Existence and Uniqueness of Solutions for Fractional Integro-Differential Equations Involving the Hadamard Derivatives

Abstract

In this paper, we study the existence and uniqueness of solutions for the following fractional boundary value problem, consisting of the Hadamard fractional derivative: ${}_H{D}^{\alpha }x\left(t\right)=Af(t,x\left(t\right))+{\sum }_{i=1}^k{C}_i{}_H{I}^{{\beta }_i}{g}_i(t,x\left(t\right)),t\in (1,e),$ supplemented with fractional Hadamard boundary conditions: ${}_H{D}^{\xi }x\left(1\right)=0,{}_H{D}^{\xi }x\left(e\right)=a{}_H{D}^{\frac{\alpha -\xi -1}{2}}\left({}_H{D}^{\xi }x\left(t\right)\right){|}_{t=\delta },\delta \in (1,e),$ where $1<\alpha \le 2$, $0<\xi \le \frac{1}{2}$, $a\in (0,\infty )$, $1<\alpha -\xi <2$, $0<{\beta }_i<1$, $A,C_i$, $1\le i\le k$, are real constants, ${}_H{D}^{\alpha }$ is the Hadamard fractional derivative of order $\alpha$ and ${}_H{I}^{{\beta }_i}$ is the Hadamard fractional integral of order ${\beta }_i$. By using some fixed point theorems, existence and uniqueness results are obtained. Finally, an example is given for demonstration.

1. Introduction

Fractional differential equations with Hadamard derivatives have been extensively investigated in recent years. In particular, the existence and uniqueness of the solution of Cauchy problems for fractional differential equations involving Hadamard derivatives was discussed in Kilbas et al. [1] in a nonsequential setting. Klimek [2], using the contraction mapping principle and a new, equivalent norm and metric, investigated the existence and uniqueness of the solution of sequential fractional differential equations with Hadamard derivative.

Recently, Wang et al. [3], using some classical methods, discussed the existence, blowing-up solutions and Ulam–Hyers stability of fractional differential equations with Hadamard derivatives. Further, Ahmad and Ntouyas [4], Ma et al. [5] and Aljoudi et al. [6] studied two-dimensional fractional differential systems with Hadamard derivatives. In [7], the authors discussed the existence at least one solution of a fractional impulsive Cauchy problem with Hadamard derivatives. In [8], by the fixed point theory and Hyers–Ulam stability, the authors discussed results of the existence, uniqueness and stability of a multi-point boundary value problem defined by a system of coupled fractional differential equations involving Hadamard derivatives.

In [9], Rezapour et al., using the topological degree and fixed point theories, investigated the existence results for solutions of a mixed multi-order Hadamard integro-derivative conditions involving the Caputo-Hadamard derivatives. The authors in [10], using Banach’s contraction mapping principle and the Leray–Schauder nonlinear alternative, studied the existence and uniqueness results of a class of impulsive boundary value problems consisting of mixed type fractional quantum and Hadamard derivatives. For more results on equations with Hadamard derivatives, we refer to [11,12,13] and the references therein.

In [14], the authors considered the following fractional boundary value problem

$$\left\{\begin{array}{c}-D^{\alpha }u\left(t\right)=Af(t,u\left(t\right))+B{I}^{\beta }g(t,u\left(t\right)),t\in (0,1),\hfill \\ D^{\delta }u\left(0\right)=0,D^{\delta +1}u\left(0\right)=0,D^{\delta }u\left(1\right)-D^{\delta }u\left(\eta \right)=a,\hfill \end{array}\right.$$

where $D_{0^{+}}^{\alpha }$ is the standard Riemann–Liouville fractional derivative of order $2<\alpha \le 3$, $0<\beta <1$, $\alpha -\delta >3$, $0<\eta <1$, $f,g$ are given continuous functions, and $A,B$ are real constants. They obtained the existence of positive solutions by means of Sadovskii’s fixed point theorem.

Zhou et al. [15] considered the existence of a positive solution for the singular nonlinear fractional differential equation boundary value problem:

$$\left\{\begin{array}{c}{D}^{\alpha }u\left(t\right)+f(t,u\left(t\right))=0,t\in (0,1),\hfill \\ u\left(0\right)=0,u\left(1\right)=a{D}^{\frac{\alpha -1}{2}}u\left(t\right){|}_{t=\xi },\hfill \end{array}\right.$$

(1)

where $1<\alpha \le 2$, $0<\xi \le \frac{1}{2}$, $a\in (0,\infty )$, $a\Gamma \left(\alpha \right){\xi }^{\frac{\alpha -1}{2}}<\Gamma \left(\frac{\alpha +1}{2}\right)$, and $f:(0,1]\times [0,+\infty )\to [0,+\infty )$ with ${lim}_{t\to 0^{+}}f(t,·)=+\infty$. Using a fixed-point theorem in a partially ordered set, it was proved that problem (1) has one positive solution.

Inspired by the above works, in this paper, we consider the following fractional boundary value problem with a Hadamard derivative:

$$\left\{\begin{array}{c}{\displaystyle {}_H{D}^{\alpha }x\left(t\right)=Af(t,x\left(t\right))+\sum _{i=1}^k{C}_i{}_H{I}^{{\beta }_i}{g}_i(t,x\left(t\right)),t\in (1,e),}\hfill \\ {}_H{D}^{\xi }x\left(1\right)=0,{}_H{D}^{\xi }x\left(e\right)=a{}_H{D}^{\frac{\alpha -\xi -1}{2}}\left({}_H{D}^{\xi }x\left(t\right)\right){|}_{t=\delta },\delta \in (1,e),\hfill \end{array}\right.$$

(2)

where $1<\alpha \le 2$, $0<\xi \le \frac{1}{2}$, $a\in \left(0,\frac{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}}\right)$, $1<\alpha -\xi <2$, $0<{\beta }_i<1$, $A,C_i$, $1\le i\le k$, are real constants, ${}_H{D}^{\alpha }$ is the Hadamard fractional derivative of order $\alpha$, and ${}_H{I}^{{\beta }_i}$ is the Hadamard fractional integral of order ${\beta }_i$.

We organize this paper as follows. In Section 2, we give some preliminaries. In Section 3, existence and uniqueness results for the problem (2) are studied via the Leray–Schauder nonlinear alternative, Krasnosel’skiĭ fixed point theorem and Banach’s fixed point theorem. Finally, in Section 4, an example is given to demonstrate the applications of our main results.

2. Preliminaries

In this section, we introduce some notations, basic definitions of fractional calculus and preliminary lemmas.

Definition 1

([1]). The Hadamard derivative of fractional order α for a function $h:[1,\infty )⟶\mathbb{R}$ is defined as:

$$\begin{array}{c}\hfill {}_H{D}^{\alpha }g\left(t\right)=\frac{1}{\Gamma (n-\alpha )}{\left(t\frac{d}{dt}\right)}^n{\int }_1^t{\left(log\frac{t}{s}\right)}^{n-\alpha -1}\frac{g\left(s\right)}{s}ds,n-1<\alpha <n,\end{array}$$

where $n=\left[\alpha \right]+1$, $\left[\alpha \right]$ denotes the integer part of the real number α and $log(·)={log}_e(·)$.

Definition 2

([1]). The Hadamard fractional integral of order α for a function h is defined as:

$$\begin{array}{c}\hfill {}_H{I}^{\alpha }h\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left(log\frac{t}{s}\right)}^{\alpha -1}\frac{h\left(s\right)}{s}ds,\alpha >0,\end{array}$$

provided the integral exists.

Furthermore, we will introduce the weighted space $C_{\gamma ,log}[a,b](0\le \gamma <1)$ of the function g on the finite interval $[a,b]$:

$$\begin{array}{c}\hfill C_{\gamma ,log}[a,b]=\left\{g:{\left(log\frac{t}{a}\right)}^{\gamma }g\left(t\right)\in C[a,b]\right\},\end{array}$$

with the norm:

$$\begin{array}{c}\hfill {\parallel g\parallel }_{C_{\gamma ,log}}=\underset{t\in [a,b]}{max}\left|{\left(log\frac{t}{a}\right)}^{\gamma }g\left(t\right)\right|.\end{array}$$

Clearly, $C_{0,log}[a,b]=C[a,b]$.

Lemma 1

([1]). For $\alpha >0$, $\beta >0$, and $0<a<b<\infty$, the following properties hold, when $f\in C_{\gamma ,log}[a,b](0\le \gamma <1)$, ${}_H{J}_{a^{+}}^{\alpha }{}_H{J}_{a^{+}}^{\beta }f{=}_H{J}_{a^{+}}^{\alpha +\beta }f$ and ${}_H{D}_{a^{+}}^{\alpha }{}_H{J}_{a^{+}}^{\alpha }f=f$.

Lemma 2

([1]). If $\alpha >0$, $\beta >0$, and $0<a<b<\infty$, then

$$\begin{array}{ccc}& & \left({}_H{I}_{a^{+}}^{\alpha }{\left(log\frac{t}{a}\right)}^{\beta -1}\right)\left(x\right)=\frac{\Gamma \left(\beta \right)}{\Gamma (\beta +\alpha )}{\left(log\frac{x}{a}\right)}^{\beta +\alpha -1},\hfill \\ & & \left({}_H{D}_{a^{+}}^{\alpha }{\left(log\frac{t}{a}\right)}^{\beta -1}\right)\left(x\right)=\frac{\Gamma \left(\beta \right)}{\Gamma (\beta -\alpha )}{\left(log\frac{x}{a}\right)}^{\beta -\alpha -1}.\hfill \end{array}$$

In particular, for $0<\alpha <1$, it holds that $\left({}_H{D}_{a^{+}}^{\alpha }{(log\frac{t}{a})}^{\alpha -1}\right)\left(x\right)=0$.

Lemma 3.

For $1<\alpha \le 2$, by the substitution $x\left(t\right)={}_H{I}^{\xi }u\left(t\right)$, problem (2) is turned into

$$\left\{\begin{array}{c}{\displaystyle {}_H{D}^{\alpha -\xi }u\left(t\right)=:g\left(t\right)=Af(t,{}_H{I}^{\xi }u\left(t\right))+\sum _{j=1}^k{C}_j{}_H{I}^{{\beta }_i}{g}_j(t,{}_H{I}^{\xi }u\left(t\right)),}\hfill \\ u\left(1\right)=0,u\left(e\right)=a{}_H{D}^{\frac{\alpha -\xi -1}{2}}u\left(t\right){|}_{t=\delta }.\hfill \end{array}\right.$$

(3)

For any $g\in C_{\gamma ,log}([1,e],\mathbb{R})$, the unique solution of the boundary value problem

$$\begin{array}{c}\hfill \left\{\begin{array}{c}{}_H{D}^{\alpha -\xi }u\left(t\right)=g\left(t\right),t\in (1,e),\hfill \\ u\left(1\right)=0,u\left(e\right)=a{}_H{D}^{\frac{\alpha -\xi -1}{2}}u\left(t\right)){|}_{t=\delta }\hfill \end{array}\right.\end{array}$$

(4)

is given by

$$\begin{array}{c}\hfill u\left(t\right)={}_H{I}^{\alpha -\xi }g\left(t\right)+\frac{{(logt)}^{\alpha -\xi -1}\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}}\left(-{}_H{I}^{\alpha -\xi }g\left(e\right)+a{}_H{I}^{\frac{\alpha -\xi +1}{2}}g\left(\delta \right)\right).\end{array}$$

Proof. 

As argued in [1], the solution of Hadamard differential equation in (4) can be written as

$$u\left(t\right)={}_H{I}^{\alpha -\xi }g\left(t\right)+c_1{(logt)}^{\alpha -\xi -1}+c_2{(logt)}^{\alpha -\xi -2},$$

(5)

for some arbitrary constants $c_1,c_2\in \mathbb{R}$.

The boundary condition $u\left(1\right)=0$ implies that $c_2=0$. Thus, one can obtain

$$u\left(e\right)={}_H{I}^{\alpha -\xi }g\left(e\right)+c_1,$$

and it follows from Lemma (2) that

$$\begin{array}{ccc}\hfill {}_H{D}^{\frac{\alpha -\xi -1}{2}}u\left(t\right)& =& {}_H{D}^{\frac{\alpha -\xi -1}{2}}{}_H{I}^{\alpha -\xi }g\left(t\right)+c_1{}_H{D}^{\frac{\alpha -\xi -1}{2}}{(logt)}^{\alpha -\xi -1}\hfill \\ & =& {}_H{I}^{\frac{\alpha -\xi +1}{2}}g\left(t\right)+c_1\frac{\Gamma (\alpha -\xi )}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{(logt)}^{\frac{\alpha -\xi -1}{2}}.\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill {}_H{D}^{\frac{\alpha -\xi -1}{2}}u\left(t\right){|}_{t=\delta }={\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{g\left(s\right)}{s}ds+c_1\frac{\Gamma (\alpha -\xi )}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{(log\delta )}^{\frac{\alpha -\xi -1}{2}}.\end{array}$$

Thus, in view of the boundary condition $u\left(e\right)=a{}_H{D}^{\frac{\alpha -\xi -1}{2}}u\left(t\right)){|}_{t=\delta }$, one has

$$\begin{array}{ccc}\hfill c_1& =& \frac{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}}[-{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{g\left(s\right)}{s}ds\hfill \\ & & +a{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{g\left(s\right)}{s}ds].\hfill \end{array}$$

Thus, the solution of problem (4) is as follows

$$\begin{array}{ccc}\hfill u\left(t\right)& =& {}_H{I}^{\alpha -\xi }g\left(t\right)+\frac{{(logt)}^{\alpha -\xi -1}\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}}\hfill \\ & & \times [-{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{g\left(s\right)}{s}ds\hfill \\ & & +a{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{g\left(s\right)}{s}ds].\hfill \end{array}$$

The proof is complete. □

Therefore, we can write the solution of the problem (2) as

$$\begin{array}{ccc}\hfill x\left(t\right)& =& {}_H{I}^{\xi }u\left(t\right)\hfill \\ & =& {}_H{I}^{\xi }[{}_H{I}^{\alpha -\xi }g\left(t\right)+\frac{{(logt)}^{\alpha -\xi -1}\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}}\hfill \\ & & \times \left(-{}_H{I}^{\alpha -\xi }g\left(e\right)+a{}_H{I}^{\frac{\alpha -\xi +1}{2}}g\left(\delta \right)\right)]\hfill \\ & =& {}_H{I}^{\alpha }g\left(t\right)+\frac{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}}\hfill \\ & & \times \left(-{}_H{I}^{\alpha -\xi }g\left(e\right)+a{}_H{I}^{\frac{\alpha -\xi +1}{2}}g\left(\delta \right)\right)\left({}_H{I}^{\xi }{(logt)}^{\alpha -\xi -1}\right),\hfill \end{array}$$

or

$$\begin{array}{ccc}\hfill x\left(t\right)& =& {}_H{I}^{\alpha }g\left(t\right)+\frac{{(logt)}^{\alpha -1}\Gamma (\alpha -\xi )\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\alpha \right)\left(\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}\right)}\hfill \\ & & \times \left(-{}_H{I}^{\alpha -\xi }g\left(e\right)+a{}_H{I}^{\frac{\alpha -\xi +1}{2}}g\left(\delta \right)\right).\hfill \end{array}$$

(6)

We can obtain the solution of problem (2) by replacing g with

$$Af(t,x\left(t\right))+\sum _{i=1}^k{C}_i{}_H{I}^{{\beta }_i}{g}_i(t,x\left(t\right)).$$

In this paper, we use the real Banach space

$$\mathcal{C}=C\left(\right[1,e],\mathbb{R}):\{f:[1,e]\to \mathbb{R}|f\text{is continuous}\}$$

endowed with the norm $\parallel x\parallel ={sup}_{t\in [1,e]}\left|x\left(t\right)\right|$.

Define an operator $\mathcal{T}:\mathcal{C}\to \mathcal{C}$ by

$$\begin{array}{ccc}\hfill \left(\mathcal{T}x\right)\left(t\right)& =& A{\int }_1^t\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t}{s}\right)}^{\alpha -1}\frac{f(s,x(s\left)\right)}{s}ds\hfill \\ & & +\sum _{i=1}^k{C}_i{\int }_1^t\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}\frac{g_i(s,x\left(s\right))}{s}ds\hfill \\ & & +{(logt)}^{\alpha -1}{A}_0[-A{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{f(s,x(s\left)\right)}{s}ds\hfill \\ & & -\sum _{i=1}^k{C}_i{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\frac{g_i(s,x\left(s\right))}{s}ds\hfill \\ & & +aA{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{f(s,x(s\left)\right)}{s}ds\hfill \\ & & +a\sum _{i=1}^k{C}_i{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}\frac{g_i(s,x\left(s\right))}{s}ds],\hfill \end{array}$$

where

$$A_0=\frac{\Gamma (\alpha -\xi )\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\Gamma \left(\alpha \right)\left(\Gamma \left(\frac{\alpha -\xi +1}{2}\right)-a\Gamma (\alpha -\xi ){(log\delta )}^{\frac{\alpha -\xi -1}{2}}\right)}.$$

Clearly, a fixed point of $\mathcal{T}$ on $\mathcal{C}$ is equivalent to a positive solution for the problem (2).

3. Main Results

In this section, by using the Leray–Schauder nonlinear alternative, Krasnosel’skiĭ fixed point theorem and Banach’s fixed point theorem, we study the existence and uniqueness of solutions of problem (2).

Theorem 1.

Let $f,g_i:[1,e]\times \mathbb{R}\to \mathbb{R}$, $i=1,\dots ,k,$ be continuous functions such that $f(t,x),g_i(t,x)\in C_{\gamma ,log}([1,e],\mathbb{R})$ with $0<\gamma <1$ and $\alpha -\xi +\gamma >0(i=1,\dots ,k)$ such that:

(A1) 

$|f(t,x_1)-f(t,x_2)|\le {\zeta }_1|x_1-x_2|$, $|g_i(t,x_1)-g_i(t,x_2)|\le {\zeta }_{i+1}|x_1-x_2|$, $t\in [1,e]$, ${\zeta }_i>0$, $i=1,\cdots ,k$, $x_1,x_2\in \mathbb{R}$ .

Then, the boundary value problem (2) has a unique solution if $\zeta <\frac{1}{\Delta }$, where $\zeta =max\{{\zeta }_i:i=1,\dots ,k+1\}$ and

$$\begin{array}{ccc}\hfill \Delta & =& \underset{t\in [1,e]}{sup}\{\left|A\right|\left[\frac{{(logt)}^{\alpha }}{\Gamma (\alpha +1)}+\left|A_0\right|{(logt)}^{\alpha -1}\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right)\right]\hfill \\ & & +\sum _{i=1}^k|C_i|[\frac{{(logt)}^{\alpha +{\beta }_i}}{\Gamma (\alpha +{\beta }_i+1)}+\left|A_0\right|{(logt)}^{\alpha -1}(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\left)\right]\}.\hfill \end{array}$$

(7)

Proof. 

Assume that $M=max\{M_i:i=1,\dots ,k+1\},$ where ${sup}_{t\in [1,e]}\left|f(t,1)\right|=M_1$, ${sup}_{t\in [1,e]}\left|g_i(t,1)\right|=M_{i+1}$. Selecting $r>\frac{\Delta M}{1-L\Delta }$ and $B_r=\{x\in \mathcal{C}:\parallel x\parallel \le r\}$, we show that $\mathcal{T}{B}_r\subset B_r$. In view of (7) and the fact $\left|f(s,x\left(s\right))\right|\le |f(s,x\left(s\right))-f(s,1)|+\left|f(s,1)\right|\le L_{1}r+M_1$, $|g_i(s,x\left(s\right))|\le |g_i(s,x\left(s\right))-g_i(s,1)|+|g_i(s,1)|\le L_{i+1}r+M_{i+1},i=1,\dots ,k,$ for $x\in B_r$, we find

$$\begin{array}{ccc}\hfill \parallel \left(\mathcal{T}x\right)\parallel & \le & (\zeta r+M)\underset{t\in [1,e]}{sup}\left\{\left|A\right|\right[\frac{{(logt)}^{\alpha }}{\Gamma (\alpha +1)}\hfill \\ & & +|A_0{|(logt)}^{\alpha -1}\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right)]+\sum _{i=1}^k|C_i|[\frac{{(logt)}^{\alpha +{\beta }_i}}{\Gamma (\alpha +{\beta }_i+1)}\hfill \\ & & +|A_0{|(logt)}^{\alpha -1}\left(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\right)\left]\right\}\hfill \\ & \le & (\zeta r+M)\Delta <r,\hfill \end{array}$$

which implies that $\mathcal{T}{B}_r\subset B_r$. Now, for $x,y\in \mathcal{C}$, we obtain:

$$\begin{array}{ccc}\hfill \parallel \mathcal{T}x-\mathcal{T}y\parallel & \le & \underset{t\in [1,e]}{sup}\{\left|A\right|{\int }_1^t\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t}{s}\right)}^{\alpha -1}\left|\frac{f(s,x(s\left)\right)}{s}-\frac{f(s,y(s\left)\right)}{s}\right|ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^t\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}\left|\frac{g_i(s,x\left(s\right))}{s}-\frac{g_i(s,y\left(s\right))}{s}\right|ds\hfill \\ & & +|A_0{|(logt)}^{\alpha -1}[\left|A\right|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\left|\frac{f(s,x(s\left)\right)}{s}-\frac{f(s,y(s\left)\right)}{s}\right|ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\left|\frac{g_i(s,x\left(s\right))}{s}-\frac{g_i(s,y\left(s\right))}{s}\right|ds\hfill \\ & & +\left|A\right|a{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\left|\frac{f(s,x(s\left)\right)}{s}-\frac{f(s,y(s\left)\right)}{s}\right|ds\hfill \\ & & +a\sum _{i=1}^k|C_i|{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi -1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i-1}\left|\frac{g_i(s,x\left(s\right))}{s}-\frac{g_i(s,y\left(s\right))}{s}\right|ds\left]\right\}\hfill \\ & \le & \zeta \underset{t\in [1,e]}{sup}\{\left|A\right|\left[\frac{{(logt)}^{\alpha }}{\Gamma (\alpha +1)}+\left|A_0\right|{(logt)}^{\alpha -1}\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right)\right]\hfill \\ & & +\sum _{i=1}^k|C_i|[\frac{{(logt)}^{\alpha +{\beta }_i}}{\Gamma (\alpha +{\beta }_i+1)}+\left|A_0\right|{(logt)}^{\alpha -1}(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\left)\right]\}\parallel x-y\parallel \hfill \\ & =& \zeta \Delta \parallel x-y\parallel .\hfill \end{array}$$

Since $\zeta <\frac{1}{\Delta }$ then $\mathcal{T}$ is a contraction. Thus, by using Banach fixed point theorem, the proof is completed. □

Theorem 2.

Assume that $f,g_i:[1,e]\times \mathbb{R}\to \mathbb{R}$, $i=1,\dots ,k,$ are continuous functions such that $f(t,x),g_i(t,x)\in C_{\gamma ,log}([1,e],\mathbb{R})$ with $0<\gamma <1$, $\alpha -\xi +\gamma >0(i=1,\dots ,k)$ and satisfy the following conditions:

(A2) 

$|f(t,x_1\left(t\right))-f(t,x_2\left(t\right))|\le \mu \left(t\right)|x_1-x_2|,|g_i(t,x_1\left(t\right))-g_i(t,x_2\left(t\right))|\le {\varrho }_i\left(t\right)|x_1-x_2|$, for $t\in [1,e]$, $x_1,x_2\in \mathbb{R}$, $\mu \left(t\right),{\varrho }_i\left(t\right)\in L^{\frac{1}{\rho }}([1,e],{\mathbb{R}}^{+})$ and $\rho \in (0,min\{1,\alpha +\beta ,\alpha -\xi ,\frac{\alpha -\xi +1}{2}+{\beta }_i\})$ for $i=1,\dots ,k;$

(A3) 

${\displaystyle \left|A\right|\parallel m\parallel Z_1+\sum _{i=1}^k|C_i|\parallel n_i\parallel Z_{i+1}<1}$, where

$$\begin{array}{ccc}\hfill Z_1& =& \frac{1}{\Gamma \left(\alpha \right)}{\left(\frac{1-\rho }{\alpha -\rho }\right)}^{1-\rho }+\frac{|A_0|}{\Gamma (\alpha -\xi )}{\left(\frac{1-\rho }{\alpha -\xi -\rho }\right)}^{1-\rho }\hfill \\ & & +\frac{a|A_0|}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(\frac{1-\rho }{\frac{\alpha -\xi +1}{2}-\rho }\right)}^{1-\rho }{(log\delta )}^{\frac{\alpha -\xi +1}{2}-\rho },\hfill \\ \hfill Z_{i+1}& =& \frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(\frac{1-\rho }{\alpha +{\beta }_i-\rho }\right)}^{1-\rho }+\frac{|A_0|}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(\frac{1-\rho }{\alpha -\xi +{\beta }_i-\gamma }\right)}^{1-\gamma }\hfill \\ & & +\left(\frac{a|A_0|}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}\right){\left(\frac{1-\rho }{\frac{\alpha -\xi +1}{2}+{\beta }_i-\rho }\right)}^{1-\rho }{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i-\rho },(i=1,\dots ,k),\hfill \end{array}$$

and $\parallel m\parallel =({\int }_0^1{\left|m\left(s\right)\right|}^{\frac{1}{\rho }}{ds)}^{\rho },m=\mu ,{\varrho }_i$.

Then, the problem (2) has a unique solution.

Proof. 

From condition (A2) and Hölder inequality, one can find

$$\begin{array}{ccc}& & \parallel \mathcal{T}x-\mathcal{T}y\parallel \hfill \\ & \le & \underset{t\in [1,e]}{sup}\left\{\left|A\right|{\int }_1^t\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t}{s}\right)}^{\alpha -1}\mu \left(s\right)\right|\frac{x\left(s\right)}{s}-\frac{y\left(s\right)}{s}|ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^t\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}{\varrho }_i\left(s\right)|\frac{x\left(s\right)}{s}-\frac{y\left(s\right)}{s}|ds\hfill \\ & & +|A_0\left|\right[\left|A\right|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\mu \left(s\right)|\frac{x\left(s\right)}{s}-\frac{y\left(s\right)}{s}|ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}{\varrho }_i\left(s\right)|\frac{x\left(s\right)}{s}-\frac{y\left(s\right)}{s}|ds\hfill \\ & & +a\left|A\right|{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\mu \left(s\right)|\frac{x\left(s\right)}{s}-\frac{y\left(s\right)}{s}|ds\hfill \\ & & +a\sum _{i=1}^k|C_i|{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi -1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}{\varrho }_i\left(s\right)|\frac{x\left(s\right)}{s}-\frac{y\left(s\right)}{s}|ds\left]\right\}\hfill \\ & \le & \underset{t\in [1,e]}{sup}\{\frac{\left|A\right|\parallel \mu \parallel }{\Gamma \left(\alpha \right)}{\left(\frac{1-\rho }{\alpha -\rho }\right)}^{1-\rho }{(logt)}^{\alpha -\rho }+\sum _{i=1}^k\frac{|C_i|\parallel {\varrho }_i\parallel }{\Gamma (\alpha +{\beta }_i)}{\left(\frac{1-\rho }{\alpha +{\beta }_i-\rho }\right)}^{1-\rho }{(logt)}^{\alpha +{\beta }_i-\rho }\hfill \\ & & +|A_0\left|\right[\frac{\left|A\right|\parallel \mu \parallel }{\Gamma (\alpha -\xi )}{\left(\frac{1-\rho }{\alpha -\xi -\rho }\right)}^{1-\rho }+\sum _{i=1}^k\frac{|C_i|\parallel {\varrho }_i\parallel }{\Gamma (\alpha -\xi +{\beta }_i)}{\left(\frac{1-\rho }{\alpha -\xi +{\beta }_i-\rho }\right)}^{1-\rho }\hfill \\ & & +\frac{a\left|A\right|\parallel \mu \parallel }{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(\frac{1-\rho }{\frac{\alpha -\xi +1}{2}-\rho }\right)}^{1-\rho }{(log\delta )}^{\frac{\alpha -\xi +1}{2}-\gamma }\hfill \\ & & +a\sum _{i=1}^k\frac{|C_i|\parallel {\varrho }_i\parallel }{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\left(\frac{1-\rho }{\frac{\alpha -\xi +1}{2}+{\beta }_i-\rho }\right)}^{1-\rho }{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i-\rho }\left]\right\}\parallel x-y\parallel \hfill \\ & \le & \left\{A\right|\parallel \mu \parallel [\frac{1}{\Gamma \left(\alpha \right)}{\left(\frac{1-\rho }{\alpha -\rho }\right)}^{1-\rho }+\frac{|A_0|}{\Gamma (\alpha -\xi )}{\left(\frac{1-\rho }{\alpha -\xi -\rho }\right)}^{1-\rho }\hfill \\ & & +\frac{a|A_0|}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(\frac{1-\rho }{\frac{\alpha -\xi +1}{2}-\rho }\right)}^{1-\rho }{(log\delta )}^{\frac{\alpha -\xi +1}{2}-\rho }]\hfill \\ & & +\sum _{i=1}^k|C_i|\parallel {\varrho }_i\parallel [\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(\frac{1-\rho }{\alpha +{\beta }_i-\rho }\right)}^{1-\rho }+\frac{|A_0|}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(\frac{1-\rho }{\alpha -\xi +{\beta }_i-\rho }\right)}^{1-\rho }\hfill \\ & & +\left(\frac{a|A_0|}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}\right){\left(\frac{1-\rho }{\frac{\alpha -\xi +1}{2}+{\beta }_i-\rho }\right)}^{1-\rho }{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i-\rho }\left]\right\}\parallel x-y\parallel \hfill \\ & =& \left[\left|A\right|\parallel \mu \parallel Z_1+\sum _{i=1}^k|C_i|\parallel {\varrho }_i\parallel Z_{i+1}\right]\parallel x-y\parallel ,\hfill \end{array}$$

where $x,y\in \mathbb{R}$ and $t\in [1,e]$.

Thus, the condition $\left(A3\right)$ implies that $\mathcal{T}$ is a contraction mapping. Thus, by using the Banach fixed point theorem, we have the desired conclusion. □

We prove the next existence result by applying the Leray–Schauder nonlinear alternative [16,17].

Lemma 4

(See [16,17]). Assume that E is a Banach space, C is a closed, convex subset of E, U is an open subset of C and $0\in U$. Let F:Ū $\to C$ be a completely continuous operator. Then, either:

(i) 

F has a fixed point in Ū, or

(ii) 

there is an element $u\in \partial U$ (the boundary of U in C) and $\lambda \in (0,1)$ with $u=\lambda F\left(u\right)$.

Theorem 3.

Suppose that $f,g_i:[1,e]\times \mathbb{R}\to \mathbb{R},i=1,\dots ,k,$ are continuous functions and $f(t,x),g_i(t,x)\in C_{\gamma ,log}([1,e],\mathbb{R})$ with $0<\gamma <1$ and $\alpha -\xi +\gamma >0(i=1,\dots ,k)$. Let the following conditions hold:

(H1) 

There exist $k+1$ functions $\theta ,{\theta }_i\in C([1,e],{\mathbb{R}}^{+})$, $1\le i\le k$, and $k+1$ nondecreasing functions $\phi ,{\phi }_i:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$, $i=1,\dots ,k$, such that

$$\left|f(t,v)\right|\le \theta \left(t\right)\phi (\parallel v\parallel ),|g_i(t,v)|\le {\theta }_i\left(t\right){\phi }_i(\parallel v\parallel ),$$

for all $(t,v)\in [1,e]\times \mathbb{R}$ and $i=1,\dots ,k;$

(H2) 

There exists a constant $M>0$ such that

$$\frac{M}{\left|A\right|\phi \left(M\right)\parallel \theta \parallel \Omega +{\sum }_{i=1}^k|C_i|{\Omega }_i{\phi }_i\left(M\right)\parallel {\theta }_i\parallel }>1,$$

where

$$\Omega =\frac{1}{\Gamma (\alpha +1)}+\left|A_0\right|\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right),$$

and

$${\Omega }_i=\frac{1}{\Gamma (\alpha +{\beta }_i+1)}+\left|A_0\right|\left(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\right),(i=1,\dots ,k).$$

Then, problem (2) has at least one solution on $[1,e]$.

Proof. 

Define the operator $\mathcal{T}:\mathcal{C}\to \mathcal{C}$ as follows:

$$\begin{array}{ccc}\hfill \left(\mathcal{T}x\right)\left(t\right)& =& A{\int }_1^t\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t}{s}\right)}^{\alpha -1}\frac{f(s,x(s\left)\right)}{s}ds\hfill \\ & & +\sum _{i=1}^k{C}_i{\int }_1^t\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}\frac{g_i(s,x\left(s\right))}{s}ds\hfill \\ & & +{(logt)}^{\alpha -1}{A}_0[-A{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{f(s,x(s\left)\right)}{s}ds\hfill \\ & & +\sum _{i=1}^k{C}_i{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\frac{g_i(s,xs))}{s}ds\hfill \\ & & +aA{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{f(s,x(s\left)\right)}{s}ds\hfill \\ & & +a\sum _{i=1}^k{C}_i{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}\frac{g_i(s,x\left(s\right))}{s}ds].\hfill \end{array}$$

We establish that $\mathcal{T}$ is uniformly bounded in $C\left(\right[1,e],\mathbb{R})$. Let $B_r=\{x\in C([1,e],\mathbb{R}):\parallel x\parallel \le r\}$ be a bounded set in $C\left(\right[1,e],\mathbb{R})$. Thus, we can find

$$\begin{array}{ccc}\hfill \left|\right(\mathcal{T}x\left)\right(t\left)\right|& \le & \left|A\right|{\int }_1^t\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t}{s}\right)}^{\alpha -1}\theta \left(s\right)\phi (\parallel x\parallel )\frac{ds}{s}\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^t\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}{\theta }_i\left(s\right){\phi }_i(\parallel x\parallel )\frac{ds}{s}\hfill \\ & & +|A_0{|(logt)}^{\alpha -1}\left[\right|A|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\theta \left(s\right)\phi (\parallel x\parallel )\frac{ds}{s}\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}{\theta }_i\left(s\right){\phi }_i(\parallel x\parallel )\frac{ds}{s}\hfill \\ & & +\left|A\right|a{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\theta \left(s\right)\phi (\parallel x\parallel )\frac{ds}{s}\hfill \\ & & +a\sum _{i=1}^k|C_i|{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi -1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}{\theta }_i\left(s\right){\phi }_i(\parallel x\parallel )\frac{ds}{s}]\hfill \\ & \le & \left|A\right|\phi \left(r\right)\parallel \theta \parallel \left[\frac{{(logt)}^{\alpha }}{\Gamma (\alpha +1)}+\left|A_0\right|{(logt)}^{\alpha -1}\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\frac{\delta }{s})}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right)\right]\hfill \\ & & +\sum _{i=1}^k|C_i|{\phi }_i\left(r\right)\parallel {\theta }_i\parallel [\frac{{(logt)}^{\alpha +{\beta }_i}}{\Gamma (\alpha +{\beta }_i+1)}+\left|A_0\right|{(logt)}^{\alpha -1}(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\left)\right].\hfill \end{array}$$

Consequently

$$\begin{array}{ccc}\hfill \parallel \mathcal{T}x\parallel & \le & \left|A\right|\phi \left(r\right)\parallel \theta \parallel \left[\frac{1}{\Gamma (\alpha +1)}+\left|A_0\right|\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right)\right]\hfill \\ & & +\sum _{i=1}^k|C_i|{\phi }_i\left(r\right)\parallel {\theta }_i\parallel [\frac{1}{\Gamma (\alpha +{\beta }_i+1)}+\left|A_0\right|(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\left)\right]\hfill \\ & =& \left|A\right|\phi \left(r\right)\parallel \theta \parallel \Omega +\sum _{i=1}^k|C_i|{\phi }_i\left(r\right)\parallel {\theta }_i\parallel {\Omega }_i.\hfill \end{array}$$

Set $K:=\left|A\right|\phi \left(r\right)\parallel \theta \parallel \Omega +{\sum }_{i=1}^k|C_i|{\phi }_i\left(r\right)\parallel {\theta }_i\parallel {\Omega }_i$; therefore, we find $\parallel \mathcal{T}x\parallel \le K$. Thus, $\mathcal{T}$ is uniformly bounded in $C\left(\right[1,e],\mathbb{R})$.

Next, we show that $\mathcal{T}$ is equicontinuous. Suppose that $t_1,t_2\in [1,e]$ with $t_1<t_2$ and $x\in B_r$; thus, we have

$$\begin{array}{ccc}& & |\left(\mathcal{T}x\right)\left(t_2\right)-\left(\mathcal{T}x\right)\left(t_1\right)|\hfill \\ & \le & \frac{\left|A\right|}{\Gamma \left(\alpha \right)}{\int }_1^{t_1}\left[{\left(log\frac{t_2}{s}\right)}^{\alpha -1}-{\left(log\frac{t_1}{s}\right)}^{\alpha -1}\right]\left|\frac{f(s,x(s\left)\right)}{s}\right|ds\hfill \\ & & +\frac{\left|A\right|}{\Gamma \left(\alpha \right)}{\int }_{t_1}^{t_2}{\left(log\frac{t_2}{s}\right)}^{\alpha -1}\left|\frac{f(s,x(s\left)\right)}{s}\right|ds\hfill \\ & & +\sum _{i=1}^k\frac{|C_i|}{\Gamma (\alpha +{\beta }_i)}{\int }_1^{t_1}\left[{\left(log\frac{t_2}{s}\right)}^{\alpha +{\beta }_i-1}-{\left(log\frac{t_1}{s}\right)}^{\alpha +{\beta }_i-1}\right]\left|\frac{g_i(s,x\left(s\right))}{s}\right|ds\hfill \\ & & +\sum _{i=1}^k\frac{|C_i|}{\Gamma (\alpha +{\beta }_i)}{\int }_{t_1}^{t_2}{\left(log\frac{t_2}{s}\right)}^{\alpha +{\beta }_i-1}\left|\frac{g_i(s,x\left(s\right))}{s}\right|ds\hfill \\ & & +\left[{(log{t}_2)}^{\alpha -1}-{(log{t}_1)}^{\alpha -1}\right]|A_0\left|\right[\left|A\right|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\left|\frac{f(s,x(s\left)\right)}{s}\right|ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\left|\frac{g_i(s,x\left(s\right))}{s}\right|ds\hfill \\ & & +a\left|A\right|{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\left|\frac{f(s,x(s\left)\right)}{s}\right|ds\hfill \\ & & +a\sum _{i=1}^k|C_i|{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}\left|\frac{g_i(s,x\left(s\right))}{s}\right|ds]\hfill \\ & \le & \frac{\left|A\right|}{\Gamma \left(\alpha \right)}{\int }_1^{t_1}\left[{\left(log\frac{t_2}{s}\right)}^{\alpha -1}-{\left(log\frac{t_1}{s}\right)}^{\alpha -1}\right]\theta \left(s\right)\phi \left(r\right)ds\hfill \\ & & +\frac{\left|A\right|}{\Gamma \left(\alpha \right)}{\int }_{t_1}^{t_2}{\left(log\frac{t_2}{s}\right)}^{\alpha -1}\theta \left(s\right)\psi \left(r\right)ds\hfill \\ & & +\sum _{i=1}^k\frac{|C_i|}{\Gamma (\alpha +{\beta }_i)}{\int }_1^{t_1}[{(log\frac{t_2}{s})}^{\alpha +{\beta }_i-1}-{(log\frac{t_1}{s})}^{\alpha +{\beta }_i-1}]{\theta }_i\left(s\right){\phi }_i\left(r\right)ds\hfill \\ & & +\sum _{i=1}^k\frac{|C_i|}{\Gamma (\alpha +{\beta }_i)}{\int }_{t_1}^{t_2}{\left(log\frac{t_2}{s}\right)}^{\alpha +{\beta }_i-1}{\theta }_i\left(s\right){\phi }_i\left(r\right)ds\hfill \\ & & +[{(log{t}_2)}^{\alpha -1}-{(log{t}_1)}^{\alpha -1}]|A_0\left|\right[\left|A\right|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\theta \left(s\right)\phi \left(r\right)ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}{\theta }_i\left(s\right){\phi }_i\left(r\right)ds\hfill \\ & & +a\left|A\right|{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\theta \left(s\right)\phi \left(r\right)ds\hfill \\ & & +a\sum _{i=1}^k|C_i|{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}{\theta }_i\left(s\right){\phi }_i\left(r\right)ds].\hfill \end{array}$$

Clearly, the right hand side of the above inequality tends to zero independently of $x\in B_r$ as $t_2-t_1\to 0$. Therefore, the Arzelá–Ascoli theorem implies that $\mathcal{T}:C\left(\right[1,e],\mathbb{R})\to C\left(\right[1,e],\mathbb{R})$ is completely continuous.

Let x be a solution of the equations $x=\lambda \mathcal{T}x$ for $\lambda \in (0,1)$. Hence, by straightforward computations, we find

$$\begin{array}{ccc}\hfill \parallel x\parallel & =& \parallel \lambda \left(\mathcal{T}x\right)\parallel \hfill \\ & \le & \left|A\right|\phi (\parallel x\parallel )\parallel \theta \parallel \left[\frac{1}{\Gamma (\alpha +1)}+\left|A_0\right|\left(\frac{1}{\Gamma (\alpha -\xi +1)}+a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}}}{\Gamma (\frac{\alpha -\xi +1}{2}+1)}\right)\right]\hfill \\ & & +\sum _{i=1}^k|C_i|{\phi }_i(\parallel x\parallel )\parallel {\theta }_i\parallel [\frac{1}{\Gamma (\alpha +{\beta }_i+1)}+\left|A_0\right|(\frac{1}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +a\frac{{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i+1)}\left)\right],\hfill \end{array}$$

so,

$$\frac{\parallel x\parallel }{\left|A\right|\phi (\parallel x\parallel )\parallel \theta \parallel \Omega +{\sum }_{i=1}^k|C_i|{\Omega }_i{\phi }_i(\parallel x\parallel )\parallel {\theta }_i\parallel }\le 1.$$

By the condition (H2), there exists M with $\parallel x\parallel \ne M$. Set

$$U=\{v\in C([1,e],\mathbb{R}):\parallel v\parallel <M\}.$$

Clearly, continuity of the functions f and $g_i(1\le i\le k)$ imply that the operators $\mathcal{T}:$ Ū $\to C\left(\right[1,e],\mathbb{R})$ is continuous. Furthermore, it is completely continuous. By the definition of U there is no $u\in \partial U$ such that $x=\lambda \mathcal{T}\left(x\right)$ for some $\lambda \in (0,1)$. Therefore, by Lemma 4, we have the desired conclusion. □

The proof of the next theorem is fully based on the Krasnosel’skiĭs fixed point Theorem [18].

Lemma 5

([18]). Let S be a nonempty, convex, closed, and bounded set such that $S\subset \mathbb{X},$ and let $\mathcal{A}:\mathbb{X}\to \mathbb{X}$ and $\mathcal{B}:S\to \mathbb{X}$ be two operators that satisfy the following:

(i) 

$\mathcal{A}$ is contraction,

(ii) 

$\mathcal{B}$ is completely continuous, and

(iii) 

$x=\mathcal{A}x+\mathcal{B}y,\forall y\in S\Rightarrow x\in S.$

Then, there exists a solution of the operator equation $x=\mathcal{A}x+\mathcal{B}x.$

Theorem 4.

Suppose that $f,g_i:[1,e]\times \mathbb{R}\to \mathbb{R},i=1,\dots ,k,$ are continuous functions such that $f(t,x),g_i(t,x)\in C_{\gamma ,log}([1,e],\mathbb{R})$ with $0<\gamma <1$ and $\alpha -\xi +\gamma >0(i=1,\dots ,k)$.

Assume that:

(H3) 

There exist positive functions $L,L_i\in C([1,e],{\mathbb{R}}^{+})$, $i=1,\dots ,k$, such that

$$|f(t,u)-f(t,v)|\le L\left(t\right)\parallel u-v\parallel ,|g_i(t,u)-g_i(t,v)|\le L_i\left(t\right)\parallel u-v\parallel ,$$

for all $(t,u),(t,v)\in [1,e]\times \mathbb{R}$ and $i=1,\dots ,k$;

(H4) 

There exist $\mu ,{\mu }_i\in C([1,e],{\mathbb{R}}^{+})$, $i=1,\dots ,k$, such that

$$\left|f\right(t,u\left)\right|\le \mu \left(t\right),\forall (t,u)\in [1,e]\times \mathbb{R},$$

$$|g_i(t,u)|\le {\mu }_i\left(t\right),\forall (t,u)\in [1,e]\times \mathbb{R}.$$

Then, the problem (2) has at least one solution provided

$$\begin{array}{ccc}\hfill {\lambda }_0& :=& A_0\left[\right|A|I^{\alpha -\xi }L\left(e\right)+\sum _{i=1}^k|C_i|I^{\alpha -\xi +{\beta }_i}{L}_i\left(e\right)+a\left|A\right|I^{\frac{\alpha -\xi +1}{2}}L\left(\delta \right)\hfill \\ & & +a\sum _{i=1}^k|C_i|I^{\frac{\alpha -\xi +1}{2}+{\beta }_i}{L}_i\left(\delta \right)]<1.\hfill \end{array}$$

Proof. 

First, we consider a subset S of $\mathcal{C}$ defined by $S=\{x\in \mathcal{C}:\parallel x\parallel \le r\}$, where r is given by

$$\begin{array}{ccc}& & r=\left[\frac{\left|A\right|}{\Gamma (\alpha +1)}+\frac{A_0\left|A\right|}{\Gamma (\alpha -\xi +1)}+a\frac{\left|A\right|}{\Gamma \left(\frac{\alpha -\xi +3}{2}\right)}{(log\delta )}^{\frac{\alpha -\xi +1}{2}}\right]\parallel \mu \parallel \hfill \\ & & +\sum _{i=1}^k\{\frac{|C_i|}{\Gamma (\alpha +{\beta }_i+1)}+\frac{A_0\left|C_i\right|}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +\frac{a{A}_0\left|C_i\right|}{\frac{\alpha -\xi +1}{2}\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}\}\parallel {\mu }_i\parallel .\hfill \end{array}$$

We now define the operators ${\mathcal{A}}_1:\mathcal{C}\to \mathcal{C}$ and ${\mathcal{A}}_2:S\to \mathcal{C}$ by

$$\begin{array}{ccc}\hfill \left({\mathcal{A}}_{1}u\right)\left(t\right)& =& {(logt)}^{\alpha -1}{A}_0[-A{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{f(s,u(s\left)\right)}{s}ds\hfill \\ & & -\sum _{i=1}^k{C}_i{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\frac{g_i(s,u\left(s\right))}{s}ds\hfill \\ & & +aA{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{f(s,u(s\left)\right)}{s}ds\hfill \\ & & +a\sum _{i=1}^k{C}_i{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{(log\frac{\delta }{s})}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}\frac{g_i(s,u\left(s\right))}{s}ds],t\in [1,e].\hfill \\ \hfill \left({\mathcal{A}}_{2}u\right)\left(t\right)& =& A{\int }_1^t\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t}{s}\right)}^{\alpha -1}\frac{f(s,u(s\left)\right)}{s}ds\hfill \\ & & +\sum _{i=1}^k{C}_i{\int }_1^t\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}\frac{g_i(s,u\left(s\right))}{s}ds,t\in [1,e].\hfill \end{array}$$

Clearly, $\mathcal{T}u={\mathcal{A}}_{1}u+{\mathcal{A}}_{2}u$. In the next steps, we show that the operators ${\mathcal{A}}_1$ and ${\mathcal{A}}_2$ fulfill all the assumptions of Lemma 5. The proof is divided into three steps:

Step 1. ${\mathcal{A}}_1$ is a contraction. For $u,v\in \mathcal{C}$ and for every $t\in [1,e]$, by (H3), we have

$$\begin{array}{ccc}& & |\left({\mathcal{A}}_{1}u\right)\left(t\right)-\left({\mathcal{A}}_{1}v\right)\left(t\right)|\hfill \\ & \le & {(logt)}^{\alpha -1}{A}_0\left[\right|A|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi )}{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{1}{s}L\left(s\right)\parallel u-v\parallel ds\hfill \\ & & +\sum _{i=1}^k|C_i|{\int }_1^e\frac{1}{\Gamma (\alpha -\xi +{\beta }_i)}{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\frac{1}{s}{L}_i\left(s\right)\parallel u-v\parallel ds\hfill \\ & & +a\left|A\right|{\int }_1^{\delta }\frac{1}{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{1}{s}L\left(s\right)\parallel u-v\parallel ds\hfill \\ & & +a\sum _{i=1}^k|C_i|{\int }_1^{\delta }\frac{1}{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}\frac{1}{s}{L}_i\left(s\right)]\hfill \\ & \le & A_0\left[\right|A|I^{\alpha -\xi }L\left(e\right)+\sum _{i=1}^k|C_i|I^{\alpha -\xi +{\beta }_i}{L}_i\left(e\right)+a\left|A\right|I^{\frac{\alpha -\xi +1}{2}}L\left(\delta \right)\hfill \\ & & +a\sum _{i=1}^k|C_i|I^{\frac{\alpha -\xi +1}{2}+{\beta }_i}{L}_i\left(\delta \right)]\parallel u-v\parallel \hfill \\ & =& {\lambda }_0\parallel u-v\parallel .\hfill \end{array}$$

Consequently $\parallel \left({\mathcal{A}}_{1}u\right)-\left({\mathcal{A}}_{1}v\right)\parallel \le {\lambda }_0\parallel u-v\parallel$. Since ${\lambda }_0<1$, therefore ${\mathcal{A}}_1$ is a contraction on $\mathcal{C}$.

Step 2. The operator ${\mathcal{A}}_2$ is completely continuous on $S$. The continuity of f and $g_i(i=1,\dots ,k)$ imply that the operator ${\mathcal{A}}_2$ is continuous. Furthermore, ${\mathcal{A}}_2$ is uniformly bounded on S as

$$\begin{array}{ccc}\hfill \parallel ({\mathcal{A}}_{2}u\parallel & \le & \frac{\left|A\right|\parallel \mu \parallel }{\Gamma (\alpha +1)}{(logt)}^{\alpha }+\sum _{i=1}^k\frac{|C_i|\parallel {\mu }_i{\parallel }_{L^1}}{\Gamma (\alpha +{\beta }_i+1)}{(logt)}^{\alpha +{\beta }_i}\hfill \\ & \le & \frac{\left|A\right|\parallel \mu \parallel }{\Gamma (\alpha +1)}+\sum _{i=1}^k\frac{|C_i|\parallel {\mu }_i\parallel }{\Gamma (\alpha +{\beta }_i+1)}.\hfill \end{array}$$

We now claim that ${\mathcal{A}}_2$ is equicontinuous on S. To this end, we have

$$\begin{array}{ccc}& & |\left({\mathcal{A}}_{2}u\right)\left(t_2\right)-\left({\mathcal{A}}_{2}u\right)\left(t_1\right)|\hfill \\ & \le & \parallel \mu \parallel \left|A\right||{\int }_1^{t_1}\frac{1}{\Gamma \left(\alpha \right)}\left({\left(log\frac{t_2}{s}\right)}^{\alpha -1}-{\left(log\frac{t_1}{s}\right)}^{\alpha -1}\right)\frac{1}{s}ds\hfill \\ & & +\sum _{i=1}^k{M}_i\left|C_i\right|{\int }_1^{t_1}\frac{1}{\Gamma (\alpha +{\beta }_i)}\left({\left(log\frac{t_2}{s}\right)}^{\alpha +{\beta }_i-1}-{\left(log\frac{t_1}{s}\right)}^{\alpha +{\beta }_i-1}\right)\frac{1}{s}ds\hfill \\ & & +\left|A\right|\parallel \mu \parallel {\int }_{t_1}^{t_2}\frac{1}{\Gamma \left(\alpha \right)}{\left(log\frac{t_2}{s}\right)}^{\alpha -1}\frac{1}{s}ds\hfill \\ & & +\sum _{i=1}^k|C_i|M_i{\int }_{t_1}^{t_2}\frac{1}{\Gamma (\alpha +{\beta }_i)}{\left(log\frac{t_2}{s}\right)}^{\alpha +{\beta }_i-1}\frac{1}{s}ds|\hfill \\ & \le & \frac{2\parallel \mu \parallel \left|A\right|}{\Gamma (\alpha +1)}|{\left(log\frac{t_2}{t_1}\right)}^{\alpha }+\sum _{i=1}^k\frac{2M_i\left|C_i\right|}{\Gamma (\alpha +{\beta }_i+1)}|{\left(log\frac{t_2}{t_1}\right)}^{\alpha +{\beta }_i}|,\hfill \end{array}$$

which yields that $|\left({\mathcal{A}}_{2}u\right)\left(t_2\right)-\left({\mathcal{A}}_{2}u\right)\left(t_1\right)|\to 0$ as $t_2\to t_1$. Then, ${\mathcal{A}}_2$ is equicontinuous and the Arzelá–Ascoli theorem implies that ${\mathcal{A}}_2$ is compact on S, and thus the operator ${\mathcal{A}}_2$ is completely continuous.

Step 3. We show that the third condition (iii) of Lemma 5 is fulfilled. For any $v\in S$, we have by (H4)

$$\begin{array}{ccc}\hfill \left|u\right(t\left)\right|& =& |{\mathcal{A}}_{1}u\left(t\right)+{\mathcal{A}}_{2}v\left(t\right)|\le |{\mathcal{A}}_{1}u\left(t\right)|+|{\mathcal{A}}_{2}v\left(t\right)|\hfill \\ & \le & \frac{\left|A\right|\parallel \mu \parallel }{\Gamma \left(\alpha \right)}{\int }_1^t{\left(log\frac{t}{s}\right)}^{\alpha -1}\frac{1}{s}ds+\sum _{i=1}^k\frac{|C_i|\parallel {\mu }_i\parallel }{\Gamma (\alpha +{\beta }_i)}{\int }_1^t{\left(log\frac{t}{s}\right)}^{\alpha +{\beta }_i-1}\frac{1}{s}ds\hfill \\ & & +{(loge)}^{\alpha -1}{A}_0[\frac{\left|A\right|\parallel \mu \parallel }{\Gamma (\alpha -\xi )}{\int }_1^e{\left(log\frac{e}{s}\right)}^{\alpha -\xi -1}\frac{1)}{s}ds\hfill \\ & & +\sum _{i=1}^k\frac{|C_i|\parallel {\mu }_i\parallel }{\Gamma (\alpha -\xi +{\beta }_i)}{\int }_1^e{\left(log\frac{e}{s}\right)}^{\alpha -\xi +{\beta }_i-1}\frac{1}{s}ds\hfill \\ & & +a\frac{\left|A\right|\parallel \mu \parallel }{\Gamma \left(\frac{\alpha -\xi +1}{2}\right)}{\int }_1^{\delta }{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}}\frac{1}{s}ds\hfill \\ & & +a\sum _{i=1}^k\frac{|C_i|\parallel {\mu }_i\parallel }{\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{\int }_1^{\delta }{\left(log\frac{\delta }{s}\right)}^{\frac{\alpha -\xi -1}{2}+{\beta }_i}\frac{1}{s}ds]\hfill \\ & \le & \left[\frac{\left|A\right|}{\Gamma (\alpha +1)}+\frac{A_0\left|A\right|}{\Gamma (\alpha -\xi +1)}+a\frac{\left|A\right|}{\Gamma \left(\frac{\alpha -\xi +3}{2}\right)}{(log\delta )}^{\frac{\alpha -\xi +1}{2}}\right]\parallel \mu \parallel \hfill \\ & & +\sum _{i=1}^k\{\frac{|C_i|}{\Gamma (\alpha +{\beta }_i+1)}+\frac{A_0\left|C_i\right|}{\Gamma (\alpha -\xi +{\beta }_i+1)}\hfill \\ & & +\frac{a{A}_0\left|C_i\right|}{\frac{\alpha -\xi +1}{2}\Gamma (\frac{\alpha -\xi +1}{2}+{\beta }_i)}{(log\delta )}^{\frac{\alpha -\xi +1}{2}+{\beta }_i}\}\parallel {\mu }_i\parallel =r,\hfill \end{array}$$

which implies $\parallel u\parallel \le r,$ and thus $u\in S.$

Thus, all the assumptions of Lemma 5 are satisfied and the conclusion of Lemma 5 implies that the boundary-value problem (2) has at least one solution on $[1,e]$. Therefore, the proof is completed. □

Remark 1.

We know that, the Hadamard derivative is exactly the Riemann–Liouville derivative with respect to the logarithm function, according to the formalism of fractional calculus with respect to functions as described in standard textbooks, such as [1,19]. This means that, after a substitution or change of variables, the fractional differential Equation (2) studied here can be rewritten as a Riemann–Liouville differential equation on the interval $[0,1]$.

4. Application

Example 1.

Consider the following fractional boundary value problem:

$$\left\{\begin{array}{c}{\displaystyle {}_H{D}^{\frac{3}{2}}x\left(t\right)=\frac{1}{40e^t}(1+cos|x\left(t\right)\left|\right)+\sum _{i=1}^2{}_H{I}^{{\beta }_i}{g}_i(t,x\left(t\right)),t\in (1,e),}\hfill \\ {\displaystyle {}_H{D}^{\frac{1}{4}}x\left(1\right)=0,{}_H{D}^{\frac{1}{4}}x\left(e\right)=\frac{1}{2}{D}^{\frac{1}{8}}\left({}_H{D}^{\frac{1}{4}}x\left(t\right)\right){|}_{t=\frac{3}{2}}.}\hfill \end{array}\right.$$

(8)

Here, ${\beta }_1=\frac{1}{2}$, ${\beta }_2=\frac{1}{3}$, $g_i(t,x)=\frac{1}{30t^2{e}^t}\left(\right|x|+cos\left|x\right|)$. With the given data, we obtain $A_0=0.4758156$, $\Delta =3.1385793$ and ${\zeta }_1=0.067957$, ${\zeta }_2=0.1812187$ as $|f(t,x_1)-f(t,x_2)|\le 0.067957|x_1-x_2|$, $|g_i(t,x_1)-g_i(t,x_2)|\le 0.1812187|x_1-x_2|$. Clearly, $\zeta =max\{{\zeta }_1,{\zeta }_2\}=0.1812187$ and $\zeta <\frac{1}{\Delta }$. Hence, Theorem 1 implies that the problem (8) has a unique solution.

5. Conclusions

In this article, we considered a fractional boundary value problem involving the Hadamard derivatives and Hadamard fractional integral. We investigated the existence criterion by using the Leray–Schauder nonlinear alternative, Krasnosel’skiĭ fixed point theorem and Banach’s fixed point theorem. In the last part of the article, we investigated the consistency of our theoretical findings by demonstrating an example.

The work accomplished in this paper is new and enriches the literature on fractional differential equations. In future works, one can extend the given fractional boundary value problem to more fractional derivatives, such as the Caputo–Hadamard derivative, $\psi$–Hilfer fractional derivatives and $(k,\psi )$–Hilfer fractional derivative.