,2. Then, there exists a unique solution of system (3) on c,d provided that(26)ℓ1C1+D1+ℓ2C2+D2+C3+C4+D3+D4<1.Proof .
It suffices to display that the operatorQ has a unique fixed point. For this aim, Banach’s theorem will be applied. Put supz∈c,df1z,0,0≔M<∞ and supz∈c,df2z,0,0≔N<∞. Now, we locate Br=u,v∈Y×Y;u,v≤r, in which(27)r≥MC1+D1+NC2+D21−ℓ1C1+D1+ℓ2C2+D2+C3+C4+D3+D4.
First, we indicate thatQBr⊆Br. Assume that u,v∈Br and z∈c,d. Due to (), we have(28)f1z,uz,vz≤f1z,uz,vz−f1z,0,0+f1z,0,0≤ℓ1uz+vz+M=ℓ1r+M.
Similarly, we have(29)f2z,uz,vz≤ℓ2r+N.
Hence, we infer that(30)Q1u,vz≤rσ1d−c+d−cγ1−1ΘΓγ1G3σ2λ1∫cds−cdH1s+σ2∑i=1nμi∫ηiξis−cds+σ1d−c+G2d−cγ1−1ΘΓγ1σ1λ2∫cds−cdH2s+σ1∑r=1pνr∫ζrθrs−cds+σ2d−c+ℓ1r+M1Γα1+1+d−cγ1−1ΘΓγ1G31Γα1+1+G2d−cγ−1ΘΓγ1λ21Γα1+1∫cds−cα1dH2s+1Γα1+1∑r=1pνr∫ζrθrs−cα1ds+ℓ2r+Nd−cγ1−1ΘΓγ1G31Γα2+1∑i=1nμi∫ηiξis−cα2ds+λ1Γα2+1∫cds−cα2dH1s+G2d−cγ1−1ΘΓγ11Γα2+1=ℓ1r+MC1+ℓ2r+NC2+rC3+C4.
Consequently,(31)Q1uv≤ℓ1r+MC1+ℓ2r+NC2+rC3+C4.
In the same way, we have(32)Q2u,v≤ℓ1r+MD1+ℓ2r+ND2+rD3+D4.
Hence,(33)Qu,v≤ℓ1r+MC1+D1+ℓ2r+NC2+D2+rC3+C4+D3+D3≤r,which yields that QBr⊆Br.
Now, it is provedQ:Y×Y⟶Y×Y is a contraction. Applying condition (25), for any u1,v1,u2,v2∈Y×Y and for each z∈c,d, we have(34)Q1u1,v1z−Q1u2,v2z≤ℓ1u1−u2+v1−v21Γα1+1+d−cγ1−1ΘΓγ1G31Γα1+1+G2d−cγ1−1ΘΓγ1λ21Γα1+1∫cds−cα1dH2s+1Γα1+2∑r=1pνr∫ζrθrs−cα1ds+ℓ2u1−u2+v1−v2d−cγ1−1ΘΓγ1G3×1Γα2+2∑i=1nμi∫ηiξis−cα2ds+λ1Γα2+1∫cds−cα2dH1s+G2d−cγ1−1ΘΓγ11Γα2+1+u1−u2σ1d−c+G3d−cγ1−1ΘΓγ1d−cσ1+G2d−cγ1−1ΘΓγ1d−c×σ1λ2∫cds−cdH2s+G2d−cγ1−1ΘΓγ1σ1∑r=1pνr∫ζrθrs−cds+v1−v2d−cγ1−1ΘΓγ1G3σ2λ1|∫cds−cdH1s+d−cγ1−1ΘΓγ1G3σ2∑i=1nμi∫ηiξis−cds+G2d−cγ1−1ΘΓγ1d−cσ2=ℓ1C1+ℓ2C2u1−u2+v1−v2+C3u1−u2+C4v1−v2≤ℓ1C1+ℓ2C2+C3+C4u1−u2+v1−v2,and hence(35)Q1u1,v1−Q1u2,v2≤ℓ1C1+ℓ2C2+C3+C4u1−u2+v1−v2.
Furthermore, we deduce that(36)Q2u1,v1−Q2u2,v2≤ℓ1D1+ℓ2D2+D3+D4u1−u2+v1−v2.
Using (25) and (33), we concluded that(37)Qu1,v1−Qu2,v2≤ℓ1C1+D1+ℓ2C2+D2+C3+C4+D3+D4×u1−u2+v1−v2.
Asℓ1C1+D1+ℓ2C2+D2+C3+C4+D3+D4<1, so the operator Q is a contraction and by applying Lemma 2, the operator Q has a unique solution which is the solution of the problem (3). The proof is finished.


## 5. Existence Results

Two existence results are proved in this section.


### 5.1. Existence Result via Leray–Schauder Alternative

The Leray–Schauder alternative (Lemma3) is used in the proof of our first existence result.Theorem 2.
LetΘ≠0 and f1, f2:c,d×ℝ2⟶ℝ be continuous functions. Assume thatH1 There exist real constants ui,vi≥0 for i=1,2 and u0,v0>0 such that for all u,v∈ℝ, we have(38)f1z,uz,vz≤u0+u1u+u2v,f2z,uz,vz≤v0+v1u+v2v.IfC1+D1u1+C2+D2v1+C3+D3<1 and C1+D1u2+C2+D2v2+C4+D4<1, where Ci, Di for i=1,2,3,4 are given by (23) and (24), respectively, then system (1.3) admits at least one solution on c,d.Proof .
The functionsf1,f2 are continuous on c,d×ℝ2. Thus, the operator Q is continuous. Now, we will show that the operator Q:Y×Y⟶Y×Y is completely continuous. Let Br⊂Y×Y be a bounded set, where Br=u,v∈Y×Y:u,v≤r. Then, for any u,v∈Br, there exist positive real numbers P1 and P2 such that f1z,ut,vz≤P1 and f2z,uz,vz≤P2.
Thus, for eachu,v∈Br, we have(39)Q1u,vz≤σ1∫czusds+Iα1f1z,uz,vz+d−cγ1−1ΘΓγ1G3σ2λ1∫cd∫csvtdtdH1s+λ1∫cdf1z,uz,vzdH1s+σ2∑i=1nμi∫ηiξi∫asvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds+Iα1f1d,ud,vd+G2σ1λ2∫cd∫cs|utdtdH2s+σ1∑r=1pνr∫ζrθr∫cs|utdtds+λ2∫cdIα1f1s,us,vsdH2s+σ2∫cdvsds+Iα2f2d,ud,vd+∑r=1pνr∫ζrθrIα1f1s,us,vsds≤rσ1d−c+d−cγ1−1ΘΓγ1G3σ2λ1∫cds−cdH1s+σ2∑i=1nμi∫ηiξis−cds+σ1d−c+G2d−cγ1−1ΘΓγ1σ1λ2∫cds−cdH2s+σ1∑r=1pνr∫ζrθrs−cds+σ2d−c+P11Γα1+1+d−cγ1−1ΘΓγ1G31Γα1+1+G2d−cγ−1ΘΓγ1λ21Γα1+1∫cds−cα1dH2s+1Γα1+1∑r=1pνr∫ζrθrs−cα1ds+P2d−cγ1−1ΘΓγ1G31Γα2+1∑i=1nμi∫ηiξis−cα2ds+λ1Γα2+1∫cds−cα2dH1s+G2d−cγ1−1ΘΓγ11Γα2+1=P1C1+P2C2+rC3+C4,which yields(40)Q1u,v≤C1P1+C2P2+rC3+C4.
Similarly, we obtain that(41)Q2u,v≤D1P1+D2P2+rD3+D4.
Hence, from the above inequalities, we get that the operatorQ is uniformly bounded, since(42)Qu,v≤C1+D1P1+C2+D2P2+rC3+C4+D3+D4.
Next, we are going to prove that the operatorQ is equicontinuous. Let τ1,τ2∈c,d with τ1<τ2. Then, we have(43)Q1u,vτ2−Q1u,vτ1≤σ1rτ2−τ1+P1∫cτ1τ2−cα1−1−τ2−cα1−1Γα1ds+P1∫τ1τ2τ2−cα1−1Γα1ds+τ2−cγ1−1−τ2−cγ1−1ΘΓγ1rG3σ2λ1∫cds−cdH1s+σ2∑i=1nμi∫ηiξis−cds+σ1d−c+G2rσ1λ2∫cds−cdH2s+σ1∑r=1pνr∫ζrθrs−cds+σ2d−c+P1G31Γα1+1+G2λ21Γα1+1∫cds−cα1dH2s+1Γα1+1∑r=1pνr∫ζrθrs−cα1ds+P2G31Γα2+1∑i=1nμi∫ηiξis−cα2ds+λ1Γα2+1∫cds−cα2dH1s+G21Γα2+1.
Therefore, we obtain(44)Q1u,vτ2−Q1u,vτ1⟶0,asτ1⟶τ2.
Analogously, we can get the following inequality:(45)Q2u,vτ2−Q2u,vτ1⟶0,asτ1⟶τ2.
Hence, the setQℬr is equicontinuous. Accordingly, Arzelá–Ascoli theorem implies that the operator Q is completely continuous.
Finally, we shall show the boundedness of the setZ=u,v∈Y×Y:u,v=μQu,v,0≤μ≤1. Let any u,v∈Z, then u,v=μQu,v. We have, for all z∈c,d,(46)uz=μQ1u,vz,vz=μQ2u,vz.
Then, we get(47)u≤u0+u1u+u2vC1+v0+v1u+v2vC2+uC3+vC4,v≤u0+u1u+u2vD1+v0+v1u+v2vD2+uD3+vD4,which imply that(48)u+v≤C1+D1u0+C2+D2v0+C1+D1u1+C2+D2v1+C3+D3u+C1+D1u2+C2+D2v2+C4+D4‖v‖.
Thus, we obtain(49)u,v≤C1+D1u0+C2+D2v0M∗,where M∗=min1−C1+D1u1−C2+D2v1−C3+D3,1−C1+D1u2−C2+D2v2−C4+D4, which shows that the set Z is bounded. Therefore, by Leray–Schauder alternative (Lemma 3), the operator Q has at least one fixed point. Hence, we deduce that problem (3) admits a solution on c,d, which completes the proof.


### 5.2. Existence Result via Krasnosel’skiĭ’s Fixed-Point Theorem

Now, Krasnosel’skiĭ’s fixed-point theorem (Lemma4) is applied to prove our second existence result.Theorem 3.
Assume thatΘ≠0 and f1,f2:c,d×ℝ2⟶ℝ are continuous functions satisfying condition (4.8) in Theorem 4.1. Furthermore, suppose that there exist positive constants R1 and R2 such that for all z∈c,d and u,v∈ℝ, we have(50)f1z,u,v≤R1,f2z,u,v≤R2.IfC3+C4<1, D3+D4<1 and d−cα1/Γα1+1l1+d−cα2/d−cα2ℓ2<1, then problem (1.3) admits a solution on c,d.Proof .
First, we decompose the operatorQ defined by (1) into four operators as(51)S1u,vz=−σ1∫czusds+z−cγ1−1ΘΓγ1G3−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2f2s,us,vsdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds−Iα1f1d,ud,vd+G2−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1f1s,us,vsdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1f1s,us,vsds+σ2∫cdvsds−Iα2f2d,ud,vd,S2u,vz=Iα1f1z,uz,vzIα1f1u,vz,S3u,vz=−σ2∫czvsds+z−cγ2−1ΘΓγ2G1−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1f1s,us,vsdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1f1s,us,vsds+σ2∫cdvsds−Iα2f2d,ud,vd+G4−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2f2s,us,vsdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds−Iα1f1d,ud,vd,S4u,vz=Iα2f2z,uz,vzIα2f2u,vz.
Accordingly,Q1u,vz=S1u,vz+S2u,vz and Q2u,vz=S3u,vz+S4u,vz. Let Bε=u,v∈Y×Y;u,v≤ε with(52)ε≥maxC1R1+C2R21−C3+C4,D1R1+D2R21−D3+D4.
First, it is showed thatQ1x,y+Q2u,v∈Bε for all x,y,u,v∈Bε. According to the proof of Theorem 1, we get(53)S2u,vzIα1f1z,uz,vz,S4u,vzIα2f2z,uz,vz.
Consequently,Q1x,y+Q2u,v∈Bε and we conclude the condition i of Lemma 4. Now, it is indicated that the operator S2,S4 is a contraction mapping. For x1,y1,x2,y2∈Bε, we infer that(54)S2x1,y1z−S2x2,y2z≤Iα1f1x1,y1−f1x2,y2z≤ℓ1x1−x2+y1−y2Iα11d≤ℓ1d−cα1Γα1+1x1−x2+y1−y2,(55)S4x1,y1z−S4x2,y2z∣≤Iα2f2x1,y1−f2x2,y2z≤ℓ2x1−x2+y1−y2Iα21d≤ℓ2d−cα2Γα2+1x1−x2+y1−y2.
Asd−cα1/Γα1+1ℓ1+d−cα2/Γα2+1ℓ2<1, the operator S2,S4 is a contraction and the condition iii of Lemma 4 is concluded. In the final step, the condition ii of Lemma 4 is verified for the operator S1,S3. As the functions f1,f2 are continuous, one can see that the operator S1,S3 is continuous. Furthermore, for u,v∈Bε, as in the proof of Theorem 1, we have(56)S1u,vz≤C1−1Γα1+1R1+C2R2+C3+C4ε=P∗,S3u,vz≤D1R1+D2−1Γα2+1R2+D3+D4ε=Q∗.
Hence,S1,S2u,v≤P∗+Q∗, which implies that S1,S3Bε is uniformly bounded. Now, we claim that the set S1,S3Bε is equicontinuous. For this aim, let τ1,τ2∈c,d with τ1<τ2. For any u,v∈Bε, similar to the proofs of equicontinuous for the operators Q1 and Q2 in 2, we can show that S1u,vτ2−S1u,vτ1, S3u,vτ2−S3u,vτ1⟶0 as τ1⟶τ2. Consequently, the set S1,S3Bε is equicontinuous and by applying Arzelá-Ascoli theorem, the operator S1,S3 will be compact on Bε. Therefore, by applying Lemma 4, problem (3) has at least one solution on c,d. This completes the proof.Remark 1.
In Theorem2, the functions f1,f2 are bounded by linear planes in three-dimension space. While, in Theorem 3, f1,f2 are bounded by fixed constants and also satisfied Lipschitz condition in (25).


## 5.1. Existence Result via Leray–Schauder Alternative

The Leray–Schauder alternative (Lemma3) is used in the proof of our first existence result.Theorem 2.
LetΘ≠0 and f1, f2:c,d×ℝ2⟶ℝ be continuous functions. Assume thatH1 There exist real constants ui,vi≥0 for i=1,2 and u0,v0>0 such that for all u,v∈ℝ, we have(38)f1z,uz,vz≤u0+u1u+u2v,f2z,uz,vz≤v0+v1u+v2v.IfC1+D1u1+C2+D2v1+C3+D3<1 and C1+D1u2+C2+D2v2+C4+D4<1, where Ci, Di for i=1,2,3,4 are given by (23) and (24), respectively, then system (1.3) admits at least one solution on c,d.Proof .
The functionsf1,f2 are continuous on c,d×ℝ2. Thus, the operator Q is continuous. Now, we will show that the operator Q:Y×Y⟶Y×Y is completely continuous. Let Br⊂Y×Y be a bounded set, where Br=u,v∈Y×Y:u,v≤r. Then, for any u,v∈Br, there exist positive real numbers P1 and P2 such that f1z,ut,vz≤P1 and f2z,uz,vz≤P2.
Thus, for eachu,v∈Br, we have(39)Q1u,vz≤σ1∫czusds+Iα1f1z,uz,vz+d−cγ1−1ΘΓγ1G3σ2λ1∫cd∫csvtdtdH1s+λ1∫cdf1z,uz,vzdH1s+σ2∑i=1nμi∫ηiξi∫asvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds+Iα1f1d,ud,vd+G2σ1λ2∫cd∫cs|utdtdH2s+σ1∑r=1pνr∫ζrθr∫cs|utdtds+λ2∫cdIα1f1s,us,vsdH2s+σ2∫cdvsds+Iα2f2d,ud,vd+∑r=1pνr∫ζrθrIα1f1s,us,vsds≤rσ1d−c+d−cγ1−1ΘΓγ1G3σ2λ1∫cds−cdH1s+σ2∑i=1nμi∫ηiξis−cds+σ1d−c+G2d−cγ1−1ΘΓγ1σ1λ2∫cds−cdH2s+σ1∑r=1pνr∫ζrθrs−cds+σ2d−c+P11Γα1+1+d−cγ1−1ΘΓγ1G31Γα1+1+G2d−cγ−1ΘΓγ1λ21Γα1+1∫cds−cα1dH2s+1Γα1+1∑r=1pνr∫ζrθrs−cα1ds+P2d−cγ1−1ΘΓγ1G31Γα2+1∑i=1nμi∫ηiξis−cα2ds+λ1Γα2+1∫cds−cα2dH1s+G2d−cγ1−1ΘΓγ11Γα2+1=P1C1+P2C2+rC3+C4,which yields(40)Q1u,v≤C1P1+C2P2+rC3+C4.
Similarly, we obtain that(41)Q2u,v≤D1P1+D2P2+rD3+D4.
Hence, from the above inequalities, we get that the operatorQ is uniformly bounded, since(42)Qu,v≤C1+D1P1+C2+D2P2+rC3+C4+D3+D4.
Next, we are going to prove that the operatorQ is equicontinuous. Let τ1,τ2∈c,d with τ1<τ2. Then, we have(43)Q1u,vτ2−Q1u,vτ1≤σ1rτ2−τ1+P1∫cτ1τ2−cα1−1−τ2−cα1−1Γα1ds+P1∫τ1τ2τ2−cα1−1Γα1ds+τ2−cγ1−1−τ2−cγ1−1ΘΓγ1rG3σ2λ1∫cds−cdH1s+σ2∑i=1nμi∫ηiξis−cds+σ1d−c+G2rσ1λ2∫cds−cdH2s+σ1∑r=1pνr∫ζrθrs−cds+σ2d−c+P1G31Γα1+1+G2λ21Γα1+1∫cds−cα1dH2s+1Γα1+1∑r=1pνr∫ζrθrs−cα1ds+P2G31Γα2+1∑i=1nμi∫ηiξis−cα2ds+λ1Γα2+1∫cds−cα2dH1s+G21Γα2+1.
Therefore, we obtain(44)Q1u,vτ2−Q1u,vτ1⟶0,asτ1⟶τ2.
Analogously, we can get the following inequality:(45)Q2u,vτ2−Q2u,vτ1⟶0,asτ1⟶τ2.
Hence, the setQℬr is equicontinuous. Accordingly, Arzelá–Ascoli theorem implies that the operator Q is completely continuous.
Finally, we shall show the boundedness of the setZ=u,v∈Y×Y:u,v=μQu,v,0≤μ≤1. Let any u,v∈Z, then u,v=μQu,v. We have, for all z∈c,d,(46)uz=μQ1u,vz,vz=μQ2u,vz.
Then, we get(47)u≤u0+u1u+u2vC1+v0+v1u+v2vC2+uC3+vC4,v≤u0+u1u+u2vD1+v0+v1u+v2vD2+uD3+vD4,which imply that(48)u+v≤C1+D1u0+C2+D2v0+C1+D1u1+C2+D2v1+C3+D3u+C1+D1u2+C2+D2v2+C4+D4‖v‖.
Thus, we obtain(49)u,v≤C1+D1u0+C2+D2v0M∗,where M∗=min1−C1+D1u1−C2+D2v1−C3+D3,1−C1+D1u2−C2+D2v2−C4+D4, which shows that the set Z is bounded. Therefore, by Leray–Schauder alternative (Lemma 3), the operator Q has at least one fixed point. Hence, we deduce that problem (3) admits a solution on c,d, which completes the proof.


## 5.2. Existence Result via Krasnosel’skiĭ’s Fixed-Point Theorem

Now, Krasnosel’skiĭ’s fixed-point theorem (Lemma4) is applied to prove our second existence result.Theorem 3.
Assume thatΘ≠0 and f1,f2:c,d×ℝ2⟶ℝ are continuous functions satisfying condition (4.8) in Theorem 4.1. Furthermore, suppose that there exist positive constants R1 and R2 such that for all z∈c,d and u,v∈ℝ, we have(50)f1z,u,v≤R1,f2z,u,v≤R2.IfC3+C4<1, D3+D4<1 and d−cα1/Γα1+1l1+d−cα2/d−cα2ℓ2<1, then problem (1.3) admits a solution on c,d.Proof .
First, we decompose the operatorQ defined by (1) into four operators as(51)S1u,vz=−σ1∫czusds+z−cγ1−1ΘΓγ1G3−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2f2s,us,vsdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds−Iα1f1d,ud,vd+G2−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1f1s,us,vsdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1f1s,us,vsds+σ2∫cdvsds−Iα2f2d,ud,vd,S2u,vz=Iα1f1z,uz,vzIα1f1u,vz,S3u,vz=−σ2∫czvsds+z−cγ2−1ΘΓγ2G1−σ1λ2∫cd∫csutdtdH2s+λ2∫cdIα1f1s,us,vsdH2s−σ1∑r=1pνr∫ζrθr∫csutdtds+∑r=1pνr∫ζrθrIα1f1s,us,vsds+σ2∫cdvsds−Iα2f2d,ud,vd+G4−σ2λ1∫cd∫csvtdtdH1s+λ1∫cdIα2f2s,us,vsdH1s−σ2∑i=1nμi∫ηiξi∫csvtdtds+∑i=1nμi∫ηiξiIα2f2s,us,vsds+σ1∫cdusds−Iα1f1d,ud,vd,S4u,vz=Iα2f2z,uz,vzIα2f2u,vz.
Accordingly,Q1u,vz=S1u,vz+S2u,vz and Q2u,vz=S3u,vz+S4u,vz. Let Bε=u,v∈Y×Y;u,v≤ε with(52)ε≥maxC1R1+C2R21−C3+C4,D1R1+D2R21−D3+D4.
First, it is showed thatQ1x,y+Q2u,v∈Bε for all x,y,u,v∈Bε. According to the proof of Theorem 1, we get(53)S2u,vzIα1f1z,uz,vz,S4u,vzIα2f2z,uz,vz.
Consequently,Q1x,y+Q2u,v∈Bε and we conclude the condition i of Lemma 4. Now, it is indicated that the operator S2,S4 is a contraction mapping. For x1,y1,x2,y2∈Bε, we infer that(54)S2x1,y1z−S2x2,y2z≤Iα1f1x1,y1−f1x2,y2z≤ℓ1x1−x2+y1−y2Iα11d≤ℓ1d−cα1Γα1+1x1−x2+y1−y2,(55)S4x1,y1z−S4x2,y2z∣≤Iα2f2x1,y1−f2x2,y2z≤ℓ2x1−x2+y1−y2Iα21d≤ℓ2d−cα2Γα2+1x1−x2+y1−y2.
Asd−cα1/Γα1+1ℓ1+d−cα2/Γα2+1ℓ2<1, the operator S2,S4 is a contraction and the condition iii of Lemma 4 is concluded. In the final step, the condition ii of Lemma 4 is verified for the operator S1,S3. As the functions f1,f2 are continuous, one can see that the operator S1,S3 is continuous. Furthermore, for u,v∈Bε, as in the proof of Theorem 1, we have(56)S1u,vz≤C1−1Γα1+1R1+C2R2+C3+C4ε=P∗,S3u,vz≤D1R1+D2−1Γα2+1R2+D3+D4ε=Q∗.
Hence,S1,S2u,v≤P∗+Q∗, which implies that S1,S3Bε is uniformly bounded. Now, we claim that the set S1,S3Bε is equicontinuous. For this aim, let τ1,τ2∈c,d with τ1<τ2. For any u,v∈Bε, similar to the proofs of equicontinuous for the operators Q1 and Q2 in 2, we can show that S1u,vτ2−S1u,vτ1, S3u,vτ2−S3u,vτ1⟶0 as τ1⟶τ2. Consequently, the set S1,S3Bε is equicontinuous and by applying Arzelá-Ascoli theorem, the operator S1,S3 will be compact on Bε. Therefore, by applying Lemma 4, problem (3) has at least one solution on c,d. This completes the proof.Remark 1.
In Theorem2, the functions f1,f2 are bounded by linear planes in three-dimension space. While, in Theorem 3, f1,f2 are bounded by fixed constants and also satisfied Lipschitz condition in (25).


## 6. Examples

Now, we present some examples to show the benefits of our results.Example 1.
Consider the following coupled system of Hilfer type sequential fractional differential equations involving Riemann–Stieltjes integral multistrip boundary conditions of the form(57)HD5/42/3+120HD1/42/3uz=f1z,uz,vz,z∈1/8,13/8,HD7/41/3+115HD3/41/3vz=f2z,uz,vz,z∈1/8,13/8,u18=0,v18=0,u138=14∫1/813/8vsde−2s+27∫1/25/8vsds+311∫9/85/4vsds,v138=15∫1/813/8usde−3s+413∫1/43/8usds+517∫3/47/8usds+619∫11/83/2usds.
Here,α1=5/4, α2=7/4, β1=2/3, β2=1/3, c=1/8, d=13/8, λ1=1/4, λ2=1/5, H1t=e−2t, H2t=e−3t, n=2, μ1=2/7, μ2=3/11, η1=1/2, η2=9/8, ξ1=5/8, ξ2=5/4, p=3, ν1=4/13, ν2=5/17, ν3=6/19, ζ1=1/4, ζ2=3/4, ζ3=11/8, θ1=3/8, θ2=7/8, and θ3=3/2. Then, we can compute that γ1=7/4, γ2=11/6, G1≈1.474766913, G2≈−0.03380798224, G3≈1.490431261, G4≈0.03704876432, Θ≈2.199291254, C1≈1.765659740, C2≈0.006128241272, C3≈0.1499796921, C4≈0.000526941741, D1≈1.242948896, D2≈0.06369559511, D3≈0.1998340850, and D4≈0.003945721441.(i)
The Lipschitzian functionsf1,f2:1/8,13/8×ℝ2⟶ℝ are given by(58)f1t,u,v=18t+4u2+2u21+u+sinv8t+5+12cos2πt,(59)f2t,u,v=18t+3tan−1u+e1−8t8t+74v2+5v51+v+14loget.From direct computation to (59)-(60), we get(60)f1t,u1,v1−f1t,u2,v2≤15u1−u2+16v1−v2,f2t,u1,v1−f2t,u2,v2≤14u1−u2+18v1−v2,foru1,u2,v1,v2∈ℝ. Setting ℓ1=1/5 and ℓ2=1/4, we obtain ℓ1C1+D1+ℓ2C2+D2+C3+C4+D3+D4≈0.9734641265<1. By application of our Theorem 1, the problem of Hilfer type sequential fractional differential system involving Riemann–Stieltjes integral multistrip boundary conditions (58) with (59)-(60) has a unique solution on 1/8,13/8.(ii)
Let the nonlinear functionsf1,f2:1/8,13/8×ℝ2⟶ℝ be defined by(61)f1t,u,v=12cos2πt+15ue−v4+v15sin4u31+v14,(62)f2t,u,v=13sin2πt+u18cos12v41+u17+vπtan−1u.We remark thatf1t,u,v≤1/2+1/5u+1/3v and f2t,u,v≤1/3+1/4u+1/2v. Now, we choose the constants as in Theorem 2 by u0=1/2, u1=1/5, u2=1/3, v0=1/3, v1=1/4, and v2=1/2. Then, we can find that C1+D1u1+C2+D2v1+C3+D3≈0.9748101164<1 and C1+D1u2+C2+D2v2+C4+D4≈0.7915367403<1. The benefit of Theorem 2 can be used to conclude that the coupled system of Hilfer type sequential fractional differential equations subject to boundary conditions (58) with (62)-(63) has at least one solution on 1/8,13/8.(iii)
Suppose that two Lipschitzian functionsf1,f2:1/8,13/8×ℝ2⟶ℝ are stated by(63)f1t,u,v=18t+1+18t+2u1+u+18t+3tan−1v,(64)f2t,u,v=18t+4+13e1−8tsinu+15e−1−8t2v1+v.
Actually, we can compute the bounds of the above two functions byf1t,u,v≤5/6+π/8, f2t,u,v≤47/60, for all u,v∈ℝ. In addition, we can find that f1t,u1,v1−f1t,u2,v2≤1/3u1−u2+1/4v1−v2 and f2t,u1,v1−f2t,u2,v2≤1/3u1−u2+1/5v1−v2, and thus we can set ℓ1=1/3 and ℓ2=1/3 satisfying condition (4.8) in Theorem 1. Then, we obtain C3+C4≈0.1505066338<1, D3+D4≈0.2037798064<1, and d−cα1/Γα1+1ℓ1+d−cα1/Γα2+1ℓ2≈0.9097463067<1 that all conditions in Theorem 3 are fulfilled. In this step, we conclude that the problem (58) with (64) has at least one solution on 1/8,13/8. Finally, we observe that the uniqueness result cannot be obtained because ℓ1C1+D1+ℓ2C2+D2+C3+C4+D3+D4≈1.380430597>1.


## 7. Conclusions

In the present research, we studied a coupled system of Hilfer type sequential fractional differential equations supplemented with Riemann–Stieltjes integral multistrip boundary conditions. First, an auxiliary lemma, concerning a linear variant of the considered problem, has been proved which is pivotal to converting the coupled system into a fixed point problem. Then, existence and uniqueness results are established via standard fixed point theorems. Thus, the classical Banach fixed point theorem is applied to obtain a uniqueness result, while Leray–Schauder alternative and Krasnosel’skiĭ’s fixed point theorem are applied to present the existence results. Numerical examples are also constructed to illustrate the obtained results. The obtained results are new and enrich the existing literature on coupled systems of Hilfer type sequential fractional differential equations.



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