# Existence Results for a Coupled System of Nonlinear Fractional Differential Equation with Four-Point Boundary Conditions

Authors: M. Gaber; M. G. Brikaa
Journal: ISRN Mathematical Analysis
 (2011)

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## 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.


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## Body

## 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 by(1.1)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 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 Section2, 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 orderq>0 of function f:(0,∞)→R is given by
(2.1)Iqf(t)=1⌈(q)∫0tf(s)(t-s)1-qds,
provided that the integral exists.Definition 2.2 (see [5]).
The Riemann-Liouville fractional derivative of orderq>0 of function f:(0,∞)→R is given by
(2.2)Dqf(t)=1Γ(n-q)(ddt)n∫0tf(s)(t-s)q-n+1ds,
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]).
Letn-1<α≤n, Dαu(t) exists for t∈(0,1). Then,
(2.3)IαDαu(t)=u(t)+C1tα-1+C2tα-2+⋯+Cntα-n.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. Let(2.4)μ1=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.LetC(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.
Lety∈C(J) be a given function and 1<α<2. Then, the unique solution of
(2.5)Dαu(t)=y(t),t∈(0,1),(2.6)u(0)=γu(ζ),u(1)=δu(η),
is given by
(2.7)u(t)=∫01K1(t,s)y(s)ds,
where K1(t,s) is Green's function given by
(2.8)K1(t,s)=1Γ(α)(ϖ1ν1+μ1λ1){K11(t,s),0≤t≤ζ,K12(t,s),ζ<t≤η,K13(t,s),η<t≤1,(2.9)K11(t,s)={[(t-s)α-1(ϖ1ν1+μ1λ1)+γtα-2(ϖ1-λ1t)(ζ-s)α-1+δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],0≤s≤t,[γtα-2(ϖ1-λ1t)(ζ-s)α-1+δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],t<s≤ζ,[δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1]ζ<s≤η,-tα-2(ν1t+μ1)(1-s)α-1,η<s≤1,K12(t,s)={[(t-s)α-1(ϖ1ν1+μ1λ1)+γtα-2(ϖ1-λ1t)(ζ-s)α-1+δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],0≤s≤ζ,[(t-s)α-1(ϖ1ν1+μ1λ1)+δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],ζ<s≤t,[δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],t<s≤η,-tα-2(ν1t+μ1)(1-s)α-1,η<s≤1,K13(t,s)={[(t-s)α-1(ϖ1ν1+μ1λ1)+γtα-2(ϖ1-λ1t)(ζ-s)α-1+δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],0≤s≤ζ,[(t-s)α-1(ϖ1ν1+μ1λ1)+δtα-2(ν1t+μ1)(η-s)α-1-tα-2(ν1t+μ1)(1-s)α-1],ζ<s≤η,[(t-s)α-1(ϖ1ν1+μ1λ1)-tα-2(ν1t+μ1)(1-s)α-1],η<s≤t,-tα-2(ν1t+μ1)(1-s)α-1,t<s≤1.Proof.
ForC1, C2∈R, the general solution of (2.5) can be written as
(2.10)u(t)=Iαy(t)+C1tα-1+C2tα-2=∫0t(t-s)α-1Γ(α)y(s)ds+C1tα-1+C2tα-2.
By the boundary condition,(2.11)u(0)=γu(ζ),C2=1ν1(γ∫0ζ(ζ-s)α-1Γ(α)y(s)ds+μ1C1).
By the boundary condition,(2.12)u(1)=δu(η),C1=1ϖ1(δ∫0η(η-s)α-1Γ(α)y(s)ds-∫01(1-s)α-1Γ(α)y(s)ds-λ1C2).
Substituting (2.11) into (2.12), we get
(2.13)C1=δν1(ϖ1ν1+μ1λ1)∫0η(η-s)α-1Γ(α)y(s)ds-ν1(ϖ1ν1+μ1λ1)∫01(1-s)α-1Γ(α)y(s)ds-γλ1(ϖ1ν1+μ1λ1)∫0ζ(ζ-s)α-1Γ(α)y(s)ds.
Substituting (2.13) into (2.11), we get
(2.14)C2=γϖ1(ϖ1ν1+μ1λ1)∫0ζ(ζ-s)α-1Γ(α)y(s)ds+δμ1(ϖ1ν1+μ1λ1)∫0η(η-s)α-1Γ(α)y(s)ds-μ1(ϖ1ν1+μ1λ1)∫01(1-s)α-1Γ(α)y(s)ds.
Thus, the unique solution of (2.5) and (2.6) is
(2.15)u(t)=∫0t(t-s)α-1Γ(α)y(s)ds+γtα-2(ϖ1-λ1t)(ϖ1ν1+μ1λ1)∫0ζ(ζ-s)α-1Γ(α)y(s)ds+δtα-2(ν1t+μ1)(ϖ1ν1+μ1λ1)∫0η(η-s)α-1Γ(α)y(s)ds-tα-2(ν1t+μ1)(ϖ1ν1+μ1λ1)∫01(1-s)α-1Γ(α)y(s)ds=∫01K1(t,s)y(s)ds,
where K1(t,s) is given by (2.8).Similarly, the general solution of(2.16)Dβv(t)=y(t),t∈(0,1),v(0)=γv(ζ),v(1)=δv(η),
is(2.17)v(t)=∫01K2(t,s)y(s)ds,
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:(2.18)u(t)=∫01K1(t,s)f(s,v(s),Dpv(s))ds,v(t)=∫01K2(t,s)g(s,u(s),Dqu(s))ds.


## 3. Main Results

Lemma 3.1.
Assume thatf,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 Lemma2.5, so we omit it.Let us define an operatorF:X×Y→X×Y as(3.1)F(u,v)(t)=(F1v(t),F2u(t)),
where(3.2)F1v(t)=∫01K1(t,s)f(s,v(s),Dpv(s))ds,F2u(t)=∫01K2(t,s)g(s,u(s),Dqu(s))ds.In view of the continuity ofK1,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 onf and g as(A1)
there exists a nonnegative functiona(t)∈L(0,1) such that
(3.3)|f(t,x,y)|≤a(t)+ϵ1|x|ρ1+ϵ2|y|ρ2,ϵ1,ϵ2>0,0<ρ1,ρ2<1,(A2)
there exist a nonnegative functionb(t)∈L(0,1) such that
(3.4)|g(t,x,y)|≤b(t)+δ1|x|σ1+δ2|y|σ2,δ1,δ2>0,0<σ1,σ2<1.Let us set the following notations for convenience:(3.5)A=1αΓ(α+1)⌈(α-q+1)(ϖ1ν1+μ1λ1)×[γζαΓ(α+1)(α-q)(ϖ1-λ1)+δηαΓ(α+1)(α-q)(ν1+μ1)+αΓ(α+1)(ν1(1+ϖ1)+μ1(1+λ1))+α⌈(α-q+1)((1-δηα)(ν1+μ1)-γζα(ϖ1-λ1))],B=1βΓ(β+1)⌈(β-p+1)(ϖ2ν2+μ2λ2)×[γζβΓ(β+1)(β-p)(ϖ2-λ2)+δηβΓ(β+1)(β-p)(ν2+μ2)+βΓ(β+1)(ν2(1+ϖ2)+μ2(1+λ2))+β⌈(β-p+1)((1-δηβ)(ν2+μ2)-γζβ(ϖ2-λ2))],m=max⁡t∈J∫01|a(s)k1(t,s)|ds+1⌈(α-q)(ϖ1ν1+μ1λ1)×[γ(ϖ1-λ1)∫0ζ(ζ-s)α-1a(s)ds+δ(ν1+μ1)∫0η(η-s)α-1a(s)ds+(ν1(1+ϖ1)+μ1(1+λ1))∫01(1-s)α-q-1a(s)ds],n=max⁡t∈J∫01|b(s)k1(t,s)|ds+1⌈(β-p)(ϖ2ν2+μ2λ2)×[γ(ϖ2-λ2)∫0ζ(ζ-s)β-1b(s)ds+δ(ν2+μ2)∫0η(η-s)β-1b(s)ds+(ν2(1+ϖ2)+μ2(1+λ2))∫01(1-s)β-p-1b(s)ds].Define a ballW in the Banach space X×Y as(3.6)W={(u(t),v(t))∣(u(t),v(t))∈X×Y,‖(u(t),v(t))‖X×Y≤R,t∈J},
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 thatF:W→W:
(3.7)|F1v(t)|=|∫01k1(t,s)f(s,v(s),Dpv(s))ds|≤∫01|a(s)k1(t,s)|ds+(ϵ1|R|ρ1+ϵ2|R|ρ2)∫01|k1(t,s)|ds=∫01|a(s)k1(t,s)|ds+(ϵ1|R|ρ1+ϵ2|R|ρ2)×(-∫0t(t-s)α-1Γ(α)ds-γtα-2(ϖ1-λ1t)(ϖ1ν1+μ1λ1)∫0ζ(ζ-s)α-1Γ(α)ds-δtα-2(ν1t+μ1)(ϖ1ν1+μ1λ1)∫0η(η-s)α-1Γ(α)ds+tα-2(ν1t+μ1)(ϖ1ν1+μ1λ1)∫01(1-s)α-1Γ(α)ds)=∫01|a(s)k1(t,s)|ds+(ϵ1|R|ρ1+ϵ2|R|ρ2)×(-tαΓ(α+1)+tα-2(ν1t+μ1)-γζαtα-2(ϖ1-λ1t)-δηαtα-2(ν1t+μ1)(ϖ1ν1+μ1λ1)Γ(α+1))≤∫01|a(s)k1(t,s)|ds+(ϵ1|R|ρ1+ϵ2|R|ρ2)×((1-δηα)(ν1+μ1)-γζα(ϖ1-λ1)(ϖ1ν1+μ1λ1)Γ(α+1)),|DqF1v(t)|=|DqIαf(t,v(t),Dpv(t))+1(ϖ1ν1+μ1λ1)×[γ(ϖ1Dqtα-2-λ1Dqtα-1)Iαf(ζ,v(ζ),Dpv(ζ))+δ(ν1Dqtα-1+μ1Dqtα-2)Iαf(η,v(η),Dpv(η))-(ν1Dqtα-1+μ1Dqtα-2)Iαf(1,v(1),Dpv(1))]1(ϖ1ν1+μ1λ1)|≤∫0t(t-s)α-q-1Γ(α-q)f(s,v(s),Dpv(s))ds+1⌈(α-q)(ϖ1ν1+μ1λ1)×[γ(ϖ1-λ1)∫0ζ(ζ-s)α-1f(s,v(s),Dpv(s))ds+δ(ν1+μ1)∫0η(η-s)α-1f(s,v(s),Dpv(s))ds+(ν1+μ1)∫01(1-s)α-1f(s,v(s),Dpv(s))ds]≤1Γ(α-q)[∫0t(t-s)α-q-1a(s)ds+(ϵ1|R|ρ1+ϵ2|R|ρ2)∫0t(t-s)α-q-1ds]+1⌈(α-q)(ϖ1ν1+μ1λ1)×[γ(ϖ1-λ1)∫0ζ(ζ-s)α-1a(s)ds+γ(ϖ1-λ1)(ϵ1|R|ρ1+ϵ2|R|ρ2)×∫0ζ(ζ-s)α-1ds+δ(ν1+μ1)∫0η(η-s)α-1a(s)ds+δ(ν1+μ1)(ϵ1|R|ρ1+ϵ2|R|ρ2)×∫0η(η-s)α-1ds+(ν1+μ1)∫01(1-s)α-1a(s)ds+(ν1+μ1)(ϵ1|R|ρ1+ϵ2|R|ρ2)×∫01(1-s)α-1ds]≤1⌈(α-q)(ϖ1ν1+μ1λ1)×[γ(ϖ1-λ1)∫0ζ(ζ-s)α-1a(s)ds+γ(ϖ1-λ1)(ϵ1|R|ρ1+ϵ2|R|ρ2)×∫0ζ(ζ-s)α-1ds+δ(ν1+μ1)∫0η(η-s)α-1a(s)ds+δ(ν1+μ1)(ϵ1|R|ρ1+ϵ2|R|ρ2)×∫0η(η-s)α-1ds+(ν1(1+ϖ1)+μ1(1+λ1))(∫01(1-s)α-1a(s)ds+(ϵ1|R|ρ1+ϵ2|R|ρ2)×∫01(1-s)α-1ds)]=1⌈(α-q)(ϖ1ν1+μ1λ1)×[γ(ϖ1-λ1)∫0ζ(ζ-s)α-1a(s)ds+δ(ν1+μ1)∫0η(η-s)α-1a(s)ds+(ν1(1+ϖ1)+μ1(1+λ1))∫01(1-s)α-q-1a(s)ds]+(ϵ1|R|ρ1+ϵ2|R|ρ2)×[γζα(α-q)(ϖ1-λ1)+δηα(α-q)(ν1+μ1)+α(ν1(1+ϖ1)+μ1(1+λ1))α⌈(α-q+1)(ϖ1ν1+μ1λ1)].
Thus,(3.8)‖F1v(t)‖x=max⁡t∈J|F1v(t)|+max⁡t∈J|DqF1v(t)|≤m+(ϵ1|R|ρ1+ϵ2|R|ρ2)A≤R3+R3+R3=R.
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.
SinceF1v(t),F2v(t),DqF1v(t),DqF2v(t) are continuous on J, therefore, F:W→W.
Now, we show thatF is a completely continuous operator. For that we fix
(3.9)M=max⁡t∈J|f(t,v(t),Dpv(t))|,N=max⁡t∈J|g(t,u(t),Dpv(t))|.
For(u,v)∈W,t,τ∈J(t<τ), we have
(3.10)|F1v(t)-F1v(τ)|=|∫01(k1(t,s)-k1(τ,s))f(s,v(s),Dpv(s))ds|≤M[∫0t|(k1(t,s)-k1(τ,s))|ds+∫tτ|(k1(t,s)-k1(τ,s))|ds+∫τζ|(k1(t,s)-k1(τ,s))|ds+∫ζη|(k1(t,s)-k1(τ,s))|ds+∫η1|(k1(t,s)-k1(τ,s))|ds]=MΓ(α)(ϖ1ν1+μ1λ1)×[∫0t{((τ-s)α-1-(t-s)α-1)[ϖ1ν1+μ1λ1]+(τα-1-tα-1)[ν1(1-s)α-1+γλ1(ζ-s)α-1-δν1(η-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1-γϖ1(ζ-s)α-1-δμ1(η-s)α-1]}ds+∫tτ{(τ-s)α-1[ϖ1ν1+μ1λ1]+(τα-1-tα-1)[ν1(1-s)α-1+γλ1(ζ-s)α-1-δν1(η-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1-γϖ1(ζ-s)α-1-δμ1(η-s)α-1]}ds+∫τζ{(τα-1-tα-1)[ν1(1-s)α-1+γλ1(ζ-s)α-1-δν1(η-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1-γϖ1(ζ-s)α-1-δμ1(η-s)α-1]}ds+∫ζη{(τα-1-tα-1)[ν1(1-s)α-1-δν1(η-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1-δμ1(η-s)α-1]}ds+∫η1{(τα-1-tα-1)[ν1(1-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1]}ds]=MΓ(α)(ϖ1ν1+μ1λ1)×[∫0τ(τ-s)α-1[ϖ1ν1+μ1λ1]ds-∫0t(t-s)α-1[ϖ1ν1+μ1λ1]ds+∫0ζ[(τα-1-tα-1)[γλ1(ζ-s)α-1]-(τα-2-tα-2)[γϖ1(ζ-s)α-1]]ds+∫0η[(τα-1-tα-1)[ν1(1-s)α-1-δν1(η-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1-δμ1(η-s)α-1]]ds+∫η1[(τα-1-tα-1)[ν1(1-s)α-1]+(τα-2-tα-2)[μ1(1-s)α-1]]ds]=MΓ(α+1)(ϖ1ν1+μ1λ1)×[(τα-tα)[ϖ1ν1+μ1λ1]+(τα-1-tα-1)[γζα(λ1-ϖ1)+(ν1+μ1)(1-δηα)]],|DqF1v(t)-DqF1v(τ)|=|DqIαf(t,v(t),Dpv(t))+1(ϖ1ν1+μ1λ1)×[γ(ϖ1Dqtα-2-λ1Dqtα-1)Iαf(ζ,v(ζ),Dpv(ζ))+δ(ν1Dqtα-1+μ1Dqtα-2)Iαf(η,v(η),Dpv(η))-(ν1Dqtα-1+μ1Dqtα-2)Iαf(1,v(1),Dpv(1))]-DqIαf(τ,v(τ),Dpv(τ))-1(ϖ1ν1+μ1λ1)×[γ(ϖ1Dqτα-2-λ1Dqτα-1)Iαf(ζ,v(ζ),Dpv(ζ))+δ(ν1Dqτα-1+μ1Dqτα-2)Iαf(η,v(η),Dpv(η))-(ν1Dqτα-1+μ1Dqτα-2)Iαf(1,v(1),Dpv(1))]1(ϖ1ν1+μ1λ1)|≤1Γ(α-q)[|∫0t(t-s)α-q-1f(s,v(s),Dpv(s))ds-∫0τ(τ-s)α-q-1f(s,v(s),Dpv(s))ds|]+M⌈(α-q)(ϖ1ν1+μ1λ1)×|(tα-q-1-τα-q-1){-γλ1∫0ζ(ζ-s)α-1ds+δν1∫0η(η-s)α-1ds-ν1∫01(1-s)α-1ds}|+Mα⌈(α-q-1)(ϖ1ν1+μ1λ1)×|(tα-q-2-τα-q-2){γϖ1∫0ζ(ζ-s)α-1ds+δμ1∫0η(η-s)α-1ds-μ1∫01(1-s)α-1ds}|≤1Γ(α-q)[|∫0t(t-s)α-q-1f(s,v(s),Dpv(s))ds-∫0τ(t-s)α-q-1f(s,v(s),Dpv(s))ds|+|∫0τ(t-s)α-q-1f(s,v(s),Dpv(s))ds-∫0τ(τ-s)α-q-1f(s,v(s),Dpv(s))ds|1(ϖ1ν1+μ1λ1)]+M(γζαλ1+ν1(1-δηα))α⌈(α-q)(ϖ1ν1+μ1λ1)(tα-q-1-τα-q-1)+M(γζαϖ1+μ1(1-δηα))α2⌈(α-q-1)(ϖ1ν1+μ1λ1)(tα-q-2-τα-q-2)≤MΓ(α-q)[∫0τ((τ-s)α-q-1-(t-s)α-q-1)ds+∫tτ(t-s)α-q-1ds]+M(γζαλ1+ν1(1-δηα))α⌈(α-q)(ϖ1ν1+μ1λ1)(tα-q-1-τα-q-1)+M(γζαϖ1+μ1(1-δηα))α2⌈(α-q-1)(ϖ1ν1+μ1λ1)(tα-q-2-τα-q-2)=MΓ(α-q+1)(τα-q-tα-q)+M(γζαλ1+ν1(1-δηα))α⌈(α-q)(ϖ1ν1+μ1λ1)(tα-q-1-τα-q-1)+M(γζαϖ1+μ1(1-δηα))α2⌈(α-q-1)(ϖ1ν1+μ1λ1)(tα-q-2-τα-q-2).
Analogously, it can be proved that(3.11)|F1u(t)-F1u(τ)|≤NΓ(β+1)(ϖ2ν2+μ2λ2)×[(τβ-tβ)[ϖ2ν2+μ2λ2]+(τβ-1-tβ-1)[γζβ(λ2-ϖ2)+(ν2+μ2)(1-δηβ)]]|DpF1u(t)-DpF1u(τ)|≤NΓ(β-p+1)(τβ-p-tβ-p)+N(γζβλ2+ν2(1-δηβ))β⌈(β-p)(ϖ2ν2+μ2λ2)(tβ-p-1-τβ-p-1)+N(γζβϖ2+μ2(1-δηβ))β2⌈(β-p-1)(ϖ2ν2+μ2λ2)(tβ-p-2-τβ-p-2).
Since the functionstα,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).



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*Source: 468346-2011-12-15.xml*