Existence and Uniqueness of Solutions for Singular Fractional Integro-Differential Equations with p-Laplacian and Two Kinds of Fractional Derivatives

Abstract

The paper is devoted to the study of a class of singular high-order fractional integro-differential equations with p-Laplacian operator, involving both the Riemann–Liouville fractional derivative and the Caputo fractional derivative. First, we investigate the problem by the method of reducing the order of fractional derivative. Then, by using the Schauder fixed point theorem, the existence of solutions is proved. The upper and lower bounds for the unique solution of the problem are established under various conditions by employing the Banach contraction mapping principle. Furthermore, four numerical examples are presented to illustrate the applications of our main results.

1. Introduction

In recent years, fractional boundary value problems have appeared widely in many fields of science, such as the characterization of the behavior for viscoelastic materials [1], the development of algorithms for signal and image processing [2], and so on. To cite a specific instance, by introducing fractional derivatives to describe the relationship between strain and strain force, in [3], the author introduced fractional models to describe the creep, stress relaxation, and hysteresis of viscoelastic materials by incorporating fractional derivatives into the relationship between strain and stress. The theoretical results obtained are closer to the actual situation, which provides an important theoretical basis for the study of the properties of actual viscoelastic materials. Many scholars have conducted in-depth and extensive research in this field [4,5,6,7,8], and they have obtained significant results for the heat equation [9], the impulsive problem [10], and the system [11].

In addition, as a crucial tool for modeling fluid mechanics, the p-Laplacian equations have been studied by many researchers [12,13,14,15,16]. For example, Turbulence is a state of fluid flow in physics, and the study of turbulent characteristics in porous media has always been a hot topic in the field. Since this process is influenced by many parameters, research on turbulent flow fields in porous media is extremely complex. Therefore, modeling this process has remained a key issue both domestically and internationally. In [12], the author introduced the following differential equation with a p-Laplacian operator to describe the mechanical phenomenon of turbulence in porous media:

$${\left({\varphi }_p\left(u^{\prime }\left(t\right)\right)\right)}^{\prime }=f(t,u\left(t\right),u^{\prime }\left(t\right)).$$

This work represented a significant step forward in the study of p-Laplacian differential equations. Specifically, the introduction of the p-Laplacian operator into fractional differential equations has been a key development, which has increasingly become a hot topic in research.

In the case of problems containing a type of fractional derivative, the existence and uniqueness of solutions have been extensively studied. In particular, the Riemann–Liouville fractional differential equations have been received considerable attention. For example, in [17] the author considered the following problem:

$$\left\{\begin{array}{l}{D}_{0+}^{\beta }{\varphi }_p\left(D_{0+}^{\alpha }u\left(t\right)\right)=f(t,u\left(t\right)),0<t<1,\\ u\left(0\right)=0,D_{0+}^{\gamma }u\left(1\right)={\displaystyle \sum _{i=1}^{m-2}}{\xi }_i{D}_{0+}^{\gamma }u\left({\eta }_i\right),\\ D_{0+}^{\alpha }u\left(0\right)=0,{\varphi }_p\left(D_{0+}^{\alpha }u\right)\left(1\right)={\displaystyle \sum _{i=1}^{m-2}}{\rho }_i{\varphi }_p\left(D_{0+}^{\alpha }u\right)\left({\eta }_i\right),\end{array}\right.$$

where $1<\alpha$, $\beta \le 2$, $3<\alpha +\beta \le 4$, $0<\gamma \le 1$, $\alpha -\gamma >1$, $0<{\xi }_i,{\rho }_i,{\eta }_i\le 1$, ${\sum }_{i=1}^{m-2}{\xi }_i{\eta }_i^{\alpha -\gamma -1}<1$, ${\sum }_{i=1}^{m-2}{\rho }_i{\eta }_i^{\alpha -\gamma -1}<1$, $f:[0,1]\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ is continuous.

By using the Banach contraction mapping principle, the author established the uniqueness result for this problem. Further, by using the fixed point theorem of mixed monotone operators, in [18], the authors obtained the existence of positive solutions to the same problem. Building on this foundation, the study in [19] extended the analysis to a class of fractional differential equations with a generalized Laplacian operator. Following a methodology similar to [19], the work in [20] further incorporated a fractional integral operator into the nonlinear term. In contrast, in [21], the authors introduced a fractional derivative in the nonlinear term, which necessitated an analysis within a function space framework distinct from that employed in the preceding works.

On the other hand, fractional differential equations involving the Caputo derivative have also been investigated. For instance, in [22], the authors investigated a class of integro-differential equations by applying the generalized contraction principle. Subsequently, the studies presented in [23,24] extended the research to the existence of positive solutions. Furthermore, the works in [25,26] made significant strides by establishing both the existence and uniqueness of solutions. While the authors of [27] investigated the generalized system of fractional differential equations, the study in [28] focused on a class of generalized Caputo fractional boundary value problems. It is worth noting that the analyses in [23,24,25] were particularly complex due to the inclusion of derivatives of functions in their nonlinear terms. Moreover, the impulsive problem addressed in [25] contributed to the analytical complexity of the study.

In contrast to the aforementioned studies, in [29], the author studied the following initial value problem:

$$\left\{\begin{array}{l}{}^{c}D_{0+}^{\beta }\left({\varphi }_p{(}^c{D}_{0+}^{\alpha }x\left(t\right))\right)=t^{-\gamma }f(t,x\left(t\right))0<t<1,\\ x\left(0\right)=x_0,{(}^c{D}_{0+}^{\alpha }x)\left(0\right)=x_{\alpha },x_0,x_{\alpha }\in \mathbb{R},\end{array}\right.$$

where $0\le \gamma$$<\alpha$, $\beta \le 1$, $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ is continuous. By means of the Schaefer’s fixed point theorem, the authors established existence results for this problem.

Specifically, the various types of multi-point boundary value problems are an active area of research [25,26,30,31,32]. For instance, in [26], the authors studied the existence and uniqueness of solutions to the following problems:

$$\left\{\begin{array}{l}\left(r{}^{c}D_{0+}^{{\sigma }_1}+(1-r)I_{0+}^{{\sigma }_2}\right)z\left(t\right)=f(t,z\left(t\right))+{}^{c}D_{0+}^{{\sigma }_3}g(t,z\left(t\right)),0\le t\le T,\\ z\left(0\right)=0,\\ r\left({}^{c}D_{0+}^{{\nu }_1}z\right)\left(T\right)+(1-r)\left({}^{c}D_{0+}^{{\nu }_2}z\right)\left(T\right)=a_0,\end{array}\right.$$

where ${\sigma }_1$, ${\sigma }_3\in (1,2]$$({\sigma }_1>{\sigma }_3)$, ${\sigma }_2\in (0,1]$, $0<{\nu }_1$, ${\nu }_2<{\sigma }_1-{\sigma }_3$, $r\in [0,1]$, $f,g:[0,T]\times \mathbb{R}\to \mathbb{R}$ are continuous. The authors provided examples and presented numerical data in tables and figures to demonstrate the theoretical results.

It is worth mentioning that in [33], the authors investigated fixed point theorems in a class of generalized metric spaces and applied them to a type of nonlinear fractional differential equation problem, which represented a highly novel field of research. Recently, in [34], the authors investigated the following problem with generalized Caputo fractional derivative:

$$\left\{\begin{array}{l}{}_c{}^{\varrho }D_{0+}^{\sigma }x\left(t\right)=-f(t,x\left(t\right)),\theta \le t\le \vartheta ,\\ x\left(\theta \right)={\lambda }_1,x\left(\vartheta \right)={\lambda }_2.\end{array}\right.$$

where $1<\sigma \le 2$, $f:[\theta ,\vartheta ]\times \mathbb{R}\to \mathbb{R}$ is continuous. The authors proved the uniqueness of the solution by applying the Banach contraction mapping principle under the condition that the nonlinear term satisfies the Lipschitz condition. It is worth noting that the authors verified the theoretical results through an approximate approach to the parameter.

Furthermore, the problems, which contain two types of fractional derivatives, have been studied recently. For instance, building upon the work of [35], in [36], the authors investigated the following problem:

$$\left\{\begin{array}{l}{D}_{0+}^{\alpha }{\varphi }_p\left({}^{c}D_{0+}^{\beta }u\left(t\right)\right)=f(t,u\left(t\right),{}^{c}D_{0+}^{\beta }u\left(t\right)),0<t<1,\\ \left({}^{c}D_{0+}^{\beta }u\right)\left(0\right)=u^{\prime }\left(0\right)=0,\\ u\left(1\right)=r_{1}u\left(\eta \right),\left({}^{c}D_{0+}^{\beta }u\right)\left(1\right)=r_2\left({}^{c}D_{0+}^{\beta }u\right)\left(\xi \right),\end{array}\right.$$

where $\alpha$, $\beta \in (1,2]$, $r_1$, $r_2>2$, $f:[0,1]\times [0,+\infty )\times (-\infty ,0]\to [0,+\infty )$ is continuous. By the method of lower and upper solutions, the authors obtained the existence of positive solutions. In [37], the authors applied the same method to a class of integral boundary value problems. Building upon [36], the work in [38] modified the boundary conditions and successfully established the existence of at least three positive solutions using the Leggett–Williams fixed point theorem. Subsequently, in [39], the authors extended the research scope significantly by generalizing the mixed derivative problem to a system of fractional differential equations. Further, in [40], the authors combined the quantum fractional derivative with the Riemann–Liouville derivative, applying the same method as in [38] to investigate multiple positive solutions.

Inspired by the above-mentioned work, we study the following fractional boundary value problem involving the p-Laplacian operator:

$$\left\{\begin{array}{l}-{}^{c}D_{0+}^{\delta }{\varphi }_p\left(D_{0+}^{\alpha }u\left(t\right)\right)+f(t,u\left(t\right),\left(Hu\right)\left(t\right),D_{0+}^{\beta }u\left(t\right))=0,0<t<1,\\ u\left(0\right)=D_{0+}^{{\gamma }_1}u\left(0\right)=\cdots =D_{0+}^{{\gamma }_{n-2}}u\left(0\right)=0,\\ D_{0+}^{\alpha }u\left(0\right)=0,{\left({\varphi }_p\left(D_{0+}^{\alpha }u\right)\right)}^{″}\left(0\right)=0,\\ {\varphi }_p\left(D_{0+}^{\alpha }u\left(1\right)\right)=k_0{\int }_0^1{b}_0\left(s\right){\varphi }_p\left(D_{0+}^{\alpha }u\left(s\right)\right)ds,\\ D_{0+}^{z_0}u\left(1\right)=k_1{\int }_0^1{b}_1\left(s\right)D_{0+}^{z_1}u\left(s\right)d{A}_1\left(s\right)+k_2{\int }_0^{\zeta }{b}_2\left(s\right)D_{0+}^{z_2}u\left(s\right)d{A}_2\left(s\right)\\ +k_3{\displaystyle \sum _{i=1}^{\infty }}{a}_i{D}_{0+}^{z_3}u\left({\xi }_i\right),\end{array}\right.$$

(1)

where $2<\delta \le 3$, $n-1<\alpha \le n$$(n\ge 3)$, $n-3<\beta \le n-2$, $1<\alpha -\beta \le 2$, $j-1<{\gamma }_j\le j$$(j=1,2,\dots ,n-2)$, $\beta -{\gamma }_j\le n-2-j$$(j=1,2,\dots ,n-2)$, $\beta \le z_l\le z_0\le \alpha -1$$(l=1,2,3)$; $k_l\ge 0$$(l=0,1,2,3)$, $0<\zeta \le 1$, $a_i\ge 0$, $0<{\xi }_i<1$$(i=1,2,\dots )$; $b_{\mathsf{ı}}\in L^1(0,1)$, $b_{\mathsf{ı}}:(0,1)\to {\mathbb{R}}_{+}$ $(\mathsf{ı}=0,1,2)$ is continuous, $\left(Hu\right)\left(t\right)={\int }_0^{t}h(t,s)u\left(s\right)ds$ with the bounded function $h\in L^1(0,1)$, $h:(0,1)\to {\mathbb{R}}_{+}$ is continuous, and ${\int }_0^1·d{A}_i\left(s\right)$ represent the standard Riemann–Stieltjes integral, in which $A_i:[0,1]\to \mathbb{R}$$(i=1,2)$ is function of bounded variation and $f:(0,1)\times {\mathbb{R}}^3\to \mathbb{R}$ is continuous.

Remark 1.

In this paper, we always use the following notation: ${}^{c}D_{0+}^{\alpha }$ denotes the Caputo fractional derivative of order α, $D_{0+}^{\beta }$ denotes the Riemann–Liouville fractional derivative of order β, ${\varphi }_p$ denotes the p-Laplacian operator as ${\varphi }_p\left(s\right)={\left|s\right|}^{p-2}s$ with $p>1$, and the inverse function of ${\varphi }_p$ is ${\varphi }_q\left(x\right)={\left|x\right|}^{q-2}x$ with $1/p+1/q=1$.

Let $x\left(t\right)=D_{0^{+}}^{\beta }u\left(t\right)$, $t\in [0,1]$. Then, the problem (1) changes to the following problem:

$$\left\{\begin{array}{l}-{}^{c}D_{0+}^{\delta }{\varphi }_p(D_{0+}^{\alpha -\beta }x(t))+f(t,I_{0+}^{\beta }x(t),(\mathcal{H}x)(t),x(t))=0,0<t<1,\\ D_{0+}^{{\gamma }_{n-2}-\beta }x(0)=0,\\ D_{0+}^{\alpha -\beta }x(0)=0,{({\varphi }_p(D_{0+}^{\alpha -\beta }x))}^{″}(0)=0,\\ {\varphi }_p(D_{0+}^{\alpha -\beta }x(1))=k_0{\int }_0^1{b}_0(s){\varphi }_p(D_{0+}^{\alpha -\beta }x(s))ds,\\ D_{0+}^{z_0-\beta }x(1)=k_1{\int }_0^1{b}_1(s)D_{0+}^{z_1-\beta }x(s)d{A}_1(s)\\ +k_2{\int }_0^{\zeta }{b}_2(s)D_{0+}^{z_2-\beta }x(s)d{A}_2(s)\\ +k_3{\displaystyle \sum _{i=1}^{\infty }}{a}_i{D}_{0+}^{z_3-\beta }x({\xi }_i),\end{array}\right.$$

(2)

where $1<\alpha -\beta \le 2$, $\left(\mathcal{H}x\right)\left(t\right)={\int }_0^{t}h(t,s)I_{0+}^{\beta }x\left(s\right)ds$.

The innovation of our paper includes the following four points. First of all, as far as we know, there has been no research focusing on the problem (1) with more complex boundary conditions. The key characteristic of the problem (1) is that it involves two kinds of fractional derivatives, and the nonlinear term contains both integral operator and fractional derivatives. In the process of reducing the order of problem (1) in Lemma 5, since two types of fractional derivatives need to be considered, and the boundary conditions simultaneously involve integral and infinite-point conditions, the difficulty of the order reduction analysis is increased. On the other hand, the complexity of the boundary conditions leads to a rather cumbersome form of the Green’s function, which also complicates the proof process of Lemma 9 and Lemma 10. It is necessary to carefully consider the relationship of the relevant parameters to ensure the acquisition of good properties of the solution.

Secondly, by the method of reduction of order, we transform the problem (1) into a new equivalent lower-order problem (2), in which the nonlinear term has no fractional derivative. Accordingly, we denote the Banach space $C[0,1]$ with the norm

$$\parallel x\parallel =\underset{0\le t\le 1}{sup}\left|x\left(t\right)\right|.$$

As a result, we discuss the problem (2) in the space $C[0,1]$, whereas the problem (1) requires consideration in a complex functional space endowed with fractional derivatives. For details of the method of reduction of order, we refer the readers to [41,42,43,44,45] and so on. If the reduced-order method is not employed, all arguments must be developed within the more complex space $Y=\left\{x\right|x\in C[0,1],D_{0^{+}}^{\beta }x\in C[0,1]\}$ with the norm

$$\parallel x\parallel =\underset{0\le t\le 1}{sup}\left|x\left(t\right)\right|+\underset{0\le t\le 1}{sup}\left|D_{0^{+}}^{\beta }x\left(t\right)\right|.$$

This necessitates the prior establishment of several key properties of the operator $\left(D_{0^{+}}^{\beta }T\right)$, whether by applying the Schauder fixed point theorem in Theorem 1 or the Banach contraction mapping principle in Theorems 2–5. Consequently, the process entails considerably heavier computations and greater theoretical difficulty.

Thirdly, as far as we know, many studies have established uniqueness results for p-Laplacian boundary value problems by using the Banach contraction mapping principle. In this case, the nonlinear term is required to satisfy the Lipschitz condition, cf. [17,35,46]. In fact, a function $g:[0,1]\times I\to \mathbb{R}$ is said to satisfy the Lipschitz condition on the interval I if there exists a constant $L>0$ such that

$$|g(t,x_1)-g(t,x_2)|\le L|x_1-x_2|,\forall x_1,x_2\in I,t\in [0,1].$$

This condition implies that g is uniformly continuous on I. This is a very strong condition, so its applications are limited in many studies. However, we obtain the uniqueness results for the problem (1) under more general conditions, which allow the nonlinear term to be singular at time variables $t=0$ and $t=1$. In the proofs of Theorems 2–5, we relax the classical Lipschitz condition by allowing the constant to be a function, thereby accommodating certain singularities in the nonlinear term. Using the integrability of the Lipschitz function, we have proved the conclusion.

Finally, it is worth pointing out that the estimation of solutions for the problem (1) has not been considered before. Especially, the validity of the results is proved by numerical examples.

Due to the complexity of the problem (1), the examples constructed in Section 5 must incorporate numerous parameters, resulting in a sophisticated case study. We hope that these examples can offer practical guidance for real-world problems.

The rest of the paper is organized as follows. In Section 2, we present some relevant definitions and lemmas, transform the problem (1) into an equivalent low-order problem (2), and obtain some significant properties of the relevant Green’s function. In Section 3, we give the main proof for the existence results of the problem (1). In Section 4, we establish the uniqueness result of the problem (1). In Section 5, we give four numerical examples to illustrate our theoretical results.

2. Preliminaries and Lemmas

Definition 1.

The Riemann–Liouville fractional integral of order $\upsilon >0$ of a function $u:(0,\infty )\to \mathbb{R}$ is given by

$$I_{0^{+}}^{\upsilon }u\left(t\right)={\displaystyle \frac{1}{\Gamma \left(\upsilon \right)}}{\int }_0^t{(t-\eta )}^{\upsilon -1}u\left(\eta \right)d\eta ,$$

where the right-hand side exists a.e. on $(0,\infty )$.

Definition 2.

The Riemann–Liouville fractional derivative of order $\upsilon >0$ of a continuous function $u:(0,\infty )\to \mathbb{R}$ is given by

$$D_{0^{+}}^{\upsilon }u\left(t\right)={\left({\displaystyle \frac{d}{dt}}\right)}^n{I}_{0^{+}}^{n-\upsilon }u\left(t\right),$$

where the right-hand side exists a.e. on $(0,\infty )$, where $n-1<\upsilon <n$ (n is a positive integer). In particular, if $\upsilon =n$, then $D_{0^{+}}^{\upsilon }u\left(t\right)=u^{\left(n\right)}\left(t\right)$.

Definition 3.

The Caputo fractional derivative of order $\upsilon >0$ of a continuous function $u:(0,\infty )\to \mathbb{R}$ is given by

$${}^{c}D_{0+}^{\upsilon }u\left(t\right)=I_{0^{+}}^{n-\upsilon }{u}^{\left(n\right)}\left(t\right),$$

with the right-hand side pointwise defined on $(0,\infty )$, $n-1<\upsilon <n$ (n is a positive integer). In particular, if $\upsilon =n$, then ${}^{c}D_{0+}^{\upsilon }u\left(t\right)=u^{\left(n\right)}\left(t\right)$.

Lemma 1

([47]). Let $u\in C(0,1)\cap L^1(0,1)$. Then, the equation

$$D_{0^{+}}^{\upsilon }u\left(t\right)=0$$

has

$$u\left(t\right)=\sum _{i=1}^n{b}_i{t}^{\upsilon -i},$$

as the unique solution, where $n-1<\upsilon \le n$ (n is a positive integer), $b_1,\cdots ,b_n\in \mathbb{R}$.

Lemma 2

([47]). Let $u,D_{0^{+}}^{\upsilon }u\in C(0,1)\cap L^1(0,1)$. Then,

$$I_{0^{+}}^{\upsilon }{D}_{0^{+}}^{\upsilon }u\left(t\right)=u\left(t\right)+\sum _{i=1}^n{b}_i{t}^{\upsilon -i},$$

where $n-1<\upsilon \le n$ (n is a positive integer), $b_1,\cdots ,b_n\in \mathbb{R}$.

Lemma 3

([4]). Let $u\in A{C}^n[0,1]$. Then,

$$I_{0^{+}}^{\upsilon }{}^{c}D_{0+}^{\upsilon }u\left(t\right)=u\left(t\right)+\sum _{i=0}^{n-1}{b}_i{t}^i,$$

where $n-1<\upsilon \le n$ (n is a positive integer), $b_0,\cdots ,b_{n-1}\in \mathbb{R}$.

Lemma 4

([4]). If $u\in L^1(0,1)$, ${\upsilon }_1\ge {\upsilon }_2>0$, then

$$D_{0^{+}}^{{\upsilon }_1}{I}_{0^{+}}^{{\upsilon }_1}u\left(t\right)=u\left(t\right),D_{0^{+}}^{{\upsilon }_2}{I}_{0^{+}}^{{\upsilon }_1}u\left(t\right)=I_{0^{+}}^{{\upsilon }_1-{\upsilon }_2}u\left(t\right).$$

Lemma 5.

Suppose that $u\in C[0,1]$ is a solution of the problem (1), then $x=D_{0^{+}}^{\beta }u$ is a solution of the problem (2). Conversely, if $x\in C[0,1]$ is a solution of the problem (2), then $u=I_{0^{+}}^{\beta }x$ is a solution of the problem (1).

Proof. 

Let $x=D_{0^{+}}^{\beta }u$, where $u\in C[0,1]$ is a solution of the problem (1). By Lemma 2, we have

$$u\left(t\right)=I_{0^{+}}^{\beta }x\left(t\right)-c_1{t}^{\beta -1}-\dots -c_{n-2}{t}^{\beta -(n-2)},c_i\in \mathbb{R},i=1,2,\cdots ,n-2.$$

Since $u\left(0\right)=D_{0+}^{{\gamma }_1}u\left(0\right)=\cdots =D_{0+}^{{\gamma }_{n-3}}u\left(0\right)=0$, we obtain $c_{n-2}=\cdots =c_1=0$. It follows that

$$u\left(t\right)=I_{0^{+}}^{\beta }x\left(t\right),t\in [0,1].$$

(3)

Consequently, it follows that

$$\left(Hu\right)\left(t\right)={\int }_0^{t}h(t,s)u\left(s\right)ds={\int }_0^{t}h(t,s)I_{0^{+}}^{\beta }x\left(s\right)ds=\left(\mathcal{H}x\right)\left(t\right),$$

(4)

$$\begin{array}{cc}\hfill D_{0^{+}}^{\alpha }u\left(t\right)& ={\displaystyle \frac{d^n}{d{t}^n}}{I}_{0^{+}}^{n-\alpha }u\left(t\right)={\displaystyle \frac{d^n}{d{t}^n}}{I}_{0^{+}}^{n-\alpha }\left(I_{0^{+}}^{\beta }x\left(t\right)\right)={\displaystyle \frac{d^n}{d{t}^n}}{I}_{0^{+}}^{n-(\alpha -\beta )}x\left(t\right)\hfill \\ \hfill & ={\displaystyle \frac{d^2}{d{t}^2}}\left({\displaystyle \frac{d^{n-2}}{d{t}^{n-2}}}{I}_{0^{+}}^{n-2}\right)I_{0^{+}}^{2-(\alpha -\beta )}x\left(t\right)\hfill \\ \hfill & ={\displaystyle \frac{d^2}{d{t}^2}}{I}_{0^{+}}^{2-(\alpha -\beta )}x\left(t\right)\hfill \\ \hfill & =D_{0^{+}}^{\alpha -\beta }x\left(t\right)\hfill \end{array}$$

(5)

and

$$D_{0+}^{{\gamma }_{n-2}}u\left(t\right)=D_{0+}^{{\gamma }_{n-2}-\beta }x\left(t\right),D_{0^{+}}^{z_i}u\left(t\right)=D_{0^{+}}^{z_i-\beta }x\left(t\right),i=0,1,2,3.$$

(6)

From (3), (4) and (5), we have

$$\begin{array}{cc}\hfill & {}^{c}D_{0+}^{\delta }{\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(t\right)\right)-f(t,I_{0+}^{\beta }x\left(t\right),\left(\mathcal{H}x\right)\left(t\right),x\left(t\right))\hfill \\ \hfill & ={}^{c}D_{0+}^{\delta }{\varphi }_p\left(D_{0+}^{\alpha }u\left(t\right)\right)-f(t,u\left(t\right),\left(Hx\right)\left(t\right),D_{0+}^{\beta }u\left(t\right))=0,\hfill \end{array}$$

(7)

$$\begin{array}{c}\hfill D_{0+}^{\alpha -\beta }x\left(0\right)=D_{0+}^{\alpha }u\left(0\right)=0,{\left({\varphi }_p\left(D_{0+}^{\alpha -\beta }x\right)\right)}^{″}\left(0\right)={\left({\varphi }_p\left(D_{0+}^{\alpha }u\right)\right)}^{″}\left(0\right)=0\end{array}$$

(8)

and

$$\begin{array}{cc}\hfill {\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(1\right)\right)& ={\varphi }_p\left(D_{0+}^{\alpha }u\left(1\right)\right)\hfill \\ \hfill & =k_0{\int }_0^1{b}_0\left(s\right){\varphi }_p\left(D_{0+}^{\alpha }u\left(s\right)\right)ds,\hfill \\ \hfill & =k_0{\int }_0^1{b}_0\left(s\right){\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(s\right)\right)ds.\hfill \end{array}$$

(9)

By using (6), we obtain

$$D_{0+}^{{\gamma }_{n-2}-\beta }x\left(0\right)=D_{0+}^{{\gamma }_{n-2}}u\left(0\right)=0$$

(10)

and

$$\begin{array}{cc}\hfill D_{0+}^{z_0-\beta }x\left(1\right)=& k_1{\int }_0^1{b}_1\left(s\right)D_{0+}^{z_1-\beta }x\left(s\right)d{A}_1\left(s\right)+k_2{\int }_0^{\eta }{b}_2\left(s\right)D_{0+}^{z_2-\beta }x\left(s\right)d{A}_2\left(s\right)\hfill \\ \hfill & +k_3{\displaystyle \sum _{i=1}^{\infty }}{a}_i{D}_{0+}^{z_3-\beta }x\left({\xi }_i\right).\hfill \end{array}$$

(11)

Combining (7)–(11), we conclude that $x=D_{0^{+}}^{\beta }u$ is a solution of the problem (2).

Conversely, let $x\in C[0,1]$ be a solution of the problem (2). Then, $u=I_{0^{+}}^{\beta }x$ is a solution of the problem (1). Since the proof is similar to that of Lemma 3 in [44], we omit it here. □

Remark 2.

According to Lemma 5, we deduce that, under the assumption that $1<\alpha -\beta \le 2$, the problem (1) is equivalent to the problem (2). Consequently, we focus on establishing the results for the problem (2) in the following.

Lemma 6.

Let $W\in C(0,1)\cap L^1(0,1)$. Then, the problem

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha -\beta }x\left(t\right)+W\left(t\right)=0,0<t<1,1<\alpha -\beta \le 2,\hfill \\ D_{0+}^{{\gamma }_{n-2}-\beta }x\left(0\right)=0,\hfill \\ D_{0+}^{z_0-\beta }x\left(1\right)=k_1{\int }_0^1{b}_1\left(s\right)D_{0+}^{z_1-\beta }x\left(s\right)d{A}_1\left(s\right)\hfill \\ +k_2{\int }_0^{\zeta }{b}_2\left(s\right)D_{0+}^{z_2-\beta }x\left(s\right)d{A}_2\left(s\right)\hfill \\ +k_3{\displaystyle \sum _{i=1}^{\infty }}{a}_i{D}_{0+}^{z_3-\beta }x\left({\xi }_i\right),\hfill \end{array}\right.$$

(12)

has a unique solution

$$x\left(t\right)={\int }_0^{1}G(t,s)W\left(s\right)ds,$$

where

$$\begin{array}{cc}\hfill G(t,s)=& P_0(t,s)+t^{\alpha -\beta -1}\left({\int }_0^1{P}_1(\tau ,s)b_1\left(\tau \right)d{A}_1\left(\tau \right)\right)\hfill \\ \hfill & +t^{\alpha -\beta -1}\left({\int }_0^{\zeta }{P}_2(\tau ,s)b_2\left(\tau \right)d{A}_2\left(\tau \right)\right)\hfill \\ \hfill & +t^{\alpha -\beta -1}{\displaystyle \sum _{i=1}^{\infty }}{a}_i{P}_3({\xi }_i,s),\hfill \end{array}$$

(13)

in which

$$P_0(t,s)={\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}\left\{\begin{array}{c}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}-{(t-s)}^{\alpha -\beta -1},s\le t,\hfill \\ t^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1},t\le s,\hfill \end{array}\right.$$

(14)

$$P_i(t,s)={\displaystyle \frac{k_i}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_i)}}\left\{\begin{array}{c}{t}^{\alpha -z_i-1}{(1-s)}^{\alpha -z_0-1}-{(t-s)}^{\alpha -z_i-1},s\le t,\hfill \\ t^{\alpha -z_i-1}{(1-s)}^{\alpha -z_0-1},t\le s,\hfill \end{array}\right.$$

(15)

(i = 1, 2, 3)

$${\varrho }_1={\displaystyle \frac{k_1}{\Gamma (\alpha -z_1)}}{\int }_0^1{s}^{\alpha -z_1-1}{b}_1\left(s\right)d{A}_1\left(s\right),$$

(16)

$${\varrho }_2={\displaystyle \frac{k_2}{\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }{s}^{\alpha -z_2-1}{b}_2\left(s\right)d{A}_2\left(s\right),$$

(17)

$${\varrho }_3={\displaystyle \frac{k_3}{\Gamma (\alpha -z_3)}}\sum _{i=1}^{\infty }{a}_i{\xi }_i^{\alpha -z_3-1},$$

(18)

$${\sigma }_1={\displaystyle \frac{1}{\Gamma (\alpha -z_0)}}-{\varrho }_1-{\varrho }_2-{\varrho }_3\ne 0.$$

(19)

Obviously, $G:[0,1]\times [0,1]\to \mathbb{R}$ is continuous.

Proof. 

By Lemma 2, Equation (12) can be written as

$$x\left(t\right)=c_1{t}^{\alpha -\beta -1}+c_2{t}^{\alpha -\beta -2}-I_{0+}^{\alpha -\beta }W\left(t\right),c_1,c_2\in \mathbb{R}.$$

Since $D_{0+}^{{\gamma }_{n-2}-\beta }x\left(0\right)=0$, we obtain $c_2=0$. Thus,

$$x\left(t\right)=c_1{t}^{\alpha -\beta -1}-I_{0+}^{\alpha -\beta }W\left(t\right).$$

(20)

Applying $D_{0+}^{z_i-\beta }$$(i=0,1,2,3)$ on both sides of (20), we have

$$D_{0+}^{z_i-\beta }x\left(t\right)=c_1{\displaystyle \frac{\Gamma (\alpha -\beta )}{\Gamma (\alpha -z_i)}}{t}^{\alpha -z_i-1}-I_{0+}^{\alpha -z_i}W\left(t\right),i=0,1,2,3.$$

(21)

Substituting (21) into the second boundary condition of (12), we derive

$$\begin{array}{cc}\hfill c_1=& \left\{{\displaystyle \frac{1}{\Gamma (\alpha -z_0)}}{\int }_0^1{(1-s)}^{\alpha -z_0-1}W\left(s\right)ds\right.\hfill \\ \hfill & -{\displaystyle \frac{k_1}{\Gamma (\alpha -z_1)}}{\int }_0^1\left({\int }_0^s{(s-\tau )}^{\alpha -z_1-1}W\left(\tau \right)d\tau \right)b_1\left(s\right)d{A}_1\left(s\right)\hfill \\ \hfill & -{\displaystyle \frac{k_2}{\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }\left({\int }_0^s{(s-\tau )}^{\alpha -z_2-1}W\left(\tau \right)d\tau \right)b_2\left(s\right)d{A}_2\left(s\right)\hfill \\ \hfill & -\left.{\displaystyle \frac{k_3}{\Gamma (\alpha -z_3)}}{\displaystyle \sum _{i=1}^{\infty }}{a}_i{\int }_0^{{\xi }_i}{({\xi }_i-s)}^{\alpha -z_3-1}W\left(s\right)ds\right\}{\displaystyle \frac{1}{{\sigma }_1\Gamma (\alpha -\beta )}},\hfill \end{array}$$

(22)

Substituting $c_1$ into (20) and simplifying, we deduce that

$$\begin{array}{cc}\hfill x\left(t\right)=& -{\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}{\int }_0^t{(t-s)}^{\alpha -\beta -1}W\left(s\right)ds\hfill \\ \hfill & +\left\{{\displaystyle \frac{1}{\Gamma (\alpha -z_0)}}{\int }_0^1{(1-s)}^{\alpha -z_0-1}W\left(s\right)ds\right.\hfill \\ \hfill & -{\displaystyle \frac{k_1}{\Gamma (\alpha -z_1)}}{\int }_0^1\left({\int }_0^s{(s-\tau )}^{\alpha -z_1-1}W\left(\tau \right)d\tau \right)b_1\left(s\right)d{A}_1\left(s\right)\hfill \\ \hfill & -{\displaystyle \frac{k_2}{\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }\left({\int }_0^s{(s-\tau )}^{\alpha -z_2-1}W\left(\tau \right)d\tau \right)b_2\left(s\right)d{A}_2\left(s\right)\hfill \\ \hfill & -\left.{\displaystyle \frac{k_3}{\Gamma (\alpha -z_3)}}{\displaystyle \sum _{i=1}^{\infty }}{a}_i{\int }_0^{{\xi }_i}{({\xi }_i-s)}^{\alpha -z_3-1}W\left(s\right)ds\right\}{\displaystyle \frac{t^{\alpha -\beta -1}}{{\sigma }_1\Gamma (\alpha -\beta )}}\hfill \\ \hfill =& {\int }_0^1{P}_0(t,s)W\left(s\right)ds\hfill \\ \hfill & +{\int }_0^1{t}^{\alpha -\beta -1}\left({\int }_0^1{P}_1(\tau ,s)b_1\left(\tau \right)d{A}_1\left(\tau \right)\right)W\left(s\right)ds\hfill \\ \hfill & +{\int }_0^1{t}^{\alpha -\beta -1}\left({\int }_0^{\zeta }{P}_2(\tau ,s)b_2\left(\tau \right)d{A}_2\left(\tau \right)\right)W\left(s\right)ds\hfill \\ \hfill & +{\int }_0^1{t}^{\alpha -\beta -1}{\displaystyle \sum _{j=1}^{\infty }}{a}_i{P}_3({\xi }_i,s)W\left(s\right)ds\hfill \\ \hfill =& {\int }_0^{1}G(t,s)W\left(s\right)ds.\hfill \end{array}$$

(23)

The proof is complete. □

Lemma 7.

Let $W\in C(0,1)\cap L^1(0,1)$. Then, the following problem

$$\left\{\begin{array}{c}{}^{c}D_{0+}^{\delta }x\left(\theta \right)+W\left(\theta \right)=0,0<\theta <1,\hfill \\ {\varphi }_q\left(x\left(0\right)\right)=x^{″}\left(0\right)=0,\hfill \\ x\left(1\right)=k_0{\int }_0^1{b}_0\left(s\right)x\left(s\right)ds,\hfill \end{array}\right.$$

(24)

has a unique solution

$$x\left(\theta \right)={\int }_0^{1}V(\theta ,s)W\left(s\right)ds,$$

where ${\varphi }_q={\varphi }_p^{-1}$,

$$V(\theta ,s)=V_1(\theta ,s)+V_2(\theta ,s),$$

(25)

in which

$$V_1(\theta ,s)={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}\left\{\begin{array}{c}\theta {(1-s)}^{\delta -1}-{(\theta -s)}^{\delta -1},s\le \theta ,\hfill \\ \theta {(1-s)}^{\delta -1},\theta \le s,\hfill \end{array}\right.$$

(26)

$$V_2(\theta ,s)={\displaystyle \frac{k_0\theta }{{\sigma }_2}}{\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau ,$$

(27)

$${\sigma }_2=1-k_0{\int }_0^1\theta b_0\left(\theta \right)d\theta \ne 0.$$

(28)

Proof. 

Obviously, the problem (24) can be written as

$$x\left(\theta \right)=b_0+b_1\theta +b_2{\theta }^2-I_{0+}^{\delta }W\left(\theta \right).$$

The boundary condition ${\varphi }_q\left(x\left(0\right)\right)=0$ implies that $x\left(0\right)=0$. Substituting this result and the condition $x^{″}\left(0\right)=0$ into the above equation yields that $b_0=b_2=0$. Thus,

$$x\left(\theta \right)=b_1\theta -I_{0+}^{\delta }W\left(\theta \right).$$

(29)

Note that the boundary condition $x\left(1\right)=k_0{\int }_0^1{b}_0\left(s\right)x\left(s\right)ds$. We infer that

$$b_1={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}{\int }_0^1{(1-s)}^{\delta -1}W\left(s\right)ds+k_0{\int }_0^1{b}_0\left(s\right)x\left(s\right)ds.$$

(30)

It follows that

$$\begin{array}{cc}\hfill x\left(\theta \right)=& {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}{\int }_0^1\theta {(1-s)}^{\delta -1}W\left(s\right)ds+k_0{\int }_0^1\theta b_0\left(s\right)x\left(s\right)ds\hfill \\ \hfill & -{\displaystyle \frac{1}{\Gamma \left(\delta \right)}}{\int }_0^{\theta }{(\theta -s)}^{\delta -1}W\left(s\right)ds\hfill \\ \hfill =& {\int }_0^1{V}_1(\theta ,s)W\left(s\right)ds+k_0\theta {\int }_0^1{b}_0\left(s\right)x\left(s\right)ds.\hfill \end{array}$$

(31)

Multiplying Equation (31) on both sides by $b_0\left(\theta \right)$ and integrating on $(0,1)$, we have

$$\begin{array}{cc}\hfill {\int }_0^1{b}_0\left(\theta \right)x\left(\theta \right)d\theta =& {\int }_0^1{b}_0\left(\theta \right)\left({\int }_0^1{V}_1(\theta ,s)W\left(s\right)ds\right)d\theta \hfill \\ \hfill & +k_0{\int }_0^1\theta b_0\left(\theta \right)\left({\int }_0^1{b}_0\left(s\right)x\left(s\right)ds\right)d\theta \hfill \\ \hfill =& {\int }_0^1\left({\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau \right)W\left(s\right)ds\hfill \\ \hfill & +\left(k_0{\int }_0^1\theta b_0\left(\theta \right)d\theta \right)\left({\int }_0^1{b}_0\left(s\right)x\left(s\right)ds\right).\hfill \end{array}$$

(32)

Then, we obtain

$${\int }_0^1{b}_0\left(\theta \right)x\left(\theta \right)d\theta ={\displaystyle \frac{1}{{\sigma }_2}}{\int }_0^1\left({\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau \right)W\left(s\right)ds.$$

(33)

Substituting (33) into (31), we have

$$\begin{array}{cc}\hfill x\left(\theta \right)& ={\int }_0^1{V}_1(\theta ,s)W\left(s\right)ds+{\displaystyle \frac{k_0\theta }{{\sigma }_2}}{\int }_0^1\left({\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau \right)W\left(s\right)ds\hfill \\ \hfill & ={\int }_0^1{V}_1(\theta ,s)W\left(s\right)ds+{\int }_0^1{V}_2(\theta ,s)W\left(s\right)ds\hfill \\ \hfill & ={\int }_0^{1}V(\theta ,s)W\left(s\right)ds.\hfill \end{array}$$

(34)

□

Lemma 8.

Let $W\in C(0,1)\cap L^1(0,1)$. Then, the following problem

$$\left\{\begin{array}{c}-{}^{c}D_{0+}^{\delta }{\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(t\right)\right)+W\left(t\right)=0,0<t<1,\hfill \\ D_{0+}^{{\gamma }_{n-2}-\beta }x\left(0\right)=D_{0+}^{\alpha -\beta }x\left(0\right)={\left({\varphi }_p\left(D_{0+}^{\alpha -\beta }x\right)\right)}^{″}\left(0\right)=0,\hfill \\ {\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(1\right)\right)=k_0{\int }_0^1{b}_0\left(s\right){\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(s\right)\right)ds,\hfill \\ D_{0+}^{z_0-\beta }x\left(1\right)=k_1{\int }_0^1{b}_1\left(s\right)D_{0+}^{z_1-\beta }x\left(s\right)d{A}_1\left(s\right)\hfill \\ +k_2{\int }_0^{\zeta }{b}_2\left(s\right)D_{0+}^{z_2-\beta }x\left(s\right)d{A}_2\left(s\right)+k_3{\displaystyle \sum _{i=1}^{\infty }}{a}_i{D}_{0+}^{z_3-\beta }x\left({\xi }_i\right),\hfill \end{array}\right.$$

(35)

has a unique solution

$$x\left(t\right)={\int }_0^{1}G(t,s){\varphi }_q\left({\int }_0^{1}V(s,\tau )W\left(\tau \right)d\tau \right)ds,$$

where $G(t,s)$ and $V(t,s)$ are defined in Lemmas 6 and 7, respectively.

Proof. 

Let $h\left(t\right)=-{\varphi }_p\left(D_{0+}^{\alpha -\beta }x\left(t\right)\right)$. Then, we can resolve the problem (35) into the following two problems:

$$\left\{\begin{array}{c}{}^{c}D_{0+}^{\delta }h\left(t\right)+W\left(t\right)=0,0<t<1,\hfill \\ {\varphi }_q\left(h\left(0\right)\right)=h^{″}\left(0\right)=0,\hfill \\ h\left(1\right)=k_0{\int }_0^1{b}_0\left(s\right)h\left(s\right)ds\hfill \end{array}\right.$$

(36)

and

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha -\beta }x\left(t\right)+{\varphi }_q\left(h\left(t\right)\right)=0,0<t<1,\hfill \\ D_{0+}^{{\gamma }_{n-2}-\beta }x\left(0\right)=0,\hfill \\ D_{0+}^{z_0-\beta }x\left(1\right)=k_1{\int }_0^1{b}_1\left(s\right)D_{0+}^{z_1-\beta }x\left(s\right)d{A}_1\left(s\right)\hfill \\ +k_2{\int }_0^{\zeta }{b}_2\left(s\right)D_{0+}^{z_2-\beta }x\left(s\right)d{A}_2\left(s\right)+k_3{\displaystyle \sum _{i=1}^{\infty }}{a}_i{D}_{0+}^{z_3-\beta }x\left({\xi }_i\right).\hfill \end{array}\right.$$

(37)

From Lemma 7, the problem (36) has a unique solution:

$$h\left(t\right)={\int }_0^{1}V(t,s)W\left(s\right)ds.$$

(38)

From Lemma 6, the problem (37) has a unique solution:

$$x\left(t\right)={\int }_0^{1}G(t,s){\varphi }_q\left(h\left(s\right)\right)ds.$$

(39)

Therefore, the problem (35) has a unique solution:

$$x\left(t\right)={\int }_0^{1}G(t,s){\varphi }_q\left({\int }_0^{1}V(s,\tau )W\left(\tau \right)d\tau \right)ds.$$

□

Lemma 9.

Let ${\int }_0^1{s}^{\alpha -z_1-1}{b}_1\left(s\right)d{A}_1\left(s\right)\ge 0,$ ${\int }_0^{\zeta }{s}^{\alpha -z_2-1}{b}_2\left(s\right)d{A}_2\left(s\right)\ge 0$, ${\sum }_{i=1}^{\infty }{a}_i{\xi }_i^{\alpha -z_3-1}$$<+\infty$, $k_i\ge 0$$(i=1,2,3)$, ${\sigma }_1>0$ (defined in Lemma 6). Then, $G(t,s)$ has the following properties:

(1) $t^{\alpha -\beta -1}J\left(s\right)\le G(t,s)\le \mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}$, where

$$J\left(s\right)={\mathsf{{\rm Y}}}_0{J}_0\left(s\right)+{\mathsf{{\rm Y}}}_1{J}_1\left(s\right)+{\mathsf{{\rm Y}}}_2{J}_2\left(s\right)+{\mathsf{{\rm Y}}}_3{J}_3\left(s\right),\mathsf{{\rm Y}}={\mathsf{{\rm Y}}}_0+{\mathsf{{\rm Y}}}_1+{\mathsf{{\rm Y}}}_2+{\mathsf{{\rm Y}}}_3,$$

in which

$${\mathsf{{\rm Y}}}_0={\displaystyle \frac{1}{\Gamma (\alpha -\beta )}},{\mathsf{{\rm Y}}}_1={\displaystyle \frac{k_1}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_1)}}{\int }_0^1{\tau }^{\alpha -z_1-1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right),$$

$${\mathsf{{\rm Y}}}_2={\displaystyle \frac{k_2}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }{\tau }^{\alpha -z_2-1}{b}_2\left(\tau \right)d{A}_2\left(\tau \right),$$

$${\mathsf{{\rm Y}}}_3={\displaystyle \frac{k_3}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_3)}}\sum _{i=1}^{\infty }{a}_i{\xi }_i^{\alpha -z_3-1},$$

$$J_0\left(s\right)={(1-s)}^{\alpha -z_0-1}\left(1-{(1-s)}^{z_0-\beta }\right),$$

$$J_i\left(s\right)={(1-s)}^{\alpha -z_0-1}\left(1-{(1-s)}^{z_0-z_i}\right),i=1,2,3.$$

(2) $G(t,s)\ge 0$, t, $s\in [0,1]\times [0,1]$.

Proof. 

(1) For $s\le t$,

$$\begin{array}{cc}\hfill P_0(t,s)& ={\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}\left(t^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}-{(t-s)}^{\alpha -\beta -1}\right)\hfill \\ & \ge {\displaystyle \frac{t^{\alpha -\beta -1}}{\Gamma (\alpha -\beta )}}{(1-s)}^{\alpha -z_0-1}\left(1-{(1-s)}^{z_0-\beta }\right)\hfill \\ & ={\mathsf{{\rm Y}}}_0{t}^{\alpha -\beta -1}{J}_0\left(s\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill P_0(t,s)& \le {\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}\hfill \\ & ={\mathsf{{\rm Y}}}_0{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}.\hfill \end{array}$$

For $t\le s$,

$$\begin{array}{cc}\hfill P_0(t,s)& ={\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}\hfill \\ & \ge {\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}\left(1-{(1-s)}^{z_0-\beta }\right)\hfill \\ & ={\mathsf{{\rm Y}}}_0{t}^{\alpha -\beta -1}{J}_0\left(s\right),\hfill \end{array}$$

$$P_0(t,s)={\mathsf{{\rm Y}}}_0{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}.$$

Similarly, we obtain

$$P_i(t,s)\ge {\displaystyle \frac{k_i{t}^{\alpha -z_i-1}{J}_i\left(s\right)}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_i)}},i=1,2,3,$$

$$P_i(t,s)\le {\displaystyle \frac{k_i{t}^{\alpha -z_i-1}{(1-s)}^{\alpha -z_0-1}}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_i)}},i=1,2,3.$$

Thus, we have

$$\begin{array}{cc}\hfill G(t,s)=& P_0(t,s)+t^{\alpha -\beta -1}\left({\int }_0^1{P}_1(\tau ,s)b_1\left(\tau \right)d{A}_1\left(\tau \right)\right)\hfill \\ & +t^{\alpha -\beta -1}\left({\int }_0^{\zeta }{P}_2(\tau ,s)b_2\left(\tau \right)d{A}_2\left(\tau \right)\right)+t^{\alpha -\beta -1}{\displaystyle \sum _{i=1}^{\infty }}{a}_i{P}_3({\xi }_i,s)\hfill \\ \hfill \ge & t^{\alpha -\beta -1}{\mathsf{{\rm Y}}}_0{J}_0\left(s\right)+t^{\alpha -\beta -1}{\mathsf{{\rm Y}}}_1{J}_1\left(s\right)+t^{\alpha -\beta -1}{\mathsf{{\rm Y}}}_2{J}_2\left(s\right)+t^{\alpha -\beta -1}{\mathsf{{\rm Y}}}_3{J}_3\left(s\right)\hfill \\ \hfill =& t^{\alpha -\beta -1}J\left(s\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill G(t,s)\le & {\mathsf{{\rm Y}}}_0{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}+{\mathsf{{\rm Y}}}_1{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}\hfill \\ & +{\mathsf{{\rm Y}}}_2{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}+{\mathsf{{\rm Y}}}_3{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}\hfill \\ \hfill =& \mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}.\hfill \end{array}$$

Moreover, from the above proof, (2) is true, clearly. □

Lemma 10.

Let ${\sigma }_2>0$ (defined in Lemma 7). Then, the following properties hold:

(1) ${\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t(1-t^{\delta -2}){(1-s)}^{\delta -1}\le V_1(t,s)\le {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}$.

(2) ${\displaystyle \frac{{\Xi }_1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}\le V_2(t,s)\le {\displaystyle \frac{{\Xi }_2}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}$, where

$${\Xi }_1={\displaystyle \frac{k_0}{{\sigma }_2}}{\int }_0^1\tau (1-{\tau }^{\delta -2})b_0\left(\tau \right)d\tau ,{\Xi }_2={\displaystyle \frac{k_0}{{\sigma }_2}}{\int }_0^1\tau b_0\left(\tau \right)d\tau .$$

(3) ${\Lambda }_{1}t(1-t^{\delta -2}){(1-s)}^{\delta -1}\le V(t,s)\le {\Lambda }_{2}t{(1-s)}^{\delta -1}$, where

$${\Lambda }_1={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}(1+{\Xi }_1),{\Lambda }_2={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}(1+{\Xi }_2).$$

(4) for $\iota >0$,

$${\int }_0^{1}V(t,s)s^{\iota }ds={\displaystyle \frac{\mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}(t-t^{\delta +\iota }+\varpi t),$$

where $\varpi ={\displaystyle \frac{k_0}{{\sigma }_2}}{\int }_0^1(\tau -{\tau }^{\delta +\iota })b_0\left(\tau \right)d\tau$.

(5) $V(t,s)\ge 0$, $t,s\in [0,1]\times [0,1]$.

Proof. 

(1) For $s\le t$,

$$\begin{array}{cc}\hfill V_1(t,s)& \ge {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}\left(t{(1-s)}^{\delta -1}-t^{\delta -1}{(1-s)}^{\delta -1}\right)\hfill \\ & ={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t(1-t^{\delta -2}){(1-s)}^{\delta -1},\hfill \end{array}$$

$$V_1(t,s)\le {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}.$$

For $t\le s$,

$$V_1(t,s)\ge {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t(1-t^{\delta -2}){(1-s)}^{\delta -1},$$

$$V_1(t,s)={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}.$$

(2) It follows from (1) that

$$\begin{array}{cc}\hfill V_2(t,s)& ={\displaystyle \frac{k_{0}t}{{\sigma }_2}}{\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau \hfill \\ & \ge {\displaystyle \frac{k_0}{{\sigma }_2\Gamma \left(\delta \right)}}\left({\int }_0^1\tau (1-{\tau }^{\delta -2})b_0\left(\tau \right)d\tau \right)t{(1-s)}^{\delta -1}\hfill \\ & ={\displaystyle \frac{{\Xi }_1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1},\hfill \end{array}$$

$$\begin{array}{cc}\hfill V_2(t,s)& ={\displaystyle \frac{k_{0}t}{{\sigma }_2}}{\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau \hfill \\ & \le {\displaystyle \frac{k_0}{{\sigma }_2\Gamma \left(\delta \right)}}\left({\int }_0^1\tau b_0\left(\tau \right)d\tau \right)t{(1-s)}^{\delta -1}\hfill \\ & ={\displaystyle \frac{{\Xi }_2}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}.\hfill \end{array}$$

(3) On the basis of (1) and (2), we obtain

$$\begin{array}{cc}\hfill V(t,s)& \ge {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t(1-t^{\delta -2}){(1-s)}^{\delta -1}+{\displaystyle \frac{{\Xi }_1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}\hfill \\ & \ge \left({\displaystyle \frac{1}{\Gamma \left(\delta \right)}}+{\displaystyle \frac{{\Xi }_1}{\Gamma \left(\delta \right)}}\right)t(1-t^{\delta -2}){(1-s)}^{\delta -1}\hfill \\ & ={\Lambda }_{1}t(1-t^{\delta -2}){(1-s)}^{\delta -1}.\hfill \end{array}$$

$$\begin{array}{cc}\hfill V(t,s)& \le {\displaystyle \frac{1}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}+{\displaystyle \frac{{\Xi }_2}{\Gamma \left(\delta \right)}}t{(1-s)}^{\delta -1}\hfill \\ & ={\Lambda }_{2}t{(1-s)}^{\delta -1}.\hfill \end{array}$$

(4) For $\iota >0$,

$$\begin{array}{cc}\hfill {\int }_0^1{V}_1(t,s)s^{\iota }ds& ={\displaystyle \frac{1}{\Gamma \left(\delta \right)}}\left({\int }_0^{1}t{(1-s)}^{\delta -1}{s}^{\iota }ds-{\int }_0^t{(t-s)}^{\delta -1}{s}^{\iota }ds\right)\hfill \\ & ={\displaystyle \frac{\mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}(t-t^{\delta +\iota }),\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\int }_0^1{V}_2(t,s)s^{\iota }ds& ={\int }_0^1\left({\displaystyle \frac{k_{0}t}{{\sigma }_2}}{\int }_0^1{V}_1(\tau ,s)b_0\left(\tau \right)d\tau \right)s^{\iota }ds\hfill \\ & ={\displaystyle \frac{k_{0}t}{{\sigma }_2}}{\int }_0^1\left({\int }_0^1{V}_1(\tau ,s)s^{\iota }ds\right)b_0\left(\tau \right)d\tau \hfill \\ & ={\displaystyle \frac{k_0\mathcal{B}(\iota +1,\delta )t}{{\sigma }_2\Gamma \left(\delta \right)}}{\int }_0^1(\tau -{\tau }^{\delta +\iota })b_0\left(\tau \right)d\tau ,\hfill \\ & ={\displaystyle \frac{\mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi t,\hfill \end{array}$$

where $\varpi ={\displaystyle \frac{k_0}{{\sigma }_2}}{\int }_0^1(\tau -{\tau }^{\delta +\iota })b_0\left(\tau \right)d\tau$. It follows that

$${\int }_0^{1}V(t,s)s^{\iota }ds={\displaystyle \frac{\mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}(t-t^{\delta +\iota }+\varpi t).$$

From the above proof, (5) is true. The proof is complete. □

Lemma 11

([35]). (1) If $1<p\le 2$, $|x_1|$, $|x_2|\ge x_0>0$, and $x_1{x}_2>0$, then

$$|{\varphi }_p\left(x_1\right)-{\varphi }_p\left(x_2\right)|\le (p-1)x_0^{p-2}|x_1-x_2|.$$

(2) If $p>2$, $\left|x_1\right|,\left|x_2\right|\le X_0$, then

$$|{\varphi }_p\left(x_1\right)-{\varphi }_p\left(x_2\right)|\le (p-1)X_0^{p-2}|x_1-x_2|.$$

(3) If $1<p\le 2$, then for $x_1$, $x_2\ge 0$,

$$2^{p-2}({\varphi }_p\left(x_1\right)+{\varphi }_p\left(x_2\right))\le {\varphi }_p(x_1+x_2)\le {\varphi }_p\left(x_1\right)+{\varphi }_p\left(x_2\right).$$

(4) If $p>2$, then for $x_1$, $x_2\ge 0$,

$${\varphi }_p\left(x_1\right)+{\varphi }_p\left(x_2\right)\le {\varphi }_p(x_1+x_2)\le 2^{p-2}({\varphi }_p\left(x_1\right)+{\varphi }_p\left(x_2\right)).$$

Remark 3.

In this paper, $\Gamma \left(t\right)$ represents the Gamma function

$$\Gamma \left(t\right)={\int }_0^{+\infty }{s}^{t-1}{e}^{-s}ds,t>0,$$

and $\mathcal{B}(t,s)$ represents the Beta function

$$\mathcal{B}(t,s)={\int }_0^1{l}^{t-1}{(1-l)}^{s-1}dl,t>0,s>0.$$

In the following, we define two operators $T_1$, $T:C[0,1]\to C[0,1]$ by

$$\left(T_{1}x\right)\left(t\right)={\int }_0^{1}V(t,s)f(s,I_{0^{+}}^{\beta }x\left(s\right),\mathcal{H}x\left(s\right),x\left(s\right))ds,t\in [0,1],$$

$$\left(Tx\right)\left(t\right)={\int }_0^{1}G(t,s){\varphi }_q\left(\left(T_{1}x\right)\left(s\right)\right)ds,t\in [0,1].$$

It is easy to verify that if T has a fixed point x in $C[0,1]$, then x is a solution of the problem (2) on $[0,1]$.

3. Existence of Solutions

In this section, we give the following hypothesis:

For $\left({\mathbf{A}}_{\mathbf{0}}\right)$, there exist non-negative functions $a_i\in L^2(0,1)$$(i=0,1,2,3)$ such that for a.e. $t\in (0,1),$ and for any $(x_1,x_2,x_3)\in {\mathbb{R}}^3$,

$$|f(t,x_1,x_2,x_3)|\le a_0\left(t\right)+\sum _{i=1}^3{a}_i\left(t\right){\left|x_i\right|}^{{\theta }_i},$$

where ${\theta }_i\in (0,p-1)$$(i=1,2,3)$.

In the following, for any $r>0$, we denote

$$\begin{array}{cc}\hfill K_r=& {\Lambda }_2\left({\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1)\parallel a_0{\parallel }_{L_2}+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(2\beta {\theta }_1+1,2\delta -1){(\Gamma (\beta +1))}^{-{\theta }_1}{\parallel a_1\parallel }_{L_2}{r}^{{\theta }_1}\right.\hfill \\ \hfill & \left.+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){(\Gamma (\beta +1))}^{-{\theta }_2}{M}_h^{{\theta }_2}\parallel a_2{\parallel }_{L_2}{r}^{{\theta }_2}+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){\parallel a_3\parallel }_{L_2}{r}^{{\theta }_3}\right),\hfill \end{array}$$

(40)

where ${∥a_i∥}_{L_2}={\left({\int }_0^1{a}_i^2\left(t\right)dt\right)}^{{\displaystyle \frac{1}{2}}}$$(i=0,1,2,3)$, $M_h={sup}_{t\in [0,1]}{\int }_0^{t}h(t,s)s^{\beta }ds$, and ${\Lambda }_2$ is defined in Lemma 10.

Lemma 12.

Let ${\mathcal{C}}_i\ge 0$$(i=1,2,3,4)$ and considering the function

$$\mathcal{C}\left(t\right)=t^{p-1}-{\mathcal{C}}_1{t}^{{\theta }_1}-{\mathcal{C}}_2{t}^{{\theta }_2}-{\mathcal{C}}_3{t}^{{\theta }_3}-{\mathcal{C}}_4,t\in [0,+\infty ).$$

Since ${\theta }_i\in (0,p-1)$$(i=1,2,3)$, we derive that $\mathcal{C}\left(t\right)\to +\infty$$(t\to +\infty )$. Thus, there exists $t_0>0$ such that

$$\mathcal{C}\left(t\right)\ge 0,\forall t\ge t_0.$$

Lemma 13.

Suppose that $\left({\mathbf{A}}_{\mathbf{0}}\right)$ holds. Then, for any $r>0,$$T:S_r\to C[0,1]$ is well defined, where $S_r=\{x\in C[0,1]:\parallel x\parallel \le r\}$.

Proof. 

For any $x\in S_r$, we have

$$\left|I_{0^{+}}^{\beta }x\left(t\right)\right|\le I_{0^{+}}^{\beta }\left|x\left(t\right)\right|\le {\displaystyle \frac{\parallel x\parallel }{\Gamma \left(\beta \right)}}{\int }_0^t{(t-s)}^{\beta -1}ds\le {\displaystyle \frac{r{t}^{\beta }}{\Gamma (\beta +1)}},t\in [0,1],$$

(41)

$$\begin{array}{cc}\hfill \left|\left(\mathcal{H}x\right)\left(t\right)\right|& \le {\int }_0^{t}h(t,s)I_{0^{+}}^{\beta }\left|x\left(s\right)\right|ds\le {\displaystyle \frac{r}{\Gamma (\beta +1)}}{\int }_0^{t}h(t,s)s^{\beta }ds\hfill \\ & \le {\displaystyle \frac{M_{h}r}{\Gamma (\beta +1)}},t\in [0,1].\hfill \end{array}$$

(42)

Based on $\left({\mathbf{A}}_{\mathbf{0}}\right)$, Lemma 10(3) and the above analysis, we derive that for any $x\in S_r$,

$$\begin{array}{cc}\hfill & |T_{1}x\left(t\right)|\hfill \\ \hfill =& \left|{\int }_0^{1}V(t,s)f(s,I_{0^{+}}^{\beta }x\left(s\right),\mathcal{H}x\left(s\right),x\left(s\right))ds\right|\hfill \\ \hfill \le & {\int }_0^{1}V(t,s)\left(a_0\left(s\right)+a_1\left(s\right)|I_{0^{+}}^{\beta }{x\left(s\right)|}^{{\theta }_1}+a_2{\left(s\right)\left|\mathcal{H}x\left(s\right)\right|}^{{\theta }_2}+a_3\left(s\right){\left|x\left(s\right)\right|}^{{\theta }_3}\right)ds\hfill \\ \hfill \le & {\int }_0^{1}V(t,s)\left(a_0\left(s\right)+{\displaystyle \frac{a_1\left(s\right)s^{\beta {\theta }_1}{r}^{{\theta }_1}}{{(\Gamma (\beta +1))}^{{\theta }_1}}}+{\displaystyle \frac{a_2\left(s\right)M_h^{{\theta }_2}{r}^{{\theta }_2}}{{(\Gamma (\beta +1))}^{{\theta }_2}}}+a_3\left(s\right)r^{{\theta }_3}\right)ds\hfill \\ \hfill \le & {\Lambda }_{2}t\left({\int }_0^1{(1-s)}^{\delta -1}{a}_0\left(s\right)ds+{\displaystyle \frac{r^{{\theta }_1}}{{(\Gamma (\beta +1))}^{{\theta }_1}}}{\int }_0^1{(1-s)}^{\delta -1}{a}_1\left(s\right)s^{\beta {\theta }_1}ds\right.\hfill \\ \hfill & +\left.{\displaystyle \frac{M_h^{{\theta }_2}{r}^{{\theta }_2}}{{(\Gamma (\beta +1))}^{{\theta }_2}}}{\int }_0^1{(1-s)}^{\delta -1}{a}_2\left(s\right)ds+r^{{\theta }_3}{\int }_0^1{(1-s)}^{\delta -1}{a}_3\left(s\right)ds\right)\hfill \\ \hfill \le & {\Lambda }_{2}t\left\{{\left({\int }_0^1{(1-s)}^{2\delta -2}ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{a}_0^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\right.\hfill \\ \hfill & +{\displaystyle \frac{r^{{\theta }_1}}{{(\Gamma (\beta +1))}^{{\theta }_1}}}{\left({\int }_0^1{s}^{2\beta {\theta }_1}{(1-s)}^{2\delta -2}ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{a}_1^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{M_h^{{\theta }_2}{r}^{{\theta }_2}}{{(\Gamma (\beta +1))}^{{\theta }_2}}}{\left({\int }_0^1{(1-s)}^{2\delta -2}ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{a}_2^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \left.+r^{{\theta }_3}{\left({\int }_0^1{(1-s)}^{2\delta -2}ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{a}_3^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\right\}\hfill \\ \hfill =& {\Lambda }_{2}t\left({\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1)\parallel a_0{\parallel }_{L_2}+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(2\beta {\theta }_1+1,2\delta -1){(\Gamma (\beta +1))}^{-{\theta }_1}{\parallel a_1\parallel }_{L_2}{r}^{{\theta }_1}\right.\hfill \\ \hfill & \left.+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){(\Gamma (\beta +1))}^{-{\theta }_2}{M}_h^{{\theta }_2}\parallel a_2{\parallel }_{L_2}{r}^{{\theta }_2}+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){\parallel a_3\parallel }_{L_2}{r}^{{\theta }_3}\right)\hfill \\ \hfill =& K_{r}t,t\in [0,1],\hfill \end{array}$$

(43)

which, together with Lemma 9(1), implies

$$\begin{array}{cc}\hfill \left|Tx\right(t\left)\right|=& \left|{\int }_0^{1}G(t,s){\varphi }_q\left(T_{1}x\left(s\right)\right)ds\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s){\varphi }_q\left(K_{r}s\right)ds\hfill \\ \hfill =& K_r^{q-1}{\int }_0^{1}G(t,s)s^{q-1}ds\hfill \\ \hfill \le & K_r^{q-1}{\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{s}^{q-1}ds\hfill \\ \hfill =& \mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)K_r^{q-1}{t}^{\alpha -\beta -1}<+\infty ,t\in [0,1].\hfill \end{array}$$

(44)

Therefore, we deduce that $T:S_r\to C[0,1]$ is well defined. □

Theorem 1.

Suppose that $\left({\mathbf{A}}_{\mathbf{0}}\right)$ holds. Then, the problem (1) has a solution $u_1$, and there exists $r_1>0$ such that

$$|u_1\left(t\right)|\le {\displaystyle \frac{r_1}{\Gamma (\beta +1)}}{t}^{\beta },t\in [0,1].$$

Proof. 

Considering the function $\mathcal{C}\left(t\right)$ (in Lemma 12) with

$${\mathcal{C}}_1={\left(\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\right)}^{p-1}{\Lambda }_2{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(2\beta {\theta }_1+1,2\delta -1){(\Gamma (\beta +1))}^{-{\theta }_1}{\parallel a_1\parallel }_{L_2},$$

$${\mathcal{C}}_2={\left(\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\right)}^{p-1}{\Lambda }_2{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){(\Gamma (\beta +1))}^{-{\theta }_2}{M}_h^{{\theta }_2}{\parallel a_2\parallel }_{L_2},$$

$${\mathcal{C}}_3={\left(\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\right)}^{p-1}{\Lambda }_2{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){\parallel a_3\parallel }_{L_2},$$

$${\mathcal{C}}_4={\left(\mathsf{{\rm Y}}B(q,\alpha -z_0)\right)}^{p-1}{\Lambda }_2{B}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){\parallel a_0\parallel }_{L_2},$$

we can infer that there exists an $r_1$ that is sufficiently large such that $\mathcal{C}\left(r_1\right)>0$, i.e.,

$$\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)K_{r_1}^{q-1}\le r_1.$$

(45)

Let $S_{r_1}=\{x\in C[0,1]:\parallel x\parallel \le r_1\}$, and we prove $T:S_{r_1}\to S_{r_1}$, firstly. Taking account of (44) and (45), we infer that for any $x\in S_{r_1}$,

$$\left|Tx\left(t\right)\right|\le \mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)K_{r_1}^{q-1}\le r_1,t\in [0,1],$$

(46)

which means $T:S_{r_1}\to S_{r_1}$. In the following, we give the proof that $T:S_{r_1}\to S_{r_1}$ is completely continuous.

Let $\left\{x_n\right\}\subset C[0,1]$ be a sequence satisfying $x_n\to x_0$, as $n\to +\infty$. Since $f\in C((0,1)\times {\mathbb{R}}^3,\mathbb{R})$, for fixed $\tau \in (0,1)$, $f(\tau ,x,y,z)$ is uniformly continuous in regard to x, y and z on ${[-r_1,r_1]}^3$. Thus, for any $\epsilon >0$, there exists $\delta =\delta (\tau ,\epsilon )>0$ such that, for any $x_i$, $y_i,z_i\in [-r_1,r_1]$$(i=1,2)$, which satisfy $|x_1-x_2|<\delta$,$|y_1-y_2|<\delta$ and $|z_1-z_2|<\delta$,

$$|f(\tau ,x_1,y_1,z_1)-f(\tau ,x_2,y_2,z_2)|<\epsilon .$$

(47)

As $x_n\to x_0$$(n\to +\infty )$, for $\tilde{\delta }=min\{\delta ,\Gamma (\beta +1)\delta ,\Gamma (\beta +1)M_h^{-1}\delta \}$, there exists $N_0>0$, such that for any $n\ge N_0$,

$$\parallel x_n-x_0\parallel <\tilde{\delta }\le \delta .$$

Evidently, we have

$$|x_n\left(\tau \right)-x_0\left(\tau \right)|<\tilde{\delta }\le \delta ,$$

(48)

$$|I_{0^{+}}^{\beta }{x}_n\left(\tau \right)-I_{0^{+}}^{\beta }{x}_0\left(\tau \right)|<{\displaystyle \frac{\tilde{\delta }}{\Gamma (\beta +1)}}\le \delta ,$$

(49)

$$|\mathcal{H}{x}_n\left(\tau \right)-\mathcal{H}{x}_0\left(\tau \right)|<{\displaystyle \frac{M_h\tilde{\delta }}{\Gamma (\beta +1)}}\le \delta .$$

(50)

Hence, by (47)–(50), we infer that for any $n\ge N_0$,

$$|f(\tau ,I_{0^{+}}^{\beta }{x}_n\left(\tau \right),\mathcal{H}{x}_n\left(\tau \right),x_n\left(\tau \right))-f(\tau ,I_{0^{+}}^{\beta }{x}_0\left(\tau \right),\mathcal{H}{x}_0\left(\tau \right),x_0\left(\tau \right))|<\epsilon .$$

(51)

That is, for any $\tau \in (0,1)$,

$$|f(\tau ,I_{0^{+}}^{\beta }{x}_n\left(\tau \right),\mathcal{H}{x}_n\left(\tau \right),x_n\left(\tau \right))-f(\tau ,I_{0^{+}}^{\beta }{x}_0\left(\tau \right),\mathcal{H}{x}_0\left(\tau \right),x_0\left(\tau \right))|\to 0,n\to +\infty .$$

(52)

Further, by $\left({\mathbf{A}}_{\mathbf{0}}\right)$, we know

$$\begin{array}{cc}& |f(\tau ,I_{0^{+}}^{\beta }{x}_n\left(\tau \right),\mathcal{H}{x}_n\left(\tau \right),x_n\left(\tau \right))-f(\tau ,I_{0^{+}}^{\beta }{x}_0\left(\tau \right),\mathcal{H}{x}_0\left(\tau \right),x_0\left(\tau \right))|\hfill \\ \hfill \le & 2a_0\left(\tau \right)+2a_1\left(\tau \right)r_1^{{\theta }_1}+2a_2\left(\tau \right)r_1^{{\theta }_2}+2a_3\left(\tau \right)r_1^{{\theta }_3},\tau \in (0,1).\hfill \end{array}$$

It is clear that $2a_0+2r_1^{{\theta }_1}{a}_1+2r_1^{{\theta }_2}{a}_2+2r_1^{{\theta }_3}{a}_3\in L^1(0,1)$, by using $\left({\mathbf{A}}_{\mathbf{0}}\right)$. Accordingly, by means of the Lebesgue dominated convergence theorem, we obtain

$$\begin{array}{cc}\hfill & |T_1{x}_n\left(s\right)-T_1{x}_0\left(s\right)|\hfill \\ \hfill =& {\int }_0^{1}V(s,\tau )\left|f(\tau ,I_{0^{+}}^{\beta }{x}_n\left(\tau \right),\mathcal{H}{x}_n\left(\tau \right),x_n\left(\tau \right))-f(\tau ,I_{0^{+}}^{\beta }{x}_0\left(\tau \right),\mathcal{H}{x}_0\left(\tau \right),x_0\left(\tau \right))\right|d\tau \hfill \\ \hfill \le & {\Lambda }_2{\int }_0^1\left|f(\tau ,I_{0^{+}}^{\beta }{x}_n\left(\tau \right),\mathcal{H}{x}_n\left(\tau \right),x_n\left(\tau \right))-f(\tau ,I_{0^{+}}^{\beta }{x}_0\left(\tau \right),\mathcal{H}{x}_0\left(\tau \right),x_0\left(\tau \right))\right|d\tau \hfill \\ \hfill \to & 0,s\in [0,1].\hfill \end{array}$$

(53)

Based on the continuity of ${\varphi }_q$, we have

$$|{\varphi }_q\left(T_1{x}_n\left(s\right)\right)-{\varphi }_q\left(T_1{x}_0\left(s\right)\right)|\to 0,s\in [0,1].$$

(54)

By (43), we know

$$|{\varphi }_q\left(T_1{x}_n\left(s\right)\right)-{\varphi }_q\left(T_1{x}_0\left(s\right)\right)|\le 2K_{r_1}^{q-1},s\in [0,1].$$

Thus, based on Lemma 9(1) and the Lebesgue control convergence theorem, we obtain

$$\begin{array}{cc}\hfill & |\left(T{x}_n\right)\left(t\right)-\left(T{x}_0\right)\left(t\right)|\hfill \\ \hfill =& \left|{\int }_0^{1}G(t,s)({\varphi }_q\left(T_1{x}_n\left(s\right)\right)-{\varphi }_q\left(T_1{x}_0\left(s\right)\right))ds\right|\hfill \\ \hfill \le & \mathsf{{\rm Y}}{\int }_0^1\left|{\varphi }_q\left(T_1{x}_n\left(s\right)\right)-{\varphi }_q\left(T_1{x}_0\left(s\right)\right)\right|ds\to 0,t\in [0,1].\hfill \end{array}$$

(55)

Therefore, we conclude that $T:S_{r_1}\to S_{r_1}$ is continuous. □

Next, we prove $T{S}_{r_1}$ is equicontinuous. With the help of the uniformly continuity of $G(t,s)$ on $[0,1]\times [0,1]$, we infer that, for any $\tilde{\epsilon }>0$, there exists $\delta =\delta \left(\tilde{\epsilon }\right)>0$ such that, for any $t_1$, $t_2\in [0,1]$ that satisfy $|t_1-t_2|<\delta$,

$$\left|G(t_1,s)-G(t_2,s)\right|<K_{r_1}^{1-q}\tilde{\epsilon },s\in [0,1].$$

Hence, for any $x\in S_{r_1}$,

$$\begin{array}{cc}\hfill & |Tx\left(t_1\right)-Tx\left(t_2\right)|\hfill \\ \hfill =& \left|{\int }_0^{1}G(t_1,s){\varphi }_q\left(T_{1}x\left(s\right)\right)ds-{\int }_0^{1}G(t_2,s){\varphi }_q\left(T_{1}x\left(s\right)\right)ds\right|\hfill \\ \hfill \le & {\int }_0^1\left|G(t_1,s)-G(t_2,s)\right|{\varphi }_q\left(T_{1}x\left(s\right)\right)ds\hfill \\ \hfill \le & K_{r_1}^{q-1}{\int }_0^1\left|G(t_1,s)-G(t_2,s)\right|ds<\tilde{\epsilon },\hfill \end{array}$$

(56)

which implies that $T{S}_{r_1}$ is equicontinuous. In addition, based on (46), we observe that $T{S}_{r_1}$ is uniformly bounded. Hence, we deduce that $T:S_{r_1}\to S_{r_1}$ is completely continuous. It follows from the Schauder fixed point theorem that T has a fixed point $x_1\in S_{r_1}$. That is, the problem (2) has a solution $x_1$, which satisfies $\parallel x_1\parallel \le r_1$. Moreover, from Lemma 5, we derive that theproblem (1) has a solution $u_1=I_{0^{+}}^{\beta }{x}_1$, which satisfies

$$|u_1\left(t\right)|\le {\displaystyle \frac{r_1}{\Gamma (\beta +1)}}{t}^{\beta },t\in [0,1].$$

□

4. Uniqueness of Solutions

In this section, we prove the uniqueness of solutions for the cases $1<q\le 2$ and $q>2$ separately. We give the following hypotheses:

$\left({\mathbf{A}}_{\mathbf{1}}\right)$ There exist $\kappa >0$ and $\iota >0$ such that

$$f(t,x,y,z)\ge \kappa t^{\iota },\forall (t,x,y,z)\in (0,1)\times {\mathbb{R}}^3.$$

$\left({\mathbf{A}}_{\mathbf{2}}\right)$ There exist $\kappa >0$ and $\iota >0$ such that

$$f(t,x,y,z)\le -\kappa t^{\iota },\forall (t,x,y,z)\in (0,1)\times {\mathbb{R}}^3.$$

$\left({\mathbf{A}}_{\mathbf{3}}\right)$ There exists non-negative function $\mathcal{F}\in C(0,1)$, which satisfies ${(1-t)}^{\delta -1}\mathcal{F}\in$ $L^1(0,1)$, such that

$$\left|f(t,x,y,z)\right|\le \mathcal{F}\left(t\right),\forall (t,x,y,z)\in (0,1)\times {\mathbb{R}}^3.$$

$\left({\mathbf{A}}_{\mathbf{4}}\right)$ There exist non-negative functions $l_i\in L^2(0,1)$$(i=1,2,3)$ such that for a.e. $t\in (0,1),$ and for any $(x_j,y_j,z_j)\in {\mathbb{R}}^3$$(j=1,2)$,

$$|f(t,x_1,y_1,z_1)-f(t,x_2,y_2,z_2)|\le l_1\left(t\right)|x_1-x_2|+l_2\left(t\right)|y_1-y_2|+l_3\left(t\right)|z_1-z_2|.$$

$\left({\mathbf{A}}_{\mathbf{5}}\right)$ For any $r>0$, there exist non-negative functions $l_{i,r}\in L^2(0,1)$$(i=1,2,3)$ such that for a.e. $t\in (0,1),$ and for any $(x_j,y_j,z_j)\in {[-r,r]}^3$$(j=1,2)$,

$$|f(t,x_1,y_1,z_1)-f(t,x_2,y_2,z_2)|\le l_{1,r}\left(t\right)|x_1-x_2|+l_{2,r}\left(t\right)|y_1-y_2|+l_{3,r}\left(t\right)|z_1-z_2|.$$

For convenience, we denote

$$M_h=\underset{t\in [0,1]}{sup}{\int }_0^{t}h(t,s)s^{\beta }ds,$$

(57)

$$M_f={\int }_0^1{(1-\tau )}^{\delta -1}f(\tau ,0,0,0)d\tau ,$$

(58)

$$M_{\mathcal{F}}={\int }_0^1{(1-\tau )}^{\delta -1}\mathcal{F}\left(\tau \right)d\tau ,$$

(59)

$$\begin{array}{cc}\hfill \mathcal{L}=& {\Lambda }_2\left({\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(2\beta +1,2\delta -1){(\Gamma (\beta +1))}^{-1}{\parallel l_1\parallel }_{L^2}\right.\hfill \\ \hfill & \left.+M_h{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){(\Gamma (\beta +1))}^{-1}\parallel l_2{\parallel }_{L^2}+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){\parallel l_3\parallel }_{L^2}\right),\hfill \end{array}$$

(60)

$$\begin{array}{cc}\hfill {\mathcal{L}}_r=& {\Lambda }_2\left({\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(2\beta +1,2\delta -1){(\Gamma (\beta +1))}^{-1}{\parallel l_{1,r}\parallel }_{L^2}\right.\hfill \\ \hfill & \left.+M_h{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){(\Gamma (\beta +1))}^{-1}\parallel l_{2,r}{\parallel }_{L^2}+{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1){\parallel l_{3,r}\parallel }_{L^2}\right)(\forall r>0),\hfill \end{array}$$

(61)

$${\mathcal{M}}_1=(q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0),$$

(62)

$${\mathcal{M}}_2=(q-1){\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-2}\mathcal{L}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0),$$

(63)

where ${\parallel x\parallel }_{L^2}={\left({\int }_0^1{x}^2\left(t\right)dt\right)}^{{\displaystyle \frac{1}{2}}}$.

Lemma 14.

Suppose that $\left({\mathbf{A}}_{\mathbf{4}}\right)$ holds. Then, for any x, $y\in C[0,1]$,

$$|\left(T_{1}x\right)\left(t\right)-\left(T_{1}y\right)\left(t\right)|\le \mathcal{L}t\parallel x-y\parallel ,t\in [0,1].$$

Proof. 

For any x, $y\in C[0,1]$, we have

$$\left|I_{0^{+}}^{\beta }x\left(t\right)-I_{0^{+}}^{\beta }y\left(t\right)\right|\le {\displaystyle \frac{t^{\beta }}{\Gamma (\beta +1)}}\parallel x-y\parallel ,t\in [0,1],$$

(64)

$$\left|\left(\mathcal{H}x\right)\left(t\right)-\left(\mathcal{H}y\right)\left(t\right)\right|\le {\displaystyle \frac{M_h}{\Gamma (\beta +1)}}\parallel x-y\parallel ,t\in [0,1],$$

(65)

which, together with $\left({\mathbf{A}}_{\mathbf{4}}\right)$ and Lemma 10(3), yields

$$\begin{array}{cc}\hfill & |\left(T_{1}x\right)\left(t\right)-\left(T_{1}y\right)\left(t\right)|\hfill \\ \hfill \le & {\int }_0^{1}V(t,s)\left|f(s,I_{0^{+}}^{\beta }x\left(s\right),\left(\mathcal{H}x\right)\left(s\right),x\left(s\right))ds-f(s,I_{0^{+}}^{\beta }y\left(s\right),\left(\mathcal{H}y\right)\left(s\right),y\left(s\right))\right|ds\hfill \\ \hfill \le & {\int }_0^{1}V(t,s)\left(l_1\left(s\right)|I_{0^{+}}^{\beta }x\left(s\right)-I_{0^{+}}^{\beta }y\left(s\right)|+l_2\left(s\right)|\left(\mathcal{H}x\right)\left(s\right)-\left(\mathcal{H}y\right)\left(s\right)|\right.\hfill \\ \hfill & +l_3\left(s\right)|x\left(s\right)-y\left(s\right)|)ds\hfill \\ \hfill \le & \left({\Lambda }_2{\int }_0^{1}t{(1-s)}^{\delta -1}\left({\displaystyle \frac{s^{\beta }{l}_1\left(s\right)}{\Gamma (\beta +1)}}+{\displaystyle \frac{M_h{l}_2\left(s\right)}{\Gamma (\beta +1)}}+l_3\left(s\right)\right)ds\right)\parallel x-y\parallel \hfill \\ \hfill \le & \left({\displaystyle \frac{1}{\Gamma (\beta +1)}}{\left({\int }_0^1{(1-s)}^{2\delta -2}{s}^{2\beta }ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{l}_1^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\right.\hfill \\ \hfill & +{\displaystyle \frac{M_h}{\Gamma (\beta +1)}}{\left({\int }_0^1{(1-s)}^{2\delta -2}ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{l}_2^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \left.+{\left({\int }_0^1{(1-s)}^{2\delta -2}ds\right)}^{{\displaystyle \frac{1}{2}}}{\left({\int }_0^1{l}_3^2\left(s\right)ds\right)}^{{\displaystyle \frac{1}{2}}}\right){\Lambda }_{2}t\parallel x-y\parallel \hfill \\ \hfill =& \left({\displaystyle \frac{{\mathcal{B}}^{\frac{1}{2}}(2\beta +1,2\delta -1){\parallel l_1\parallel }_{L^2}}{\Gamma (\beta +1)}}+{\displaystyle \frac{M_h{\mathcal{B}}^{\frac{1}{2}}(1,2\delta -1){\parallel l_2\parallel }_{L^2}}{\Gamma (\beta +1)}}\right.\hfill \\ \hfill & +{\mathcal{B}}^{{\displaystyle \frac{1}{2}}}(1,2\delta -1)\parallel l_3{\parallel }_{L^2}){\Lambda }_{2}t\parallel x-y\parallel \hfill \\ \hfill =& \mathcal{L}t\parallel x-y\parallel ,t\in [0,1].\hfill \end{array}$$

(66)

□

4.1. Uniqueness of Solutions for $1<q\le 2$

Lemma 15.

Suppose that $\left({\mathbf{A}}_{\mathbf{1}}\right)$ and $\left({\mathbf{A}}_{\mathbf{4}}\right)$ hold. Then, $T:C[0,1]\to C[0,1]$ is well defined.

Proof. 

By means of $\left({\mathbf{A}}_{\mathbf{1}}\right)$ and Lemma 10(4), we know, for any $x\in C[0,1]$,

$$\begin{array}{cc}\hfill \left(T_{1}x\right)\left(t\right)& ={\int }_0^{1}V(t,s)f(s,I_{0^{+}}^{\beta }x\left(s\right),\left(\mathcal{H}x\right)\left(s\right),x\left(s\right))ds\ge \kappa {\int }_0^{1}V(t,s)s^{\iota }ds\hfill \\ \hfill & ={\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}(t-t^{\delta +\iota }+\varpi t),t\in [0,1],\hfill \end{array}$$

(67)

which, together with (66) and Lemma 11(1), implies that for any x, $y\in C[0,1]$,

$$\begin{array}{cc}\hfill & \left|{\varphi }_q\left(\left(T_{1}x\right)\left(t\right)\right)-{\varphi }_q\left(\left(T_{1}y\right)\left(t\right)\right)\right|\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}(t-t^{\delta +\iota }+\varpi t)\right)}^{q-2}|\left(T_{1}x\right)\left(t\right)-\left(T_{1}y\right)\left(t\right)|\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi t\right)}^{q-2}\mathcal{L}t\parallel x-y\parallel \hfill \\ \hfill =& (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}{t}^{q-1}\parallel x-y\parallel ,t\in (0,1].\hfill \end{array}$$

(68)

Meanwhile,

$$\begin{array}{cc}\hfill & \left|{\varphi }_q\left(\left(T_{1}x\right)\left(0\right)\right)-{\varphi }_q\left(\left(T_{1}y\right)\left(0\right)\right)\right|\hfill \\ \hfill =& \left|{\varphi }_q\left({\int }_0^{1}V(0,s)f(s,I_{0^{+}}^{\beta }x\left(s\right),\left(\mathcal{H}x\right)\left(s\right),x\left(s\right))ds\right)\right.\hfill \\ \hfill & \left.-{\varphi }_q\left({\int }_0^{1}V(0,s)f(s,I_{0^{+}}^{\beta }y\left(s\right),\left(\mathcal{H}y\right)\left(s\right),y\left(s\right))ds\right)\right|=0\hfill \end{array}$$

(69)

follows from $V(0,s)=0$, for $s\in [0,1]$. Thus, combining with (66), we obtain

$$\begin{array}{cc}\hfill \left|Tx\left(t\right)\right|=& \left|{\int }_0^{1}G(t,s){\varphi }_q\left(\left(T_{1}x\right)\left(s\right)\right)ds\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s)\left|{\varphi }_q\left({\int }_0^{1}V(s,\tau )f(\tau ,I_{0^{+}}^{\beta }x\left(\tau \right),\left(\mathcal{H}x\right)\left(\tau \right),x\left(\tau \right))d\tau \right)\right.\hfill \\ \hfill & -\left.{\varphi }_q\left({\int }_0^{1}V(s,\tau )f(\tau ,0,0,0)d\tau \right)\right|ds\hfill \\ \hfill & +{\int }_0^{1}G(t,s)\left|{\varphi }_q\left({\int }_0^{1}V(s,\tau )f(\tau ,0,0,0)d\tau \right)\right|ds\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\parallel x\parallel {\int }_0^{1}G(t,s)s^{q-1}ds\hfill \\ \hfill & +{\int }_0^{1}G(t,s){\varphi }_q\left({\int }_0^1{\Lambda }_{2}s{(1-\tau )}^{\delta -1}f(\tau ,0,0,0)d\tau \right)ds\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\parallel x\parallel {\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{s}^{q-1}ds\hfill \\ \hfill & +{\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{\left({\Lambda }_2{M}_f\right)}^{q-1}{s}^{q-1}ds\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\parallel x\parallel \hfill \\ \hfill & +\mathsf{{\rm Y}}{\left({\Lambda }_2{M}_f\right)}^{q-1}\mathcal{B}(q,\alpha -z_0)<+\infty ,t\in [0,1].\hfill \end{array}$$

(70)

Thus, we deduce that $T:C[0,1]\to C[0,1]$ is well defined. □

Theorem 2.

Suppose that $\left({\mathbf{A}}_{\mathbf{1}}\right)$ and $\left({\mathbf{A}}_{\mathbf{4}}\right)$ hold. Then, the problem (1) has a unique solution $u_2$ provided that ${\mathcal{M}}_1<1$. Moreover, the solution $u_2$ satisfies

$$u_2\left(t\right)\ge \Theta \left(t\right),t\in [0,1],$$

where $\Theta \left(t\right)$ defined in (75).

Proof. 

Based on Lemma 15, we know, for any x, $y\in C[0,1]$,

$$\begin{array}{cc}\hfill & \left|Tx\left(t\right)-Ty\left(t\right)\right|\hfill \\ \hfill =& \left|{\int }_0^{1}G(t,s){\varphi }_q\left(\left(T_{1}x\right)\left(s\right)\right)ds-{\int }_0^{1}G(t,s){\varphi }_q\left(\left(T_{1}y\right)\left(s\right)\right)ds\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s)\left|{\varphi }_q\left(\left(T_{1}x\right)\left(s\right)\right)-{\varphi }_q\left(\left(T_{1}y\right)\left(s\right)\right)\right|ds\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\parallel x-y\parallel {\int }_0^{1}G(t,s)s^{q-1}ds\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\parallel x-y\parallel \hfill \\ \hfill & \times {\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{s}^{q-1}ds\hfill \\ \hfill \le & (q-1){\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\varpi \right)}^{q-2}\mathcal{L}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\parallel x-y\parallel \hfill \\ \hfill =& {\mathcal{M}}_1\parallel x-y\parallel ,t\in [0,1].\hfill \end{array}$$

(71)

That is,

$$\parallel Tx-Ty\parallel \le {\mathcal{M}}_1\parallel x-y\parallel .$$

Since ${\mathcal{M}}_1<1$, we can infer that $T:C[0,1]\to C[0,1]$ is a contraction mapping. It follows from the Banach contraction mapping principle that T has a unique fixed point $x_2$ in $C[0,1]$, i.e., the problem (2) has a unique solution $x_2$. Furthermore, by means of Lemma 5, we deduce that the problem (1) has a unique solution $u_2=I_{0^{+}}^{\beta }{x}_2$.

In the following, we prove that the solution $u_2$ has a lower bound given by (74). Based on (67) and Lemma 11(3), we have

$$\begin{array}{cc}\hfill x_2\left(t\right)& =\left(T{x}_2\right)\left(t\right)={\int }_0^{1}G(t,s){\varphi }_q\left(\left(T_1{x}_2\right)\left(s\right)\right)ds\hfill \\ \hfill & \ge {\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\right)}^{q-1}{\int }_0^{1}G(t,s){\varphi }_q\left(s-s^{\delta +\iota }+\varpi s\right)ds\hfill \\ \hfill & \ge 2^{q-2}{\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\right)}^{q-1}{\int }_0^{1}G(t,s)\left({\varphi }_q(s-s^{\delta +\iota })+{\varphi }_q\left(\varpi s\right)\right)ds\hfill \\ \hfill & \ge 2^{q-2}{\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\right)}^{q-1}{\int }_0^{1}G(t,s)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill & =\mu \left(t\right),t\in [0,1],\hfill \end{array}$$

(72)

where

$$\begin{array}{cc}\hfill \mu \left(t\right)=& 2^{q-2}{\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\right)}^{q-1}\left\{M_{\mathsf{{\rm Y}}}{t}^{\alpha -\beta -1}-{\mathsf{{\rm Y}}}_0\left(\mathcal{B}(q,\alpha -\beta )t^{q+\alpha -\beta -1}\right.\right.\hfill \\ \hfill & -\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -\beta )t^{(\delta +\iota )(q-1)+\alpha -\beta }\hfill \\ \hfill & \left.\left.+{\varpi }^{q-1}\mathcal{B}(q,\alpha -\beta )t^{q+\alpha -\beta -1}\right)\right\},\hfill \end{array}$$

(73)

with

$$\begin{array}{cc}\hfill M_{\mathsf{{\rm Y}}}=& \left(\mathcal{B}(q,\alpha -z_0)-\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_0)+{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_0)\right)\mathsf{{\rm Y}}\hfill \\ \hfill & -(1+{\varpi }^{q-1})\mathcal{B}(q,\alpha -z_1){\mathsf{{\rm Y}}}_{1,1}+\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_1){\mathsf{{\rm Y}}}_{1,2}\hfill \\ \hfill & -(1+{\varpi }^{q-1})\mathcal{B}(q,\alpha -z_2){\mathsf{{\rm Y}}}_{2,1}+\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_2){\mathsf{{\rm Y}}}_{2,2}\hfill \\ \hfill & -(1+{\varpi }^{q-1})\mathcal{B}(q,\alpha -z_3){\mathsf{{\rm Y}}}_{3,1}+\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_3){\mathsf{{\rm Y}}}_{3,2},\hfill \end{array}$$

$${\mathsf{{\rm Y}}}_{1,1}={\displaystyle \frac{k_1}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_1)}}{\int }_0^1{\tau }^{q+\alpha -z_1-1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right),$$

$${\mathsf{{\rm Y}}}_{1,2}={\displaystyle \frac{k_1}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_1)}}{\int }_0^1{\tau }^{(\delta +\iota )(q-1)+\alpha -z_1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right),$$

$${\mathsf{{\rm Y}}}_{2,1}={\displaystyle \frac{k_2}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }{\tau }^{q+\alpha -z_2-1}{b}_2\left(\tau \right)d{A}_2\left(\tau \right),$$

$${\mathsf{{\rm Y}}}_{2,2}={\displaystyle \frac{k_2}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }{\tau }^{(\delta +\iota )(q-1)+\alpha -z_2}{b}_2\left(\tau \right)d{A}_2\left(\tau \right),$$

$${\mathsf{{\rm Y}}}_{3,1}={\displaystyle \frac{k_3}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_3)}}\sum _{i=1}^{\infty }{a}_i{\xi }_i^{q+\alpha -z_3-1},$$

$${\mathsf{{\rm Y}}}_{3,2}={\displaystyle \frac{k_3}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_3)}}\sum _{i=1}^{\infty }{a}_i{\xi }_i^{(\delta +\iota )(q-1)+\alpha -z_3}.$$

Consequently, we obtain

$$\begin{array}{cc}\hfill u_2\left(t\right)=& I_{0^{+}}^{\beta }{x}_2\left(t\right)\ge {\displaystyle \frac{1}{\Gamma \left(\beta \right)}}{\int }_0^t{(t-s)}^{\beta -1}\mu \left(s\right)ds\hfill \\ \hfill =& {\displaystyle \frac{2^{q-2}}{\Gamma \left(\beta \right)}}{\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\right)}^{q-1}\left(M_{\mathsf{{\rm Y}}}{\int }_0^t{(t-s)}^{\beta -1}{s}^{\alpha -\beta -1}ds\right.\hfill \\ \hfill & -{\mathsf{{\rm Y}}}_0\mathcal{B}(q,\alpha -\beta ){\int }_0^t{(t-s)}^{\beta -1}{s}^{q+\alpha -\beta -1}ds\hfill \\ \hfill & +{\mathsf{{\rm Y}}}_0\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -\beta ){\int }_0^t{(t-s)}^{\beta -1}{s}^{(\delta +\iota )(q-1)+\alpha -\beta }ds\hfill \\ \hfill & \left.-{\mathsf{{\rm Y}}}_0{\varpi }^{q-1}\mathcal{B}(q,\alpha -\beta ){\int }_0^t{(t-s)}^{\beta -1}{s}^{q+\alpha -\beta -1}ds\right)\hfill \\ \hfill =& \Theta \left(t\right),t\in [0,1],\hfill \end{array}$$

(74)

where

$$\begin{array}{cc}\hfill \Theta \left(t\right)=& 2^{q-2}{\left({\displaystyle \frac{\kappa \mathcal{B}(\iota +1,\delta )}{\Gamma \left(\delta \right)}}\right)}^{q-1}\left(M_{\mathsf{{\rm Y}}}{\displaystyle \frac{\Gamma (\alpha -\beta )}{\Gamma \left(\alpha \right)}}{t}^{\alpha -1}-{\mathsf{{\rm Y}}}_0{\displaystyle \frac{\Gamma (q+\alpha -\beta )\mathcal{B}(q,\alpha -\beta )}{\Gamma (q+\alpha )}}{t}^{q+\alpha -1}\right.\hfill \\ \hfill & +{\mathsf{{\rm Y}}}_0{\displaystyle \frac{\Gamma \left(\right(\delta +\iota \left)\right(q-1)+\alpha -\beta +1)\mathcal{B}\left(\right(\delta +\iota \left)\right(q-1)+1,\alpha -\beta )}{\Gamma \left(\right(\delta +\iota \left)\right(q-1)+\alpha +1)}}{t}^{(\delta +\iota )(q-1)+\alpha }\hfill \\ \hfill & \left.-{\mathsf{{\rm Y}}}_0{\varpi }^{q-1}{\displaystyle \frac{\Gamma (q+\alpha -\beta )\mathcal{B}(q,\alpha -\beta )}{\Gamma (q+\alpha )}}\mathcal{B}(q,\alpha -\beta )t^{q+\alpha -1}\right).\hfill \end{array}$$

(75)

Remark 4.

Given that the lengthy expression of $G(t,s)$ and the resulting elaborate computation for

$${\int }_0^{1}G(t,s)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds,$$

we provide the detailed calculation below for completeness. By calculation, we obtain

$$\begin{array}{cc}\hfill & {\int }_0^1{P}_0(t,s)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill =& {\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}{\int }_0^1{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill & -{\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}{\int }_0^t{(t-s)}^{\alpha -\beta -1}\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill =& {\displaystyle \frac{t^{\alpha -\beta -1}}{\Gamma (\alpha -\beta )}}\left(\mathcal{B}(q,\alpha -z_0)-\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_0)+{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_0)\right)\hfill \\ \hfill & -{\displaystyle \frac{1}{\Gamma (\alpha -\beta )}}\left(t^{q+\alpha -\beta -1}\mathcal{B}(q,\alpha -\beta )-t^{(\delta +\iota )(q-1)+\alpha -\beta }\mathcal{B}((\delta +\iota )(q-1)+1,\right.\hfill \\ \hfill & \left.\alpha -\beta )+{\varpi }^{q-1}{t}^{q+\alpha -\beta -1}\mathcal{B}(q,\alpha -\beta )\right),\hfill \end{array}$$

(76)

$$\begin{array}{cc}\hfill & {\int }_0^1\left({\int }_0^1{P}_1(\tau ,s)b_1\left(\tau \right)d{A}_1\left(\tau \right)\right)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill =& {\int }_0^1\left({\int }_0^1{P}_1(\tau ,s)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\right)b_1\left(\tau \right)d{A}_1\left(\tau \right)\hfill \\ \hfill =& {\displaystyle \frac{k_1}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_1)}}{\int }_0^1{\tau }^{\alpha -z_1-1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right)\hfill \\ \hfill & \times \left(\mathcal{B}(q,\alpha -z_0)-\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_0)+{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_0)\right)\hfill \\ \hfill & -{\displaystyle \frac{k_1}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_1)}}(\mathcal{B}(q,\alpha -z_1){\int }_0^1{\tau }^{q+\alpha -z_1-1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right)\hfill \\ \hfill & -\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_1){\int }_0^1{\tau }^{(\delta +\iota )(q-1)+\alpha -z_1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right)\hfill \\ \hfill & +{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_1){\int }_0^1{\tau }^{q+\alpha -z_1-1}{b}_1\left(\tau \right)d{A}_1\left(\tau \right)),\hfill \end{array}$$

(77)

$$\begin{array}{cc}\hfill & {\int }_0^1\left({\int }_0^{\zeta }{P}_2(\tau ,s)b_2\left(\tau \right)d{A}_2\left(\tau \right)\right)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill =& {\int }_0^{\zeta }\left({\int }_0^1{P}_2(\tau ,s)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\right)b_2\left(\tau \right)d{A}_2\left(\tau \right)\hfill \\ \hfill =& {\displaystyle \frac{k_2}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_2)}}{\int }_0^{\zeta }{\tau }^{\alpha -z_2-1}{b}_2\left(\tau \right)d{A}_2\left(\tau \right)\hfill \\ \hfill & \times \left(\mathcal{B}(q,\alpha -z_0)-\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_0)+{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_0)\right)\hfill \\ \hfill & -{\displaystyle \frac{k_2}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_2)}}(\mathcal{B}(q,\alpha -z_2){\int }_0^{\zeta }{\tau }^{q+\alpha -z_2-1}{b}_2\left(\tau \right)d{A}_2\left(\tau \right)\hfill \\ \hfill & -\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_2){\int }_0^{\zeta }{\tau }^{(\delta +\iota )(q-1)+\alpha -z_2}{b}_2\left(\tau \right)d{A}_2\left(\tau \right)\hfill \\ \hfill & +{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_2){\int }_0^{\zeta }{\tau }^{q+\alpha -z_2-1}{b}_2\left(\tau \right)d{A}_2\left(\tau \right)),\hfill \end{array}$$

(78)

and

$$\begin{array}{cc}\hfill & {\int }_0^1{\displaystyle \sum _{i=1}^{\infty }}{a}_i{P}_3({\xi }_i,s)\left(s^{q-1}-s^{(\delta +\iota )(q-1)}+{\left(\varpi s\right)}^{q-1}\right)ds\hfill \\ \hfill =& {\displaystyle \frac{k_3}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_3)}}(\mathcal{B}(q,\alpha -z_0)-\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_0)\hfill \\ \hfill & +{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_0)){\displaystyle \sum _{i=1}^{\infty }}{a}_i{\xi }_i^{\alpha -z_3-1}\hfill \\ \hfill & -{\displaystyle \frac{k_3}{{\sigma }_1\Gamma (\alpha -\beta )\Gamma (\alpha -z_3)}}(\mathcal{B}(q,\alpha -z_3){\displaystyle \sum _{i=1}^{\infty }}{a}_i{\xi }_i^{q+\alpha -z_3-1}\hfill \\ \hfill & -\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -z_3){\displaystyle \sum _{i=1}^{\infty }}{a}_i{\xi }_i^{(\delta +\iota )(q-1)+\alpha -z_3}\hfill \\ \hfill & +{\varpi }^{q-1}\mathcal{B}(q,\alpha -z_3){\displaystyle \sum _{i=1}^{\infty }}{a}_i{\xi }_i^{q+\alpha -z_3-1}).\hfill \end{array}$$

(79)

From (76)–(79), we deduce that

$$\begin{array}{cc}\hfill & {\int }_0^{1}G(t,s)\left({\varphi }_q\left(s\right)-{\varphi }_q\left(s^{\delta +\iota }\right)+{\varphi }_q\left(\varpi s\right)\right)ds\hfill \\ \hfill =& M_{\mathsf{{\rm Y}}}{t}^{\alpha -\beta -1}-{\mathsf{{\rm Y}}}_0\left(\mathcal{B}(q,\alpha -\beta )t^{q+\alpha -\beta -1}\right.\hfill \\ \hfill & -\mathcal{B}((\delta +\iota )(q-1)+1,\alpha -\beta )t^{(\delta +\iota )(q-1)+\alpha -\beta }\left.+{\varpi }^{q-1}\mathcal{B}(q,\alpha -\beta )t^{q+\alpha -\beta -1}\right).\hfill \end{array}$$

(80)

□

Theorem 3.

Suppose that $\left({\mathbf{A}}_{\mathbf{2}}\right)$ and $\left({\mathbf{A}}_{\mathbf{4}}\right)$ hold. Then, the problem (1) has a unique solution $u_3$ provided that ${\mathcal{M}}_1<1$. Moreover, the solution $u_3$ satisfies

$$u_3\left(t\right)\le -\Theta \left(t\right),t\in [0,1].$$

Proof. 

Let $g(t,x,y,z)=-f(t,x,y,z)$. Then the proof is similar to Theorem 2. So, we omit the details here. □

4.2. Uniqueness of Solutions for $q>2$

Theorem 4.

Suppose that $\left({\mathbf{A}}_{\mathbf{3}}\right)$ and $\left({\mathbf{A}}_{\mathbf{4}}\right)$ hold. Then the problem (1) has a unique solution $u_4$ provided that ${\mathcal{M}}_2<1$. Moreover, the solution $u_4$ satisfies

$$|u_4\left(t\right)|\le {\displaystyle \frac{{\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-1}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\Gamma (\alpha -\beta )}{\Gamma \left(\alpha \right)}}{t}^{\alpha -1},t\in [0,1].$$

Proof. 

By using $\left({\mathbf{A}}_{\mathbf{3}}\right)$ and Lemma 10(3), we infer that for any $x\in C[0,1]$,

$$\begin{array}{cc}\hfill |\left(T_{1}x\right)\left(t\right)|& \le {\int }_0^{1}V(t,s)\left|f(s,I_{0^{+}}^{\beta }x\left(s\right),\left(\mathcal{H}x\right)\left(s\right),x\left(s\right))\right|ds\hfill \\ & \le {\int }_0^{1}V(t,s)\mathcal{F}\left(s\right)ds\hfill \\ & \le {\Lambda }_{2}t{\int }_0^1{(1-s)}^{\delta -1}\mathcal{F}\left(s\right)ds\hfill \\ & ={\Lambda }_2{M}_{\mathcal{F}}t,t\in [0,1],\hfill \end{array}$$

(81)

which, together with Lemma 11(2) and Lemma 14, implies that for any x, $y\in C[0,1]$,

$$\begin{array}{cc}\hfill & \left|{\varphi }_q\left(\left(T_{1}x\right)\left(t\right)\right)-{\varphi }_q\left(\left(T_{1}y\right)\left(t\right)\right)\right|\hfill \\ \hfill \le & (q-1){\left({\Lambda }_2{M}_{\mathcal{F}}t\right)}^{q-2}\left|\left(T_{1}x\right)\left(t\right)-\left(T_{1}y\right)\left(t\right)\right|\hfill \\ \hfill \le & (q-1){\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-2}\mathcal{L}{t}^{q-1}\parallel x-y\parallel ,t\in [0,1].\hfill \end{array}$$

(82)

Hence, for any x, $y\in C[0,1]$,

$$\begin{array}{cc}\hfill & \left|Tx\left(t\right)-Ty\left(t\right)\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s)\left|{\varphi }_q\left(\left(T_{1}x\right)\left(s\right)\right)-{\varphi }_q\left(\left(T_{1}y\right)\left(s\right)\right)\right|ds\hfill \\ \hfill \le & (q-1){\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-2}\mathcal{L}\parallel x-y\parallel {\int }_0^{1}G(t,s)s^{q-1}ds\hfill \\ \hfill \le & (q-1){\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-2}\mathcal{L}\parallel x-y\parallel {\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{s}^{q-1}ds\hfill \\ \hfill \le & (q-1){\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-2}\mathcal{L}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\parallel x-y\parallel \hfill \\ \hfill =& {\mathcal{M}}_2\parallel x-y\parallel ,t\in [0,1].\hfill \end{array}$$

(83)

That is,

$$\parallel Tx-Ty\parallel \le {\mathcal{M}}_2\parallel x-y\parallel .$$

(84)

Since ${\mathcal{M}}_2<1$, we deduce that $T:C[0,1]\to C[0,1]$ is a contraction mapping. Thus, by the Banach contraction mapping principle, we know that T has a unique fixed point $x_4$ in $C[0,1]$, i.e., the problem (2) has a unique solution $x_4$. Moreover, by using (81) and Lemma 9, we find an upper bound of $x_4$ on $[0,1]$:

$$\begin{array}{cc}\hfill |x_4\left(t\right)|=& \left|T{x}_4\left(t\right)\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s){\varphi }_q\left(\right|\left(T_1{x}_4\right)\left(s\right)\left|\right)ds\hfill \\ \hfill \le & {\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-1}{\int }_0^{1}G(t,s){\varphi }_q\left(s\right)ds\hfill \\ \hfill \le & {\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-1}{\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{s}^{q-1}ds\hfill \\ \hfill =& {\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-1}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)t^{\alpha -\beta -1},t\in [0,1].\hfill \end{array}$$

(85)

Finally, by using Lemma 5, we deduce that the problem (1) has a unique solution $u_4=I_{0^{+}}^{\beta }{x}_4$, which satisfies

$$|u_4\left(t\right)|\le {\displaystyle \frac{{\left({\Lambda }_2{M}_{\mathcal{F}}\right)}^{q-1}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\Gamma (\alpha -\beta )}{\Gamma \left(\alpha \right)}}{t}^{\alpha -1},t\in [0,1].$$

(86)

□

Lemma 16.

Suppose that $1\le M_h\le \Gamma (\beta +1)$ and $\left({\mathbf{A}}_{\mathbf{5}}\right)$ hold. Then, for any x, $y\in S_r$,

$$|T_{1}x\left(t\right)-T_{1}y\left(t\right)|\le {\mathcal{L}}_{r}t\parallel x-y\parallel ,t\in [0,1].$$

Proof. 

Obviously, for any $x\in S_r$,

$$\left|I_{0^{+}}^{\beta }x\left(t\right)\right|\le {\displaystyle \frac{r{t}^{\beta }}{\Gamma (\beta +1)}}\le r,t\in [0,1],$$

(87)

$$\left|\left(\mathcal{H}x\right)\left(t\right)\right|\le {\displaystyle \frac{r}{\Gamma (\beta +1)}}{\int }_0^{t}h(t,s)s^{\beta }ds\le {\displaystyle \frac{M_{h}r}{\Gamma (\beta +1)}}\le r,t\in [0,1].$$

(88)

Then, similar to the proof of Lemma 14, we can prove that

$$|\left(T_{1}x\right)\left(t\right)-\left(T_{1}y\right)\left(t\right)|\le {\mathcal{L}}_{r}t\parallel x-y\parallel ,t\in [0,1].$$

(89)

□

Lemma 17.

Suppose that $\left({\mathbf{A}}_{\mathbf{5}}\right)$ and the following conditions hold:

$\left(i\right)$ $1\le M_h\le \Gamma (\beta +1)$;

$\left(ii\right)$ There exists $r>0$ such that the function

$$U\left(r\right)=r-Z\left(r\right){(q-1)}^{-1}{\mathcal{L}}_r^{-1}({\mathcal{L}}_{r}r+{\Lambda }_2{M}_f)\ge 0,$$

(90)

where

$$Z\left(r\right)=(q-1){({\mathcal{L}}_{r}r+{\Lambda }_2{M}_f)}^{q-2}{\mathcal{L}}_r\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0).$$

Then, $T:S_r\to S_r$ is well defined.

Proof. 

From (87) and (88), we know, for any $x\in S_r$,

$$\left|I_{0^{+}}^{\beta }x\left(t\right)\right|\le r,\left|\left(\mathcal{H}x\right)\left(t\right)\right|\le r,t\in [0,1].$$

(91)

Thus, based on Lemma 16 and Lemma 10(3), we deduce that for any $x\in S_r$,

$$\begin{array}{cc}\hfill |T_{1}x\left(t\right)|=& \left|{\int }_0^{1}V(t,s)f(s,I_{0^{+}}^{\beta }x\left(s\right),\left(\mathcal{H}x\right)\left(s\right),x\left(s\right))ds\right|\hfill \\ \hfill \le & \left|{\int }_0^{1}V(t,s)\left(f(s,I_{0^{+}}^{\beta }x\left(s\right),\left(\mathcal{H}x\right)\left(s\right),x\left(s\right))-f(s,0,0,0)\right)ds\right|\hfill \\ \hfill & +\left|{\int }_0^{1}V(t,s)f(s,0,0,0)ds\right|\hfill \\ \hfill \le & {\mathcal{L}}_{r}rt+{\Lambda }_2{M}_{f}t,t\in [0,1].\hfill \end{array}$$

(92)

Hence, we have

$$\begin{array}{cc}\hfill \left|Tx\right(t\left)\right|=& \left|{\int }_0^{1}G(t,s){\varphi }_q\left(T_{1}x\left(s\right)\right)ds\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s){\varphi }_q\left({\mathcal{L}}_{r}rs+{\Lambda }_2{M}_{f}s\right)ds\hfill \\ \hfill \le & {({\mathcal{L}}_{r}r+{\Lambda }_2{M}_f)}^{q-1}{\int }_0^1\mathsf{{\rm Y}}{t}^{\alpha -\beta -1}{(1-s)}^{\alpha -z_0-1}{s}^{q-1}ds\hfill \\ \hfill \le & {({\mathcal{L}}_{r}r+{\Lambda }_2{M}_f)}^{q-1}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\le r,\hfill \end{array}$$

(93)

which means $T:S_r\to S_r$ is well defined. □

Theorem 5.

Suppose that $\left({\mathbf{A}}_{\mathbf{5}}\right)$ and the following conditions hold:

$\left(i\right)$ $1\le M_h\le \Gamma (\beta +1)$;

$\left(ii\right)$ There exists $r_{\ast }>0$ such that $Z\left(r_{\ast }\right)<1$ and $U\left(r_{\ast }\right)\ge 0$.

Then, the problem (1) has a unique solution $u_5$, which satisfies

$$|u_5\left(t\right)|\le {\displaystyle \frac{r_{\ast }}{\Gamma (\beta +1)}}{t}^{\beta },t\in [0,1].$$

Proof. 

By Lemma 17, we know $T:S_{r_{\ast }}\to S_{r_{\ast }}$ is well defined. In the following, we prove that T is a contraction mapping. Indeed, based on (92), Lemma 16 and Lemma 11(2), we know, for any x, $y\in S_{r_{\ast }}$,

$$\begin{array}{cc}\hfill & \left|{\varphi }_q\left(T_{1}x\left(t\right)\right)-{\varphi }_q\left(T_{1}y\left(t\right)\right)\right|\hfill \\ \hfill \le & (q-1){({\mathcal{L}}_{r_{\ast }}{r}_{\ast }t+{\Lambda }_2{M}_{f}t)}^{q-2}|T_{1}x\left(t\right)-T_{1}y\left(t\right)|\hfill \\ \hfill \le & (q-1){({\mathcal{L}}_{r_{\ast }}{r}_{\ast }t+{\Lambda }_2{M}_{f}t)}^{q-2}{\mathcal{L}}_{r_{\ast }}t\parallel x-y\parallel \hfill \\ \hfill =& (q-1){({\mathcal{L}}_{r_{\ast }}{r}_{\ast }+{\Lambda }_2{M}_f)}^{q-2}{\mathcal{L}}_{r_{\ast }}{t}^{q-1}\parallel x-y\parallel ,t\in [0,1].\hfill \end{array}$$

(94)

Thus, for any x, $y\in S_{r_{\ast }}$,

$$\begin{array}{cc}\hfill & \left|Tx\left(t\right)-Ty\left(t\right)\right|\hfill \\ \hfill \le & {\int }_0^{1}G(t,s)\left|{\varphi }_q\left(T_{1}x\left(s\right)\right)ds-{\varphi }_q\left(T_{1}y\left(s\right)\right)\right|ds\hfill \\ \hfill \le & (q-1){({\mathcal{L}}_{r_{\ast }}{r}_{\ast }+{\Lambda }_2{M}_f)}^{q-2}{\mathcal{L}}_{r_{\ast }}\parallel x-y\parallel {\int }_0^{1}G(t,s)s^{q-1}ds\hfill \\ \hfill \le & (q-1){({\mathcal{L}}_{r_{\ast }}{r}_{\ast }+{\Lambda }_2{M}_f)}^{q-2}{\mathcal{L}}_{r_{\ast }}\mathsf{{\rm Y}}\mathcal{B}(q,\alpha -z_0)\parallel x-y\parallel \hfill \\ \hfill =& Z\left(r_{\ast }\right)\parallel x-y\parallel ,t\in [0,1].\hfill \end{array}$$

(95)

Since $Z\left(r_{\ast }\right)<1$, $T:S_{r_{\ast }}\to S_{r_{\ast }}$ is a contraction mapping. In view of the Banach contraction mapping principle, we deduce that T has a unique fixed point $x_5$ in $S_{r_{\ast }}$, i.e., the problem (2) has a unique solution $x_5$ satisfying $\parallel x_5\parallel \le r_{\ast }$. Moreover, by using Lemma 5, we deduce that the problem (1) has a unique solution $u_5=I_{0^{+}}^{\beta }{x}_5$, which satisfies

$$|u_5\left(t\right)|\le {\displaystyle \frac{r_{\ast }}{\Gamma (\beta +1)}}{t}^{\beta },t\in [0,1].$$

(96)

□

5. Examples

Example 1.

We consider the following problem:

$$\left\{\begin{array}{l}-{}^{c}D_{0+}^{{\displaystyle \frac{5}{2}}}{\varphi }_5\left(u^{‴}\left(t\right)\right)+{\displaystyle \frac{4.1472\times {10}^8}{\sqrt{\pi }}}\left(t^{{\displaystyle \frac{3}{2}}}-8t^{{\displaystyle \frac{5}{2}}}+{\displaystyle \frac{144}{7}}{t}^{{\displaystyle \frac{7}{2}}}\right.\\ \left.-{\displaystyle \frac{64}{3}}{t}^{{\displaystyle \frac{9}{2}}}+{\displaystyle \frac{256}{33}}{t}^{{\displaystyle \frac{11}{2}}}\right)=0,0<t<1,\\ u\left(0\right)=u^{\prime }\left(0\right)=u^{‴}\left(0\right)={\left({\varphi }_5\left(u^{‴}\left(0\right)\right)\right)}^{″}={\varphi }_5\left(u^{‴}\left(1\right)\right)=0,\\ u^{\prime }\left(1\right)={\displaystyle \frac{3(2^{\frac{7}{4}}-1)}{3\times 2^{\frac{17}{5}}-2^{\frac{99}{20}}+2^{\frac{16}{5}}}}{\displaystyle \sum _{i=1}^{\infty }}{2}^{-i}{u}^{\prime }(2^{{\displaystyle \frac{16-5i}{20}}}).\end{array}\right.$$

(97)

Proof. 

Let $p=5$, $\delta ={\displaystyle \frac{5}{2}}$, $\alpha =3$$(n=3)$, $\beta =1$, ${\gamma }_1=1$, $z_i=1$$(i=0,1,2,3)$, $\zeta ={\displaystyle \frac{1}{7}}$, $k_0=0$, $k_1=0$, $k_2={\displaystyle \frac{1}{3}}$, $k_3=3(2^{{\displaystyle \frac{7}{4}}}-1){(3\times 2^{{\displaystyle \frac{17}{5}}}-2^{{\displaystyle \frac{99}{20}}}+2^{{\displaystyle \frac{16}{5}}})}^{-1}$, $b_0\left(s\right)=s^{-{\displaystyle \frac{1}{2}}}$, $b_1\left(s\right)=s^{-{\displaystyle \frac{1}{2}}}$, $b_2\left(s\right)\equiv 0$, $a_i=2^{-i}$, ${\xi }_i=2^{{\displaystyle \frac{16-5i}{20}}}$$(i=1,2,\dots )$, $\left(Hu\right)\left(t\right)={\int }_0^{t}u\left(s\right)ds$,

$$A_1\left(t\right)=\left\{\begin{array}{l}\Gamma \left({\displaystyle \frac{6}{5}}\right),t\in \left[0,{\displaystyle \frac{10}{11}}\right),\\ \Gamma \left({\displaystyle \frac{11}{5}}\right),t\in \left[{\displaystyle \frac{10}{11}},1\right],\end{array}\right.A_2\left(t\right)=\left\{\begin{array}{l}\Gamma \left({\displaystyle \frac{8}{7}}\right),t\in \left[0,{\displaystyle \frac{2}{7}}\right),\\ 3,t\in \left[{\displaystyle \frac{2}{7}},1\right].\end{array}\right.$$

$$f(t,u,v,w)={\displaystyle \frac{4.1472\times {10}^8}{\sqrt{\pi }}}\left(t^{{\displaystyle \frac{3}{2}}}-8t^{{\displaystyle \frac{5}{2}}}+{\displaystyle \frac{3856}{3465}}{t}^{{\displaystyle \frac{13}{2}}}-{\displaystyle \frac{464}{105}}{t}^{{\displaystyle \frac{1}{2}}}u+{\displaystyle \frac{7712}{1155}}{t}^{{\displaystyle \frac{1}{2}}}v+{\displaystyle \frac{72}{35}}{t}^{{\displaystyle \frac{1}{2}}}w\right).$$

Then, we can regard the problem (97) as a specific case of the problem (1). Let ${\theta }_1={\theta }_2={\theta }_3=1,$ and

$$a_0\left(t\right)={\displaystyle \frac{4.1472\times {10}^8}{\sqrt{\pi }}}\left(t^{{\displaystyle \frac{3}{2}}}-8t^{{\displaystyle \frac{5}{2}}}+{\displaystyle \frac{3856}{3465}}{t}^{{\displaystyle \frac{13}{2}}}\right),a_1\left(t\right)={\displaystyle \frac{1282.8672\times {10}^7}{7\sqrt{\pi }}}{t}^{{\displaystyle \frac{1}{2}}},$$

$$a_2\left(t\right)={\displaystyle \frac{21322.1376\times {10}^7}{77\sqrt{\pi }}}{t}^{{\displaystyle \frac{1}{2}}},a_3\left(t\right)={\displaystyle \frac{597.1968\times {10}^7}{7\sqrt{\pi }}}{t}^{{\displaystyle \frac{1}{2}}}.$$

Obviously, $a_i\in L^2(0,1)$$(i=0,1,2,3)$ and f satisfies the condition $\left({\mathbf{A}}_{\mathbf{0}}\right)$. Let $r_1={\displaystyle \frac{3}{2}}$. Then, all the assumptions of Theorem 1 are fulfilled. With these data, we obtain a solution $u_1$ to the problem (97) and establish an upper bound for $u_1$. Figure 1 and

Table 1. The solution $u_1$ with its upper bound.

t	0	0.2	0.4	0.6	0.8	1
$u_1\left(t\right)$	0	0.0037	0.0538	0.2462	0.6963	1.5000
The upper bound of $u_1\left(t\right)$	0	0.3000	0.6000	0.9000	1.2000	1.5000

show that the solution $u_1$ has an upper bound. This thereby verifies the validity and effectiveness of Theorem 1. □

Example 2.

We consider the following problem:

$$\left\{\begin{array}{l}-{}^{c}D_{0+}^3{\varphi }_2\left(D_{0+}^{{\displaystyle \frac{5}{2}}}u\left(t\right)\right)+{\displaystyle \frac{315\sqrt{\pi }}{8}}=0,0<t<1,\\ u\left(0\right)=u^{\prime }\left(0\right)=D_{0+}^{{\displaystyle \frac{5}{2}}}u\left(0\right)={\left({\varphi }_2\left(D_{0+}^{{\displaystyle \frac{5}{2}}}u\left(0\right)\right)\right)}^{″}=0,\\ {\varphi }_2\left(D_{0+}^{{\displaystyle \frac{5}{2}}}u\left(1\right)\right)=0,u^{\prime }\left(1\right)=0.\end{array}\right.$$

(98)

Proof. 

Let $p=2$, $\delta =3$, $\alpha ={\displaystyle \frac{5}{2}}$$(n=3)$, $\beta ={\displaystyle \frac{3}{4}}$, ${\gamma }_1=1$, $z_0=1$, $z_1={\displaystyle \frac{4}{5}}$, $z_2={\displaystyle \frac{5}{6}}$, $z_3={\displaystyle \frac{6}{7}}$, $\zeta ={\displaystyle \frac{13}{14}}$, $k_0=k_1=0$, $k_2=2$, $k_3=1$, $a_i\equiv 0$, ${\xi }_i=2^{-i}$$(i=1,2,\dots )$, $b_0\left(s\right)=s^{{\displaystyle \frac{1}{3}}}$, $b_1\left(s\right)=s^{-{\displaystyle \frac{1}{2}}}$, $b_2\left(s\right)\equiv 0$, $\left(Hu\right)\left(t\right)={\int }_0^{t}u\left(s\right)ds$,

$$A_1\left(t\right)=\left\{\begin{array}{l}{\displaystyle \frac{5}{7}},t\in \left[0,{\displaystyle \frac{1}{7}}\right),\\ {\displaystyle \frac{6}{7}},t\in \left[{\displaystyle \frac{1}{7}},1\right],\end{array}\right.A_2\left(t\right)=\left\{\begin{array}{l}{\displaystyle \frac{5}{8}},t\in \left[0,{\displaystyle \frac{1}{3}}\right),\\ {\displaystyle \frac{8}{9}},t\in \left[{\displaystyle \frac{1}{3}},1\right].\end{array}\right.$$

$$\begin{array}{cc}\hfill f(t,u,v,w)=& \left|u\right|+\left|v\right|+\left|w\right|+{\displaystyle \frac{315\sqrt{\pi }}{8}}-{\displaystyle \frac{13}{9}}{t}^{{\displaystyle \frac{3}{2}}}+t^{{\displaystyle \frac{7}{2}}}-{\displaystyle \frac{8}{33}}{t}^{{\displaystyle \frac{11}{2}}}-{\displaystyle \frac{26}{45}}{t}^{{\displaystyle \frac{5}{2}}}+{\displaystyle \frac{2}{9}}{t}^{{\displaystyle \frac{9}{2}}}\hfill \\ & -{\displaystyle \frac{16}{429}}{t}^{{\displaystyle \frac{13}{2}}}-{\displaystyle \frac{\sqrt{\pi }}{\Gamma (3/4)}}\left({\displaystyle \frac{13}{9}}{t}^{{\displaystyle \frac{3}{4}}}-{\displaystyle \frac{20}{11}}{t}^{{\displaystyle \frac{11}{4}}}+{\displaystyle \frac{128}{209}}{t}^{{\displaystyle \frac{19}{4}}}\right).\hfill \end{array}$$

Then, we can regard the problem (98) as the problem (1). It is easy to check that f satisfies $\left({\mathbf{A}}_{\mathbf{1}}\right)$ and $\left({\mathbf{A}}_{\mathbf{4}}\right)$ with $\kappa =10$, $\iota =2$, $l_1\left(t\right)=l_2\left(t\right)=l_3\left(t\right)=1$. And, ${\mathcal{M}}_1=0.1258<1$. Thus, all the assumptions of Theorem 2 are fulfilled. With these data, the numerical results in Figure 2 and Figure 3 and

Table 2. The solution $u_2$ with its lower bound.

t	0	0.2	0.4	0.6	0.8	1
$u_2\left(t\right)$	0	0.1257	0.3265	0.5186	0.6467	0.6869
The lower bound of $u_2\left(t\right)$	0	0.0023	0.0061	0.0098	0.0123	0.0120

show that the solution $u_2$ has a lower bound. Among them, Figure 2 illustrates the relationship between the solution $u_2$ and its lower bound. For a clearer visualization of the lower bound’s trajectory, we provide Figure 3. Thus, this numerical example confirms Theorem 2. □

Example 3.

We consider the following equation:

$$\left\{\begin{array}{l}-{}^{c}D_{0+}^{{\displaystyle \frac{21}{10}}}{\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{11}{5}}}u\left(t\right)\right)+{\displaystyle \frac{792\Gamma \left(\frac{1}{5}\right)}{125000\Gamma \left(\frac{1}{10}\right)}}{\left({\displaystyle \frac{897\Gamma \left(\frac{3}{5}\right)}{17\Gamma \left(\frac{2}{5}\right)}}\right)}^{{\displaystyle \frac{1}{2}}}{t}^{{\displaystyle \frac{1}{10}}}=0,0<t<1,\\ u\left(0\right)=u^{\prime }\left(0\right)=D_{0+}^{{\displaystyle \frac{11}{5}}}u\left(0\right)={\left({\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{11}{5}}}u\left(0\right)\right)\right)}^{″}=0,\\ {\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{11}{5}}}u\left(1\right)\right)=3{\int }_0^1{s}^{-{\displaystyle \frac{1}{5}}}{\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{11}{5}}}u\left(s\right)\right)ds,\\ D_{0+}^{{\displaystyle \frac{6}{5}}}u\left(1\right)={\displaystyle \frac{8}{\Gamma \left(\frac{32}{5}\right)}}{\int }_0^1{s}^{-{\displaystyle \frac{1}{2}}}{D}_{0+}^{{\displaystyle \frac{11}{10}}}u\left(s\right)d{A}_1\left(s\right)\\ +{\displaystyle \frac{8}{\Gamma \left(\frac{32}{5}\right)}}{\int }_0^{{\displaystyle \frac{10}{11}}}{s}^{-{\displaystyle \frac{4}{5}}}{D}_{0+}^{{\displaystyle \frac{4}{5}}}u\left(s\right)d{A}_2\left(s\right)+{\displaystyle \frac{1}{2\Gamma \left(\frac{32}{5}\right)}}{\displaystyle \sum _{i=1}^{\infty }}\Gamma \left({\displaystyle \frac{67}{10}}\right)2^{-{\displaystyle \frac{1}{2}}i}{D}_{0+}^{{\displaystyle \frac{9}{10}}}u\left(2^{-{\displaystyle \frac{5}{57}}i}\right),\end{array}\right.$$

(99)

where

$$A_1\left(t\right)=\left\{\begin{array}{l}1,t\in \left[0,{\displaystyle \frac{1}{2}}\right),\\ 1+\Gamma \left({\displaystyle \frac{13}{2}}\right),t\in \left[{\displaystyle \frac{1}{2}},1\right],\end{array}\right.A_2\left(t\right)=\left\{\begin{array}{l}1,t\in \left[0,{\displaystyle \frac{1}{2}}\right),\\ 1+\Gamma \left({\displaystyle \frac{34}{5}}\right),t\in \left[{\displaystyle \frac{1}{2}},1\right].\end{array}\right.$$

Proof. 

Let $p={\displaystyle \frac{3}{2}}$, $\delta ={\displaystyle \frac{21}{10}}$, $\alpha ={\displaystyle \frac{11}{5}}$$(n=3)$, $\beta =1$, ${\gamma }_1=1$, $z_0={\displaystyle \frac{6}{5}}$, $z_1={\displaystyle \frac{11}{10}}$, $z_2={\displaystyle \frac{4}{5}}$, $z_3={\displaystyle \frac{9}{10}}$, $k_0=3$, $k_1={\displaystyle \frac{8}{\Gamma \left(\frac{32}{5}\right)}}$, $k_2={\displaystyle \frac{8}{\Gamma \left(\frac{32}{5}\right)}}$, $k_3={\displaystyle \frac{1}{2\Gamma \left(\frac{32}{5}\right)}}$, $\zeta ={\displaystyle \frac{10}{11}}$, $b_0\left(s\right)=s^{-{\displaystyle \frac{1}{5}}}$, $b_1\left(s\right)=s^{-{\displaystyle \frac{1}{2}}}$, $b_2\left(s\right)=s^{-{\displaystyle \frac{4}{5}}}$, $\left(Hu\right)\left(t\right)={\int }_0^{t}u\left(s\right)ds$,

$$f(t,u,v,w)={\displaystyle \frac{792\Gamma \left(\frac{1}{5}\right)}{125000\Gamma \left(\frac{1}{10}\right)}}{\left({\displaystyle \frac{897\Gamma \left(\frac{3}{5}\right)}{17\Gamma \left(\frac{2}{5}\right)}}\right)}^{{\displaystyle \frac{1}{2}}}{\displaystyle \frac{t^{\frac{1}{10}}(\frac{t^{\frac{33}{5}}}{{10}^6}+1)(\frac{5t^{\frac{38}{5}}}{38\times {10}^6}+1)(\frac{33t^{\frac{28}{5}}}{5\times {10}^6}+1)}{\left(\right|u|+1)\left(\right|v|+1)\left(\right|w|+1)}}.$$

Then, we can regard the problem (99) as the problem (1). Let

$$\mathcal{F}\left(t\right)={\displaystyle \frac{792\Gamma \left(\frac{1}{5}\right)}{125000\Gamma \left(\frac{1}{10}\right)}}{\left({\displaystyle \frac{897\Gamma \left(\frac{3}{5}\right)}{17\Gamma \left(\frac{2}{5}\right)}}\right)}^{{\displaystyle \frac{1}{2}}}{t}^{{\displaystyle \frac{1}{10}}}({\displaystyle \frac{t^{\frac{33}{5}}}{{10}^6}}+1)({\displaystyle \frac{5t^{\frac{38}{5}}}{38\times {10}^6}}+1)({\displaystyle \frac{33t^{\frac{28}{5}}}{5\times {10}^6}}+1)$$

and

$$l_i\left(t\right)={\displaystyle \frac{792\Gamma \left(\frac{1}{5}\right)}{125000\Gamma \left(\frac{1}{10}\right)}}{\left({\displaystyle \frac{897\Gamma \left(\frac{3}{5}\right)}{17\Gamma \left(\frac{2}{5}\right)}}\right)}^{{\displaystyle \frac{1}{2}}}{t}^{{\displaystyle \frac{1}{10}}}({\displaystyle \frac{t^{\frac{33}{5}}}{{10}^6}}+1)({\displaystyle \frac{5t^{\frac{38}{5}}}{38\times {10}^6}}+1)({\displaystyle \frac{33t^{\frac{28}{5}}}{5\times {10}^6}}+1)(i=1,2,3).$$

It is easy to check that f satisfies conditions $\left({\mathbf{A}}_{\mathbf{3}}\right)$ and $\left({\mathbf{A}}_{\mathbf{4}}\right)$. Moreover, by using the above data, we obtain ${\mathcal{M}}_2\le 0.0029<1$. Therefore, all the assumptions of Theorem 4 are fulfilled. With these data, the numerical results in Figure 4 and Figure 5 as well as

Table 3. The solution $u_4$ with its upper bound.

t	0	0.2	0.4	0.6	0.8	1
$u_4\left(t\right)$$(\times {10}^{-6})$	0	0.0000	0.0024	0.0343	0.2293	1.0000
The upper bound of $u_4\left(t\right)$ $(\times {10}^{-4})$	0	0.7665	1.7610	2.8646	4.0457	5.2879

show that the solution $u_4$ has an upper bound. Among them, Figure 4 illustrates the relationship between the solution $u_4$ and its upper bound. For a clearer visualization of the trajectory of solution $u_4$, we provide Figure 5. Therefore, this numerical example demonstrates the validity of Theorem 4. □

Example 4.

We consider the following problem:

$$\left\{\begin{array}{l}-{}^{c}D_{0+}^3{\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{7}{2}}}u\left(t\right)\right)+{\displaystyle \frac{3}{5}}=0,0<t<1,\\ u\left(0\right)=D_{0+}^{{\displaystyle \frac{6}{7}}}u\left(0\right)=D_{0+}^{{\displaystyle \frac{13}{7}}}u\left(0\right)=0,\\ D_{0+}^{{\displaystyle \frac{7}{2}}}u\left(0\right)=0,{\left({\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{7}{2}}}u\left(0\right)\right)\right)}^{″}=0,\\ {\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{7}{2}}}u\left(1\right)\right)={\displaystyle \frac{7}{2}}{\int }_0^1{s}^{-{\displaystyle \frac{1}{2}}}{\varphi }_{{\displaystyle \frac{3}{2}}}\left(D_{0+}^{{\displaystyle \frac{7}{2}}}u\left(s\right)\right)ds,\\ D_{0+}^{{\displaystyle \frac{5}{2}}}u\left(1\right)=1152{\int }_0^1{D}_{0+}^{{\displaystyle \frac{1}{2}}}u\left(s\right)d{A}_1\left(s\right)\\ +64{\int }_0^{{\displaystyle \frac{10}{11}}}{D}_{0+}^{{\displaystyle \frac{3}{2}}}u\left(s\right)d{A}_2\left(s\right)+4{\displaystyle \sum _{i=1}^{\infty }}{2}^{-{\displaystyle \frac{1}{2}}i}{D}_{0+}^{{\displaystyle \frac{3}{2}}}u\left(2^{-{\displaystyle \frac{1}{16}}i}\right),\end{array}\right.$$

(100)

where

$$A_1\left(t\right)=\left\{\begin{array}{l}1,t\in \left[0,{\displaystyle \frac{1}{2}}\right),\\ 9,t\in \left[{\displaystyle \frac{1}{2}},1\right],\end{array}\right.A_2\left(t\right)=\left\{\begin{array}{l}3,t\in \left[0,{\displaystyle \frac{1}{2}}\right),\\ 11,t\in \left[{\displaystyle \frac{1}{2}},1\right].\end{array}\right.$$

Proof. 

Let $p={\displaystyle \frac{3}{2}}$, $\delta =3$, $\alpha ={\displaystyle \frac{7}{2}}$$(n=4)$, $\beta ={\displaystyle \frac{3}{2}}$, ${\gamma }_1={\displaystyle \frac{6}{7}}$, ${\gamma }_2={\displaystyle \frac{13}{7}}$, $z_0={\displaystyle \frac{5}{2}}$, $z_1={\displaystyle \frac{1}{2}}$, $z_2=z_3={\displaystyle \frac{3}{2}}$, $k_0={\displaystyle \frac{72}{9}}$, $k_1=1152$, $k_2=64$, $k_3=4$, $\zeta ={\displaystyle \frac{10}{11}}$, $a_i=2^{-{\displaystyle \frac{1}{2}}i}$, ${\xi }_i=2^{-{\displaystyle \frac{1}{16}}i}$$(i=1,2,\dots )$, $b_0\left(s\right)=s^{-{\displaystyle \frac{1}{2}}}$, $b_1\left(s\right)\equiv 1$, $b_2\left(s\right)\equiv 1$, $\left(Hu\right)\left(t\right)=\Gamma \left({\displaystyle \frac{7}{2}}\right){\int }_0^{t}u\left(s\right)ds$,

$$f(t,u,v,w)={\displaystyle \frac{3\left(\frac{2^{12}}{363738375\sqrt{\pi }}{t}^{\frac{19}{2}}+1\right)\left(\frac{2^{10}}{509233725}{t}^{\frac{21}{2}}+1\right)\left(\frac{1}{5600}{t}^8+1\right)}{5\left(\right|u|+1)\left(\right|v|+1)\left(\right|w|+1)}}.$$

Then, we can regard the problem (100) as the problem (1). By using the above data, it is easy to check that f satisfies the condition $\left({\mathbf{A}}_{\mathbf{5}}\right)$ with

$$l_{i,r}\left(t\right)={\displaystyle \frac{3}{5}}\left({\displaystyle \frac{2^{12}{t}^{\frac{19}{2}}}{363738375\sqrt{\pi }}}+1\right)\left({\displaystyle \frac{2^{10}{t}^{\frac{21}{2}}}{509233725}}+1\right)\left({\displaystyle \frac{t^8}{5600}}+1\right)(i=1,2,3).$$

Let $r_{\ast }=1$. Then, by calculation, we have $U\left(r_{\ast }\right)>0,$$Z\left(r_{\ast }\right)\le 0.0245<1,$ and $1<M_h=\Gamma (\beta +1)=\Gamma \left(2.5\right)$. Thus, all the assumptions of Theorem 5 are fulfilled. With these data, the numerical results in Figure 6 and Figure 7 and

Table 4. The solution $u_5$ with its upper bound.

t	0	0.2	0.4	0.6	0.8	1
$u_5\left(t\right)$$(\times {10}^{-5})$	0	0.0000	0.0001	0.0050	0.0763	0.6353
The upper bound of $u_5\left(t\right)$	0	0.0673	0.1903	0.3496	0.5383	0.7523

show that the solution $u_5$ has an upper bound. Among them, Figure 6 illustrates the relationship between the solution $u_5$ and its upper bound. For a clearer visualization of the trajectory of solution $u_5$, we provide Figure 7. Therefore, these numerical results validate the validity of Theorem 5. □

6. Conclusions

In this paper, we have addressed the problem of fractional differential equations with the p-Laplacian operator. By the method of reducing the order of fractional derivative, even when the nonlinear term involves fractional derivatives, we can still study the problem in the function space $C[0,1]$, which significantly simplifies our analysis. By employing the Schauder fixed point theorem and the Banach contraction principle, we established the existence and uniqueness of solutions. We have systematically validated our theoretical results through numerical examples, tabulated data, and graphical analyses. Furthermore, through analysis of the properties of Green’s function, we developed and applied a novel approach to approximate solutions. This approach demonstrated effectiveness and practicality in estimating solutions. Notably, the generalized boundary conditions adopted in this work provided a powerful modeling tool for physics phenomena.