Table of Contents
ISRN Mathematical Analysis
VolumeΒ 2011, Article IDΒ 468346, 14 pages
http://dx.doi.org/10.5402/2011/468346
Research Article

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

1Faculty of Education Al-Arish, Suez Canal University, Ismailia 41522, Egypt
2Faculty of Computers and Informatics, Suez Canal University, Ismailia 41522, Egypt

Received 30 August 2011; Accepted 20 October 2011

Academic Editor: J.-L.Β Wu

Copyright Β© 2011 M. Gaber and M. G. Brikaa. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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: 𝐷𝛼𝑒(𝑑)=𝑓(𝑑,𝑣(𝑑),π·π‘šπ‘£(𝑑)), π‘‘βˆˆ(0,1), 𝐷𝛽𝑣(𝑑)=𝑓(𝑑,𝑒(𝑑),𝐷𝑛𝑒(𝑑)), π‘‘βˆˆ(0,1), 𝑒(0)=𝛾𝑒(𝜁), 𝑒(1)=𝛿𝑒(πœ‚), 𝑣(0)=𝛾𝑣(𝜁), 𝑣(1)=𝛿𝑣(πœ‚), where 𝛼,𝛽,π‘š,𝑛,𝜁,πœ‚,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 by𝐷𝛼𝑒(𝑑)=𝑓(𝑑,𝑣(𝑑),π·π‘šπ·π‘£(𝑑)),π‘‘βˆˆ(0,1),𝛽𝑣(𝑑)=𝑓(𝑑,𝑒(𝑑),𝐷𝑛𝑒(𝑑)),π‘‘βˆˆ(0,1),𝑒(0)=𝛾𝑒(𝜁),𝑒(1)=𝛿𝑒(πœ‚),𝑣(0)=𝛾𝑣(𝜁),𝑣(1)=𝛿𝑣(πœ‚),(1.1) where 1<𝛼,𝛽<2, π‘š,𝑛,𝛾,𝛿>0, 0<𝜁<πœ‚<1, π›Όβˆ’π‘›β‰₯1, π›½βˆ’π‘šβ‰₯1, π›Ώπœ‚π›Όβˆ’1<1, π›Ώπœ‚π›Όβˆ’2<1, π›Ύπœπ›Όβˆ’1<1, π›Ύπœπ›Όβˆ’2<1, π›Ώπœ‚π›½βˆ’1<1, π›Ώπœ‚π›½βˆ’2<1, π›Ύπœπ›½βˆ’1<1, π›Ύπœπ›½βˆ’2<1, 𝐷 is the standard Riemann-Liouville fractional derivative, and 𝑓,π‘”βˆΆ[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 π‘ž>0 of function π‘“βˆΆ(0,∞)→𝑅 is given by πΌπ‘ž1𝑓(𝑑)=ξ€œβŒˆ(π‘ž)𝑑0𝑓(𝑠)(π‘‘βˆ’π‘ )1βˆ’π‘žπ‘‘π‘ ,(2.1) provided that the integral exists.

Definition 2.2 (see [5]). The Riemann-Liouville fractional derivative of order π‘ž>0 of function π‘“βˆΆ(0,∞)→𝑅 is given by π·π‘ž1𝑓(𝑑)=𝑑Γ(π‘›βˆ’π‘ž)ξ‚π‘‘π‘‘π‘›ξ€œπ‘‘0𝑓(𝑠)(π‘‘βˆ’π‘ )π‘žβˆ’π‘›+1𝑑𝑠,(2.2) where 𝑛=[π‘ž]+1 and [π‘ž] denotes the integral part of number π‘ž, provided that the right side is pointwise defined on (0,∞).

Lemma 2.3 ([5]). Let π‘›βˆ’1<𝛼≀𝑛, 𝐷𝛼𝑒(𝑑) exists for π‘‘βˆˆ(0,1). Then, 𝐼𝛼𝐷𝛼𝑒(𝑑)=𝑒(𝑑)+𝐢1π‘‘π›Όβˆ’1+𝐢2π‘‘π›Όβˆ’2+β‹―+πΆπ‘›π‘‘π›Όβˆ’π‘›.(2.3)

Remark 2.4. The following properties are useful for our discussion: 𝐼𝛼𝐼𝛽𝑓(𝑑)=𝐼𝛼+𝛽𝑓(𝑑), 𝐷𝛼𝐼𝛼𝑓(𝑑)=𝑓(𝑑), 𝛼>0, 𝛽>0, π‘“βˆˆπΏ(0,1); 𝐼𝛼𝐷𝛼𝑓(𝑑)=𝑓(𝑑), 0<𝛼<1, 𝑓(𝑑)∈𝐢[0,1] and 𝐷𝛼𝑓(𝑑)∈𝐢(0,1)∩𝐿(0,1);πΌπ›ΌβˆΆπΆ[0,1]→𝐢[0,1],𝛼>0.

For convenience, we introduce the following notation. Letπœ‡1=π›Ύπœπ›Όβˆ’1,𝜈1=1βˆ’π›Ύπœπ›Όβˆ’2,πœ›1=1βˆ’π›Ώπœ‚π›Όβˆ’1,πœ†1=1βˆ’π›Ώπœ‚π›Όβˆ’2,πœ‡2=π›Ύπœπ›½βˆ’1,𝜈2=1βˆ’π›Ύπœπ›½βˆ’2,πœ›2=1βˆ’π›Ώπœ‚π›½βˆ’1,πœ†2=1βˆ’π›Ώπœ‚π›½βˆ’2.(2.4)

Let 𝐢(𝐽) Denote the space of all continuous functions defined on 𝐽=[0,1]. Let 𝑋={𝑒(𝑑)βˆΆπ‘’βˆˆπΆ(𝐽)andπ·π‘›π‘’βˆˆπΆ(𝐽)} be a Banach space endowed with the norm ‖𝑒‖π‘₯=maxπ‘‘βˆˆπ½|𝑒(𝑑)|+maxπ‘‘βˆˆπ½|𝐷𝑛𝑒(𝑑)|, where 1<𝛼<2, 0<π‘›β‰€π›Όβˆ’1, see [11] Lemma 3.2, and let π‘Œ={𝑣(𝑑)βˆΆπ‘£βˆˆπΆ(𝐽)andπ·π‘šπ‘£βˆˆπΆ(𝐽)} be a Banach space equipped with the norm β€–π‘£β€–π‘Œ=maxπ‘‘βˆˆπ½|𝑣(𝑑)|+maxπ‘‘βˆˆπ½|π·π‘šπ‘£(𝑑)|, where 1<𝛽<2, 0<π‘šβ‰€π›½βˆ’1. Thus, (π‘‹Γ—π‘Œ,β€–β‹…β€–π‘‹Γ—π‘Œ) is a Banach with the norm defined by β€–(𝑒,𝑣)β€–π‘‹Γ—π‘Œ=max{‖𝑒‖π‘₯,β€–π‘£β€–π‘Œ} for (𝑒,𝑣)βˆˆπ‘‹Γ—π‘Œ.

Lemma 2.5. Let π‘¦βˆˆπΆ(𝐽) be a given function and 1<𝛼<2. Then, the unique solution of 𝐷𝛼𝑒(𝑑)=𝑦(𝑑),π‘‘βˆˆ(0,1),(2.5)𝑒(0)=𝛾𝑒(𝜁),𝑒(1)=𝛿𝑒(πœ‚),(2.6) is given by ξ€œπ‘’(𝑑)=10𝐾1(𝑑,𝑠)𝑦(𝑠)𝑑𝑠,(2.7) where 𝐾1(𝑑,𝑠) is Green's function given by 𝐾1(1𝑑,𝑠)=ξ€·πœ›Ξ“(𝛼)1𝜈1+πœ‡1πœ†1ξ€ΈβŽ§βŽͺ⎨βŽͺ⎩𝐾11𝐾(𝑑,𝑠),0β‰€π‘‘β‰€πœ,12𝐾(𝑑,𝑠),𝜁<π‘‘β‰€πœ‚,13(𝑑,𝑠),πœ‚<𝑑≀1,(2.8)𝐾11⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©ξ‚ƒ(𝑑,𝑠)=(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έ+π›Ύπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1𝑑(πœβˆ’π‘ )π›Όβˆ’1+π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,0≀𝑠≀𝑑,π›Ύπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1𝑑(πœβˆ’π‘ )π›Όβˆ’1+π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,𝑑<π‘ β‰€πœ,π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1ξ‚„πœ<π‘ β‰€πœ‚,βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1𝐾,πœ‚<𝑠≀1,12⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©ξ‚ƒ(𝑑,𝑠)=(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έ+π›Ύπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1𝑑(πœβˆ’π‘ )π›Όβˆ’1+π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,0β‰€π‘ β‰€πœ,(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έ+π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,𝜁<𝑠≀𝑑,π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1ξ‚„,𝑑<π‘ β‰€πœ‚,βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1𝐾,πœ‚<𝑠≀1,13(⎧βŽͺβŽͺβŽͺβŽͺ⎨βŽͺβŽͺβŽͺβŽͺβŽ©ξ‚ƒπ‘‘,𝑠)=(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έ+π›Ύπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1𝑑(πœβˆ’π‘ )π›Όβˆ’1+π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,0β‰€π‘ β‰€πœ,(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έ+π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,𝜁<π‘ β‰€πœ‚,(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έβˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1ξ‚„,πœ‚<𝑠≀𝑑,βˆ’π‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έ(1βˆ’π‘ )π›Όβˆ’1,𝑑<𝑠≀1.(2.9)

Proof. For 𝐢1, 𝐢2βˆˆπ‘…, the general solution of (2.5) can be written as 𝑒(𝑑)=𝐼𝛼𝑦(𝑑)+𝐢1π‘‘π›Όβˆ’1+𝐢2π‘‘π›Όβˆ’2=ξ€œπ‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)𝑦(𝑠)𝑑𝑠+𝐢1π‘‘π›Όβˆ’1+𝐢2π‘‘π›Όβˆ’2.(2.10)
By the boundary condition, 𝐢𝑒(0)=𝛾𝑒(𝜁),2=1𝜈1ξ‚΅π›Ύξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)𝑦(𝑠)𝑑𝑠+πœ‡1𝐢1ξ‚Ά.(2.11)
By the boundary condition, 𝐢𝑒(1)=𝛿𝑒(πœ‚),1=1πœ›1ξ‚΅π›Ώξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€œΞ“(𝛼)𝑦(𝑠)π‘‘π‘ βˆ’10(1βˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)𝑦(𝑠)π‘‘π‘ βˆ’πœ†1𝐢2ξ‚Ά.(2.12)
Substituting (2.11) into (2.12), we get 𝐢1=π›Ώπœˆ1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’πœˆΞ“(𝛼)𝑦(𝑠)𝑑𝑠1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œ10(1βˆ’π‘ )π›Όβˆ’1βˆ’Ξ“(𝛼)𝑦(𝑠)π‘‘π‘ π›Ύπœ†1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)𝑦(𝑠)𝑑𝑠.(2.13)
Substituting (2.13) into (2.11), we get 𝐢2=π›Ύπœ›1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1+Ξ“(𝛼)𝑦(𝑠)π‘‘π‘ π›Ώπœ‡1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’πœ‡Ξ“(𝛼)𝑦(𝑠)𝑑𝑠1ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œ10(1βˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)𝑦(𝑠)𝑑𝑠.(2.14)
Thus, the unique solution of (2.5) and (2.6) is ξ€œπ‘’(𝑑)=𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’1Ξ“+(𝛼)𝑦(𝑠)π‘‘π‘ π›Ύπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1π‘‘ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1+Ξ“(𝛼)𝑦(𝑠)π‘‘π‘ π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1βˆ’π‘‘Ξ“(𝛼)𝑦(𝑠)π‘‘π‘ π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œ10(1βˆ’π‘ )π›Όβˆ’1=ξ€œΞ“(𝛼)𝑦(𝑠)𝑑𝑠10𝐾1(𝑑,𝑠)𝑦(𝑠)𝑑𝑠,(2.15) where 𝐾1(𝑑,𝑠) is given by (2.8).

Similarly, the general solution of 𝐷𝛽𝑣(𝑑)=𝑦(𝑑),π‘‘βˆˆ(0,1),𝑣(0)=𝛾𝑣(𝜁),𝑣(1)=𝛿𝑣(πœ‚),(2.16) isξ€œπ‘£(𝑑)=10𝐾2(𝑑,𝑠)𝑦(𝑠)𝑑𝑠,(2.17) where 𝐾2(𝑑,𝑠), can be obtained from 𝐾1(𝑑,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:ξ€œπ‘’(𝑑)=10𝐾1(𝑑,𝑠)𝑓(𝑠,𝑣(𝑠),π·π‘π‘£ξ€œπ‘£(𝑠))𝑑𝑠,(𝑑)=10𝐾2(𝑑,𝑠)𝑔(𝑠,𝑒(𝑠),π·π‘žπ‘’(𝑠))𝑑𝑠.(2.18)

3. Main Results

Lemma 3.1. Assume that 𝑓,π‘”βˆΆπ½Γ—π‘…Γ—π‘…β†’π‘… are continuous functions. Then, (𝑒,𝑣)βˆˆπ‘‹Γ—π‘Œ is a solution of (1.1) if and only if (𝑒,𝑣)βˆˆπ‘‹Γ—π‘Œ is a solution of (2.18).

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

Let us define an operator πΉβˆΆπ‘‹Γ—π‘Œβ†’π‘‹Γ—π‘Œ as𝐹𝐹(𝑒,𝑣)(𝑑)=1𝑣(𝑑),𝐹2𝑒(𝑑),(3.1) where𝐹1ξ€œπ‘£(𝑑)=10𝐾1(𝑑,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝𝑣(𝑠))𝑑𝑠,𝐹2ξ€œπ‘’(𝑑)=10𝐾2(𝑑,𝑠)𝑔(𝑠,𝑒(𝑠),π·π‘žπ‘’(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 π‘Ž(𝑑)∈𝐿(0,1) such that ||||𝑓(𝑑,π‘₯,𝑦)β‰€π‘Ž(𝑑)+πœ–1|π‘₯|𝜌1+πœ–2||𝑦||𝜌2,πœ–1,πœ–2>0,0<𝜌1,𝜌2<1,(3.3)(A2)there exist a nonnegative function 𝑏(𝑑)∈𝐿(0,1) such that ||||𝑔(𝑑,π‘₯,𝑦)≀𝑏(𝑑)+𝛿1|π‘₯|𝜎1+𝛿2||𝑦||𝜎2,𝛿1,𝛿2>0,0<𝜎1,𝜎2<1.(3.4)

Let us set the following notations for convenience∢1𝐴=ξ›ξ€·πœ›π›ΌΞ“(𝛼+1)(π›Όβˆ’π‘ž+1)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ€Ίπ›Ύπœπ›Όξ€·πœ›Ξ“(𝛼+1)(π›Όβˆ’π‘ž)1βˆ’πœ†1ξ€Έ+π›Ώπœ‚π›Όξ€·πœˆΞ“(𝛼+1)(π›Όβˆ’π‘ž)1+πœ‡1ξ€Έξ€·πœˆ+𝛼Γ(𝛼+1)1ξ€·1+πœ›1ξ€Έ+πœ‡1ξ€·1+πœ†1ξ€Έξ€Έ+π›ΌβŒˆ(π›Όβˆ’π‘ž+1)ξ€·ξ€·1βˆ’π›Ώπœ‚π›Όπœˆξ€Έξ€·1+πœ‡1ξ€Έβˆ’π›Ύπœπ›Όξ€·πœ›1βˆ’πœ†1,1𝐡=ξ›ξ€·πœ›π›½Ξ“(𝛽+1)(π›½βˆ’π‘+1)2𝜈2+πœ‡2πœ†2ξ€ΈΓ—ξ€Ίπ›Ύπœπ›½ξ€·πœ›Ξ“(𝛽+1)(π›½βˆ’π‘)2βˆ’πœ†2ξ€Έ+π›Ώπœ‚π›½ξ€·πœˆΞ“(𝛽+1)(π›½βˆ’π‘)2+πœ‡2ξ€Έξ€·πœˆ+𝛽Γ(𝛽+1)2ξ€·1+πœ›2ξ€Έ+πœ‡2ξ€·1+πœ†2ξ‚€ξ€·ξ€Έξ€Έ+π›½βŒˆ(π›½βˆ’π‘+1)1βˆ’π›Ώπœ‚π›½πœˆξ€Έξ€·2+πœ‡2ξ€Έβˆ’π›Ύπœπ›½ξ€·πœ›2βˆ’πœ†2ξ€Έ,ξ‚ξ‚„π‘š=maxπ‘‘βˆˆπ½ξ€œ10||π‘Ž(𝑠)π‘˜1||1(𝑑,𝑠)𝑑𝑠+ξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έπ›Ύξ€·πœ›1βˆ’πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1ξ€·πœˆπ‘Ž(𝑠)𝑑𝑠+𝛿1+πœ‡1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€·πœˆπ‘Ž(𝑠)𝑑𝑠1ξ€·1+πœ›1ξ€Έ+πœ‡1ξ€·1+πœ†1ξ€œξ€Έξ€Έ10(1βˆ’π‘ )π›Όβˆ’π‘žβˆ’1ξ‚Ή,π‘Ž(𝑠)𝑑𝑠𝑛=maxπ‘‘βˆˆπ½ξ€œ10||𝑏(𝑠)π‘˜1||1(𝑑,𝑠)𝑑𝑠+ξ›ξ€·πœ›(π›½βˆ’π‘)2𝜈2+πœ‡2πœ†2ξ€ΈΓ—ξ‚Έπ›Ύξ€·πœ›2βˆ’πœ†2ξ€Έξ€œπœ0(πœβˆ’π‘ )π›½βˆ’1ξ€·πœˆπ‘(𝑠)𝑑𝑠+𝛿2+πœ‡2ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›½βˆ’1+ξ€·πœˆπ‘(𝑠)𝑑𝑠2ξ€·1+πœ›2ξ€Έ+πœ‡2ξ€·1+πœ†2ξ€œξ€Έξ€Έ10(1βˆ’π‘ )π›½βˆ’π‘βˆ’1ξ‚Ή.𝑏(𝑠)𝑑𝑠(3.5)

Define a ball π‘Š in the Banach space π‘‹Γ—π‘Œ asξ€½β€–β€–π‘Š=(𝑒(𝑑),𝑣(𝑑))∣(𝑒(𝑑),𝑣(𝑑))βˆˆπ‘‹Γ—π‘Œ,(𝑒(𝑑),𝑣(𝑑))π‘‹Γ—π‘Œξ€Ύβ‰€π‘…,π‘‘βˆˆπ½,(3.6) where 𝑅β‰₯max{(3π΄πœ–1)1/(1βˆ’πœŒ1),(3π΄πœ–2)1/(1βˆ’πœŒ2),(3𝐡𝛿1)1/(1βˆ’πœŽ1),(3𝐡𝛿2)1/(1βˆ’πœŽ2),3π‘š,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 πΉβˆΆπ‘Šβ†’π‘Š: ||𝐹1||=||||ξ€œπ‘£(𝑑)10π‘˜1(𝑑,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||β‰€ξ€œπ‘£(𝑠))𝑑𝑠10||π‘Ž(𝑠)π‘˜1||ξ€·πœ–(𝑑,𝑠)𝑑𝑠+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€Έξ€œ10||π‘˜1||=ξ€œ(𝑑,𝑠)𝑑𝑠10||π‘Ž(𝑠)π‘˜1(||ξ€·πœ–π‘‘,𝑠)𝑑𝑠+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξƒ©βˆ’ξ€œπ‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’1Ξ“(𝛼)π‘‘π‘ βˆ’π›Ύπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1π‘‘ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1βˆ’Ξ“(𝛼)π‘‘π‘ π›Ώπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1𝑑Γ(𝛼)𝑑𝑠+π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξ€œ10(1βˆ’π‘ )π›Όβˆ’1ξƒͺ=ξ€œΞ“(𝛼)𝑑𝑠10||π‘Ž(𝑠)π‘˜1(||ξ€·πœ–π‘‘,𝑠)𝑑𝑠+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξƒ©βˆ’π‘‘π›Ό+𝑑Γ(𝛼+1)π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έβˆ’π›Ύπœπ›Όπ‘‘π›Όβˆ’2ξ€·πœ›1βˆ’πœ†1π‘‘ξ€Έβˆ’π›Ώπœ‚π›Όπ‘‘π›Όβˆ’2ξ€·πœˆ1𝑑+πœ‡1ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξƒͺβ‰€ξ€œΞ“(𝛼+1)10||π‘Ž(𝑠)π‘˜1(||ξ€·πœ–π‘‘,𝑠)𝑑𝑠+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2×1βˆ’π›Ώπœ‚π›Όπœˆξ€Έξ€·1+πœ‡1ξ€Έβˆ’π›Ύπœπ›Όξ€·πœ›1βˆ’πœ†1ξ€Έξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€Έξƒͺ,||𝐷Γ(𝛼+1)π‘žπΉ1𝑣||=||||𝐷(𝑑)π‘žπΌπ›Όπ‘“(𝑑,𝑣(𝑑),𝐷𝑝𝑣1(𝑑))+ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚ƒπ›Ύξ‚€πœ›1π·π‘žπ‘‘π›Όβˆ’2βˆ’πœ†1π·π‘žπ‘‘π›Όβˆ’1𝐼𝛼𝑓(𝜁,𝑣(𝜁),π·π‘ξ‚€πœˆπ‘£(𝜁))+𝛿1π·π‘žπ‘‘π›Όβˆ’1+πœ‡1π·π‘žπ‘‘π›Όβˆ’2𝐼𝛼𝑓(πœ‚,𝑣(πœ‚),π·π‘βˆ’ξ‚€πœˆπ‘£(πœ‚))1π·π‘žπ‘‘π›Όβˆ’1+πœ‡1π·π‘žπ‘‘π›Όβˆ’2𝐼𝛼𝑓(1,𝑣(1),𝐷𝑝||||β‰€ξ€œπ‘£(1))𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1Ξ“(π›Όβˆ’π‘ž)𝑓(𝑠,𝑣(𝑠),𝐷𝑝1𝑣(𝑠))𝑑𝑠+ξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έπ›Ύξ€·πœ›1βˆ’πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1𝑓(𝑠,𝑣(𝑠),π·π‘ξ€·πœˆπ‘£(𝑠))𝑑𝑠+𝛿1+πœ‡1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1𝑓(𝑠,𝑣(𝑠),𝐷𝑝+ξ€·πœˆπ‘£(𝑠))𝑑𝑠1+πœ‡1ξ€Έξ€œ10(1βˆ’π‘ )π›Όβˆ’1𝑓(𝑠,𝑣(𝑠),𝐷𝑝≀1𝑣(𝑠))π‘‘π‘ ξ‚Έξ€œΞ“(π›Όβˆ’π‘ž)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1ξ€·πœ–π‘Ž(𝑠)𝑑𝑠+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€Έξ€œπ‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1ξ‚Ή+1𝑑𝑠(ξ€·πœ›π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έπ›Ύξ€·πœ›1βˆ’πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1π‘Žξ€·πœ›(𝑠)𝑑𝑠+𝛾1βˆ’πœ†1πœ–ξ€Έξ€·1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1ξ€·πœˆπ‘‘π‘ +𝛿1+πœ‡1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€·πœˆπ‘Ž(𝑠)𝑑𝑠+𝛿1+πœ‡1πœ–ξ€Έξ€·1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€·πœˆπ‘‘π‘ 1+πœ‡1ξ€Έξ€œ10(1βˆ’π‘ )π›Όβˆ’1ξ€·πœˆπ‘Ž(𝑠)𝑑𝑠+1+πœ‡1πœ–ξ€Έξ€·1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξ€œ10(1βˆ’π‘ )π›Όβˆ’1≀1π‘‘π‘ ξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έπ›Ύξ€·πœ›1βˆ’πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1ξ€·πœ›π‘Ž(𝑠)𝑑𝑠+𝛾1βˆ’πœ†1πœ–ξ€Έξ€·1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1ξ€·πœˆπ‘‘π‘ +𝛿1+πœ‡1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€·πœˆπ‘Ž(𝑠)𝑑𝑠+𝛿1+πœ‡1πœ–ξ€Έξ€·1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€·πœˆπ‘‘π‘ 1ξ€·1+πœ›1ξ€Έ+πœ‡1ξ€·1+πœ†1ξ‚΅ξ€œξ€Έξ€Έ10(1βˆ’π‘ )π›Όβˆ’1π‘Žξ€·πœ–(𝑠)𝑑𝑠+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξ€œ10(1βˆ’π‘ )π›Όβˆ’1=1𝑑𝑠(ξ€·πœ›π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έπ›Ύξ€·πœ›1βˆ’πœ†1ξ€Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1π‘Žξ€·πœˆ(𝑠)𝑑𝑠+𝛿1+πœ‡1ξ€Έξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1π‘Ž+ξ€·πœˆ(𝑠)𝑑𝑠1ξ€·1+πœ›1ξ€Έ+πœ‡1ξ€·1+πœ†1ξ€œξ€Έξ€Έ10(1βˆ’π‘ )π›Όβˆ’π‘žβˆ’1ξ‚Ή+ξ€·πœ–π‘Ž(𝑠)𝑑𝑠1||𝑅||𝜌1+πœ–2||𝑅||𝜌2ξ€ΈΓ—ξƒ¬π›Ύπœπ›Όξ€·πœ›(π›Όβˆ’π‘ž)1βˆ’πœ†1ξ€Έ+π›Ώπœ‚π›Όξ€·πœˆ(π›Όβˆ’π‘ž)1+πœ‡1ξ€Έξ€·πœˆ+𝛼1ξ€·1+πœ›1ξ€Έ+πœ‡1ξ€·1+πœ†1ξ€Έξ€Έξ€·πœ›π›ΌβŒˆ(π›Όβˆ’π‘ž+1)1𝜈1+πœ‡1πœ†1ξ€Έξƒ­.(3.7)
Thus, ‖‖𝐹1𝑣‖‖(𝑑)π‘₯=maxπ‘‘βˆˆπ½||𝐹1𝑣||(𝑑)+maxπ‘‘βˆˆπ½||π·π‘žπΉ1𝑣||ξ€·πœ–(𝑑)β‰€π‘š+1||𝑅||𝜌1+πœ–2||𝑅||𝜌2𝐴≀𝑅3+𝑅3+𝑅3=𝑅.(3.8)
Similarly, it can be shown that ‖𝐹2𝑣(𝑑)‖𝑦≀𝑛+(πœ–1|𝑅|𝜌1+πœ–2|𝑅|𝜌2)𝐡≀𝑅. Hence, we conclude that ‖𝐹1𝑣(𝑑)β€–π‘₯×𝑦≀𝑅.
Since 𝐹1𝑣(𝑑),𝐹2𝑣(𝑑),π·π‘žπΉ1𝑣(𝑑),π·π‘žπΉ2𝑣(𝑑) are continuous on 𝐽, therefore, πΉβˆΆπ‘Šβ†’π‘Š.
Now, we show that 𝐹 is a completely continuous operator. For that we fix 𝑀=maxπ‘‘βˆˆπ½||𝑓(𝑑,𝑣(𝑑),𝐷𝑝||𝑣(𝑑)),𝑁=maxπ‘‘βˆˆπ½||𝑔(𝑑,𝑒(𝑑),𝐷𝑝||𝑣(𝑑)).(3.9)
For (𝑒,𝑣)βˆˆπ‘Š,𝑑,𝜏∈𝐽(𝑑<𝜏), we have ||𝐹1𝑣(𝑑)βˆ’πΉ1||=||||ξ€œπ‘£(𝜏)10ξ€·π‘˜1(𝑑,𝑠)βˆ’π‘˜1ξ€Έ(𝜏,𝑠)𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||ξ‚Έξ€œπ‘£(𝑠))𝑑𝑠≀𝑀𝑑0||ξ€·π‘˜1(𝑑,𝑠)βˆ’π‘˜1ξ€Έ||ξ€œ(𝜏,𝑠)𝑑𝑠+πœπ‘‘||ξ€·π‘˜1(𝑑,𝑠)βˆ’π‘˜1ξ€Έ||ξ€œ(𝜏,𝑠)𝑑𝑠+𝜁𝜏||ξ€·π‘˜1(𝑑,𝑠)βˆ’π‘˜1ξ€Έ||+ξ€œ(𝜏,𝑠)π‘‘π‘ πœ‚πœ||ξ€·π‘˜1(𝑑,𝑠)βˆ’π‘˜1ξ€Έ||ξ€œ(𝜏,𝑠)𝑑𝑠+1πœ‚||ξ€·π‘˜1(𝑑,𝑠)βˆ’π‘˜1ξ€Έ||ξ‚Ή=𝑀(𝜏,𝑠)π‘‘π‘ ξ€·πœ›Ξ“(𝛼)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έξ€œπ‘‘0ξ€½ξ€·(πœβˆ’π‘ )π›Όβˆ’1βˆ’(π‘‘βˆ’π‘ )π›Όβˆ’1πœ›ξ€Έξ€Ί1𝜈1+πœ‡1πœ†1ξ€»+ξ€·πœπ›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1+π›Ύπœ†1(πœβˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœˆ1(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ύπœ›1(πœβˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœ‡1(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€œξ€»ξ€Ύπ‘‘π‘ πœπ‘‘ξ€½(πœβˆ’π‘ )π›Όβˆ’1ξ€Ίπœ›1𝜈1+πœ‡1πœ†1ξ€»+ξ€·πœπ›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1+π›Ύπœ†1(πœβˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœˆ1(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ύπœ›1(πœβˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœ‡1(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€œξ€»ξ€Ύπ‘‘π‘ πœπœπœξ€½ξ€·π›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1+π›Ύπœ†1(πœβˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœˆ1(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ύπœ›1(πœβˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœ‡1(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€œξ€»ξ€Ύπ‘‘π‘ πœ‚πœπœξ€½ξ€·π›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœˆ1(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœ‡1(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€œξ€»ξ€Ύπ‘‘π‘ 1πœ‚πœξ€½ξ€·π›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1ξ‚Ή=π‘€ξ€»ξ€Ύπ‘‘π‘ Ξ“ξ€·πœ›(𝛼)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚Έξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1ξ€Ίπœ›1𝜈1+πœ‡1πœ†1ξ€»ξ€œπ‘‘π‘ βˆ’π‘‘0(π‘‘βˆ’π‘ )π›Όβˆ’1ξ€Ίπœ›1𝜈1+πœ‡1πœ†1ξ€»+ξ€œπ‘‘π‘ πœ0πœξ€Ίξ€·π›Όβˆ’1βˆ’π‘‘π›Όβˆ’1ξ€Έξ€Ίπ›Ύπœ†1(πœβˆ’π‘ )π›Όβˆ’1ξ€»βˆ’ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2ξ€Έξ€Ίπ›Ύπœ›1(πœβˆ’π‘ )π›Όβˆ’1+ξ€œξ€»ξ€»π‘‘π‘ πœ‚0πœξ€Ίξ€·π›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœˆ1(πœ‚βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1βˆ’π›Ώπœ‡1(πœ‚βˆ’π‘ )π›Όβˆ’1+ξ€œξ€»ξ€»π‘‘π‘ 1πœ‚πœξ€Ίξ€·π›Όβˆ’1βˆ’π‘‘π›Όβˆ’1πœˆξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1ξ€»+ξ€·πœπ›Όβˆ’2βˆ’π‘‘π›Όβˆ’2πœ‡ξ€Έξ€Ί1(1βˆ’π‘ )π›Όβˆ’1ξ‚Ή=π‘€ξ€»ξ€»π‘‘π‘ ξ€·πœ›Ξ“(𝛼+1)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ€Ί(πœπ›Όβˆ’π‘‘π›Ό)ξ€Ίπœ›1𝜈1+πœ‡1πœ†1ξ€»+ξ€·πœπ›Όβˆ’1βˆ’π‘‘π›Όβˆ’1ξ€Έξ€Ίπ›Ύπœπ›Όξ€·πœ†1βˆ’πœ›1ξ€Έ+ξ€·πœˆ1+πœ‡1ξ€Έξ€·1βˆ’π›Ώπœ‚π›Ό,||π·ξ€Έξ€»ξ€»π‘žπΉ1𝑣(𝑑)βˆ’π·π‘žπΉ1𝑣||=||||𝐷(𝜏)π‘žπΌπ›Όπ‘“(𝑑,𝑣(𝑑),𝐷𝑝𝑣1(𝑑))+ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ‚ƒπ›Ύξ‚€πœ›1π·π‘žπ‘‘π›Όβˆ’2βˆ’πœ†1π·π‘žπ‘‘π›Όβˆ’1𝐼𝛼𝑓(𝜁,𝑣(𝜁),π·π‘ξ‚€πœˆπ‘£(𝜁))+𝛿1π·π‘žπ‘‘π›Όβˆ’1+πœ‡1π·π‘žπ‘‘π›Όβˆ’2𝐼𝛼𝑓(πœ‚,𝑣(πœ‚),π·π‘βˆ’ξ‚€πœˆπ‘£(πœ‚))1π·π‘žπ‘‘π›Όβˆ’1+πœ‡1π·π‘žπ‘‘π›Όβˆ’2𝐼𝛼𝑓(1,𝑣(1),𝐷𝑝𝑣(1))βˆ’π·π‘žπΌπ›Όπ‘“(𝜏,𝑣(𝜏),𝐷𝑝1𝑣(𝜏))βˆ’ξ€·πœ›1𝜈1+πœ‡1πœ†1ξ€ΈΓ—ξ€Ίπ›Ύξ€·πœ›1π·π‘žπœπ›Όβˆ’2βˆ’πœ†1π·π‘žπœπ›Όβˆ’1𝐼𝛼𝑓(𝜁,𝑣(𝜁),π·π‘ξ€·πœˆπ‘£(𝜁))+𝛿1π·π‘žπœπ›Όβˆ’1+πœ‡1π·π‘žπœπ›Όβˆ’2𝐼𝛼𝑓(πœ‚,𝑣(πœ‚),π·π‘βˆ’ξ€·πœˆπ‘£(πœ‚))1π·π‘žπœπ›Όβˆ’1+πœ‡1π·π‘žπœπ›Όβˆ’2𝐼𝛼𝑓(1,𝑣(1),𝐷𝑝||||≀1𝑣(1))||||ξ€œΞ“(π›Όβˆ’π‘ž)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1𝑓(𝑠,𝑣(𝑠),π·π‘ξ€œπ‘£(𝑠))π‘‘π‘ βˆ’πœ0(πœβˆ’π‘ )π›Όβˆ’π‘žβˆ’1𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||ξƒ­+𝑀𝑣(𝑠))π‘‘π‘ ξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—||||ξ‚€π‘‘π›Όβˆ’π‘žβˆ’1βˆ’πœπ›Όβˆ’π‘žβˆ’1ξ‚ξ‚»βˆ’π›Ύπœ†1ξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1𝑑𝑠+π›Ώπœˆ1ξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1π‘‘π‘ βˆ’πœˆ1ξ€œ10(1βˆ’π‘ )π›Όβˆ’1ξ‚Ό||||+π‘€π‘‘π‘ π›Όξ›ξ€·πœ›(π›Όβˆ’π‘žβˆ’1)1𝜈1+πœ‡1πœ†1ξ€ΈΓ—||||ξ‚€π‘‘π›Όβˆ’π‘žβˆ’2βˆ’πœπ›Όβˆ’π‘žβˆ’2ξ‚ξ‚»π›Ύπœ›1ξ€œπœ0(πœβˆ’π‘ )π›Όβˆ’1𝑑𝑠+π›Ώπœ‡1ξ€œπœ‚0(πœ‚βˆ’π‘ )π›Όβˆ’1π‘‘π‘ βˆ’πœ‡1ξ€œ10(1βˆ’π‘ )π›Όβˆ’1ξ‚Ό||||≀1𝑑𝑠||||ξ€œΞ“(π›Όβˆ’π‘ž)𝑑0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1𝑓(𝑠,𝑣(𝑠),π·π‘ξ€œπ‘£(𝑠))π‘‘π‘ βˆ’πœ0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||+||||ξ€œπ‘£(𝑠))π‘‘π‘ πœ0(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1𝑓(𝑠,𝑣(𝑠),π·π‘ξ€œπ‘£(𝑠))π‘‘π‘ βˆ’πœ0(πœβˆ’π‘ )π›Όβˆ’π‘žβˆ’1𝑓(𝑠,𝑣(𝑠),𝐷𝑝||||ξƒ­+𝑀𝑣(𝑠))π‘‘π‘ π›Ύπœπ›Όπœ†1+𝜈1ξ€·1βˆ’π›Ώπœ‚π›Όξ€Έξ€Έπ›Όξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€Έξ‚€π‘‘π›Όβˆ’π‘žβˆ’1βˆ’πœπ›Όβˆ’π‘žβˆ’1+π‘€ξ€·π›Ύπœπ›Όπœ›1+πœ‡1ξ€·1βˆ’π›Ώπœ‚π›Όξ€Έξ€Έπ›Ό2ξ›ξ€·πœ›(π›Όβˆ’π‘žβˆ’1)1𝜈1+πœ‡1πœ†1ξ€Έξ‚€π‘‘π›Όβˆ’π‘žβˆ’2βˆ’πœπ›Όβˆ’π‘žβˆ’2ξ‚β‰€π‘€ξ‚Έξ€œΞ“(π›Όβˆ’π‘ž)𝜏0ξ€·(πœβˆ’π‘ )π›Όβˆ’π‘žβˆ’1βˆ’(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1ξ€Έξ€œπ‘‘π‘ +πœπ‘‘(π‘‘βˆ’π‘ )π›Όβˆ’π‘žβˆ’1ξ‚Ή+π‘€ξ€·π‘‘π‘ π›Ύπœπ›Όπœ†1+𝜈1ξ€·1βˆ’π›Ώπœ‚π›Όξ€Έξ€Έπ›Όξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€Έξ‚€π‘‘π›Όβˆ’π‘žβˆ’1βˆ’πœπ›Όβˆ’π‘žβˆ’1+π‘€ξ€·π›Ύπœπ›Όπœ›1+πœ‡1ξ€·1βˆ’π›Ώπœ‚π›Όξ€Έξ€Έπ›Ό2ξ›ξ€·πœ›(π›Όβˆ’π‘žβˆ’1)1𝜈1+πœ‡1πœ†1ξ€Έξ‚€π‘‘π›Όβˆ’π‘žβˆ’2βˆ’πœπ›Όβˆ’π‘žβˆ’2=𝑀Γ(π›Όβˆ’π‘ž+1)(πœπ›Όβˆ’π‘žβˆ’π‘‘π›Όβˆ’π‘ž)+π‘€ξ€·π›Ύπœπ›Όπœ†1+𝜈1ξ€·1βˆ’π›Ώπœ‚π›Όξ€Έξ€Έπ›Όξ›ξ€·πœ›(π›Όβˆ’π‘ž)1𝜈1+πœ‡1πœ†1ξ€Έξ‚€π‘‘π›Όβˆ’π‘žβˆ’1βˆ’πœπ›Όβˆ’π‘žβˆ’1+π‘€ξ€·π›Ύπœπ›Όπœ›1+πœ‡1ξ€·1βˆ’π›Ώπœ‚π›Όξ€Έξ€Έπ›Ό2(ξ€·πœ›π›Όβˆ’π‘žβˆ’1)1𝜈1+πœ‡1πœ†1ξ€Έξ‚€π‘‘π›Όβˆ’π‘žβˆ’2βˆ’πœπ›Όβˆ’π‘žβˆ’2.(3.10)
Analogously, it can be proved that ||𝐹1𝑒(𝑑)βˆ’πΉ1||≀𝑁𝑒(𝜏)ξ€·πœ›Ξ“(𝛽+1)2𝜈2+πœ‡2πœ†2ξ€ΈΓ—πœξ€Ίξ€·π›½βˆ’π‘‘π›½πœ›ξ€Έξ€Ί2𝜈2+πœ‡2πœ†2ξ€»+ξ€·πœπ›½βˆ’1βˆ’π‘‘π›½βˆ’1ξ€Έξ€Ίπ›Ύπœπ›½ξ€·πœ†2βˆ’πœ›2ξ€Έ+ξ€·πœˆ2+πœ‡2ξ€Έξ€·1βˆ’π›Ώπœ‚π›½||𝐷𝑝𝐹1𝑒(𝑑)βˆ’π·π‘πΉ1||≀𝑁𝑒(𝜏)ξ€·πœΞ“(π›½βˆ’π‘+1)π›½βˆ’π‘βˆ’π‘‘π›½βˆ’π‘ξ€Έ+π‘ξ‚€π›Ύπœπ›½πœ†2+𝜈2ξ€·1βˆ’π›Ώπœ‚π›½ξ€Έξ‚π›½ξ›ξ€·πœ›(π›½βˆ’π‘)2𝜈2+πœ‡2πœ†2ξ€Έξ‚€π‘‘π›½βˆ’π‘βˆ’1βˆ’πœπ›½βˆ’π‘βˆ’1+π‘ξ‚€π›Ύπœπ›½πœ›2+πœ‡2ξ‚€1βˆ’π›Ώπœ‚π›½ξ‚ξ‚π›½2ξ›ξ€·πœ›(π›½βˆ’π‘βˆ’1)2𝜈2+πœ‡2πœ†2ξ€Έξ‚€π‘‘π›½βˆ’π‘βˆ’2βˆ’πœπ›½βˆ’π‘βˆ’2.(3.11)
Since the functions 𝑑𝛼,π‘‘π›Όβˆ’1,𝑑𝛽,π‘‘π›½βˆ’1,π‘‘π›Όβˆ’π‘ž,π‘‘π›Όβˆ’π‘žβˆ’1,π‘‘π›½βˆ’π‘,π‘‘π›½βˆ’π‘βˆ’1 are uniformly continuous on 𝐽, therefore, it follows from the above estimates that πΉπ‘Š is an equicontinuous set. Also, it is uniformly bounded as πΉπ‘ŠβŠ‚π‘Š. 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).

References

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