A Study for Coupled Systems of Nonlinear Boundary Value Problem

Abstract

This chapter deals with the existence and uniqueness of solutions for a coupled system of fractional differential equations with coupled nonlocal and integral boundary conditions and for the system of two-point boundary value problem when we take the case of integer derivative. The existence results for the fist problem are obtained by using Leray-Shauder nonlinear alternative and Banach contraction principle and for the second problem, we derive explicit eigenvalue intervals of λ for the existence of at least one positive solution by using Krasnosel’skii fixed point theorem. An illustrative examples is presented at the end for each problem to illustrate the validity of our results.

1. Introduction

In this chapter, we are interested in the existence of solutions for the nonlinear fractional boundary value problem (BVP)

$$\left\{\begin{array}{ll}{}^c{D}^{\alpha }u\left(t\right)=f\left(t,u\left(t\right),v\left(t\right)\right),& t\in \left[0,1\right],2<\alpha \le 3,\\ {}^c{D}^{\beta }v\left(t\right)=g\left(t,u\left(t\right),v\left(t\right)\right),& t\in \left[0,1\right],2<\beta \le 3,\\ \mathit{\lambda u}\left(0\right)+\mathit{\gamma u}\left(1\right)=u\left(\eta \right),& \mathit{\lambda v}\left(0\right)+\mathit{\gamma v}\left(1\right)=v\left(\eta \right),\\ u\left(0\right)={\int }_0^{\eta }u\left(s\right)\mathit{ds},& v\left(0\right)={\int }_0^{\eta }v\left(s\right)\mathit{ds},\\ {\lambda }^c{D}^{p}u\left(0\right)+{\gamma }^c{D}^{p}u\left(1\right){=}^c{D}^{p}u\left(\eta \right),& 1<p\le 2.\end{array}\right.$$E1

We also study the integer case of problem

$$\left\{\begin{array}{c}\begin{array}{ll}{u}^{\left(4\right)}\left(t\right)=\mathit{\lambda a}\left(t\right)f\left(v\left(t\right)\right),& 0<t<1,\\ v^{\left(4\right)}\left(t\right)=\mathit{\lambda b}\left(t\right)g\left(u\left(t\right)\right),& 0<t<1,\\ u\left(0\right)=0,u^{\prime }\left(0\right)=0,u^{″}\left(1\right)=0,& u^{″\prime }\left(1\right)=0,\\ v\left(0\right)=0,v^{\prime }\left(0\right)=0,v^{″}\left(1\right)=0,& v^{″\prime }\left(1\right)=0.\end{array}\end{array}\right.$$E2

where $D_{0^{+}}^{\alpha },D_{0^{+}}^{\beta }$ are the standard Riemann-Liouville fractional derivative of order $\alpha$ and $\beta$, the functions $f,g\in C\left(\left(0,1\right)\times {\mathbb{R}}^2,\mathbb{R}\right)$, the functions $f,g\in C\left(\left(0,1\right)\times \mathbb{R},\mathbb{R}\right)$ in the second problem and $\lambda >0,a,b\in C\left(\left[0,1\right],\left[0,\infty \right)\right)$.

The first definition of fractional derivative was introduced at the end of the nineteenth century by Liouville and Riemann, but the concept of non-integer derivative and integral, as a generalization of the traditional integer order differential and integral calculus, was mentioned already in 1695 by Leibniz and L’Hospital. In fact, fractional derivatives provide an excellent tool for the description of memory and hereditary properties of various materials and processes. The mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of complex medium, polymer rheology, Bode’s analysis of feedback amplifiers, capacitor theory, electrical circuits, electro-analytical chemistry, biology, control theory, fitting of experimental data, involves derivatives of fractional order. In consequence, the subject of fractional differential equations is gaining much importance and attention. For more details we refer the reader to [1, 2, 3, 4, 5, 6] and the references cited therein.

Boundary value problems for nonlinear differential equations arise in a variety of areas of applied mathematics, physics and variational problems of control theory. A point of central importance in the study of nonlinear boundary value problems is to understand how the properties of nonlinearity in a problem influence the nature of the solutions to the boundary value problems. The multi-point boundary conditions are important in various physical problems of applied science when the controllers at the end points of the interval (under consideration) dissipate or add energy according to the sensors located, at intermediate points, see [7, 8] and the references therein. We quote also that realistic problems arising from economics, optimal control, stochastic analysis can be modelled as differential inclusion. The study of fractional differential inclusions was initiated by EL-Sayad and Ibrahim [9]. Also, recently, several qualitative results for fractional differential inclusion were obtained in [10, 11, 12, 13] and the references therein.

The techniques of nonlinear analysis, as the main method to deal with the problems of nonlinear differential equations (DEs), nonlinear fractional differential equations (FDEs), nonlinear partial differential equations (PDEs), nonlinear fractional partial differential equations (FPDEs), nonlinear stochastic fractional partial differential equations (SFPDEs), plays an essential role in the research of this field, such as establishing the existence, uniqueness and multiplicity of solutions (or positive solutions) and mild solutions for nonlinear of different kinds of FPDEs, FPDEs, SFPDEs, inclusion differential equations and inclusion fractional differential equations with various boundary conditions, by using different techniques (approaches). For more details, see [14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37] and the references therein. For example, iterative method is an important tool for solving linear and nonlinear Boundary Value Problems. It has been used in the research areas of mathematics and several branches of science and other fields. However, Many authors showed the existence of positive solutions for a class of boundary value problem at resonance case. Some recent devolopment for resonant case can be found in [38, 39]. Let us cited few papers. Zhang et al. [40] studied the existence of two positive solutions of following singular fractional boundary value problems:

$$\left\{\begin{array}{c}\begin{array}{ll}{D}_{0+}^{\alpha }u\left(t\right)+f\left(t,u\left(t\right)\right)=0,& t\in \left(0,1\right)\\ u\left(0\right)=0,D_{0+}^{\beta }u\left(0\right)=0,& D_{0+}^{\beta }u\left(1\right)={\sum }_{j=1}^{\infty }{D}_{0+}^{\beta }u\left({\eta }_j\right),\end{array}\end{array}\right.$$E3

where $D_{0+}^{\alpha },D_{0+}^{\beta }$ are the stantard Riemann-Liouville fractional derivative of order $\alpha \in \left(2,3\right],\beta \in \left[1,2\right],f\in C\left(\left[0,1\right]\times \mathbb{R},\mathbb{R}\right)$ and $a_j,{\eta }_j\in \left(0,1\right),\alpha -\beta \ge 1$ with ${\sum }_{\infty }^{i=0}{a}_j{\eta }_j^{\alpha -\beta -1}<1$.

In [41], the authors studied the boundary value problems of the fractional order differential equation:

$$\left\{\begin{array}{c}\begin{array}{l}{D}_{0+}^{\alpha }u\left(t\right)=f\left(t,u\left(t\right)\right)=0,t\in \left(0,1\right),\\ u\left(0\right)=0,D_{0+}^{\beta }u\left(1\right)={\mathit{aD}}_{0+}^{\beta }u\left(\eta \right),\end{array}\end{array}\right.$$E4

where $1<\alpha \le 2,0<\eta <1,0<a,\beta <1,$ $f\in C\left(\left[0,1\right]\times {\mathbb{R}}^2,\mathbb{R}\right)$ and $D_{0+}^{\alpha },D_{0+}^{\beta }$ are the stantard Riemann-Liouville fractional derivative of order $\alpha$. They obtained the multiple positive solutions by the Leray-Schauder nonlinear alternative and the fixed point theorem on cones.

In 2015, Alsulami et al. [42] studied the existence of solutions of the following nonlinear third-order ordinary differential inclusion with multi-strip boundary conditions

$$\left\{\begin{array}{c}\begin{array}{l}{u}^{\left(3\right)}\left(t\right)\in F\left(t,u\left(t\right)\right),t\in \left(0,1\right),\\ u\left(0\right)=0,u^{\prime }\left(0\right)=0,\\ u\left(1\right)=\underset{i=1}{\sum ^{n-2}}{{\alpha }_i}_{{\zeta }_i}^{{\eta }_i}u\left(s\right)\mathit{ds},\\ 0<{\zeta }_i<{\eta }_i<1,i=1,2,\dots ,n-2,n\ge 3.\end{array}\end{array}\right.$$E5

In 2017, Resapour et al. [43] investigated a Caputo fractional inclusion with integral boundary condition for the following problem

$$\left\{\begin{array}{c}\begin{array}{l}{}^c{D}^{\alpha }u\left(t\right)\in F\left(t,u\left(t\right){,}^c{D}^{\beta }u\left(t\right),u^{\prime }\left(t\right)\right),\\ u\left(0\right)+u^{\prime }\left(0\right){+}^c{D}^{\beta }u\left(0\right)={\int }_0^{\eta }u\left(s\right)\mathit{ds},\\ u\left(1\right)+u^{\prime }\left(1\right){+}^c{D}^{\beta }u\left(1\right)={\int }_0^{\nu }u\left(s\right)\mathit{ds},\end{array}\end{array}\right.$$E6

where $1<\alpha \le 2,\eta ,\nu ,\beta \in \left(0,1\right),$ $F:\left[0,1\right]\times \mathbb{R}\times \mathbb{R}\times \mathbb{R}\to 2^{\mathbb{R}}$ is a compact valued multifunction and ${}^c{D}^{\alpha }$ denotes the Caputo fractional derivative of order $\alpha$.

Inspired and motivated by the works mentioned above, The goal of this chapter is to establish the existence and uniqueness results for the nonlocal boundary value problem system (1) by using some well-known tools of fixed point theory such as Banach contraction principle and Leray-Shauder nonlinear alternative and the existence of at least one positive solution for the system of two-point boundary value problem (2) by using Krasnosel’skii fixed point theorem. The aim of the last results is to establish some simple criteria for the existence of single positive solutions of the BVPs (2) in explicit intervals for $\lambda$. The chapter is organized as follows. In Section 2, we recall some preliminary facts that we need in the sequel, for more details; see [44] and we give main results of problem (1). Finally, we give an example to illustrate our result. In Section 3, deals with main results of problem (2) and we give an example to illustrate our results.

2. Existence and uniqueness results for problem (1)

2.1 Preliminaries

In this section, we introduce some definitions and lemmas, see [2, 4, 44, 45, 46].

Definition 2.1. Let $\alpha >0,n-1<\alpha <n,n=\left[\alpha \right]+1$ and $u\in C\left(\left[0,\infty \right),\mathbb{R}\right)$. The Caputo derivative of fractional order $\alpha$ for the function $u$ is defined by

$${}^c{D}^{\alpha }u\left(t\right)={\frac{1}{\mathrm{\Gamma }\left(n-\alpha \right)}}_0^t{\left(t-s\right)}^{n-\alpha -1}{u}^{\left(n\right)}\left(s\right)\mathit{ds},$$E7

where $\mathrm{\Gamma }\left(\cdot \right)$ is the Eleur Gamma function.

Definition 2.2. The Riemann-Liouville fractional integral of order $\alpha >0$ of a function $u:\left(0,\infty \right)\to \mathbb{R}$ is given by

$$I^{\alpha }u\left(t\right)={\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}}_0^t{\left(t-s\right)}^{\alpha -1}u\left(s\right)\mathit{ds},t>0,$$E8

where $\mathrm{\Gamma }\left(\cdot \right)$ is the Eleur Gamma function, provided that the right side is pointwise defined on $\left(0,\infty \right)$.Lemma 2.1. Let $\alpha >0,n-1<\alpha <n$ and the function $g:\left[0,T\right]\to \mathbb{R}$ be continuous for each $T>0$. Then, the general solution of the fractional differential equation ${}^c{D}^{\alpha }g\left(t\right)=0$ is given by

$$g\left(t\right)=c_0+c_{1}t+\cdots +c_{n-1}{t}^{n-1},$$E9

where $c_0,c_1,\dots ,c_{n-1}$ are real constants and $n=\left[\alpha \right]+1$.

Also, in [19], authors have been proved that for each $T>0$ and $u\in C\left(\left[0,T\right]\right)$ we have

$$I^{\mathit{\alpha c}}{D}^{\alpha }u\left(t\right)=u\left(t\right)+c_0+c_{1}t+\cdots +c_{n-1}{t}^{n-1},$$E10

where $c_0,c_1,\dots ,c_{n-1}$ are real constants and $n=\left[\alpha \right]+1$.

2.2 Existence results

Let $X=\left\{u\left(t\right):u\left(t\right)\in C\left(\left[0,1\right],\mathbb{R}\right)\right\}$ endowed with the norm $‖u‖=\underset{t\in \left[0,1\right]}{sup}\left|u\left(t\right)\right|$ such that $‖u‖<\infty$. Then $\left(X,‖.‖\right)$ is a Banach space and the product space $\left(X\times X,‖\left(u,v\right)‖\right)$ is also a Banach space equipped with the norm $‖\left(u,v\right)‖=‖u‖+‖v‖$.

Throughout the first section, we let

$$\begin{array}{l}M=\frac{\mathrm{\Gamma }\left(3-p\right)}{\left|\gamma -{\eta }^{2-p}\right|}\ne 0,\left|\lambda +\gamma -1\right|\ne 0,\left|\gamma -{\eta }^2\right|\ne 0,Q=\left|2\left(1-\eta \right)\left(\gamma -\eta \right)+{\eta }^2\left|\lambda +\gamma -1\right|\right|\ne 0,\\ A\left(t\right)=\left|{\mathrm{\Lambda }}_1\left(t\right)\right|=\left|\lambda +\gamma -1\right|\left({\eta }^2+2\left(1-\eta \right)t\right),\\ \begin{array}{rc}B\left(t\right)=\left|{\mathrm{\Lambda }}_2\left(t\right)\right|=& \left({\eta }^3\left|\lambda +\gamma -1\right|+3\left|\gamma -{\eta }^2\right|\left(1-\eta \right)\right)\left({\eta }^2+2\left(1-\eta \right)t\right)-Q\left({\eta }^3+3\left(1-\eta \right)t^2\right),\end{array}\end{array}$$E11

and

$$Q=2\left(1-\eta \right)\left(\gamma -\eta \right)+{\eta }^2\left(\lambda +\gamma -1\right)\ne 0.$$E12

Lemma 2.2. Let $y\in C\left(\left[0,1\right],\mathbb{R}\right)$. Then the solution of the linear differential system

$$\left\{\begin{array}{c}\begin{array}{l}{}^c{D}^{\alpha }u\left(t\right)=y\left(t\right),{}^c{D}^{\beta }v\left(t\right)=h\left(t\right),t\in \left[0,1\right],2<\alpha ,\beta \le 3\\ \mathit{\lambda u}\left(0\right)+\mathit{\gamma u}\left(1\right)=v\left(\eta \right),\mathit{\lambda v}\left(0\right)+\mathit{\gamma v}\left(1\right)=u\left(\eta \right),\\ u\left(0\right)={\int }_0^{\eta }v\left(s\right)\mathit{ds},v\left(0\right)={\int }_0^{\eta }u\left(s\right)\mathit{ds},\\ \begin{array}{c}{\lambda }^c{D}^{p}u\left(0\right)+{\gamma }^c{D}^{p}u\left(1\right){=}^c{D}^{p}v\left(\eta \right),1<p\le 2,\\ {\lambda }^c{D}^{p}v\left(0\right)+{\gamma }^c{D}^{p}v\left(1\right){=}^c{D}^{p}u\left(\eta \right),1<p\le 2,\end{array}\end{array}\end{array}\right.$$E13

is equivalent to the system of integral equations

$$\begin{array}{l}u\left(t\right)=\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}y\left(s\right)\mathit{ds}+{\frac{1}{1-\eta }}_0^{\eta }\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{{\mathrm{\Lambda }}_1\left(t\right)}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{{\mathrm{\Lambda }}_2\left(t\right)M}{6\left(1-\eta \right)Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}h\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}y\left(s\right)\mathit{ds}\right]\\ +\frac{{\mathrm{\Lambda }}_1\left(t\right)}{Q\left(\lambda +\gamma -1\right)}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}h\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}y\left(s\right)\mathit{ds}\right],\end{array}$$E14

and

$$\begin{array}{l}v\left(t\right)=\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}y\left(s\right)\mathit{ds}+\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{{\mathrm{\Lambda }}_1\left(t\right)}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{{\mathrm{\Lambda }}_2\left(t\right)M}{6\left(1-\eta \right)Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}h\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}y\left(s\right)\mathit{ds}\right]\\ +\frac{{\mathrm{\Lambda }}_1\left(t\right)}{Q\left(\lambda +\gamma -1\right)}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}h\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\alpha \right)}y\left(s\right)\mathit{ds}\right],\end{array}$$E15

where

$${\mathrm{\Lambda }}_1\left(t\right)=\left(\lambda +\gamma -1\right)\left({\eta }^2+2\left(1-\eta \right)t\right),$$E16

and

$${\mathrm{\Lambda }}_2\left(t\right)=\left({\eta }^3\left(\lambda +\gamma -1\right)+3\left(\gamma -{\eta }^2\right)\left(1-\eta \right)\right)\left({\eta }^2+2\left(1-\eta \right)t\right)-Q\left({\eta }^3+3\left(1-\eta \right)t^2\right).$$

Proof. It is well known that the solution of equation ${}^c{D}^{\alpha }u\left(t\right)=y\left(t\right)$ can be written as

$$u\left(t\right)=I^{\alpha }y\left(t\right)+c_0+c_{1}t+c_2{t}^2,$$E17

$$v\left(t\right)=I^{\beta }h\left(t\right)+d_0+d_{1}t+d_2{t}^2,$$E18

where $c_0,c_{1,}{c}_2\in \mathbb{R}$ and and $d_0,d_{1,}{d}_2\in \mathbb{R}$ are arbitrary constants.

Then, from (68) we have

$$u^{\prime }\left(t\right)=I^{\alpha -1}y\left(t\right)+c_1+2c_{2}t,$$E19

and

$${}^c{D}^{p}u\left(t\right)=I^{\alpha -p}y\left(t\right)+c_2\frac{2t^{2-p}}{\mathrm{\Gamma }\left(3-p\right)},1<p\le 2.$$E20

By using the three-point boundary conditions, we obtain.

$$\begin{array}{l}{c}_2=\frac{M}{2}\left(I^{\beta -p}y\left(\eta \right)-\gamma I^{\alpha -p}y\left(1\right)\right),\\ c_0=\begin{array}{l}-\frac{2{\eta }^2\left(\lambda +\gamma -1\right)}{2\left(1-\eta \right)Q}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{Г\left(\beta \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}+\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{Г\left(\beta \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{\left({\eta }^2\left[{\eta }^3\left(\lambda +\gamma -1\right)+3\left(\gamma -{\eta }^2\right)\left(1-\eta \right)\right]-{\eta }^{3}Q\right)M}{2\left(1-\eta \right)Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{Г\left(\beta -p\right)}h\left(s\right)\mathit{ds}\right.\\ \left.-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{Г\left(\alpha -p\right)}y\left(s\right)\mathit{ds}\right]+\frac{{\eta }^2}{Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{Г\left(\beta \right)}h\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{Г\left(\alpha \right)}y\left(s\right)\mathit{ds}\right],\end{array}\end{array}$$E21

and

$$\begin{array}{l}{c}_1=\frac{-2\left(\lambda +\gamma -1\right)}{Q}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}h\left(\tau \right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{\left({\eta }^3\left(\lambda +\gamma -1\right)+3\left(\gamma -{\eta }^2\right)\left(1-\eta \right)\right)M}{3Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}h\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}y\left(s\right)\mathit{ds}\right]\\ +\frac{2\left(1-\eta \right)}{Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}y\left(s\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}y\left(s\right)\mathit{ds}\right].\end{array}$$E22

Substituting the values of constants $c_0,c_1$ and $c_2$ in (68), we get solution (64). Similarly, we obtain solution (65). The proof is complete.

The following rolations hold:

$$\left|A\left(t\right)\right|\le \left|\beta +\gamma -1\right|\left({\eta }^2+2\left(1-\eta \right)\right)=A_1,$$E23

and

$$\left|B\left(t\right)\right|\le \left|\left({\eta }^3\left|\beta +\gamma -1\right|+3\left|\gamma -{\eta }^2\right|\left(1-\eta \right)\right)\left({\eta }^2+2\left(1-\eta \right)\right)-Q\left({\eta }^3+3\left(1-\eta \right)\right)\right|=B_1,$$E24

For the sake of brevity, we set

$$\begin{array}{l}{\Delta }_1=\frac{{\eta }^{\beta +1}}{\left(1-\eta \right)\mathrm{\Gamma }\left(\beta +2\right)}+\frac{A_1{\eta }^{\beta +1}}{Q\left(1-\eta \right)\mathrm{\Gamma }\left(\beta +2\right)}+\frac{{\mathit{MB}}_1{\eta }^{\beta -p}}{\left(1-\eta \right)Q\mathrm{\Gamma }\left(\lambda -p+1\right)}\\ +\frac{A_1{\eta }^{\beta }}{Q\left|\beta +\gamma -1\right|\mathrm{\Gamma }\left(\beta +1\right)}\\ {\Delta }_2=\frac{{\mathit{MB}}_1\gamma }{6\left(1-\eta \right)Q\mathrm{\Gamma }\left(\alpha -p+1\right)}+\frac{A_1\gamma }{Q\left|\lambda +\gamma -1\right|\mathrm{\Gamma }\left(\alpha +1\right)}+\frac{1}{\mathrm{\Gamma }\left(\alpha +1\right)}\\ {\Delta }_3=\frac{{\eta }^{\alpha +1}}{\left(1-\eta \right)\mathrm{\Gamma }\left(\alpha +2\right)}+\frac{A_1{\eta }^{\alpha +1}}{Q\left(1-\eta \right)\mathrm{\Gamma }\left(\alpha +2\right)}+\frac{{\mathit{MB}}_1{\eta }^{\alpha -p}}{\left(1-\eta \right)Q\mathrm{\Gamma }\left(\alpha -p+1\right)}\\ +\frac{A_1{\eta }^{\alpha }}{Q\left|\lambda +\gamma -1\right|\mathrm{\Gamma }\left(\alpha +1\right)},\end{array}$$E25

and

$${\Delta }_4=\frac{{\mathit{MB}}_1\gamma }{6\left(1-\eta \right)Q\mathrm{\Gamma }\left(\beta -p+1\right)}+\frac{A_1\gamma }{Q\left|\lambda +\gamma -1\right|\mathrm{\Gamma }\left(\beta +1\right)}+\frac{1}{\mathrm{\Gamma }\left(\beta +1\right)}.$$E26

In view of Lemma 2, we define the operator $T:X\times X\to X\times X$ by

$$T\left(u,v\right)\left(t\right)=\left(\begin{array}{c}{T}_1\left(u,v\right)\left(t\right)\\ T_2\left(u,v\right)\left(t\right)\end{array}\right),$$E27

where

$$\begin{array}{l}{T}_1\left(u,v\right)\left(t\right)=\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}+\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}g\left(\tau ,u\left(\tau \right),v\left(\tau \right)\right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{B\left(t\right)M}{6\left(1-\eta \right)Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}g\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}f\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}\right]\\ +\frac{A\left(t\right)}{Q\left|\beta +\gamma -1\right|}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}g\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}.\right]\\ -\frac{A\left(t\right)}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}g\left(\tau ,u\left(\tau \right),v\left(\tau \right)\right)\mathit{d\tau }\right)\mathit{ds},\end{array}$$E28

and

$$\begin{array}{l}{T}_2\left(u,v\right)\left(t\right)=\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}g\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}+\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(\tau ,u\left(\tau \right),v\left(\tau \right)\right)\mathit{d\tau }\right)\mathit{ds}\\ -\frac{B\left(t\right)M}{6\left(1-\eta \right)Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}f\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}g\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}\right]\\ +\frac{A\left(t\right)}{Q\left|\beta +\gamma -1\right|}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}-\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}g\left(s,u\left(s\right),v\left(s\right)\right)\mathit{ds}.\right]\\ -\frac{A\left(t\right)}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}f\left(\tau ,u\left(\tau \right),v\left(\tau \right)\right)\mathit{d\tau }\right)\mathit{ds}.\end{array}$$E29

Observe that the boundary value problem (1) has solutions if the operator equation $\left(u,v\right)=T\left(u,v\right)$ has fixed points.

Now we are in a position to present the first main results of this paper. The existence results is based on Leray-Shauder nonlinear alternative.

Lemma 2.3. [44] (Leray-Schauder alternative). Let $E$ be a Banach space and $T:E\to E$ be a completely continuous operator (i.e., a map restricted to any bounded set in $E$ is compact). Let

$$\epsilon =\left\{\left(u,v\right)\in X\times X:\left(u,v\right)=\mathit{\lambda T}\left(u,v\right),\mathit{\text{for\hspace{0.17em}some}}0<\lambda <1\right\}.$$E30

Then either the $\epsilon \left(T\right)$ is unbounded or $T$ has at least one fixed point.

Theorem 1.1 Assume that $f,g:\left[0,1\right]\times \mathbb{R}\times \mathbb{R}\to \mathbb{R}$ are a continuous function and.

$\left(H_1\right)$ there exist a function $k_i\ge 0m_i\ge 0,i=1,2$ and $k_0>0,m_0>0$ such that $\forall u\in \mathbb{R},\forall v\in \mathbb{R},i=1,2$, we have

$$\left|f\left(t,u,v\right)\right|\le k_0+k_1\left|u\right|+k_2\left|v\right|,$$E31

and

$$\left|g\left(t,u,v\right)\right|\le m_0+m_1\left|u\right|+m_2\left|v\right|.$$E32

If $\left({\Delta }_2+{\Delta }_3\right)k_1+\left({\Delta }_1+{\Delta }_4\right)m_1<1$ and $\left({\Delta }_2+{\Delta }_3\right)k_2+\left({\Delta }_1+{\Delta }_4\right),m_3<1$, where ${\Delta }_i,i=\mathrm{1,2,3,4}$ are given above. Then the boundary value problem (1)–(58) has at least one solution on [0,1].

Proof. It is clear that $T$ is a continuous operator where $T:X\times X\to X\times X$ is defined above. Now, we show that $T$ is completely continuous. Let $\mathrm{\Omega }\subset X\times X$ be bounded. Then there exist positive constants $L_1$ and $L_2$ such that

$$\left|f\left(t,u\left(t\right),v\left(t\right)\right)\right|\le L_1,\left|g\left(t,u\left(t\right),v\left(t\right)\right)\right|\le L_2,\forall \left(u,v\right)\in \mathrm{\Omega }.$$E33

Then for any $\left(u,v\right)\in \mathrm{\Omega }$, we have

$$\begin{array}{l}\left|T_1\left(u,v\right)\left(t\right)\right|\le \frac{L_2}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}\\ +\frac{\left|A\left(t\right)\right|L_2}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}+L_1\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\\ +\frac{M\left|B\left(t\right)\right|}{6\left(1-\eta \right)Q}\left[L_2\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}+\gamma L_1\underset{0}{\overset{0}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\mathit{ds}\right]\\ +\frac{\left|A\left(t\right)\right|}{Q\left|\lambda +\gamma -1\right|}\left[L_2\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}+\gamma L_1\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right],\\ \le L_2\left\{\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}+\frac{A_1}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\mathit{\alpha \beta }\right)}\mathit{d\tau }\right)\mathit{ds}\right.\\ \left.+\frac{{\mathit{MB}}_1}{6\left(1-\eta \right)Q}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}+\frac{A_1}{6\left|\lambda +\gamma -1\right|}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}\right\}\\ +L_1\left\{\frac{\mathit{M\gamma }{B}_1}{6\left(1-\eta \right)Q}\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\mathit{ds}+\frac{A_1\gamma }{Q\left|\lambda +\gamma -1\right|}\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right.\left.+\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right\},\\ \le L_2{\Delta }_1+L_1{\mathrm{\Delta }}_2.\end{array}$$E34

Hence

$$‖T_1\left(u,v\right)‖\le L_2{\Delta }_1+L_1{\mathrm{\Delta }}_2.$$E35

In the same way, we can obtain that

$$‖T_2\left(u,v\right)‖\le L_1{\Delta }_3+L_2{\mathrm{\Delta }}_4.$$E36

Thus, it follows from (78) and (95) that the operator $T$ is uniformly bounded, since $‖T\left(u,v\right)‖\le L_1\left({\Delta }_1+{\Delta }_3\right)+L_2\left({\mathrm{\Delta }}_2+{\Delta }_4\right)$. Now, we show that $T$ is equicontinuous. Let $t_1,t_2\in \left[0,1\right]$ with $t_1<t_2$. Then we have

$$\begin{array}{l}\left|T_1\left(u\left(t_2\right),v\left(t_2\right)\right)-T_1\left(u\left(t_1\right),v\left(t_1\right)\right)\right|\le L_1\underset{0}{\overset{t_1}{\int }}\frac{{\left(t_2-s\right)}^{\alpha -1}-{\left(t_1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\\ +L_1\underset{t_1}{\overset{t_2}{\int }}\frac{{\left(t_2-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}+\frac{\left|A\left(t_2\right)-A\left(t_1\right)\right|L_2}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}\\ +\frac{\left(B\left(t_2\right)-B\left(t_1\right)\right)M}{6\left(1-\eta \right)Q}\left[L_2\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}+\gamma L_1\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\mathit{ds}\right]\\ +\frac{A\left(t_2\right)-A\left(t_1\right)}{Q\left|\lambda +\gamma -1\right|}\left[L_2\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}-\gamma L_1\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right].\end{array}$$E37

Obviously, the right-hand side of the above inequality tends to zero as $t_2\to t_1$. Similarly, we have

$$\begin{array}{l}\left|T_2\left(u\left(t_2\right),v\left(t_2\right)\right)-T_2\left(u\left(t_1\right),v\left(t_1\right)\right)\right|\le L_2\underset{0}{\overset{t_1}{\int }}\frac{{\left(t_2-s\right)}^{\beta -1}-{\left(t_1-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}\\ +L_2\underset{t_1}{\overset{t_2}{\int }}\frac{{\left(t_2-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}+\frac{\left|A\left(t_2\right)-A\left(t_1\right)\right|L_1}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{d\tau }\right)\mathit{ds}\\ +\frac{\left(B\left(t_2\right)-B\left(t_1\right)\right)M}{6\left(1-\eta \right)Q}\left[L_1\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\mathit{ds}+\gamma L_2\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}\right]\\ +\frac{A\left(t_2\right)-A\left(t_1\right)}{Q\left|\lambda +\gamma -1\right|}\left[L_1\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}-\gamma L_2\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}\right].\end{array}$$E38

Again, it is seen that the right-hand side of the above inequality tends to zero as $t_2\to t_1$. Thus, the operator $T$ is equicontinuous.

Therefore, the operator $T$ is completely continuous.

Finally, it will be verified that the set $\epsilon =\left\{\left(u,v\right)\in X\times X:\left(u,v\right)=\mathit{\lambda T}\left(u,v\right),0\le \lambda \le 1\right\}$ is bounded. Let $\left(u,v\right)\in \epsilon$, with $\left(u,v\right)=\mathit{\lambda T}\left(u,v\right)$ for any $t\in \left[0,1\right]$, we have

$$u\left(t\right)=\lambda T_1\left(u,v\right)\left(t\right),v\left(t\right)=\lambda T_2\left(u,v\right)\left(t\right).$$E39

Then

$$\begin{array}{l}\left|u\left(t\right)\right|\le {\Delta }_2\left(k_0+k_1\left|u\right|+k_2\left|v\right|\right)+{\Delta }_1\left(m_0+m_1\left|u\right|+m_2\left|v\right|\right),\\ ={\Delta }_2{k}_0+{\Delta }_1{m}_0+\left({\Delta }_2{k}_1+{\Delta }_1{m}_1\right)\left|u\right|+\left({\Delta }_2{k}_2+{\Delta }_1{m}_2\right)\left|v\right|,\end{array}$$E40

and

$$\begin{array}{l}\left|v\left(t\right)\right|\le {\Delta }_3\left(k_0+k_1\left|u\right|+k_2\left|v\right|\right)+{\Delta }_4\left(m_0+m_1\left|u\right|+m_2\left|v\right|\right),\\ ={\Delta }_3{k}_0+{\Delta }_4{m}_0+\left({\Delta }_3{k}_1+{\Delta }_4{m}_1\right)\left|u\right|+\left({\Delta }_3{k}_2+{\Delta }_4{m}_2\right)\left|v\right|.\end{array}$$E41

Hence we have

$$‖u‖={\Delta }_2{k}_0+{\Delta }_1{m}_0+\left({\Delta }_2{k}_1+{\Delta }_1{m}_1\right)‖u‖+\left({\Delta }_2{k}_2+{\Delta }_1{m}_2\right)‖v‖,$$E42

and

$$‖v‖={\Delta }_3{k}_0+{\Delta }_4{m}_0+\left({\Delta }_3{k}_1+{\Delta }_4{m}_1\right)\left|u\right|+\left({\Delta }_3{k}_2+{\Delta }_4{m}_2\right)\left|v\right|,$$E43

which imply that

$$\begin{array}{c}‖u‖+‖v‖=\left({\Delta }_2+{\Delta }_3\right)k_0+\left({\Delta }_1+{\Delta }_4\right)m_0+\left[\left({\Delta }_2+{\Delta }_3\right)k_1+\left({\Delta }_1+{\Delta }_4\right)m_1\right]‖u‖\\ +\left[\left({\Delta }_2+{\Delta }_3\right)k_2+\left({\Delta }_1+{\Delta }_4\right)m_2\right]‖v‖.\end{array}$$E44

Consequently,

$$‖\left(u,v\right)‖=\frac{\left({\Delta }_2+{\Delta }_3\right)k_0+\left({\Delta }_1+{\Delta }_4\right)m_0}{{\Delta }_0},$$E45

where

$${\Delta }_0=\mathit{min}\left\{1-\left[\left({\Delta }_2+{\Delta }_3\right)k_1+\left({\Delta }_1+{\Delta }_4\right)m_1\right],1-\left[\left({\Delta }_2+{\Delta }_3\right)k_2+\left({\Delta }_1+{\Delta }_4\right)m_2\right]\right\},$$E46

which proves that $\epsilon$ is bounded. Thus, by Lemma 15, the operator $T$ has at least one fixed point. Hence boundary value problem (1) has at least one solution. The proof is complete.

Now, we are in a position to present the second main results of this paper.

Theorem 1.2 Assume that $f,g:\left[0,1\right]\times {\mathbb{R}}^2\to \mathbb{R}$ are continuous functions and there exist positive constants $L_1$ and $L_2$ such that for all $t\in \left[0,1\right]$ and $u_i,v_i\in \mathbb{R},i=1,2$, we have.

1. $\left|f\left(t,u_1,u_2\right)-f\left(t,v_1,v_2\right)\right|\le L_1\left(\left|u_1-v_1\right|+\left|u_2-v_2\right|\right)$,

2. $\left|g\left(t,u_1,u_2\right)-g\left(t,v_1,v_2\right)\right|\le L_2\left(\left|u_1-v_1\right|+\left|u_2-v_2\right|\right)$.

Then the boundary value problem (1) has a unique solution on [0,1] provided

$$\left({\Delta }_1+{\Delta }_3\right)L_1+\left({\mathrm{\Delta }}_2+{\mathrm{\Delta }}_4\right)L_2<1.$$E47

Proof. Let us set $\underset{t\in \left[0,1\right]}{\mathit{sup}}\left|f\left(t\mathrm{,0,0}\right)\right|=N_1<\infty$ and $\underset{t\in \left[0,1\right]}{\mathit{sup}}\left|g\left(t\mathrm{,0,0}\right)\right|=N_2<\infty$. For $u\in X$, we observe that

$$\begin{array}{c}\left|f\left(t,u\left(t\right),v\left(t\right)\right)\right|\le \left|f\left(t,u\left(t\right)\right)-f\left(t\mathrm{,0,0}\right)\right|+\left|f\left(t\mathrm{,0,0}\right)\right|,\\ \le L_1\left(\left|u\left(t\right)\right|+\left|v\left(t\right)\right|\right)+N_1,\\ \le L_1\left(‖u‖+‖v‖\right)+N_1,\end{array}$$E48

and

$$\left|g\left(t,u\left(t\right),v\left(t\right)\right)\right|\le \left|g\left(t,u\left(t\right)\right)-g\left(t\mathrm{,0,0}\right)\right|+\left|g\left(t\mathrm{,0,0}\right)\right|\le L_2‖u‖+N_2.$$E49

Then for $u\in X$, we have

$$\begin{array}{l}\left|T_1\left(u,v\right)\left(t\right)\right|\le \frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\left[L_2‖\left(u,v\right)‖+N_2\right]\mathit{d\tau }\right)\mathit{ds}\\ +\frac{\left|A\left(t\right)\right|}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\left[L_2‖\left(u,v\right)‖+N_2\right]\mathit{d\tau }\right)\mathit{ds}+\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\left[L_1‖\left(u,v\right)‖+N_1\right]\mathit{ds}\\ +\frac{M\left|B\left(t\right)\right|}{6\left(1-\eta \right)Q}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\left[L_2‖\left(u,v\right)‖+N_2\right]\mathit{ds}+\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\left[L_1‖\left(u,v\right)‖+N_1\right]\mathit{ds}\right]\\ +\frac{\left|A\left(t\right)\right|}{Q\left|\lambda +\gamma -1\right|}\left[\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\left[L_2‖\left(u,v\right)‖+N_2\right]\mathit{ds}+\gamma \underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\left[L_1‖\left(u,v\right)‖+N_1\right]\mathit{ds},\right]\end{array}$$

$$\begin{array}{l}\le \left(L_2‖\left(u,v\right)‖+N_2\right)\left\{\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}+\frac{A_1}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}\right.\\ \left.+\frac{{\mathit{MB}}_1}{6\left(1-\eta \right)Q}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}+\frac{A_1}{6\left|\lambda +\gamma -1\right|}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}\right\}\\ \left.+\frac{{\mathit{MB}}_1}{6\left(1-\eta \right)Q}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}+\frac{A_1}{6\left|\lambda +\gamma -1\right|}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}\right\}\\ +\left(L_1‖\left(u,v\right)‖+N_1\right)\left\{\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right.\\ \left.+\frac{\mathit{M\gamma }{B}_1}{6\left(1-\eta \right)Q}\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\mathit{ds}+\frac{A_1\gamma }{Q\left|\lambda +\gamma -1\right|}\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right\},\\ \le \left(L_{2}r+N_2\right){\Delta }_1+\left(L_{1}r+N_1\right){\mathrm{\Delta }}_2\end{array}$$E50

Hence

$$‖T_1\left(u,v\right)‖\le \left(L_2{\Delta }_1+L_1{\Delta }_2\right)r+N_2{\Delta }_1+N_1{\mathrm{\Delta }}_2$$E51

In the same way, we can obtain that

$$‖T_2\left(u,v\right)‖\le \left(L_1{\Delta }_3+L_2{\Delta }_4\right)r+N_2{\Delta }_4+N_1{\mathrm{\Delta }}_3.$$E52

Consequently,

$$‖T\left(u,v\right)‖\le \left(\left({\Delta }_2+{\Delta }_3\right)L_1+\left({\Delta }_1+{\Delta }_4\right)L_2\right)r+N_2\left({\Delta }_1+{\Delta }_4\right)+N_1\left({\mathrm{\Delta }}_2+{\Delta }_3\right)\le r.$$E53

Now, for $\left(u_1,v_1\right),\left(u_2,v_2\right)\in X\times X$ and for each $t\in \left[0,1\right]$, it follows from assumption $\left(H_3\right)$ that

$$\begin{array}{l}\left|T_1\left(u_2,v_2\right)\left(t\right)-T_1\left(u_1,v_1\right)\left(t\right)\right|\le L_2\left(‖u_2-u_1‖+‖v_2-v_1‖\right)\left\{\frac{1}{1-\eta }\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}\right.\\ +\frac{A_1}{Q\left(1-\eta \right)}\underset{0}{\overset{\eta }{\int }}\left(\underset{0}{\overset{s}{\int }}\frac{{\left(s-\tau \right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{d\tau }\right)\mathit{ds}\\ \left.+{\frac{{\mathit{MB}}_1}{6\left(1-\eta \right)Q}}_0^{\eta }\frac{{\left(\eta -s\right)}^{\beta -p-1}}{\mathrm{\Gamma }\left(\beta -p\right)}\mathit{ds}+\frac{A_1}{6\left|\lambda +\gamma -1\right|}\underset{0}{\overset{\eta }{\int }}\frac{{\left(\eta -s\right)}^{\beta -1}}{\mathrm{\Gamma }\left(\beta \right)}\mathit{ds}\right\}\\ +L_1\left(‖u_2-u_1‖+‖v_2-v_1‖\right)\left\{\underset{0}{\overset{t}{\int }}\frac{{\left(t-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right.\\ \left.+\frac{\mathit{M\gamma }{B}_1}{6\left(1-\eta \right)Q}\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -p-1}}{\mathrm{\Gamma }\left(\alpha -p\right)}\mathit{ds}+\frac{A_1\gamma }{Q\left|\lambda +\gamma -1\right|}\underset{0}{\overset{1}{\int }}\frac{{\left(1-s\right)}^{\alpha -1}}{\mathrm{\Gamma }\left(\alpha \right)}\mathit{ds}\right\}\\ \le \left(L_2{\Delta }_1+L_1{\Delta }_2\right)\left(‖u_2-u_1‖+‖v_2-v_1‖\right).\end{array}$$E54

Thus

$$‖T_1\left(u_2,v_2\right)-T_1\left(u_1,v_1\right)‖\le \left(L_2{\Delta }_1+L_1{\Delta }_2\right)\left(‖u_2-u_1‖+‖v_2-v_1‖\right).$$E55

Similarly,

$$‖T_2\left(u_2,v_2\right)-T_2\left(u_1,v_1\right)‖\le \left(L_2{\Delta }_3+L_1{\Delta }_4\right)\left(‖u_2-u_1‖+‖v_2-v_1‖\right).$$E56

It follows from (101) and (110) that

$$‖T\left(u_2,v_2\right)-T\left(u_1,v_1\right)‖\le \left(L_2\left({\Delta }_1+{\Delta }_3\right)+L_1\left({\Delta }_2+{\Delta }_4\right)\right)\left(‖u_2-u_1‖+‖v_2-v_1‖\right).$$E57

Since $L_2\left({\Delta }_1+{\Delta }_3\right)+L_1\left({\Delta }_2+{\Delta }_4\right)<1$, thus $T$ is a contraction operator. Hence it follows by Banach’s contraction principle that the boundary value problem (1) has a unique solution on $\left[0,1\right]$.

We construct an example to illustrate the applicability of the results presented.

Example 2.1. Consider the following system fractional differential equation

$$\left\{\begin{array}{c}\begin{array}{l}{}^c{D}^{3}u\left(t\right)=\frac{t}{8}\left(\left(cos\left(t\right)\right)sin\left(\frac{\left|u\left(t\right)\right|+\left|v\left(t\right)\right|}{2}\right)\right)+\frac{e^{-{\left(u\left(t\right)+v\left(t\right)\right)}^2}}{1+t^2},t\in \left[0,1\right],\\ {}^c{D}^{3}v\left(t\right)=\frac{1}{32}sin\left(2\mathit{\pi u}\left(t\right)\right)+\frac{\left|v\left(t\right)\right|}{16\left(1+\left|v\left(t\right)\right|\right)}+\frac{1}{2},t\in \left[0,1\right],\end{array}\end{array}\right.$$E58

subject to the three-point coupled boundary conditions

$$\left\{\begin{array}{c}\begin{array}{c}\frac{1}{100}u\left(0\right)+\frac{1}{10}u\left(1\right)=u\left(\frac{1}{2}\right),\\ u\left(0\right)={\int }_0^{0,5}u\left(s\right)\mathit{ds},\\ {\frac{1}{100}}^c{D}^{\frac{3}{2}}u\left(0\right)+{\frac{1}{10}}^c{D}^{\frac{3}{2}}u\left(1\right){=}^c{D}^{\frac{3}{2}}u\left(\frac{1}{2}\right),\end{array}\end{array}\right.$$E59

where $f\left(t,u,v\right)=\frac{t}{8}\left(\left(cos\left(t\right)\right)sin\left(\frac{\left|u\right|+\left|v\right|}{2}\right)\right)+\frac{e^{-{\left(u+v\right)}^2}}{1+t^2},t\in \left[0,1\right],\eta =0,5,\lambda =0,01,\gamma =0,1,p=1,5$ and $g\left(t,u,v\right)=\frac{1}{32\pi }sin\left(2\mathit{\pi u}\left(t\right)\right)+\frac{\left|v\left(t\right)\right|}{16\left(1+\left|v\left(t\right)\right|\right)}+\frac{1}{2}$.

It can be easily found that $M=\frac{20}{3}$ and $Q=\frac{9}{400}$.

Furthermore, by simple computation, for every $u_i,v_i\in \mathbb{R},i=1,2$, we have

$$\left|f\left(t,u_1,u_2\right)-f\left(t,v_1,v_2\right)\right|\le L\left(\left|u_1-v_1\right|+\left|u_2-v_2\right|\right),$$E60

and

$$\left|g\left(t,u_1,u_2\right)-g\left(t,v_1,v_2\right)\right|\le L\left(\left|u_1-v_1\right|+\left|u_2-v_2\right|\right),$$E61

where $L_1=L_2=L=\frac{1}{16}$. It can be easily found that ${\mathrm{\Delta }}_1={\Delta }_3\cong 0,799562,{\mathrm{\Delta }}_2={\Delta }_4\cong 1,182808$.

Finally, since $L_1\left({\Delta }_1+{\Delta }_3\right)+L_2\left({\Delta }_2+{\Delta }_4\right)=2L\left({\Delta }_1+{\Delta }_2\right)\cong 0,247796<1$, thus all assumptions and conditions of Theorem 1.2 are satisfied. Hence, Theorem implies that the three-point boundary value problem (58, 59) has a unique solution.

3. Existence result for second problem (2)

We provide another results about the existence of solutions for the problem (2) by using the assumption.

We shall consider the Banach space $B=C\left(\left[0,1\right]\right)$ equipped with usual supermum norm and $B^{+}=C^{+}\left(\left[0,1\right]\right)$. In arriving our results, we present some notation and preliminary lemmas. The first is well known.

Lemma 3.1. Let $y\left(t\right)\in C\left[0,1\right]$. If $u\in C^4\left[0,1\right]$, then the BVP

$$\left\{\begin{array}{c}\begin{array}{c}{u}^{\left(4\right)}\left(t\right)=y\left(t\right),0\le t\le 1,\\ u\left(0\right)=u^{\prime }\left(0\right)=u^{″}\left(1\right)=u^{″}{}^{\prime }\left(1\right)=0,\end{array}\end{array}\right.$$E62

has a unique solution

$$u\left(t\right)=\underset{0}{\overset{1}{\int }}G\left(t,s\right)y\left(s\right)\mathit{ds},$$E63

where

$$G\left(t,s\right)=\left\{\begin{array}{c}\begin{array}{c}\frac{1}{6}{t}^2\left(3s-t\right),0\le t\le s\le 1,\\ \frac{1}{6}{s}^2\left(3t-s\right),0\le s\le t\le 1.\end{array}\end{array}\right.$$E64

Lemma 3.2. For any $\left(t,s\right)\in \left[0,1\right]\times \left[0,1\right]$, we have

$$0\le G\left(t,s\right)\le G\left(1,s\right)=\frac{1}{6}{s}^2\left(3-s\right)=\psi \left(s\right).$$E65

Proof. The derivatives of the function $G$ with respect to $t$ is

$$\frac{\partial }{\partial t}G\left(t,s\right)=\left\{\begin{array}{c}\begin{array}{c}\frac{1}{2}{s}^2-\frac{1}{2}{\left(s-t\right)}^2,0\le t\le s\le 1\\ \frac{1}{2}{s}^2,0\le s\le t\le 1.\end{array}\end{array}\right.$$E66

Since the derivative of the function $G$ with respect to $t$ is nonnegative for all $t\in \left[0,1\right]$, $G$ is nondecreasing function of $t$ that attaints its maximum when $t=1$. Then

$$\underset{0\le t\le 1}{max}G\left(t,s\right)=G\left(1,s\right)=\frac{1}{2}{s}^2-\frac{1}{6}{s}^3.$$E67

Lemma 3.3. Let $0<\theta <1$. Then for $y\left(t\right)\in C^{+}\left[0,1\right]$, the unique solution $u\left(t\right)$ of BVP $\left(14\right)$ is nonnegative and satisfies

$$\underset{t\in \left[\theta ,1\right]}{min}u\left(t\right)\ge \frac{2{\theta }^3}{3}‖u‖.$$E68

Proof. Let $y\left(t\right)\in C^{+}\left[0,1\right]$, then from $G\left(t,s\right)\ge 0$ we know $u\in C^{+}\left[0,1\right]$. Set $u\left(t_0\right)=‖u‖$, $t_0\in \left(0,1\right]$. We first prove that

$$\frac{G\left(t,s\right)}{G\left(t_0,s\right)}\ge \frac{2}{3}{t}^3,t,t_0,s\in \left(0,1\right].$$E69

In fact, we can consider four cases:

1. if $0<t,t_0\le s\le 1$, then

   $$\frac{G\left(t,s\right)}{G\left(t_0,s\right)}=\frac{t^2\left(3s-t\right)}{t_0^2\left(3s-t_0\right)}\ge \frac{t^2\left(2s\right)}{3s-t_0}\ge \frac{t^2\left(2s\right)}{3}\ge \frac{t^2\left(2t\right)}{3}=\frac{2t^3}{3},$$E70

2. if $0<t\le s\le t_0\le 1$, then

   $$\frac{G\left(t,s\right)}{G\left(t_0,s\right)}=\frac{t^2\left(3s-t\right)}{t_0^2\left(3s-t_0\right)}\ge \frac{t^2\left(2s\right)}{3s-t_0}\ge \frac{t^2\left(2s\right)}{3}\ge \frac{t^2\left(2t\right)}{3}=\frac{2t^3}{3},$$E71

3. if $0<s\le t,t_0\le 1$, then

   $$\frac{G\left(t,s\right)}{G\left(t_0,s\right)}=\frac{s^2\left(3t-s\right)}{s^2\left(3t_0-s\right)}=\frac{3t-s}{3t_0-s}\ge \frac{3t-s}{3t_0}\ge \frac{3t-s}{3}\ge \frac{2t+t-s}{3}\ge \frac{2t}{3}\ge \frac{2t^3}{3},$$E72

4. if $0<t_0\le s\le t\le 1$, then

   $$\frac{G\left(t,s\right)}{G\left(t_0,s\right)}=\frac{s^2\left(3t-s\right)}{t_0^2\left(3s-t_0\right)}\ge \frac{t_0^2\left(3t-s\right)}{t_0^2\left(3t-t_0\right)}\ge \frac{3t-s}{3t}\ge \frac{3t-t}{3t}\ge \frac{2t}{3}\ge \frac{2t^3}{3},$$E73

Therefore, for $t\in \left[\theta ,1\right]$, we have

$$u\left(t\right)=\underset{0}{\overset{1}{\int }}G\left(t,s\right)y\left(s\right)\mathit{ds}=\underset{0}{\overset{1}{\int }}\frac{G\left(t,s\right)}{G\left(t_0,s\right)}G\left(t_0,s\right)y\left(s\right)\mathit{ds}\ge \frac{2t^3}{3}u\left(t_0\right)\ge \frac{2{\theta }^3}{3}‖u‖.$$E74

The proof is complete.

If we let

$$K=\left\{x\in B:x\left(t\right)\ge 0\mathit{on}\left[0,1\right],\mathit{\text{and}}\underset{t\in \left[\theta ,1\right]}{min}x\left(t\right)\ge \frac{2{\theta }^3}{3}‖x‖\right\}$$E75

then it is easy to see that $K$ a cone in $B$. We not that a pair $\left(u\left(t\right),v\left(t\right)\right)$ is a solution of BVPs (2) if, and only if

$$u\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)a\left(s\right)f\left(\lambda \underset{0}{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds},t\in \left[0,1\right],$$E76

and

$$v\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds},t\in \left[0,1\right].$$E77

Now, we define an integral operator $T:K\to B$ by

$$\left(\mathit{Tu}\right)\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)a\left(s\right)f\left(\lambda \underset{0}{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds},u\in K.$$E78

We adopt the following assumptions:

$\left(H_1\right)$ $a,b\in C\left(\left(0,1\right),\left[0,\infty \right)\right)$ and each does not vanish identically on any subinterval.

$\left(H_2\right)$ $f,g\in C\left(\left[0,\infty \right),\left[0,\infty \right)\right)$ and each to be singular at $t=0\mathit{\text{or}}t=1$.

$\left(H_3\right)$ All of $f_0=\underset{x\to 0^{+}}{lim}\frac{f\left(x\right)}{x},g_0=\underset{x\to 0^{+}}{lim}\frac{g\left(x\right)}{x},f_{\infty }=\underset{x\to \infty }{lim}\frac{f\left(x\right)}{x},\mathit{\text{and}}{g}_{\infty }=\underset{x\to \infty }{lim}\frac{g\left(x\right)}{x}$ exist as real numbers.

$\left(H_4\right)$ $g\left(0\right)=0$ and $f$ is increasing function.

Lemma 3.4 Let $\lambda$ be positive number and $K$ be the cone defined above.

1. If $u\in B^{+}$ and $v:\left[0,1\right]\to \left[0,\infty \right)$ is defined by (77), then $v\in K$.

2. If $T$ is the integral operator defined by (78), then $T\left(K\right)\subset K$.

3. Assume that $\left(H_1\right),\left(H_2\right)$ hold. Then $T:K\to B$ is completely continuous.

Proof. Let $u\in B^{+}$ and $v$ be defined by (77).

1. By the nonnegativity of $G,b$ and $g$ it follows that $v\left(t\right)\ge 0$, $t\in \left[0,1\right]$. In view of $\left(H_1\right),\left(H_2\right)$, we have

   $$\underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds}\ge \underset{0}{\overset{1}{\int }}\underset{t\in \left[\theta ,1\right]}{min}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds},$$E79

   from which, we take

   $$\underset{t\in \left[\theta ,1\right]}{min}\underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds}\ge \underset{0}{\overset{1}{\int }}\underset{t\in \left[\theta ,1\right]}{min}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds}.$$E80

   Consequently, employing (68) and for $\lambda >0$, we have

   $$\begin{array}{c}\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds}\ge \lambda \underset{0}{\overset{1}{\int }}\underset{t\in \left[\theta ,1\right]}{min}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{0}{\overset{1}{\int }}G\left(t_0,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}v\left(t_0\right),t_0\in \left(0,1\right]\\ \ge \frac{2{\theta }^3}{3}‖v‖.\end{array}$$E81

   Therefore

   $$\underset{t\in \left[\theta ,1\right]}{min}v\left(t\right)\ge \frac{2{\theta }^3}{3}‖v‖.$$E82

   Which give that $v\in K$.

2. Obviously, for $v\in K,T\left(u\right)\in C^{+}\left[0,1\right]$. For $t\in \left[0,1\right]$, we have

   $$\begin{array}{c}‖\mathit{Tu}\left(t\right)‖=\underset{0\le t\le 1}{max}\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)a\left(s\right)f\left(v\left(s\right)\right)\mathit{ds}\\ \le \lambda \underset{0}{\overset{1}{\int }}G\left(1,s\right)a\left(s\right)f\left(v\left(s\right)\right)\mathit{ds},\end{array}$$E83

   and

   $$\begin{array}{c}\mathit{Tu}\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)a\left(s\right)f\left(v\left(s\right)\right)\mathit{ds}\\ =\lambda \underset{0}{\overset{1}{\int }}\frac{G\left(t,s\right)}{G\left(1,s\right)}G\left(1,s\right)a\left(s\right)f\left(v\left(s\right)\right)\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{0}{\overset{1}{\int }}G\left(1,s\right)a\left(s\right)f\left(v\left(s\right)\right)\mathit{ds}\\ \ge \frac{2}{3}{\theta }^3‖\mathit{Tu}\left(t\right)‖.\end{array}$$E84

   Which give that $\mathit{Tu}\in K$. Therefore $T:K\to K$.

3. By using standard arguments it is not difficult to show that the operator $T:K\to B$ is completely continuous.

The key tool in our approach is the following Krasnoselskii’s fixed point theorem of cone expansion-compression type.

Theorem 1.3 (See [47]) Let $B$ be a Banach space and $K\subset B$ be a cone in $B$. Assume ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$ are open subset of $B$ with $0\in {\mathrm{\Omega }}_1$ and $\overline{{\mathrm{\Omega }}_1}\subset {\mathrm{\Omega }}_2$,

$T:K\cap \left({\overline{\mathrm{\Omega }}}_2\backslash {\mathrm{\Omega }}_1\right)\to K$ be a completely continuous operator such that.

1. $‖\mathit{Tu}‖\le ‖u‖$, $\forall u\in K\cap \partial {\mathrm{\Omega }}_1$ and $‖\mathit{Tu}‖\ge ‖u‖$, $\forall u\in K\cap \partial {\mathrm{\Omega }}_2$; or.

2. $‖\mathit{Tu}‖\ge ‖u‖$, $\forall u\in K\cap \partial {\mathrm{\Omega }}_1$ and $‖\mathit{Tu}‖\le ‖u‖$, $\forall u\in K\cap \partial {\mathrm{\Omega }}_2$.

Then, $T$ has a fixed point in $K\cap \left({\overline{\mathrm{\Omega }}}_2\backslash {\mathrm{\Omega }}_1\right)$. Throughout this section, we shall use the following notations:

$$L_1=max\left\{{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)f_{\infty }\mathit{dr}\right]}^{-1},{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)g_{\infty }\mathit{dr}\right]}^{-1}\right\}$$E85

and

$$L_2=min\left\{{\left[\underset{0}{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)f_0\mathit{dr}\right]}^{-1},{\left[\underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g_0\mathit{dr}\right]}^{-1}\right\}.$$E86

$$L_3=max\left\{{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)f_0\mathit{dr}\right]}^{-1},{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)g_0\mathit{dr}\right]}^{-1}\right\}$$E87

and

$$L_4=min\left\{{\left[\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)f_{\infty }\mathit{dr}\right]}^{-1},{\left[\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g_{\infty }\mathit{dr}\right]}^{-1}\right\}.$$E88

4. Existence results

In this section, we discuss the existence of at least one positive solution for BVPs (2). We obtain the following existence results, by applying the positivity of Green’s function $G\left(t,s\right)$ and the fixed-point of cone expansion-compression type.

Theorem 1.4 Assume conditions $\left(H_1\right),\left(H_2\right)$ and $\left(H_3\right)$ are satisfied. Then, for each $\lambda$ satisfying $L_1<\lambda <L_2$ there exists a pair $\left(u,v\right)$ satisfying BVPs (2) such that $u\left(t\right)>0$ and $v\left(t\right)>0$ on $\left(0,1\right)$.

Proof. Let $L_1<\lambda <L_2$. And let $\epsilon >0$ be chosen such that

$$max\left\{{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_{\infty }-\epsilon \right)\mathit{dr}\right]}^{-1},{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(g_{\infty }-\epsilon \right)\mathit{dr}\right]}^{-1}\right\}\le \lambda ,$$E89

and

$$\lambda \le min\left\{{\left[\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_0+\epsilon \right)\mathit{dr}\right]}^{-1},{\left[\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\left(g_0+\epsilon \right)\mathit{dr}\right]}^{-1}\right\}.$$E90

From the definitions of $f_0$ and $g_0$ there exists an $R_1>0$ such that

$$f\left(u\right)\le \left(f_0+\epsilon \right)u,0<u\le R_1,$$E91

and

$$g\left(u\right)\le \left(g_0+\epsilon \right)u,0<u\le R_1,$$E92

Let $u\in K$ with $‖u‖=R_1$. From (65) and choice of $\epsilon$, we have

$$\begin{array}{c}\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(r\right)g\left(u\left(r\right)\right)\le \lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\\ \le \lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\left(g_0+\epsilon \right)u\left(r\right)\mathit{dr}\\ \le ‖u‖\lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\mathit{dr}\left(g_0+\epsilon \right)\mathit{dr}\\ \le R_1=‖u‖.\end{array}$$E93

Consequently, from (65) and choice of $\epsilon$, we have

$$\begin{array}{c}\mathit{Tu}\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)a\left(s\right)f\left(\lambda \underset{0}{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds}\\ \le \lambda \underset{0}{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)f\left(\lambda \underset{0}{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds}\\ \le \lambda \underset{0}{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\left(f_0+\epsilon \right)\left[\lambda \underset{0}{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right]\mathit{ds}\\ \le \lambda \underset{0}{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\left(f_0+\epsilon \right)R_1\mathit{ds}\\ \le R_1=‖u‖.\end{array}$$E94

So, $‖\mathit{Tu}‖\le ‖u‖$. If we set ${\mathrm{\Omega }}_1=\left\{u\in B:‖u‖<R_1\right\}$, then

$$‖\mathit{Tu}‖\le ‖u‖,\mathit{\text{for}}u\in K\cap \partial {\mathrm{\Omega }}_1$$E95

Considering the definitions of $f_{\infty }$ and $g_{\infty }$ there exists an ${\overline{R}}_2>0$ such that

$$f\left(u\right)\ge \left(f_{\infty }-\epsilon \right)u,0<u\le {\overline{R}}_2,$$E96

and

$$g\left(u\right)\ge \left(g_{\infty }-\epsilon \right)u,0<u\le {\overline{R}}_2.$$E97

Let $u\in K$ and $R_2=max\left\{2R_1,\frac{3{\overline{R}}_2}{2{\theta }^3}\right\}$ with $‖u‖=R_2$, then

$$\underset{s\in \left[\theta ,1\right]}{min}u\left(s\right)\ge \frac{2}{3}{\theta }^3‖u‖\ge {\overline{R}}_2$$E98

Thus, from (68) and choice of $\epsilon$, we have

$$\begin{array}{c}\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(r\right)g\left(u\left(r\right)\right)\ge \frac{2{\theta }^3}{3}\lambda \underset{0}{\overset{1}{\int }}G\left(1,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\left(g_{\infty }-\epsilon \right)u\left(r\right)\mathit{dr}\\ \ge ‖u‖{\left(\frac{2{\theta }^3}{3}\right)}^2\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\mathit{dr}\left(g_{\infty }-\epsilon \right)\mathit{dr}\\ \ge R_2=‖u‖.\end{array}$$E99

Consequently, from (77) and choice of $\epsilon$, we have

$$\begin{array}{c}\mathit{Tu}\left(t\right)\ge \frac{2\theta }{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)f\left(\lambda \underset{\theta }{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\left(f_{\infty }-\epsilon \right)\left[\lambda \underset{\theta }{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right]\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\left(f_{\infty }-\epsilon \right)H_2\mathit{ds}\\ \ge {\left(\frac{2{\theta }^3}{3}\right)}^2\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\left(f_{\infty }-\epsilon \right)H_2\mathit{ds}\\ \ge R_2=‖u‖.\end{array}$$E100

So, $‖\mathit{Tu}‖\ge ‖u‖$. If we set ${\mathrm{\Omega }}_2=\left\{u\in B:‖u‖<R_2\right\}$, then

$$‖\mathit{Tu}‖\ge ‖u‖,\mathit{\text{for}}u\in K\cap \partial {\mathrm{\Omega }}_2.$$E101

Applying (i) of Theorem 3.1 to (95) and (101), yields that $T$ has a fixed point $u^{\ast }\in K\cap \left({\overline{\mathrm{\Omega }}}_2/{\mathrm{\Omega }}_1\right)$. As such and with $v$ defined by

$$v\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(s\right)g\left(u\left(s\right)\right)\mathit{ds},$$E102

the pair $\left(u,v\right)$ is a desired solution of BVPs (2) for the given $\lambda$. The proof is complete.

Theorem 1.5 Assume conditions $\left(H_1\right),\left(H_2\right),\left(H_3\right)$ and $\left(H_4\right)$ are satisfied. Then, for each $\lambda$ satisfying $L_3<\lambda <L_4$ there exists a pair $\left(u,v\right)$ satisfying BVPs (2) such that $u\left(t\right)>0$ and $v\left(t\right)>0$ on $\left(0,1\right)$.

Proof. Let $L_3<\lambda <L_4$ and $\epsilon >0$ be chosen such that

$$\mathit{max}\left\{{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_0-\epsilon \right)\mathit{dr}\right]}^{-1},{\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(g_0-\epsilon \right)\mathit{dr}\right]}^{-1}\right\}\le \lambda ,$$E103

and

$$\lambda \le \mathit{min}\left\{{\left[\underset{0}{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_{\infty }+\epsilon \right)\mathit{dr}\right]}^{-1},{\left[\underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\left(g_{\infty }+\epsilon \right)\mathit{dr}\right]}^{-1}\right\}.$$E104

From the definitions of $f_0$ and $g_0$ there exists an $R_1>0$ such that

$$f\left(u\right)\ge \left(f_0-\epsilon \right)u,0<u\le R_1,$$E105

and

$$g\left(u\right)\ge \left(g_0-\epsilon \right)u,0<u\le R_1,$$E106

Now $g\left(0\right)=0$ and so there exists $0<R_2\le R_1$ such that

$$\mathit{\lambda g}\left(u\right)\le \frac{R_1}{\underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\mathit{dr}},0\le u\le R_2.$$E107

Let $u\in K$ with $‖u‖=R_2$. Then

$$\begin{array}{c}\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(r\right)g\left(u\left(r\right)\right)\le \lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\\ \le \frac{\underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)R_1\mathit{dr}}{\underset{0}{\overset{1}{\int }}\psi \left(s\right)b\left(s\right)\mathit{ds}}\\ \le R_1=‖u‖.\end{array}$$E108

Therefore, by (68), we have

$$\begin{array}{l}\mathit{Tu}\left(t\right)=\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)a\left(s\right)f\left(\lambda \underset{0}{\overset{1}{\int }}G\left(s,r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)f\left(\frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\right)\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\left(f_0-\epsilon \right)\left[{\left(\frac{2{\theta }^3}{3}\right)}^2\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\left(g_0-\epsilon \right)‖u‖\mathit{dr}\right]\mathit{ds}\\ \ge \frac{2{\theta }^3}{3}\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_0-\epsilon \right)‖u‖\\ \ge {\left(\frac{2{\theta }^3}{3}\right)}^2\lambda \underset{\theta }{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_0-\epsilon \right)‖u‖\\ \ge ‖u‖.\end{array}$$E109

So, $‖\mathit{Tu}‖\ge ‖u‖$. If we set ${\mathrm{\Omega }}_1=\left\{u\in B:‖u‖<R_2\right\}$, then

$$‖\mathit{Tu}‖\ge ‖u‖,u\in K\cap \left({\overline{\mathrm{\Omega }}}_2\backslash {\mathrm{\Omega }}_1\right)$$E110

Considering the definitions of $f_{\infty }$ and $g_{\infty }$ there exists ${\overline{R}}_1>0$ such that

$$f\left(u\right)\le \left(f_{\infty }+\epsilon \right)u,u\ge {\overline{R}}_1,$$E111

and

$$g\left(u\right)\le \left(g_{\infty }+\epsilon \right)u,u\ge {\overline{R}}_1.$$E112

We consider two cases: $g$ is bounded or $g$ is unbounded.

Case $\left(i\right)$. Suppose $g$ is bounded, say $g\left(u\right)\le N,N>0$ for all $0<u<\infty$. Then, for $u\in K$

$$\begin{array}{c}\lambda \underset{0}{\overset{1}{\int }}G\left(t,s\right)b\left(r\right)g\left(u\left(r\right)\right)\le \lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g\left(u\left(r\right)\right)\mathit{dr}\\ M=max\left\{f\left(u\right):0\le u\le \mathit{N\lambda }\underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)\mathit{dr}\right\}\end{array}$$E113

and let

$$R_3>max\left\{2R_2,\mathit{M\lambda }\underset{0}{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\mathit{ds}\right\}.$$E114

Then, for $u\in K$ with $‖u‖=R_3$, we have

$$\begin{array}{c}\mathit{Tu}\left(t\right)\le \lambda \underset{0}{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)\mathit{Mds}\\ \le R_3=‖u‖.\end{array}$$E115

So that $‖\mathit{Tu}‖\le ‖u‖$. If we set ${\mathrm{\Omega }}_2=\left\{u\in B:‖u‖\le R_3\right\}$, then, for $u\in K\cap \partial {\mathrm{\Omega }}_2$:

$$‖\mathit{Tu}‖\le ‖u‖,u\in K\cap \partial {\mathrm{\Omega }}_2$$E116

Case$\left(\mathit{ii}\right)$. $g$ is unbounded, there exists $R_3>\mathit{max}\left\{2R_2,{\overline{R}}_1\right\}$ such that $g\left(u\right)\le g\left(R_3\right)$, for $0<u\le R_3$.

Similarly, there exists $R_4>\mathit{max}\left\{2R_3,\mathit{M\lambda }{\int }_0^1\psi \left(r\right)b\left(r\right)g\left(R_3\right)\mathit{ds}\right\}$ such that $f\left(u\right)\le f\left(R_4\right)$, for $0<u\le R_4$.

Let $u\in K$ with $‖u‖=R_4$, from $\left(H_4\right)$, we have

$$\begin{array}{c}\mathit{Tu}\left(t\right)\le \lambda \underset{0}{\overset{1}{\int }}\psi \left(s\right)a\left(s\right)f\left(\lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)b\left(r\right)g\left(R_3\right)\mathit{dr}\right)\mathit{ds}\\ \le \lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)f\left(R_4\right)\mathit{ds}\\ \le \lambda \underset{0}{\overset{1}{\int }}\psi \left(r\right)a\left(r\right)\left(f_{\infty }+\epsilon \right)R_4\mathit{ds}\\ \le R_4=‖u‖.\end{array}$$E117

So, $‖\mathit{Tu}‖\le ‖u‖$. If we set ${\mathrm{\Omega }}_2=\left\{x\in C\left[0,1\right]|‖x‖\le R_4\right\}$, then

$$‖\mathit{Tu}‖\le ‖u‖,\mathit{\text{for}}u\in K\cap \partial {\mathrm{\Omega }}_2.$$E118

In either of cases, application of part $\left(\mathit{ii}\right)$ of Theorem 3.1 yields a fixed point $u^{\ast }$ of $T$ belonging to $K\cap \left({\overline{\mathrm{\Omega }}}_2/{\mathrm{\Omega }}_1\right)$, which in turn yields a pair $\left(u,v\right)$ satisfying BVPs (2) for the chosen value of $\lambda$. The proof is complete.

We construct an example to illustrate the applicability of the results presented.

Example 4.1. Consider the two-point boundary value problem

$$\left\{\begin{array}{c}\begin{array}{c}{u}^{\left(4\right)}\left(t\right)=\mathit{\lambda tv}\left(t\right)\left(v\left(t\right)e^{-v\left(t\right)}+\frac{v\left(t\right)+K}{1+\mathit{\eta v}\left(t\right)}\right),0<t<1,\\ v^{\left(4\right)}\left(t\right)=\mathit{\lambda tu}\left(t\right)\left(u\left(t\right)e^{-u\left(t\right)}+\frac{u\left(t\right)+K}{1+\mathit{\eta u}\left(t\right)}\right),0<t<1,\end{array}\end{array}\right.$$E119

and satisfying two-point boundary conditions

$$\left\{\begin{array}{c}\begin{array}{c}u\left(0\right)=0,u^{\prime }\left(0\right)=0,u^{″}\left(1\right)=0,u^{″\prime }\left(1\right)=0,\\ v\left(0\right)=0,v^{\prime }\left(0\right)=0,v^{″}\left(1\right)=0,v^{″\prime }\left(1\right)=0,\end{array}\end{array}\right.$$E120

where $a\left(t\right)=b\left(t\right)=t,f\left(v\right)=v\left({\mathit{ve}}^{-v}+\frac{v+K}{1+\mathit{\eta v}}\right),g\left(u\right)=u\left(1+\frac{u+K}{1+\mathit{\eta u}}\right)$.

By simple calculations, we find $g\left(0\right)=0,f_{\infty }=g_{\infty }=\frac{1}{\eta },f_0=g_0=K$.

Choosing $\theta =\frac{1}{3},\eta =100,\mathit{\text{and}}K={10}^4$, we obtain $L_3\cong 1,1817237,L_4\cong 9,1666667$.

By Theorem 4, it follows that for every $\lambda$ such that $1,1817237<\lambda <9,1666667$, there exists a pair $\left(u,v\right)$ satisfying BVPs (25–2526).

5. Conclusions

This chapter concerns the boundary value problem of a class of fractional differential equations involving the Caputo fractional derivative with nonlocal boundary conditions. By using the Leray-Shauder nonlinear alternative and Banach contraction principle, one shows that the problem has at least one positive solutions and has unique solution. Secondly, we derive explicit eigenvalue intervals of $\lambda$ for the existence of at least one positive solution for the second problem by using Krasnosel’skii fixed point theorem. The results of the present chapter are significantly contribute to the existing literature on the topic.

Acknowledgments

The authors want to thank the anonymous referee for the thorough reading of the manuscript and several suggestions that help us improve the presentation of the chapter.

Conflict of interest

The authors declare no conflict of interest.