Abstract

This paper studies a coupled system of nonlinear fractional differential equation with four-point boundary conditions. Applying the Schauder fixed-point theorem, an existence result is proved for the following system: 𝐷𝛼𝑢(𝑡)=𝑓(𝑡,𝑣(𝑡),𝐷𝑚𝑣(𝑡)), 𝑡∈(0,1), 𝐷𝛽𝑣(𝑡)=𝑓(𝑡,𝑢(𝑡),𝐷𝑛𝑢(𝑡)), 𝑡∈(0,1), 𝑢(0)=𝛾𝑢(𝜁), 𝑢(1)=𝛿𝑢(𝜂), 𝑣(0)=𝛾𝑣(𝜁), 𝑣(1)=𝛿𝑣(𝜂), where 𝛼,𝛽,𝑚,𝑛,𝜁,𝜂,𝛾,𝛿 satisfy certain conditions.

1. Introduction

Fractional differential equations arise in many engineering and scientific disciplines as the mathematical modelling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymerrheology, and so forth involves derivatives of fractional order. Fractional differential equations also serve as an excellent tool for the description of hereditary properties of various materials and processes. In consequence, the subject of fractional differential equations is gaining much importance and attention. For details, see [1–8] and the refrences therein.

On the other hand, the study of coupled systems involving fractional differential equations is also important as such systems occur in various problems of applied nature, for instance, see [9, 10]. Recently, in [11], the existence of nontrivial solutions was investigated for a coupled system of nonlinear fractional differential equations with two-point boundary conditions by using Schauder's fixed-point theorem. Reference [12] established the existence of a positive solution to a singular coupled system of fractional order. The existence of nontrivial solutions for a coupled system of nonlinear fractional differential equations with three-point boundary conditions was investigated in [13] by using Schauder's fixed point theorem.

In this paper, we consider a four-point boundary value problem for a coupled system of nonlinear fractional differential equation given by𝐷𝛼𝑢(𝑡)=𝑓(𝑡,𝑣(𝑡),𝐷𝑚𝐷𝑣(𝑡)),𝑡∈(0,1),𝛽𝑣(𝑡)=𝑓(𝑡,𝑢(𝑡),𝐷𝑛𝑢(𝑡)),𝑡∈(0,1),𝑢(0)=𝛾𝑢(𝜁),𝑢(1)=𝛿𝑢(𝜂),𝑣(0)=𝛾𝑣(𝜁),𝑣(1)=𝛿𝑣(𝜂),(1.1) where 1<𝛼,𝛽<2, 𝑚,𝑛,𝛾,𝛿>0, 0<𝜁<𝜂<1, 𝛼−𝑛≥1, 𝛽−𝑚≥1, 𝛿𝜂𝛼−1<1, 𝛿𝜂𝛼−2<1, 𝛾𝜁𝛼−1<1, 𝛾𝜁𝛼−2<1, 𝛿𝜂𝛽−1<1, 𝛿𝜂𝛽−2<1, 𝛾𝜁𝛽−1<1, 𝛾𝜁𝛽−2<1, 𝐷 is the standard Riemann-Liouville fractional derivative, and 𝑓,𝑔∶[0,1]×𝑅×𝑅→𝑅 are given continuous function.

The organization of this paper is as follows. In Section 2, we present some necessary definition and preliminary results that will be used to prove our main results. The proofs of our main results are given in Section 3.

2. Preliminaries

For the convenience of the reader, we present here the necessary definition from fractional calculus theory and preliminary results.

Definition 2.1 (see [5]). The Riemann-Liouville fractional integral of order 𝑞>0 of function 𝑓∶(0,∞)→𝑅 is given by 𝐼𝑞1𝑓(𝑡)=⌈(𝑞)𝑡0𝑓(𝑠)(𝑡−𝑠)1−𝑞𝑑𝑠,(2.1) provided that the integral exists.

Definition 2.2 (see [5]). The Riemann-Liouville fractional derivative of order 𝑞>0 of function 𝑓∶(0,∞)→𝑅 is given by 𝐷𝑞1𝑓(𝑡)=𝑑Γ(𝑛−𝑞)𝑑𝑡𝑛𝑡0𝑓(𝑠)(𝑡−𝑠)𝑞−𝑛+1𝑑𝑠,(2.2) where 𝑛=[𝑞]+1 and [𝑞] denotes the integral part of number 𝑞, provided that the right side is pointwise defined on (0,∞).

Lemma 2.3 ([5]). Let 𝑛−1<𝛼≤𝑛, 𝐷𝛼𝑢(𝑡) exists for 𝑡∈(0,1). Then, 𝐼𝛼𝐷𝛼𝑢(𝑡)=𝑢(𝑡)+𝐶1𝑡𝛼−1+𝐶2𝑡𝛼−2+⋯+𝐶𝑛𝑡𝛼−𝑛.(2.3)

Remark 2.4. The following properties are useful for our discussion: 𝐼𝛼𝐼𝛽𝑓(𝑡)=𝐼𝛼+𝛽𝑓(𝑡), 𝐷𝛼𝐼𝛼𝑓(𝑡)=𝑓(𝑡), 𝛼>0, 𝛽>0, 𝑓∈𝐿(0,1); 𝐼𝛼𝐷𝛼𝑓(𝑡)=𝑓(𝑡), 0<𝛼<1, 𝑓(𝑡)∈𝐶[0,1] and 𝐷𝛼𝑓(𝑡)∈𝐶(0,1)∩𝐿(0,1);𝐼𝛼∶𝐶[0,1]→𝐶[0,1],𝛼>0.

For convenience, we introduce the following notation. Let𝜇1=𝛾𝜁𝛼−1,𝜈1=1−𝛾𝜁𝛼−2,𝜛1=1−𝛿𝜂𝛼−1,𝜆1=1−𝛿𝜂𝛼−2,𝜇2=𝛾𝜁𝛽−1,𝜈2=1−𝛾𝜁𝛽−2,𝜛2=1−𝛿𝜂𝛽−1,𝜆2=1−𝛿𝜂𝛽−2.(2.4)

Let 𝐶(𝐽) Denote the space of all continuous functions defined on 𝐽=[0,1]. Let 𝑋={𝑢(𝑡)∶𝑢∈𝐶(𝐽)and𝐷𝑛𝑢∈𝐶(𝐽)} be a Banach space endowed with the norm ‖𝑢‖𝑥=max𝑡∈𝐽|𝑢(𝑡)|+max𝑡∈𝐽|𝐷𝑛𝑢(𝑡)|, where 1<𝛼<2, 0<𝑛≤𝛼−1, see [11] Lemma 3.2, and let 𝑌={𝑣(𝑡)∶𝑣∈𝐶(𝐽)and𝐷𝑚𝑣∈𝐶(𝐽)} be a Banach space equipped with the norm ‖𝑣‖𝑌=max𝑡∈𝐽|𝑣(𝑡)|+max𝑡∈𝐽|𝐷𝑚𝑣(𝑡)|, where 1<𝛽<2, 0<𝑚≤𝛽−1. Thus, (𝑋×𝑌,‖⋅‖𝑋×𝑌) is a Banach with the norm defined by ‖(𝑢,𝑣)‖𝑋×𝑌=max{‖𝑢‖𝑥,‖𝑣‖𝑌} for (𝑢,𝑣)∈𝑋×𝑌.

Lemma 2.5. Let 𝑦∈𝐶(𝐽) be a given function and 1<𝛼<2. Then, the unique solution of 𝐷𝛼𝑢(𝑡)=𝑦(𝑡),𝑡∈(0,1),(2.5)𝑢(0)=𝛾𝑢(𝜁),𝑢(1)=𝛿𝑢(𝜂),(2.6) is given by 𝑢(𝑡)=10𝐾1(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,(2.7) where 𝐾1(𝑡,𝑠) is Green's function given by 𝐾1(1𝑡,𝑠)=𝜛Γ(𝛼)1𝜈1+𝜇1𝜆1⎧⎪⎨⎪⎩𝐾11𝐾(𝑡,𝑠),0≤𝑡≤𝜁,12𝐾(𝑡,𝑠),𝜁<𝑡≤𝜂,13(𝑡,𝑠),𝜂<𝑡≤1,(2.8)𝐾11⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩(𝑡,𝑠)=(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝛾𝑡𝛼−2𝜛1−𝜆1𝑡(𝜁−𝑠)𝛼−1+𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,0≤𝑠≤𝑡,𝛾𝑡𝛼−2𝜛1−𝜆1𝑡(𝜁−𝑠)𝛼−1+𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,𝑡<𝑠≤𝜁,𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1𝜁<𝑠≤𝜂,−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1𝐾,𝜂<𝑠≤1,12⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩(𝑡,𝑠)=(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝛾𝑡𝛼−2𝜛1−𝜆1𝑡(𝜁−𝑠)𝛼−1+𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,0≤𝑠≤𝜁,(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,𝜁<𝑠≤𝑡,𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,𝑡<𝑠≤𝜂,−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1𝐾,𝜂<𝑠≤1,13(⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩𝑡,𝑠)=(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝛾𝑡𝛼−2𝜛1−𝜆1𝑡(𝜁−𝑠)𝛼−1+𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,0≤𝑠≤𝜁,(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝛿𝑡𝛼−2𝜈1𝑡+𝜇1(𝜂−𝑠)𝛼−1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,𝜁<𝑠≤𝜂,(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,𝜂<𝑠≤𝑡,−𝑡𝛼−2𝜈1𝑡+𝜇1(1−𝑠)𝛼−1,𝑡<𝑠≤1.(2.9)

Proof. For 𝐶1, 𝐶2∈𝑅, the general solution of (2.5) can be written as 𝑢(𝑡)=𝐼𝛼𝑦(𝑡)+𝐶1𝑡𝛼−1+𝐶2𝑡𝛼−2=𝑡0(𝑡−𝑠)𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝐶1𝑡𝛼−1+𝐶2𝑡𝛼−2.(2.10)
By the boundary condition, 𝐶𝑢(0)=𝛾𝑢(𝜁),2=1𝜈1𝛾𝜁0(𝜁−𝑠)𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠+𝜇1𝐶1.(2.11)
By the boundary condition, 𝐶𝑢(1)=𝛿𝑢(𝜂),1=1𝜛1𝛿𝜂0(𝜂−𝑠)𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠−10(1−𝑠)𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠−𝜆1𝐶2.(2.12)
Substituting (2.11) into (2.12), we get 𝐶1=𝛿𝜈1𝜛1𝜈1+𝜇1𝜆1𝜂0(𝜂−𝑠)𝛼−1−𝜈Γ(𝛼)𝑦(𝑠)𝑑𝑠1𝜛1𝜈1+𝜇1𝜆110(1−𝑠)𝛼−1−Γ(𝛼)𝑦(𝑠)𝑑𝑠𝛾𝜆1𝜛1𝜈1+𝜇1𝜆1𝜁0(𝜁−𝑠)𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠.(2.13)
Substituting (2.13) into (2.11), we get 𝐶2=𝛾𝜛1𝜛1𝜈1+𝜇1𝜆1𝜁0(𝜁−𝑠)𝛼−1+Γ(𝛼)𝑦(𝑠)𝑑𝑠𝛿𝜇1𝜛1𝜈1+𝜇1𝜆1𝜂0(𝜂−𝑠)𝛼−1−𝜇Γ(𝛼)𝑦(𝑠)𝑑𝑠1𝜛1𝜈1+𝜇1𝜆110(1−𝑠)𝛼−1Γ(𝛼)𝑦(𝑠)𝑑𝑠.(2.14)
Thus, the unique solution of (2.5) and (2.6) is 𝑢(𝑡)=𝑡0(𝑡−𝑠)𝛼−1Γ+(𝛼)𝑦(𝑠)𝑑𝑠𝛾𝑡𝛼−2𝜛1−𝜆1𝑡𝜛1𝜈1+𝜇1𝜆1𝜁0(𝜁−𝑠)𝛼−1+Γ(𝛼)𝑦(𝑠)𝑑𝑠𝛿𝑡𝛼−2𝜈1𝑡+𝜇1𝜛1𝜈1+𝜇1𝜆1𝜂0(𝜂−𝑠)𝛼−1−𝑡Γ(𝛼)𝑦(𝑠)𝑑𝑠𝛼−2𝜈1𝑡+𝜇1𝜛1𝜈1+𝜇1𝜆110(1−𝑠)𝛼−1=Γ(𝛼)𝑦(𝑠)𝑑𝑠10𝐾1(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,(2.15) where 𝐾1(𝑡,𝑠) is given by (2.8).

Similarly, the general solution of 𝐷𝛽𝑣(𝑡)=𝑦(𝑡),𝑡∈(0,1),𝑣(0)=𝛾𝑣(𝜁),𝑣(1)=𝛿𝑣(𝜂),(2.16) is𝑣(𝑡)=10𝐾2(𝑡,𝑠)𝑦(𝑠)𝑑𝑠,(2.17) where 𝐾2(𝑡,𝑠), can be obtained from 𝐾1(𝑡,𝑠) by replacing 𝛼 with 𝛽. Let (𝐾1,𝐾2) denote Green's function for the boundary value problem (1.1).

Consider the coupled system of integral equation:𝑢(𝑡)=10𝐾1(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝑣𝑣(𝑠))𝑑𝑠,(𝑡)=10𝐾2(𝑡,𝑠)𝑔(𝑠,𝑢(𝑠),𝐷𝑞𝑢(𝑠))𝑑𝑠.(2.18)

3. Main Results

Lemma 3.1. Assume that 𝑓,𝑔∶𝐽×𝑅×𝑅→𝑅 are continuous functions. Then, (𝑢,𝑣)∈𝑋×𝑌 is a solution of (1.1) if and only if (𝑢,𝑣)∈𝑋×𝑌 is a solution of (2.18).

Proof. The proof is immediate from Lemma 2.5, so we omit it.

Let us define an operator 𝐹∶𝑋×𝑌→𝑋×𝑌 as𝐹𝐹(𝑢,𝑣)(𝑡)=1𝑣(𝑡),𝐹2𝑢(𝑡),(3.1) where𝐹1𝑣(𝑡)=10𝐾1(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝑣(𝑠))𝑑𝑠,𝐹2𝑢(𝑡)=10𝐾2(𝑡,𝑠)𝑔(𝑠,𝑢(𝑠),𝐷𝑞𝑢(𝑠))𝑑𝑠.(3.2)

In view of the continuity of 𝐾1,𝐾2,𝑓,𝑔, it follows that 𝐹 is continuous. Moreover, by Lemma 3.1, the fixed-point of the operator 𝐹 coincides with the solution of (1.1).

For the forthcoming analysis, we introduce the growth condition on 𝑓 and 𝑔 as(A1)there exists a nonnegative function 𝑎(𝑡)∈𝐿(0,1) such that ||||𝑓(𝑡,𝑥,𝑦)≤𝑎(𝑡)+𝜖1|𝑥|𝜌1+𝜖2||𝑦||𝜌2,𝜖1,𝜖2>0,0<𝜌1,𝜌2<1,(3.3)(A2)there exist a nonnegative function 𝑏(𝑡)∈𝐿(0,1) such that ||||𝑔(𝑡,𝑥,𝑦)≤𝑏(𝑡)+𝛿1|𝑥|𝜎1+𝛿2||𝑦||𝜎2,𝛿1,𝛿2>0,0<𝜎1,𝜎2<1.(3.4)

Let us set the following notations for convenience∶1𝐴=𝜛𝛼Γ(𝛼+1)(𝛼−𝑞+1)1𝜈1+𝜇1𝜆1×𝛾𝜁𝛼𝜛Γ(𝛼+1)(𝛼−𝑞)1−𝜆1+𝛿𝜂𝛼𝜈Γ(𝛼+1)(𝛼−𝑞)1+𝜇1𝜈+𝛼Γ(𝛼+1)11+𝜛1+𝜇11+𝜆1+𝛼⌈(𝛼−𝑞+1)1−𝛿𝜂𝛼𝜈1+𝜇1−𝛾𝜁𝛼𝜛1−𝜆1,1𝐵=𝜛𝛽Γ(𝛽+1)(𝛽−𝑝+1)2𝜈2+𝜇2𝜆2×𝛾𝜁𝛽𝜛Γ(𝛽+1)(𝛽−𝑝)2−𝜆2+𝛿𝜂𝛽𝜈Γ(𝛽+1)(𝛽−𝑝)2+𝜇2𝜈+𝛽Γ(𝛽+1)21+𝜛2+𝜇21+𝜆2+𝛽⌈(𝛽−𝑝+1)1−𝛿𝜂𝛽𝜈2+𝜇2−𝛾𝜁𝛽𝜛2−𝜆2,𝑚=max𝑡∈𝐽10||𝑎(𝑠)𝑘1||1(𝑡,𝑠)𝑑𝑠+𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1×𝛾𝜛1−𝜆1𝜁0(𝜁−𝑠)𝛼−1𝜈𝑎(𝑠)𝑑𝑠+𝛿1+𝜇1𝜂0(𝜂−𝑠)𝛼−1+𝜈𝑎(𝑠)𝑑𝑠11+𝜛1+𝜇11+𝜆110(1−𝑠)𝛼−𝑞−1,𝑎(𝑠)𝑑𝑠𝑛=max𝑡∈𝐽10||𝑏(𝑠)𝑘1||1(𝑡,𝑠)𝑑𝑠+𝜛(𝛽−𝑝)2𝜈2+𝜇2𝜆2×𝛾𝜛2−𝜆2𝜁0(𝜁−𝑠)𝛽−1𝜈𝑏(𝑠)𝑑𝑠+𝛿2+𝜇2𝜂0(𝜂−𝑠)𝛽−1+𝜈𝑏(𝑠)𝑑𝑠21+𝜛2+𝜇21+𝜆210(1−𝑠)𝛽−𝑝−1.𝑏(𝑠)𝑑𝑠(3.5)

Define a ball 𝑊 in the Banach space 𝑋×𝑌 as‖‖𝑊=(𝑢(𝑡),𝑣(𝑡))∣(𝑢(𝑡),𝑣(𝑡))∈𝑋×𝑌,(𝑢(𝑡),𝑣(𝑡))𝑋×𝑌≤𝑅,𝑡∈𝐽,(3.6) where 𝑅≥max{(3𝐴𝜖1)1/(1−𝜌1),(3𝐴𝜖2)1/(1−𝜌2),(3𝐵𝛿1)1/(1−𝜎1),(3𝐵𝛿2)1/(1−𝜎2),3𝑚,3𝑛}.

Theorem 3.2. Assume that the assumptions (A1) and (A2) hold. Then, there exists a solution for the four-point boundary value problem (1.1).

Proof. As a first step, we prove that 𝐹∶𝑊→𝑊: ||𝐹1||=||||𝑣(𝑡)10𝑘1(𝑡,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||≤𝑣(𝑠))𝑑𝑠10||𝑎(𝑠)𝑘1||𝜖(𝑡,𝑠)𝑑𝑠+1||𝑅||𝜌1+𝜖2||𝑅||𝜌210||𝑘1||=(𝑡,𝑠)𝑑𝑠10||𝑎(𝑠)𝑘1(||𝜖𝑡,𝑠)𝑑𝑠+1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×−𝑡0(𝑡−𝑠)𝛼−1Γ(𝛼)𝑑𝑠−𝛾𝑡𝛼−2𝜛1−𝜆1𝑡𝜛1𝜈1+𝜇1𝜆1𝜁0(𝜁−𝑠)𝛼−1−Γ(𝛼)𝑑𝑠𝛿𝑡𝛼−2𝜈1𝑡+𝜇1𝜛1𝜈1+𝜇1𝜆1𝜂0(𝜂−𝑠)𝛼−1𝑡Γ(𝛼)𝑑𝑠+𝛼−2𝜈1𝑡+𝜇1𝜛1𝜈1+𝜇1𝜆110(1−𝑠)𝛼−1=Γ(𝛼)𝑑𝑠10||𝑎(𝑠)𝑘1(||𝜖𝑡,𝑠)𝑑𝑠+1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×−𝑡𝛼+𝑡Γ(𝛼+1)𝛼−2𝜈1𝑡+𝜇1−𝛾𝜁𝛼𝑡𝛼−2𝜛1−𝜆1𝑡−𝛿𝜂𝛼𝑡𝛼−2𝜈1𝑡+𝜇1𝜛1𝜈1+𝜇1𝜆1≤Γ(𝛼+1)10||𝑎(𝑠)𝑘1(||𝜖𝑡,𝑠)𝑑𝑠+1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×1−𝛿𝜂𝛼𝜈1+𝜇1−𝛾𝜁𝛼𝜛1−𝜆1𝜛1𝜈1+𝜇1𝜆1,||𝐷Γ(𝛼+1)𝑞𝐹1𝑣||=||||𝐷(𝑡)𝑞𝐼𝛼𝑓(𝑡,𝑣(𝑡),𝐷𝑝𝑣1(𝑡))+𝜛1𝜈1+𝜇1𝜆1×𝛾𝜛1𝐷𝑞𝑡𝛼−2−𝜆1𝐷𝑞𝑡𝛼−1𝐼𝛼𝑓(𝜁,𝑣(𝜁),𝐷𝑝𝜈𝑣(𝜁))+𝛿1𝐷𝑞𝑡𝛼−1+𝜇1𝐷𝑞𝑡𝛼−2𝐼𝛼𝑓(𝜂,𝑣(𝜂),𝐷𝑝−𝜈𝑣(𝜂))1𝐷𝑞𝑡𝛼−1+𝜇1𝐷𝑞𝑡𝛼−2𝐼𝛼𝑓(1,𝑣(1),𝐷𝑝||||≤𝑣(1))𝑡0(𝑡−𝑠)𝛼−𝑞−1Γ(𝛼−𝑞)𝑓(𝑠,𝑣(𝑠),𝐷𝑝1𝑣(𝑠))𝑑𝑠+𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1×𝛾𝜛1−𝜆1𝜁0(𝜁−𝑠)𝛼−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝜈𝑣(𝑠))𝑑𝑠+𝛿1+𝜇1𝜂0(𝜂−𝑠)𝛼−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝+𝜈𝑣(𝑠))𝑑𝑠1+𝜇110(1−𝑠)𝛼−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝≤1𝑣(𝑠))𝑑𝑠Γ(𝛼−𝑞)𝑡0(𝑡−𝑠)𝛼−𝑞−1𝜖𝑎(𝑠)𝑑𝑠+1||𝑅||𝜌1+𝜖2||𝑅||𝜌2𝑡0(𝑡−𝑠)𝛼−𝑞−1+1𝑑𝑠(𝜛𝛼−𝑞)1𝜈1+𝜇1𝜆1×𝛾𝜛1−𝜆1𝜁0(𝜁−𝑠)𝛼−1𝑎𝜛(𝑠)𝑑𝑠+𝛾1−𝜆1𝜖1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×𝜁0(𝜁−𝑠)𝛼−1𝜈𝑑𝑠+𝛿1+𝜇1𝜂0(𝜂−𝑠)𝛼−1𝜈𝑎(𝑠)𝑑𝑠+𝛿1+𝜇1𝜖1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×𝜂0(𝜂−𝑠)𝛼−1+𝜈𝑑𝑠1+𝜇110(1−𝑠)𝛼−1𝜈𝑎(𝑠)𝑑𝑠+1+𝜇1𝜖1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×10(1−𝑠)𝛼−1≤1𝑑𝑠𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1×𝛾𝜛1−𝜆1𝜁0(𝜁−𝑠)𝛼−1𝜛𝑎(𝑠)𝑑𝑠+𝛾1−𝜆1𝜖1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×𝜁0(𝜁−𝑠)𝛼−1𝜈𝑑𝑠+𝛿1+𝜇1𝜂0(𝜂−𝑠)𝛼−1𝜈𝑎(𝑠)𝑑𝑠+𝛿1+𝜇1𝜖1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×𝜂0(𝜂−𝑠)𝛼−1+𝜈𝑑𝑠11+𝜛1+𝜇11+𝜆110(1−𝑠)𝛼−1𝑎𝜖(𝑠)𝑑𝑠+1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×10(1−𝑠)𝛼−1=1𝑑𝑠(𝜛𝛼−𝑞)1𝜈1+𝜇1𝜆1×𝛾𝜛1−𝜆1𝜁0(𝜁−𝑠)𝛼−1𝑎𝜈(𝑠)𝑑𝑠+𝛿1+𝜇1𝜂0(𝜂−𝑠)𝛼−1𝑎+𝜈(𝑠)𝑑𝑠11+𝜛1+𝜇11+𝜆110(1−𝑠)𝛼−𝑞−1+𝜖𝑎(𝑠)𝑑𝑠1||𝑅||𝜌1+𝜖2||𝑅||𝜌2×𝛾𝜁𝛼𝜛(𝛼−𝑞)1−𝜆1+𝛿𝜂𝛼𝜈(𝛼−𝑞)1+𝜇1𝜈+𝛼11+𝜛1+𝜇11+𝜆1𝜛𝛼⌈(𝛼−𝑞+1)1𝜈1+𝜇1𝜆1.(3.7)
Thus, ‖‖𝐹1𝑣‖‖(𝑡)𝑥=max𝑡∈𝐽||𝐹1𝑣||(𝑡)+max𝑡∈𝐽||𝐷𝑞𝐹1𝑣||𝜖(𝑡)≤𝑚+1||𝑅||𝜌1+𝜖2||𝑅||𝜌2𝐴≤𝑅3+𝑅3+𝑅3=𝑅.(3.8)
Similarly, it can be shown that ‖𝐹2𝑣(𝑡)‖𝑦≤𝑛+(𝜖1|𝑅|𝜌1+𝜖2|𝑅|𝜌2)𝐵≤𝑅. Hence, we conclude that ‖𝐹1𝑣(𝑡)‖𝑥×𝑦≤𝑅.
Since 𝐹1𝑣(𝑡),𝐹2𝑣(𝑡),𝐷𝑞𝐹1𝑣(𝑡),𝐷𝑞𝐹2𝑣(𝑡) are continuous on 𝐽, therefore, 𝐹∶𝑊→𝑊.
Now, we show that 𝐹 is a completely continuous operator. For that we fix 𝑀=max𝑡∈𝐽||𝑓(𝑡,𝑣(𝑡),𝐷𝑝||𝑣(𝑡)),𝑁=max𝑡∈𝐽||𝑔(𝑡,𝑢(𝑡),𝐷𝑝||𝑣(𝑡)).(3.9)
For (𝑢,𝑣)∈𝑊,𝑡,𝜏∈𝐽(𝑡<𝜏), we have ||𝐹1𝑣(𝑡)−𝐹1||=||||𝑣(𝜏)10𝑘1(𝑡,𝑠)−𝑘1(𝜏,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||𝑣(𝑠))𝑑𝑠≤𝑀𝑡0||𝑘1(𝑡,𝑠)−𝑘1||(𝜏,𝑠)𝑑𝑠+𝜏𝑡||𝑘1(𝑡,𝑠)−𝑘1||(𝜏,𝑠)𝑑𝑠+𝜁𝜏||𝑘1(𝑡,𝑠)−𝑘1||+(𝜏,𝑠)𝑑𝑠𝜂𝜁||𝑘1(𝑡,𝑠)−𝑘1||(𝜏,𝑠)𝑑𝑠+1𝜂||𝑘1(𝑡,𝑠)−𝑘1||=𝑀(𝜏,𝑠)𝑑𝑠𝜛Γ(𝛼)1𝜈1+𝜇1𝜆1×𝑡0(𝜏−𝑠)𝛼−1−(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1+𝛾𝜆1(𝜁−𝑠)𝛼−1−𝛿𝜈1(𝜂−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1−𝛾𝜛1(𝜁−𝑠)𝛼−1−𝛿𝜇1(𝜂−𝑠)𝛼−1+𝑑𝑠𝜏𝑡(𝜏−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1+𝛾𝜆1(𝜁−𝑠)𝛼−1−𝛿𝜈1(𝜂−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1−𝛾𝜛1(𝜁−𝑠)𝛼−1−𝛿𝜇1(𝜂−𝑠)𝛼−1+𝑑𝑠𝜁𝜏𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1+𝛾𝜆1(𝜁−𝑠)𝛼−1−𝛿𝜈1(𝜂−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1−𝛾𝜛1(𝜁−𝑠)𝛼−1−𝛿𝜇1(𝜂−𝑠)𝛼−1+𝑑𝑠𝜂𝜁𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1−𝛿𝜈1(𝜂−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1−𝛿𝜇1(𝜂−𝑠)𝛼−1+𝑑𝑠1𝜂𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1=𝑀𝑑𝑠Γ𝜛(𝛼)1𝜈1+𝜇1𝜆1×𝜏0(𝜏−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1𝑑𝑠−𝑡0(𝑡−𝑠)𝛼−1𝜛1𝜈1+𝜇1𝜆1+𝑑𝑠𝜁0𝜏𝛼−1−𝑡𝛼−1𝛾𝜆1(𝜁−𝑠)𝛼−1−𝜏𝛼−2−𝑡𝛼−2𝛾𝜛1(𝜁−𝑠)𝛼−1+𝑑𝑠𝜂0𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1−𝛿𝜈1(𝜂−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1−𝛿𝜇1(𝜂−𝑠)𝛼−1+𝑑𝑠1𝜂𝜏𝛼−1−𝑡𝛼−1𝜈1(1−𝑠)𝛼−1+𝜏𝛼−2−𝑡𝛼−2𝜇1(1−𝑠)𝛼−1=𝑀𝑑𝑠𝜛Γ(𝛼+1)1𝜈1+𝜇1𝜆1×(𝜏𝛼−𝑡𝛼)𝜛1𝜈1+𝜇1𝜆1+𝜏𝛼−1−𝑡𝛼−1𝛾𝜁𝛼𝜆1−𝜛1+𝜈1+𝜇11−𝛿𝜂𝛼,||𝐷𝑞𝐹1𝑣(𝑡)−𝐷𝑞𝐹1𝑣||=||||𝐷(𝜏)𝑞𝐼𝛼𝑓(𝑡,𝑣(𝑡),𝐷𝑝𝑣1(𝑡))+𝜛1𝜈1+𝜇1𝜆1×𝛾𝜛1𝐷𝑞𝑡𝛼−2−𝜆1𝐷𝑞𝑡𝛼−1𝐼𝛼𝑓(𝜁,𝑣(𝜁),𝐷𝑝𝜈𝑣(𝜁))+𝛿1𝐷𝑞𝑡𝛼−1+𝜇1𝐷𝑞𝑡𝛼−2𝐼𝛼𝑓(𝜂,𝑣(𝜂),𝐷𝑝−𝜈𝑣(𝜂))1𝐷𝑞𝑡𝛼−1+𝜇1𝐷𝑞𝑡𝛼−2𝐼𝛼𝑓(1,𝑣(1),𝐷𝑝𝑣(1))−𝐷𝑞𝐼𝛼𝑓(𝜏,𝑣(𝜏),𝐷𝑝1𝑣(𝜏))−𝜛1𝜈1+𝜇1𝜆1×𝛾𝜛1𝐷𝑞𝜏𝛼−2−𝜆1𝐷𝑞𝜏𝛼−1𝐼𝛼𝑓(𝜁,𝑣(𝜁),𝐷𝑝𝜈𝑣(𝜁))+𝛿1𝐷𝑞𝜏𝛼−1+𝜇1𝐷𝑞𝜏𝛼−2𝐼𝛼𝑓(𝜂,𝑣(𝜂),𝐷𝑝−𝜈𝑣(𝜂))1𝐷𝑞𝜏𝛼−1+𝜇1𝐷𝑞𝜏𝛼−2𝐼𝛼𝑓(1,𝑣(1),𝐷𝑝||||≤1𝑣(1))||||Γ(𝛼−𝑞)𝑡0(𝑡−𝑠)𝛼−𝑞−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝑣(𝑠))𝑑𝑠−𝜏0(𝜏−𝑠)𝛼−𝑞−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||+𝑀𝑣(𝑠))𝑑𝑠𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1×||||𝑡𝛼−𝑞−1−𝜏𝛼−𝑞−1−𝛾𝜆1𝜁0(𝜁−𝑠)𝛼−1𝑑𝑠+𝛿𝜈1𝜂0(𝜂−𝑠)𝛼−1𝑑𝑠−𝜈110(1−𝑠)𝛼−1||||+𝑀𝑑𝑠𝛼𝜛(𝛼−𝑞−1)1𝜈1+𝜇1𝜆1×||||𝑡𝛼−𝑞−2−𝜏𝛼−𝑞−2𝛾𝜛1𝜁0(𝜁−𝑠)𝛼−1𝑑𝑠+𝛿𝜇1𝜂0(𝜂−𝑠)𝛼−1𝑑𝑠−𝜇110(1−𝑠)𝛼−1||||≤1𝑑𝑠||||Γ(𝛼−𝑞)𝑡0(𝑡−𝑠)𝛼−𝑞−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝑣(𝑠))𝑑𝑠−𝜏0(𝑡−𝑠)𝛼−𝑞−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||+||||𝑣(𝑠))𝑑𝑠𝜏0(𝑡−𝑠)𝛼−𝑞−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝑣(𝑠))𝑑𝑠−𝜏0(𝜏−𝑠)𝛼−𝑞−1𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||+𝑀𝑣(𝑠))𝑑𝑠𝛾𝜁𝛼𝜆1+𝜈11−𝛿𝜂𝛼𝛼𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1𝑡𝛼−𝑞−1−𝜏𝛼−𝑞−1+𝑀𝛾𝜁𝛼𝜛1+𝜇11−𝛿𝜂𝛼𝛼2𝜛(𝛼−𝑞−1)1𝜈1+𝜇1𝜆1𝑡𝛼−𝑞−2−𝜏𝛼−𝑞−2≤𝑀Γ(𝛼−𝑞)𝜏0(𝜏−𝑠)𝛼−𝑞−1−(𝑡−𝑠)𝛼−𝑞−1𝑑𝑠+𝜏𝑡(𝑡−𝑠)𝛼−𝑞−1+𝑀𝑑𝑠𝛾𝜁𝛼𝜆1+𝜈11−𝛿𝜂𝛼𝛼𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1𝑡𝛼−𝑞−1−𝜏𝛼−𝑞−1+𝑀𝛾𝜁𝛼𝜛1+𝜇11−𝛿𝜂𝛼𝛼2𝜛(𝛼−𝑞−1)1𝜈1+𝜇1𝜆1𝑡𝛼−𝑞−2−𝜏𝛼−𝑞−2=𝑀Γ(𝛼−𝑞+1)(𝜏𝛼−𝑞−𝑡𝛼−𝑞)+𝑀𝛾𝜁𝛼𝜆1+𝜈11−𝛿𝜂𝛼𝛼𝜛(𝛼−𝑞)1𝜈1+𝜇1𝜆1𝑡𝛼−𝑞−1−𝜏𝛼−𝑞−1+𝑀𝛾𝜁𝛼𝜛1+𝜇11−𝛿𝜂𝛼𝛼2(𝜛𝛼−𝑞−1)1𝜈1+𝜇1𝜆1𝑡𝛼−𝑞−2−𝜏𝛼−𝑞−2.(3.10)
Analogously, it can be proved that ||𝐹1𝑢(𝑡)−𝐹1||≤𝑁𝑢(𝜏)𝜛Γ(𝛽+1)2𝜈2+𝜇2𝜆2×𝜏𝛽−𝑡𝛽𝜛2𝜈2+𝜇2𝜆2+𝜏𝛽−1−𝑡𝛽−1𝛾𝜁𝛽𝜆2−𝜛2+𝜈2+𝜇21−𝛿𝜂𝛽||𝐷𝑝𝐹1𝑢(𝑡)−𝐷𝑝𝐹1||≤𝑁𝑢(𝜏)𝜏Γ(𝛽−𝑝+1)𝛽−𝑝−𝑡𝛽−𝑝+𝑁𝛾𝜁𝛽𝜆2+𝜈21−𝛿𝜂𝛽𝛽𝜛(𝛽−𝑝)2𝜈2+𝜇2𝜆2𝑡𝛽−𝑝−1−𝜏𝛽−𝑝−1+𝑁𝛾𝜁𝛽𝜛2+𝜇21−𝛿𝜂𝛽𝛽2𝜛(𝛽−𝑝−1)2𝜈2+𝜇2𝜆2𝑡𝛽−𝑝−2−𝜏𝛽−𝑝−2.(3.11)
Since the functions 𝑡𝛼,𝑡𝛼−1,𝑡𝛽,𝑡𝛽−1,𝑡𝛼−𝑞,𝑡𝛼−𝑞−1,𝑡𝛽−𝑝,𝑡𝛽−𝑝−1 are uniformly continuous on 𝐽, therefore, it follows from the above estimates that 𝐹𝑊 is an equicontinuous set. Also, it is uniformly bounded as 𝐹𝑊⊂𝑊. Thus, we conclude that 𝐹 is a completely continuous operator. Hence, by Schauder's fixed-point theorem, there exists a solution for the four-point boundary value problem (1.1).