Existence of Solutions for a Hadamard Fractional Boundary Value Problem at Resonance

Abstract

We explore the existence of solutions for a Hadamard fractional differential equation, subject to nonlocal boundary conditions, which contain Hadamard fractional derivatives and Riemann–Stieltjes integrals. This problem is a resonant one in the sense that the corresponding homogeneous boundary-value problem has nontrivial solutions. In the proof of the main result, we use the Mawhin continuation theorem.

1. Introduction

We consider the fractional differential equation

$${}^H{D}_{1+}^{\alpha }u\left(t\right)=f(t,u\left(t\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(t\right){,}^H{D}_{1+}^{\alpha -n+2}u\left(t\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(t\right)),t\in (1,e),$$

(1)

supplemented with the nonlocal boundary conditions

$$\left\{\begin{array}{c}{\displaystyle u^{\left(i\right)}\left(1\right)=0,i=0,\dots ,n-3,{}^H{D}_{1+}^{\alpha -1}u\left(1\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dP\left(s\right),}\hfill \\ {\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(e\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dQ\left(s\right),}\hfill \end{array}\right.$$

(2)

where $\alpha \in (n-1,n]$, $n\in \mathbb{N}$, $n\ge 3$, ${}^H{D}_{1+}^{\kappa }$ is the Hadamard fractional derivative of order $\kappa$, for $\kappa =\alpha ,\alpha -1,\dots ,\alpha -n+1$; $f:(1,e)\times {\mathbb{R}}^n\to \mathbb{R}$ is a Carathéodory function, and the integrals from the boundary conditions (2) are Riemann–Stieltjes integrals with $P,Q$ bounded variation functions.

Under some assumptions on the data of Problem (1),(2) (see assumption $\left(A2\right)$ below), this problem is a resonant one, in the sense that the corresponding homogeneous boundary-value problem has nontrivial solutions. In this paper, we prove the existence of solutions for (1) and (2) using the coincidence degree theory of J. Mawhin (more precisely, the Mawhin continuation theorem (see {[1], Theorem IV.13}, [2])). We present next some papers that relate to our problem. In [3], the authors investigate the Hadamard fractional differential equation at resonance

$$\left\{\begin{array}{c}{(}^H{D}_{1+}^{\gamma }\mathrm{u})\left(t\right)+f(t,\mathrm{u}\left(t\right))=0,t\in (1,e),\hfill \\ {\displaystyle \mathrm{u}\left(1\right)=0,\mathrm{u}\left(e\right)={\int }_1^e\mathrm{u}\left(t\right)dA\left(t\right),}\hfill \end{array}\right.$$

(3)

where $\gamma \in (1,2)$, $f:(1,e)\times \mathbb{R}\to \mathbb{R}$ satisfies Carathéodory conditions, and the integral from the boundary conditions is a Riemann-Stieltjes integral. By applying the Mawhin continuation theorem, they show the existence of solutions for Problem (3). In [4], the authors study the system of Hadamard fractional differential equations

$$\left\{\begin{array}{c}{}^H{D}_{1+}^{\alpha }\mathrm{u}\left(t\right)+f(t,\mathrm{u}\left(t\right),\mathrm{v}\left(t\right){,}^H{I}_{1+}^{{\gamma }_1}\mathrm{u}\left(t\right){,}^H{I}_{1+}^{{\delta }_1}\mathrm{v}\left(t\right))=0,t\in [1,e],\hfill \\ {}^H{D}_{1+}^{\beta }\mathrm{v}\left(t\right)+g(t,\mathrm{u}\left(t\right),\mathrm{v}\left(t\right){,}^H{I}_{1+}^{{\gamma }_2}\mathrm{u}\left(t\right){,}^H{I}_{1+}^{{\delta }_2}\mathrm{v}\left(t\right))=0,t\in [1,e],\hfill \end{array}\right.$$

(4)

subject to the coupled nonlocal boundary conditions

$$\left\{\begin{array}{c}{\mathrm{u}}^{\left(i\right)}\left(1\right)=0,i=0,\dots ,n-2,{\mathrm{v}}^{\left(i\right)}\left(1\right)=0,i=0,\dots ,m-2,\hfill \\ {\displaystyle {}^H{D}_{1+}^{\varsigma }\mathrm{u}\left(e\right)={\displaystyle \sum _{i=1}^a}{\int }_1^e{}^{H}D_{1+}^{{\varrho }_i}\mathrm{u}\left(s\right)d{H}_i\left(s\right)+{\displaystyle \sum _{i=1}^b}{\int }_1^e{}^H{D}_{1+}^{{\sigma }_i}\mathrm{v}\left(s\right)d{K}_i\left(s\right),}\hfill \\ {\displaystyle {}^H{D}_{1+}^{\vartheta }\mathrm{v}\left(e\right)={\displaystyle \sum _{i=1}^c}{\int }_1^e{}^{H}D_{1+}^{{\eta }_i}\mathrm{u}\left(s\right)d{P}_i\left(s\right)+{\displaystyle \sum _{i=1}^d}{\int }_1^e{}^H{D}_{1+}^{{\theta }_i}\mathrm{v}\left(s\right)d{Q}_i\left(s\right),}\hfill \end{array}\right.$$

(5)

where $\alpha \in (n-1,n]$, $n\in \mathbb{N}$, $n\ge 3$, $\beta \in (m-1,m]$, $m\in \mathbb{N}$, $m\ge 3$; $a,b,c,d\in \mathbb{N}$; $\varsigma \ge 1$, $0\le {\varrho }_i\le \varsigma <\alpha -1$, $i=1,\dots ,a$, $0\le {\eta }_k\le \varsigma$, $k=1,\dots ,c$, $\vartheta \ge 1$, $0\le {\sigma }_j\le \vartheta <\beta -1$, $j=1,\dots ,b$, $0\le {\theta }_{\iota }\le \vartheta$, $\iota =1,\dots ,d$; ${\gamma }_i,{\delta }_i>0$, $i=1,2$; and ${}^H{I}_{1+}^{\kappa }$ (for $\kappa ={\gamma }_i,{\delta }_i,i=1,2$) is the Hadamard fractional integral of order $\kappa$. The integrals from the boundary conditions (5) are Riemann-Stieltjes integrals with $H_i,K_j,P_k,Q_{\iota }$, $i=1,\dots ,a$, $j=1,\dots ,b$, $k=1,\dots ,c$, $\iota =1,\dots ,d$, bounded variation functions. By employing various fixed-point theorems, they prove the existence of positive solutions for Problem (4),(5). In Refs [5,6], the authors examine the system of nonlinear Hadamard fractional differential equations on an infinite interval

$$\left\{\begin{array}{c}{}^H{D}_{1+}^{\alpha }\mathrm{x}\left(t\right)+\mathfrak{a}\left(t\right)\mathfrak{f}(\mathrm{x}\left(t\right),\mathrm{y}\left(t\right))=0,t\in (1,\infty ),\hfill \\ {}^H{D}_{1+}^{\beta }\mathrm{y}\left(t\right)+\mathfrak{b}\left(t\right)\mathfrak{g}(\mathrm{x}\left(t\right),\mathrm{y}\left(t\right))=0,t\in (1,\infty ),\hfill \end{array}\right.$$

(6)

supplemented with the nonlocal boundary conditions

$$\left\{\begin{array}{c}{\displaystyle \mathrm{x}\left(1\right)={\mathrm{x}}^{\prime }\left(1\right)=\cdots ={\mathrm{x}}^{(n-2)}\left(1\right)=0,{}^H{D}_{1+}^{\alpha -1}\mathrm{x}(\infty )={\int }_1^{\infty }\mathrm{x}\left(s\right)d{\mathfrak{H}}_1\left(s\right)+{\int }_1^{\infty }\mathrm{y}\left(s\right)d{\mathfrak{H}}_2\left(s\right),}\hfill \\ {\displaystyle \mathrm{y}\left(1\right)={\mathrm{y}}^{\prime }\left(1\right)=\cdots ={\mathrm{y}}^{(m-2)}\left(1\right)=0,{}^H{D}_{1+}^{\beta -1}\mathrm{y}(\infty )={\int }_1^{\infty }\mathrm{x}\left(s\right)d{\mathfrak{K}}_1\left(s\right)+{\int }_1^{\infty }\mathrm{y}\left(s\right)d{\mathfrak{K}}_2\left(s\right),}\hfill \end{array}\right.$$

(7)

where $\alpha \in (n-1,n]$, $n\in \mathbb{N}$, $n\ge 2$, $\beta \in (m-1,m]$, $m\in \mathbb{N}$, $m\ge 2$, the functions $\mathfrak{a},\mathfrak{b}:(1,\infty )\to {\mathbb{R}}_{+}$ and $\mathfrak{f},\mathfrak{g}:{\mathbb{R}}_{+}\times {\mathbb{R}}_{+}\to {\mathbb{R}}_{+}$ verify some assumptions, ( ${\mathbb{R}}_{+}=[0,\infty )$), and the integrals from (7) are Riemann–Stieltjes integrals with ${\mathfrak{H}}_1,{\mathfrak{H}}_2,{\mathfrak{K}}_1,{\mathfrak{K}}_2:[1,\infty )\to \mathbb{R}$ functions of bounded variation. By using the fixed-point theory, they show the existence of positive solutions to Problem (6),(7), in two cases: the nonlinearities $\mathfrak{f},\mathfrak{g}$ are bounded or unbounded functions. In the paper [7], by applying the Mawhin coincidence theorem, the authors analyze the existence of at least one solution for the Riemann–Liouville fractional differential equation with Riemann–Stieltjes integral boundary conditions at resonance

$$\left\{\begin{array}{c}\left(D_{0+}^{\alpha }\mathrm{x}\right)\left(t\right)+f(t,\mathrm{x}\left(t\right),D_{0+}^{\alpha -1}\mathrm{x}\left(t\right)),t\in (0,1),\hfill \\ {\displaystyle \mathrm{x}\left(0\right)={\mathrm{x}}^{\prime }\left(0\right)=\dots ={\mathrm{x}}^{(n-2)}\left(0\right)=0,{\mathrm{x}}^{\left(k\right)}\left(1\right)={\int }_0^1{\mathrm{x}}^{\left(k\right)}\left(t\right)dA\left(t\right),}\hfill \end{array}\right.$$

where $\alpha \in (n-1,n]$, k is any integer between 0 and $n-1$, $f:(0,1)\times {\mathbb{R}}^2\to \mathbb{R}$ satisfies Carathéodory conditions, and $D_{0+}^{\kappa }$ is the Riemann-Liouville fractional derivative of order $\kappa$, for $\kappa \in \{\alpha -1,\alpha \}$. In [8], the authors investigate the resonant boundary-value problem for the Riemann-Liouville fractional differential equation

$$\left\{\begin{array}{c}\left(D_{a+}^{\alpha }\mathrm{u}\right)\left(t\right)+f(t,\mathrm{u}\left(t\right))=0,t\in (a,b),\hfill \\ {\displaystyle \mathrm{u}\left(a\right)=0,\mathrm{u}\left(b\right)=\delta {\int }_a^{\eta }\mathrm{u}\left(s\right)ds,}\hfill \end{array}\right.$$

(8)

where $\eta \in (a,b)$, and the function $f:[a,b]\times {\mathbb{R}}_{+}\to \mathbb{R}$ is continuous. By employing the fixed-point index theory and the spectral theory of linear operators, they prove the existence of positive solutions for Problem (8). In the paper [9], the authors study the existence of solutions for the infinite point Riemann-Liouville fractional boundary-value problem at resonance

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha }\mathrm{x}\left(t\right)=f(t,\mathrm{x}\left(t\right),D_{0+}^{\alpha -2}\mathrm{x}\left(t\right),D_{0+}^{\alpha -1}\mathrm{x}\left(t\right)),t\in (0,1),\hfill \\ {\displaystyle \mathrm{x}\left(0\right)=0,D_{0+}^{\alpha -1}\mathrm{x}\left(0\right)={\displaystyle \sum _{i=1}^{\infty }}{\alpha }_i{D}_{0+}^{\alpha -1}\mathrm{x}\left({\xi }_i\right),}\hfill \\ {\displaystyle D_{0+}^{\alpha -1}\mathrm{x}\left(1\right)={\displaystyle \sum _{i=1}^{\infty }}{\beta }_i{D}_{0+}^{\alpha -1}\mathrm{x}\left({\gamma }_i\right),}\hfill \end{array}\right.$$

where $\alpha \in (2,3]$, $f:[0,1]\times {\mathbb{R}}^3\to \mathbb{R}$ is a Carathéodory function, ${\xi }_i{\gamma }_i\in (0,1)$, and ${\left\{{\xi }_i\right\}}_{i=1}^{\infty },{\left\{{\gamma }_i\right\}}_{i=1}^{\infty }$ are two monotonic sequences with ${lim}_{i\to \infty }{\xi }_i=a$, ${lim}_{i\to \infty }{\gamma }_i=b$, with $a,b\in (0,1)$, ${\alpha }_i,{\beta }_i\in \mathbb{R}$. In the proof of the main theorem, they use the Mawhin continuation theorem. In [10], by applying the coincidence degree theory, the authors examine the existence of solutions for the Riemann-Liouville fractional boundary-value problem at resonance on an infinite interval.

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha }\mathrm{u}\left(t\right)=f(t,\mathrm{u}\left(t\right),D_{0+}^{\alpha -2}\mathrm{u}\left(t\right),D_{0+}^{\alpha -1}\mathrm{u}\left(t\right)),t\in (0,\infty ),\hfill \\ {\displaystyle \mathrm{u}\left(0\right)=0,D_{0+}^{\alpha -2}\mathrm{u}\left(0\right)={\displaystyle \sum _{i=1}^m}{\alpha }_i{D}_{0+}^{\alpha -2}\mathrm{u}\left({\xi }_i\right),}\hfill \\ {\displaystyle D_{0+}^{\alpha -1}\mathrm{u}(\infty )={\displaystyle \sum _{j=1}^n}{\beta }_j{D}_{0+}^{\alpha -1}\mathrm{u}\left({\eta }_j\right),}\hfill \end{array}\right.$$

where $\alpha \in (2,3]$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_m<\infty$, $0<{\eta }_1<{\eta }_2<\cdots <{\eta }_n<\infty$, ${\alpha }_i,{\beta }_j\in \mathbb{R}$ for $i=1,\dots ,m$, $j=1,\dots ,n$, and $f:(0,\infty )\times {\mathbb{R}}^3\to \mathbb{R}$ is a Carathéodory function. Various resonant boundary-value problems for Riemann-Liouville or Caputo fractional differential equations and systems, on finite or infinite intervals, are investigated in the papers [11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28].

Hadamard fractional calculus finds applications across multiple disciplines, including rheology and probability theory (see the papers [29,30] with their references). For instance, extensions of the Lomnitz logarithmic creep law are utilized to model geophysical behaviors in materials that exhibit fluid-like properties, such as igneous rocks. For additional recent studies on Hadamard fractional differential equations and systems with diverse applications, we recommend the monograph [31] and its references. The novelties of our Problem (1),(2) consist of the consideration of the resonant case for a Hadamard fractional differential equation of arbitrary order, also containing different fractional derivatives and the general boundary conditions with Riemann-Stieltjes integrals which incorporate finite or infinite multi-point boundary conditions (when P and Q are step functions with finite or infinite values), classical integral conditions (when P and Q are derivable functions), and combination of them.

The structure of our paper is as follows. In Section 2, we present some notions from the coincidence degree theory, the Mawhin continuation theorem, and some definitions and results from the Hadamard fractional calculus. In Section 3, we give the auxiliary results and the main theorem for our Problem (1),(2). Section 4 contains an example that illustrates our results, and in Section 5, we present the conclusions for this paper.

2. Preliminaries

In this section, we first present some definitions from the coincidence degree theory and the Mawhin continuation theorem from [1] (see also [2]) that we will apply in the next section.

Let $\mathcal{X}$ and $\mathcal{Y}$ be two real normed spaces. A linear operator $\mathcal{L}:D\left(\mathcal{L}\right)\subset \mathcal{X}\to \mathcal{Y}$ is called a Fredholm operator if $Ker\mathcal{L}$ has a finite dimension, and $Im\mathcal{L}$ is closed and has a finite codimension. Here $D\left(\mathcal{L}\right)$ is the domain of operator $\mathcal{L}$, $Ker\mathcal{L}={\mathcal{L}}^{-1}\left\{0\right\}$ is the kernel of $\mathcal{L}$, and $Im\mathcal{L}=\mathcal{L}\left(D\right(\mathcal{L}\left)\right)$ is the image of operator $\mathcal{L}$. The codimension of $Im\mathcal{L}$ is the dimension of $\mathcal{Y}/Im\mathcal{L}$, i.e., the dimension of the cokernel $Coker\mathcal{L}$ of $\mathcal{L}$. When $\mathcal{L}$ is a Fredholm operator, its Fredholm index is the integer $Ind\mathcal{L}=dimKer\mathcal{L}-codimIm\mathcal{L}$.

For a Fredholm operator $\mathcal{L}:\mathcal{X}\to \mathcal{Y}$ of index zero, from the basic results of linear functional analysis, we deduce that there exist continuous projectors $\mathcal{H}:\mathcal{X}\to \mathcal{X}$ and $\mathcal{K}:\mathcal{Y}\to \mathcal{Y}$ such that $Im\mathcal{H}=Ker\mathcal{L}$ and $Ker\mathcal{K}=Im\mathcal{L}$, so that $\mathcal{X}=Ker\mathcal{L}\oplus Ker\mathcal{H}$ and $\mathcal{Y}=Im\mathcal{L}\oplus Im\mathcal{K}$ as topological direct sums. Then the restriction ${\mathcal{L}}_{\mathcal{H}}$ of $\mathcal{L}$ to $D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$ is one-to-one and onto, so that its algebraic inverse ${\mathcal{T}}_{\mathcal{H}}:Im\mathcal{L}\to D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$ is defined. We denote by ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}:\mathcal{Y}\to D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$ the generalized inverse of $\mathcal{L}$ defined by ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}={\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})$.

Let $\mathcal{E}$ be a metric space and $\mathcal{N}:\mathcal{E}\to \mathcal{Y}$ be an operator (not necessarily linear). The operator $\mathcal{N}$ is $\mathcal{L}$-compact on $\mathcal{E}$ if the operators $\mathcal{K}\mathcal{N}:\mathcal{E}\to \mathcal{Y}$ and ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}:\mathcal{E}\to \mathcal{X}$ are compact on $\mathcal{E}$, i.e., continuous on $\mathcal{E}$ and such that $\mathcal{K}\mathcal{N}\left(\mathcal{E}\right)$ and ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}\left(\mathcal{E}\right)$ are relatively compact.

Theorem 1.

([1]). Let $\Omega \subset \mathcal{X}$ be an open bounded subset, $\mathcal{L}:D\left(\mathcal{L}\right)\subset \mathcal{X}\to \mathcal{Y}$ be a linear Fredholm operator of index zero, and $\mathcal{N}:\overline{\Omega }\to \mathcal{Y}$ be an $\mathcal{L}$-compact operator on $\overline{\Omega }$ which satisfy the following conditions:

(i) 

$\mathcal{L}u\ne \lambda \mathcal{N}u$ for every $(u,\lambda )\in \left(\right(D\left(\mathcal{L}\right)\setminus Ker\mathcal{L})\cap \partial \Omega )\times (0,1)$;

(ii) 

$\mathcal{N}u\notin Im\mathcal{L}$ for every $u\in Ker\mathcal{L}\cap \partial \Omega$;

(iii) 

$deg(\mathcal{K}\mathcal{N}{|}_{Ker\mathcal{L}\cap \overline{\Omega }},\Omega \cap Ker\mathcal{L},0)\ne 0$, with $\mathcal{K}:\mathcal{Y}\to \mathcal{Y}$ a continuous projector such that $Ker\mathcal{K}=Im\mathcal{L}$.

Then the equation $\mathcal{L}u=\mathcal{N}u$ has at least one solution in $D\left(\mathcal{L}\right)\cap \overline{\Omega }$.

In Theorem 1, $\partial \Omega$ is the boundary of the set $\Omega$, and $\overline{\Omega }$ is the closure of $\Omega$. In addition, in condition $\left(iii\right)$ from Theorem 1, $deg(\mathcal{K}\mathcal{N}{|}_{Ker\mathcal{L}\cap \overline{\Omega }},\Omega \cap Ker\mathcal{L},0)$ is the degree of operator ${\mathcal{K}\mathcal{N}|}_{Ker\mathcal{L}\cap \overline{\Omega }}$ with respect to $\Omega \cap Ker\mathcal{L}$ and 0 (see [1], Chapter II).

Now, we recall some definitions and lemmas from the Hadamard fractional calculus (see the monograph [32]).

Definition 1.

For a function $u:[a,b]\to \mathbb{R}$, ( $b>a\ge 1$), the left-sided Hadamard fractional integral of order $\alpha >0$ with lower limit a is defined by

$${\displaystyle {(}^H{I}_{a+}^{\alpha }u)\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_a^t{\left({\displaystyle ln\frac{t}{\tau }}\right)}^{\alpha -1}\frac{u\left(\tau \right)}{\tau }d\tau ,t>0.}$$

Definition 2.

For a function $u:[a,b]\to \mathbb{R}$, ( $b>a\ge 1$), the left-sided Hadamard fractional derivative of order $\alpha \ge 0$ with lower limit a is defined by

$${\displaystyle {}^H{D}_{a+}^{\alpha }u\left(t\right)={\left({\displaystyle t\frac{d}{dt}}\right)}^n{}^H{I}_{a+}^{n-\alpha }u\left(t\right)=\frac{1}{\Gamma (n-\alpha )}{\left({\displaystyle t\frac{d}{dt}}\right)}^n{\int }_a^t{\left({\displaystyle ln\frac{t}{\tau }}\right)}^{n-\alpha -1}\frac{u\left(\tau \right)}{\tau }d\tau ,}$$

where $n=\left[\alpha \right]+1$.

For $\alpha =m\in \mathbb{N}$, then ${}^H{D}_{a+}^{m}u\left(t\right)=\left({\delta }^{m}u\right)\left(t\right),t>a$, where $\delta =t{\displaystyle \frac{d}{dt}}$ is the $\delta$-derivative.

Lemma 1.

If $\alpha ,\beta >0$, and $a\ge 1$, then

$$\begin{array}{c}{\displaystyle \left({}^H{I}_{a+}^{\alpha }{\left({\displaystyle ln\frac{t}{a}}\right)}^{\beta -1}\right)\left(x\right)=\frac{\Gamma \left(\beta \right)}{\Gamma (\beta +\alpha )}{\left({\displaystyle ln\frac{x}{a}}\right)}^{\beta +\alpha -1},}\hfill \\ {\displaystyle \left({}^H{D}_{a+}^{\alpha }{\left({\displaystyle ln\frac{t}{a}}\right)}^{\beta -1}\right)\left(x\right)=\frac{\Gamma \left(\beta \right)}{\Gamma (\beta -\alpha )}{\left({\displaystyle ln\frac{x}{a}}\right)}^{\beta -\alpha -1},}\hfill \end{array}$$

and in particular ${(}^H{D}_{a+}^{\alpha }{(ln{\displaystyle \frac{t}{a}})}^{\alpha -j})\left(x\right)=0$, for $j=1,\dots ,\left[\alpha \right]+1$.

Lemma 2.

Let $\alpha >0$ and $z,u\in L^1(a,b)$, ( $b>a\ge 1$). Then the Hadamard fractional differential equation ${}^H{D}_{a+}^{\alpha }z\left(t\right)=0$ has the solutions

$${\displaystyle z\left(t\right)={\displaystyle \sum _{i=1}^n}{c}_i{(lnt)}^{\alpha -i},t\in (a,b),}$$

and the following formula holds

$${\displaystyle {}^H{I}_{a+}^{\alpha }{}^H{D}_{a+}^{\alpha }u\left(t\right)=u\left(t\right)+{\displaystyle \sum _{i=1}^n}{d}_i{(lnt)}^{\alpha -i},t\in (a,b),}$$

where $c_i,d_i\in \mathbb{R}$, $i=1,\dots ,n$ and $n=\left[\alpha \right]+1$.

Lemma 3.

Let $u\in L^1(a,b)$, ( $b>a\ge 1$). Then

(a) 

${}^{H}I{}_{a+}{}^{\alpha }^H{I}_{a+}^{\beta }u\left(t\right){=}^H{I}_{a+}^{\alpha +\beta }u\left(t\right),\forall \alpha ,\beta >0$;

(b) 

${}^H{D}_{a+}^{\beta }{}^H{I}_{a+}^{\alpha }u\left(t\right){=}^H{I}_{a+}^{\alpha -\beta }u\left(t\right),\forall \alpha >\beta >0$; ${}^H{D}_{a+}^{\alpha }{}^H{I}_{a+}^{\alpha }u\left(t\right)=u\left(t\right),\forall \alpha >0$.

Lemma 4.

Let $\alpha >0$, $p\in \mathbb{N}$ and $\delta =t{\displaystyle \frac{d}{dt}}$. If the fractional derivatives ${(}^H{D}_{a+}^{\alpha }u)\left(t\right)$ and ${(}^H{D}_{a+}^{\alpha +p}u)\left(t\right)$ exist, then

$$\left({\delta }^p{}^H{D}_{a+}^{\alpha }u\right)\left(t\right)={(}^H{D}_{a+}^{\alpha +p}u)\left(t\right),t\in (a,b).$$

Remark 1.

Using Lemma 4 with $p=1$, we obtain

$${\displaystyle \delta {}^H{D}_{a+}^{\alpha }u\left(t\right)={(}^H{D}_{a+}^{\alpha +1}u)\left(t\right)\iff \frac{d}{dt}{(}^H{D}_{a+}^{\alpha }u\left(t\right))=\frac{1}{t}{}^H{D}_{a+}^{\alpha +1}u\left(t\right),t\in (a,b).}$$

Then

$${\displaystyle {}^H{D}_{a+}^{\alpha }u\left(t\right){=}^H{D}_{a+}^{\alpha }u\left(t_1\right)+{\int }_{t_1}^t{}^H{D}_{a+}^{\alpha +1}u\left(s\right)\frac{ds}{s},t,t_1\in (a,b).}$$

(9)

3. Main Result

In this section, we present the auxiliary results and our main theorem for the existence of solutions to Problem (1),(2).

We denote by

$$\begin{array}{c}{\displaystyle a={\int }_1^e(P\left(e\right)-P\left(s\right))\frac{ds}{s},b={\int }_1^e(P\left(e\right)-P\left(s\right))\frac{lns}{s}ds,}\hfill \\ {\displaystyle c={\int }_1^e(Q\left(1\right)-Q\left(s\right))\frac{ds}{s},d={\int }_1^e(Q\left(1\right)-Q\left(s\right))\frac{lns}{s}ds,}\hfill \\ \Delta =ad-bc.\hfill \end{array}$$

(10)

We give firstly our main assumptions:

(A1)

The function $f:(1,e)\times {\mathbb{R}}^n\to \mathbb{R}$ is a Carathéodory function, satisfying the conditions

(a)

$f(·,u_1,u_2,\dots ,u_n)$ is measurable for each $(u_1,u_2,\dots ,u_n)\in {\mathbb{R}}^n$;

(b)

$f(t,·,·,\dots ,·)$ is continuous (in the last n variables) for a.e. $t\in (1,e)$;

(c)

For each $l>0$ there exists ${\omega }_l\in L^{\infty }(1,e)$ such that

$$|f(t,u_1,u_2,\dots ,u_n)|\le {\omega }_l\left(t\right),$$

for all $u_i\in \mathbb{R}$ with $|u_i|\le l,i=1,\dots ,n$, and a.e. $t\in (1,e)$.

(A2)

The functions $P,Q:[1,e]\to \mathbb{R}$ have bounded variations, and satisfy the resonant conditions ${\int }_1^{e}dP\left(s\right)=1$ (that is, $P\left(e\right)-P\left(1\right)=1$), and ${\int }_1^{e}dQ\left(s\right)=1$ (that is, $Q\left(e\right)-Q\left(1\right)=1$); in addition, $\Delta \ne 0$, where $\Delta$ is given by (10).

Remark 2.

If $\left(A2\right)$ holds, then Problem (1),(2) is at resonance, because the fractional differential equation

$${}^H{D}_{1+}^{\alpha }u\left(t\right)=0,t\in (1,e),$$

with the boundary conditions (2) has the nontrivial solutions $u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},$ for $t\in [1,e]$, with $a_1,a_2\in \mathbb{R}$ (see the proof of Lemma 5 below).

We introduce the Banach space

$$\mathcal{X}=\left\{u;u{,}^H{D}_{1+}^{\alpha -n+1}u,\dots {,}^H{D}_{1+}^{\alpha -1}u\in C[1,e]\right\},$$

with the norm

$${\parallel u\parallel }_{\mathcal{X}}=max\left\{{\parallel u\parallel }_{\infty }{,\parallel }^H{D}_{1+}^{\alpha -n+1}{u\parallel }_{\infty },\dots ,{{\parallel }^H{D}_{1+}^{\alpha -1}u\parallel }_{\infty }\right\},$$

where ${\parallel u\parallel }_{\infty }={sup}_{t\in [1,e]}|u\left(t\right)|$, and the Banach space $\mathcal{Y}=L^1(1,e)$ with the norm ${\parallel v\parallel }_{\mathcal{Y}}={\int }_1^e|v\left(t\right)|dt$.

We define the linear operator $\mathcal{L}:D\left(\mathcal{L}\right)\subset \mathcal{X}\to \mathcal{Y}$, with

$$D\left(\mathcal{L}\right)=\left\{u\in \mathcal{X};{}^H{D}_{1+}^{\alpha }u\in \mathcal{Y},\mathrm{and}u\mathrm{satisfies}\mathrm{the}\mathrm{boundary}\mathrm{conditions}\left(2\right)\right\},$$

given by

$$\mathcal{L}u\left(t\right){=}^H{D}_{1+}^{\alpha }u\left(t\right),t\in (1,e),u\in D\left(\mathcal{L}\right),$$

and the nonlinear operator $\mathcal{N}:\mathcal{X}\to \mathcal{Y}$, given by

$$\mathcal{N}u\left(t\right)=f(t,u\left(t\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(t\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(t\right)),t\in (1,e),u\in \mathcal{X}.$$

Using the above operators $\mathcal{L}$ and $\mathcal{N}$, Problem (1),(2) is equivalent to the operator equation

$$\mathcal{L}u=\mathcal{N}u,u\in D\left(\mathcal{L}\right).$$

(11)

By Remark 2, we see that operator $\mathcal{L}$ is non-invertible.

Lemma 5.

We suppose that $\left(A2\right)$ holds. Then operator $\mathcal{L}:D\left(\mathcal{L}\right)\subset \mathcal{X}\to \mathcal{Y}$ has the properties

$$Ker\mathcal{L}=\{u\in D\left(\mathcal{L}\right);u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},t\in [1,e],a_1,a_2\in \mathbb{R}\}\subset \mathcal{X},$$

(12)

$$Im\mathcal{L}=\{v\in \mathcal{Y};\left({\mathcal{S}}_{1}v\right)\left(t\right)=\left({\mathcal{S}}_{2}v\right)\left(t\right)=0,\forall t\in [1,e]\}\subset \mathcal{Y},$$

(13)

where

$$\begin{array}{c}{\displaystyle \left({\mathcal{S}}_{1}v\right)\left(t\right)={\int }_1^e(P\left(e\right)-P\left(s\right))v\left(s\right)\frac{ds}{s},\forall t\in [1,e],}\hfill \\ {\displaystyle \left({\mathcal{S}}_{2}v\right)\left(t\right)={\int }_1^e(Q\left(1\right)-Q\left(s\right))v\left(s\right)\frac{ds}{s},\forall t\in [1,e].}\hfill \end{array}$$

(14)

Proof. 

Let $u\in Ker\mathcal{L}$, i.e., $u\in D\left(\mathcal{L}\right)$ and $\mathcal{L}u=0$. So ${}^H{D}_{1+}^{\alpha }u\left(t\right)=0$. Then, by Lemma 2, we obtain

$$u\left(t\right)=a_1{(lnt)}^{\alpha -1}+\cdots +a_n{(lnt)}^{\alpha -n},t\in [1,e],$$

(15)

with $a_i\in \mathbb{R},i=1,\dots ,n$. By conditions $u^{\left(i\right)}\left(1\right)=0$, for $i=0,\dots ,n-3$, we deduce that $a_j=0$, for $j=3,\dots ,n$. Therefore, by (15), we find

$$u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},t\in [1,e].$$

Because ${}^H{D}_{1+}^{\alpha -1}u\left(t\right)=a_1\Gamma \left(\alpha \right)$, then, by $\left(A2\right)$, the conditions

$$\begin{array}{c}{\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(1\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dP\left(s\right)\iff a_1\Gamma \left(\alpha \right)\left({\displaystyle 1-{\int }_1^{e}dP\left(s\right)}\right)=0,\mathrm{and}}\hfill \\ {\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(e\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dQ\left(s\right)\iff a_1\Gamma \left(\alpha \right)\left({\displaystyle 1-{\int }_1^{e}dQ\left(s\right)}\right)=0,}\hfill \end{array}$$

are satisfied. So, we deduce

$$Ker\mathcal{L}\subset \left\{u\in D\left(L\right);u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},a_1,a_2\in \mathbb{R}\right\}.$$

Conversely, let $u\in D\left(\mathcal{L}\right)$, $u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},t\in [1,e],$ with $a_1,a_2\in \mathbb{R}$. We easily see that ${}^H{D}_{1+}^{\alpha }u\left(t\right)=0,t\in [1,e]$, and u verifies the conditions (2), i.e., $u\in Ker\mathcal{L}$. Hence,

$$\left\{u\in D\left(\mathcal{L}\right);u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},a_1,a_2\in \mathbb{R}\right\}\subset Ker\mathcal{L}.$$

Therefore Relationship (12) holds. We remark that $dimKer\mathcal{L}=2$.

Now we prove Relationship (13). Let $v\in Im\mathcal{L}$. Then there exists $u\in D\left(\mathcal{L}\right)$ such that $\mathcal{L}u=v$ or

$${}^H{D}_{1+}^{\alpha }u\left(t\right)=v\left(t\right),t\in (1,e),u\in D\left(\mathcal{L}\right).$$

(16)

We apply the integral operator ${}^H{I}_{1+}^{\alpha }$ to Equation (16), and we obtain

$$\begin{array}{c}u\left(t\right){=}^H{I}_{1+}^{\alpha }v\left(t\right)+a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2}+\cdots +a_n{(lnt)}^{\alpha -n}\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}\frac{v\left(s\right)}{s}ds+a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2}+\cdots +a_n{(lnt)}^{\alpha -n},}\hfill \end{array}$$

(17)

with $a_i\in \mathbb{R},i=1,\dots ,n$. By conditions $u^{\left(i\right)}\left(1\right)=0$, $i=0,\dots ,n-3$, we deduce $a_3=a_4=\cdots =a_n=0$. Therefore, by (17), we find

$${\displaystyle u\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}\frac{v\left(s\right)}{s}ds+a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2}.}$$

Then

$${\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(t\right){=}^H{I}_{1+}^{1}v\left(t\right)+a_1\Gamma \left(\alpha \right)={\int }_1^t\frac{v\left(s\right)}{s}ds+a_1\Gamma \left(\alpha \right).}$$

So conditions ${}^H{D}_{1+}^{\alpha -1}u\left(1\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dP\left(s\right)$ and ${}^H{D}_{1+}^{\alpha -1}u\left(e\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dQ\left(s\right)$, give us

$$\left\{\begin{array}{c}{\displaystyle a_1\Gamma \left(\alpha \right)={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }+a_1\Gamma \left(\alpha \right)}\right)dP\left(s\right),}\hfill \\ {\displaystyle {\int }_1^{e}v\left(s\right)\frac{ds}{s}+a_1\Gamma \left(\alpha \right)={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }+a_1\Gamma \left(\alpha \right)}\right)dQ\left(s\right),}\hfill \end{array}\right.$$

or

$$\left\{\begin{array}{c}{\displaystyle a_1\Gamma \left(\alpha \right)\left({\displaystyle 1-{\int }_1^{e}dP\left(s\right)}\right)={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dP\left(s\right),}\hfill \\ {\displaystyle a_1\Gamma \left(\alpha \right)\left({\displaystyle 1-{\int }_1^{e}dQ\left(s\right)}\right)=-{\int }_1^{e}v\left(s\right)\frac{ds}{s}+{\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dQ\left(s\right).}\hfill \end{array}\right.$$

By the resonant conditions from $\left(A2\right)$, we obtain

$$\left\{\begin{array}{c}{\displaystyle {\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dP\left(s\right)=0,}\hfill \\ {\displaystyle {\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dQ\left(s\right)-{\int }_1^{e}v\left(s\right)\frac{ds}{s}=0,}\hfill \end{array}\right.$$

and so

$$\left\{\begin{array}{c}{\displaystyle {\int }_1^e(P\left(e\right)-P\left(s\right))v\left(s\right)\frac{ds}{s}=0,}\hfill \\ {\displaystyle {\int }_1^e(Q\left(1\right)-Q\left(s\right))v\left(s\right)\frac{ds}{s}=0.}\hfill \end{array}\right.$$

Thus, $\left({\mathcal{S}}_{1}v\right)\left(t\right)=\left({\mathcal{S}}_{2}v\right)\left(t\right)=0$, for all $t\in [1,e]$, i.e.,

$$Im\mathcal{L}\subset \{v\in \mathcal{Y};\left({\mathcal{S}}_{1}v\right)\left(t\right)=\left({\mathcal{S}}_{2}v\right)\left(t\right)=0,\forall t\in [1,e]\}.$$

(18)

Conversely, let $v\in \mathcal{Y}$ with $\left({\mathcal{S}}_{1}v\right)\left(t\right)=\left({\mathcal{S}}_{2}v\right)\left(t\right)=0$ for all $t\in [1,e]$. We take $u\left(t\right)=$${}^H{I}_{1+}^{\alpha }v\left(t\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^t{(ln{\displaystyle \frac{t}{s}})}^{\alpha -1}v\left(s\right){\displaystyle \frac{ds}{s}},t\in [1,e]$. Obviously, we deduce that $u\in D\left(\mathcal{L}\right)$. Indeed we have u, ${}^H{D}_{1+}^{\alpha -n+1}u{=}^H{I}_{1+}^{n-1}u,\dots {,}^H{D}_{1+}^{\alpha -1}u{=}^H{I}_{1+}^{1}u\in C[1,e]$ and ${}^H{D}_{1+}^{\alpha }u=v\in \mathcal{Y}$. In addition, u satisfies the conditions (2), i.e., $u^{\left(i\right)}\left(1\right)=0,i=0,\dots ,n-3$. Because ${}^H{D}_{1+}^{\alpha -1}u\left(t\right){=}^H{I}_{1+}^{1}v\left(t\right)={\int }_1^{t}v\left(s\right){\displaystyle \frac{ds}{s}}$, then the conditions

$$\begin{array}{c}{\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(1\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dP\left(s\right)\iff 0={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dP\left(s\right)}\hfill \\ {\displaystyle \iff 0={\int }_1^e\left({\displaystyle {\int }_s^{e}dP\left(\tau \right)}\right)v\left(s\right)\frac{ds}{s}\iff 0={\int }_1^e(P\left(e\right)-P\left(s\right))v\left(s\right)\frac{ds}{s},}\hfill \end{array}$$

and

$$\begin{array}{c}{\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(e\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dQ\left(s\right)\iff {\int }_1^{e}v\left(s\right)\frac{ds}{s}={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dQ\left(s\right)}\hfill \\ {\displaystyle \iff {\int }_1^{e}v\left(s\right)\frac{ds}{s}={\int }_1^e\left({\displaystyle {\int }_{\tau }^{e}dQ\left(s\right)}\right)v\left(\tau \right)\frac{d\tau }{\tau }\iff {\int }_1^{e}v\left(s\right)\frac{ds}{s}={\int }_1^e\left({\displaystyle {\int }_s^{e}dQ\left(\tau \right)}\right)v\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle \iff {\int }_1^{e}v\left(s\right)\frac{ds}{s}={\int }_1^e(Q\left(e\right)-Q\left(s\right))v\left(s\right)\frac{ds}{s}\iff {\int }_1^e(Q\left(1\right)-Q\left(s\right))v\left(s\right)\frac{ds}{s}=0,}\hfill \end{array}$$

are satisfied. In addition, we have $\mathcal{L}u\left(t\right)=v\left(t\right)\iff {}^H{D}_{1+}^{\alpha }{(}^H{I}_{1+}^{\alpha }v\left(t\right))=v\left(t\right)$ for $t\in (1,e)$. Then we conclude

$$\{v\in \mathcal{Y};\left({\mathcal{S}}_{1}v\right)\left(t\right)=\left({\mathcal{S}}_{2}v\right)\left(t\right)=0,\forall t\in [1,e]\}\subset Im\mathcal{L}.$$

(19)

Therefore, by (18) and (19) we obtain Relationship (13).  □

Remark 3.

The kernel of operator $\mathcal{L}$ is characterized by logarithmic terms due to the Hadamard integral and derivative, which are involved in our problem.

Now, we define the linear operators ${\mathcal{E}}_1,{\mathcal{E}}_2:\mathcal{Y}\to \mathcal{Y}$ by

$${\displaystyle {\mathcal{E}}_{1}v=\frac{1}{\Delta }(d{\mathcal{S}}_{1}v-b{\mathcal{S}}_{2}v),{\mathcal{E}}_{2}v=\frac{1}{\Delta }(a{\mathcal{S}}_{2}v-c{\mathcal{S}}_{1}v),v\in \mathcal{Y}.}$$

(20)

Lemma 6.

We suppose that $\left(A2\right)$ holds. Then $\mathcal{L}:D\left(\mathcal{L}\right)\subset \mathcal{X}\to \mathcal{Y}$ is a Fredholm operator of index zero. Moreover, the operators $\mathcal{H}:\mathcal{X}\to \mathcal{X}$ and $\mathcal{K}:\mathcal{Y}\to \mathcal{Y}$ given by

$$\begin{array}{c}{\displaystyle \mathcal{H}u\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{\alpha -2},t\in [1,e],u\in \mathcal{X},}\hfill \\ \mathcal{K}v\left(t\right)={\mathcal{E}}_{1}v\left(t\right)+\left({\mathcal{E}}_{2}v\left(t\right)\right)lnt,t\in [1,e],v\in \mathcal{Y},\hfill \end{array}$$

(21)

are linear projectors and satisfy the relationships

$$Im\mathcal{H}=Ker\mathcal{L},Ker\mathcal{K}=Im\mathcal{L},\mathcal{X}=Ker\mathcal{H}\oplus Ker\mathcal{L},\mathcal{Y}=Im\mathcal{L}\oplus Im\mathcal{K}.$$

(22)

Proof. 

By definition of operator $\mathcal{H}$, we see that it is a continuous and linear projector operator. Indeed, we have

$${\displaystyle {\mathcal{H}}^{2}u\left(t\right)=\mathcal{H}\left(\mathcal{H}u\left(t\right)\right)=\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}\left(\mathcal{H}u\right)\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}\left(\mathcal{H}u\right)\left(1\right){(lnt)}^{\alpha -2}.}$$

(23)

We obtain

$${}^H{D}_{1+}^{\alpha -1}\left(\mathcal{H}u\right)\left(t\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\Gamma \left(\alpha \right),$$

and so ${}^H{D}_{1+}^{\alpha -1}\left(\mathcal{H}u\right)\left(1\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\Gamma \left(\alpha \right)$. In addition, we find

$${}^H{D}_{1+}^{\alpha -2}\left(\mathcal{H}u\right)\left(t\right){=}^H{D}_{1+}^{\alpha -1}u\left(1\right)(lnt){+}^H{D}_{1+}^{\alpha -2}u\left(1\right),$$

and then ${}^H{D}_{1+}^{\alpha -2}\left(\mathcal{H}u\right)\left(1\right){=}^H{D}_{1+}^{\alpha -2}u\left(1\right)$. Therefore Relationship (23) gives us ${\mathcal{H}}^{2}u\left(t\right)=\mathcal{H}u\left(t\right)$ for all $t\in [1,e]$ and $u\in \mathcal{X}$.

Moreover, $\mathcal{H}$ satisfies the relationship $Im\mathcal{H}=Ker\mathcal{L}$. To prove this, we consider first $v\in Im\mathcal{H}$; then there exists $u\in \mathcal{X}$ such that $\mathcal{H}\left(u\right)=v$, or equivalently

$${\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{\alpha -2}=v.}$$

From the above relationship, we see that $v\in D\left(\mathcal{L}\right)$, and evidently ${}^H{D}_{1+}^{\alpha }v\left(t\right)=0$; so $v\in Ker\mathcal{L}$. Therefore, we deduce that $Im\mathcal{H}\subset Ker\mathcal{L}$. In what follows, we consider $v\in Ker\mathcal{L}$, namely $v\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},$$t\in [1,e]$, $a_1,a_2\in \mathbb{R}.$ Then there exists $u\in \mathcal{X}$ such that $\mathcal{H}\left(u\right)=v$. Indeed for $u=v$, we obtain $\mathcal{H}\left(u\right)=v$, i.e., $v\in Im\mathcal{H}$. We derive that $Ker\mathcal{L}\subset Im\mathcal{H}$. Therefore, we conclude that $Im\mathcal{H}=Ker\mathcal{L}$.

Based on this last relationship, ( $Im\mathcal{H}=Ker\mathcal{L}$), we obtain $\mathcal{X}=Ker\mathcal{H}\oplus Ker\mathcal{L}$. To show this, we firstly see that

$$Ker\mathcal{H}=\{u\in \mathcal{X},{}^H{D}_{1+}^{\alpha -1}u\left(1\right)=0,{}^H{D}_{1+}^{\alpha -2}u\left(1\right)=0\}.$$

Then for $u\in \mathcal{X}$, we have $u=(u-\mathcal{H}u)+\mathcal{H}u$; so we find $\mathcal{X}=Ker\mathcal{H}+Im\mathcal{H}=Ker\mathcal{H}+Ker\mathcal{L}$. We also have the relationship $Ker\mathcal{H}\cap Ker\mathcal{L}=\left\{0\right\}$. Indeed, we easily see that $\left\{0\right\}\subset Ker\mathcal{H}\cap Ker\mathcal{L}$. In addition $Ker\mathcal{H}\cap Ker\mathcal{L}\subset \left\{0\right\}$. To prove this last inclusion, we consider $u\in Ker\mathcal{H}\cap Ker\mathcal{L}$, i.e., $\mathcal{H}u=0$ and $\mathcal{L}u=0$. Then $u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},$ with $a_1,a_2\in \mathbb{R}$, and ${}^H{D}_{1+}^{\alpha -1}u\left(1\right)=0$, and ${}^H{D}_{1+}^{\alpha -2}u\left(1\right)=0$. By the last two relationships, we deduce that $u\left(t\right)=0,$ for all $t\in [1,e]$. Therefore, we conclude that $Ker\mathcal{H}\cap Ker\mathcal{L}\subset \left\{0\right\}$. Therefore, we infer that $Ker\mathcal{H}\cap Ker\mathcal{L}=\left\{0\right\}$, and then $\mathcal{X}=Ker\mathcal{H}\oplus Ker\mathcal{L}$.

Next, we remark that the operator $\mathcal{K}$ is continuous and linear, and $dimIm\mathcal{K}=2$. By definitions of ${\mathcal{E}}_1$ and ${\mathcal{E}}_2$ (see (20)), we obtain for any $v\in \mathcal{Y}$ and $t\in [1,e]$

$$\begin{array}{c}{\displaystyle {\mathcal{E}}_1\left({\mathcal{E}}_{1}v\right)\left(t\right)=\frac{1}{\Delta }(d{\mathcal{S}}_1\left({\mathcal{E}}_{1}v\right)-b{\mathcal{S}}_2\left({\mathcal{E}}_{1}v\right))}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left[{\displaystyle d{\int }_1^e(P\left(e\right)-P\left(s\right))\left({\mathcal{E}}_{1}v\right)\left(s\right)\frac{ds}{s}-b{\int }_1^e(Q\left(1\right)-Q\left(s\right))\left({\mathcal{E}}_{1}v\right)\left(s\right)\frac{ds}{s}}\right]}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left({\mathcal{E}}_{1}v\right)\left(t\right)\left({\displaystyle d{\int }_1^e(P\left(e\right)-P\left(s\right))\frac{ds}{s}-b{\int }_1^e(Q\left(1\right)-Q\left(s\right))\frac{ds}{s}}\right)=\left({\mathcal{E}}_{1}v\right)\left(t\right),}\hfill \end{array}$$

$$\begin{array}{c}{\displaystyle {\mathcal{E}}_2(\left({\mathcal{E}}_{2}v\right)lnt)\left(t\right)=\frac{2}{\Delta }(a{\mathcal{S}}_2(\left({\mathcal{E}}_{2}v\right)lnt)-c{\mathcal{S}}_1(\left({\mathcal{E}}_{2}v\right)lnt))}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left[{\displaystyle a{\int }_1^e(Q\left(1\right)-Q\left(s\right))\left({\mathcal{E}}_{2}v\right)\left(s\right)lns\frac{ds}{s}-c{\int }_1^e(P\left(e\right)-P\left(s\right))\left({\mathcal{E}}_{2}v\right)\left(s\right)lns\frac{ds}{s}}\right]}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left({\mathcal{E}}_{2}v\right)\left(t\right)\left[{\displaystyle a{\int }_1^e(Q\left(1\right)-Q\left(s\right))lns\frac{ds}{s}-c{\int }_1^e(P\left(e\right)-P\left(s\right))lns\frac{ds}{s}}\right]=\left({\mathcal{E}}_{2}v\right)\left(t\right),}\hfill \\ {\displaystyle {\mathcal{E}}_1(\left({\mathcal{E}}_{2}v\right)lnt)\left(t\right)=\frac{1}{\Delta }\left(d{\mathcal{S}}_1(\left({\mathcal{E}}_{2}v\right)lnt)-b{\mathcal{S}}_2(\left({\mathcal{E}}_{2}v\right)lnt)\right)}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left[{\displaystyle d{\int }_1^e(P\left(e\right)-P\left(s\right))\left({\mathcal{E}}_{2}v\right)\left(s\right)lns\frac{ds}{s}-b{\int }_1^e(Q\left(1\right)-Q\left(s\right))\left({\mathcal{E}}_{2}v\right)\left(s\right)lns\frac{ds}{s}}\right]}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left[{\displaystyle d\left({\mathcal{E}}_{2}v\right)\left(t\right){\int }_1^e(P\left(e\right)-P\left(s\right))lns\frac{ds}{s}-b\left({\mathcal{E}}_{2}v\right)\left(t\right){\int }_1^e(Q\left(1\right)-Q\left(s\right))lns\frac{ds}{s}}\right]}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left({\mathcal{E}}_{2}v\right)\left(t\right)(db-bd)=0,}\hfill \\ {\displaystyle {\mathcal{E}}_2\left({\mathcal{E}}_{1}v\right)\left(t\right)=\frac{1}{\Delta }(a{\mathcal{S}}_2\left({\mathcal{E}}_{1}v\right)-c{\mathcal{S}}_1\left({\mathcal{E}}_{1}v\right))}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left[{\displaystyle a{\int }_1^e(Q\left(1\right)-Q\left(s\right))\left({\mathcal{E}}_{1}v\right)\left(s\right)\frac{ds}{s}-c{\int }_1^e(P\left(e\right)-P\left(s\right))\left({\mathcal{E}}_{1}v\right)\left(s\right)\frac{ds}{s}}\right]}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left({\mathcal{E}}_{1}v\right)\left(t\right)\left[{\displaystyle a{\int }_1^e(Q\left(1\right)-Q\left(s\right))\frac{ds}{s}-c{\int }_1^e(P\left(e\right)-P\left(s\right))\frac{ds}{s}}\right]}\hfill \\ {\displaystyle =\frac{1}{\Delta }\left({\mathcal{E}}_{1}v\right)\left(t\right)(ac-ca)=0.}\hfill \end{array}$$

Then, by the above relationships, we find

$$\begin{array}{c}{\mathcal{K}}^{2}v\left(t\right)=\mathcal{K}\left(\mathcal{K}v\right)\left(t\right)={\mathcal{E}}_1\left(\mathcal{K}v\right)\left(t\right)+{\mathcal{E}}_2\left(\mathcal{K}v\right)\left(t\right)lnt\hfill \\ ={\mathcal{E}}_1({\mathcal{E}}_{1}v\left(t\right)+\left({\mathcal{E}}_{2}v\left(t\right)\right)lnt)+{\mathcal{E}}_2({\mathcal{E}}_{1}v\left(t\right)+\left({\mathcal{E}}_{2}v\left(t\right)\right)lnt)lnt\hfill \\ ={\mathcal{E}}_1\left({\mathcal{E}}_{1}v\right)\left(t\right)+{\mathcal{E}}_1({\mathcal{E}}_{2}v\left(t\right)lnt)+{\mathcal{E}}_2\left({\mathcal{E}}_{1}v\left(t\right)\right)lnt\hfill \\ +{\mathcal{E}}_2({\mathcal{E}}_{2}v\left(t\right)lnt)lnt={\mathcal{E}}_{1}v\left(t\right)+{\mathcal{E}}_{2}v\left(t\right)lnt=\mathcal{K}v\left(t\right),\hfill \end{array}$$

for all $v\in \mathcal{Y}$ and $t\in [1,e]$. So, we deduce that $\mathcal{K}$ is a projector operator.

By Lemma 5 we have $Im\mathcal{L}\subset Ker\mathcal{K}$. We will prove next that $Ker\mathcal{K}\subset Im\mathcal{L}$. For this, let $v\in Ker\mathcal{K}$ be arbitrary but fixed for the moment. Then $\mathcal{K}v=0$, or ${\mathcal{E}}_{1}v\left(t\right)+\left({\mathcal{E}}_{2}v\left(t\right)\right)lnt=0$ for all $t\in [1,e]$. So, we obtain

$${\displaystyle \frac{1}{\Delta }(d{\mathcal{S}}_{1}v-b{\mathcal{S}}_{2}v)+\frac{1}{\Delta }(a{\mathcal{S}}_{2}v-c{\mathcal{S}}_{1}v)lnt=0.}$$

Therefore, we find the following homogeneous system of linear equations in the unknowns ${\mathcal{S}}_{1}v$ and ${\mathcal{S}}_{2}v$

$$\left\{\begin{array}{c}d{\mathcal{S}}_{1}v-b{\mathcal{S}}_{2}v=0,\hfill \\ a{\mathcal{S}}_{2}v-c{\mathcal{S}}_{1}v=0.\hfill \end{array}\right.$$

(24)

Because the determinant of the system (24) is $\Delta =ad-bc\ne 0$, we deduce that this system has only the zero solution, i.e., ${\mathcal{S}}_{1}v=0$ and ${\mathcal{S}}_{2}v=0$. So $v\in Im\mathcal{L}$, and then $Ker\mathcal{K}\subset Im\mathcal{L}$. Therefore, we conclude that $Ker\mathcal{K}=Im\mathcal{L}$.

Based on this last relationship ( $Ker\mathcal{K}=Im\mathcal{L}$), we obtain that $\mathcal{Y}=Im\mathcal{L}\oplus Im\mathcal{K}$. Indeed, for a function $v\in \mathcal{Y}$, we can write $v=(v-\mathcal{K}v)+\mathcal{K}v$, and then $v-\mathcal{K}v\in Ker\mathcal{K}=Im\mathcal{L}$. So $\mathcal{K}v\in Im\mathcal{K}$. Hence, we deduce that $v\in Im\mathcal{L}+Im\mathcal{K}$. In addition, for any $v\in Im\mathcal{L}\cap Im\mathcal{K}$ we obtain ${\mathcal{S}}_{1}v=0$, ${\mathcal{S}}_{2}v=0$, and there exists $w\in D\left(\mathcal{K}\right)$ such that $\mathcal{K}w=v$. Therefore, there exist $c_1,c_2\in \mathbb{R}$ such that $v\left(t\right)=c_1+c_{2}lnt$, $t\in [1,e]$. We deduce the system

$$\left\{\begin{array}{c}{\displaystyle {\int }_1^e(P\left(e\right)-P\left(s\right))(c_1+c_{2}lns)\frac{ds}{s}=0,}\hfill \\ {\displaystyle {\int }_1^e(Q\left(1\right)-Q\left(s\right))(c_1+c_{2}lns)\frac{ds}{s}=0,}\hfill \end{array}\right.$$

or

$$\left\{\begin{array}{c}{\displaystyle c_1{\int }_1^e(P\left(e\right)-P\left(s\right))\frac{ds}{s}+c_2{\int }_1^e(P\left(e\right)-P\left(s\right))lns\frac{ds}{s}=0,}\hfill \\ {\displaystyle c_1{\int }_1^e(Q\left(1\right)-Q\left(s\right))\frac{ds}{s}+c_2{\int }_1^s(Q\left(1\right)-Q\left(s\right))lns\frac{ds}{s}=0.}\hfill \end{array}\right.$$

Then, we find the system

$$\left\{\begin{array}{c}a{c}_1+b{c}_2=0,\hfill \\ c{c}_1+d{c}_2=0,\hfill \end{array}\right.$$

which has the unique solution $c_1=c_2=0$. Hence, we deduce that $v=0$, i.e., $Im\mathcal{L}\cap Im\mathcal{K}=\left\{0\right\}$. So $\mathcal{Y}=Im\mathcal{L}\oplus Im\mathcal{K}$, and all the relationships from (22) hold.

Therefore, by (22) we obtain $dimKer\mathcal{L}=2$ and $codimIm\mathcal{L}=2$. Because $Im\mathcal{L}$ is closed, we conclude that $\mathcal{L}$ is a Fredholm operator with the index $Ind\mathcal{L}=0$.  □

Lemma 7.

Assume that $\left(A2\right)$ holds. Then the operator ${\mathcal{T}}_{\mathcal{H}}:Im\mathcal{L}\subset \mathcal{Y}\to D\left(\mathcal{L}\right)\cap Ker\mathcal{H}\subset \mathcal{X}$ defined by

$${\displaystyle \left({\mathcal{T}}_{\mathcal{H}}v\right)\left(t\right){=}^H{I}_{1+}^{\alpha }v\left(t\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}v\left(s\right)\frac{ds}{s},v\in Im\mathcal{L},}$$

(25)

is the inverse of operator ${\mathcal{L}|}_{D\left(\mathcal{L}\right)\cap Ker\mathcal{H}}$, and $\parallel {\mathcal{T}}_{\mathcal{H}}{v\parallel }_{\mathcal{X}}\le {\parallel v\parallel }_{\mathcal{Y}}$ for all $v\in Im\mathcal{L}$.

Proof. 

We will prove that ${\mathcal{T}}_{\mathcal{H}}$ is well defined on $Im\mathcal{L}$. For $v\in Im\mathcal{L}$, we show firstly that ${\mathcal{T}}_{\mathcal{H}}v\in D\left(\mathcal{L}\right)$. Indeed, we have ${\mathcal{T}}_{\mathcal{H}}v\in \mathcal{X}$, because: ${\mathcal{T}}_{\mathcal{H}}v\in C[1,e]$, ${}^H{D}_{1+}^{\alpha -n+1}{\mathcal{T}}_{\mathcal{H}}v{=}^H{I}_{1+}^{n-1}v\in C[1,e]$, …, ${}^H{D}_{1+}^{\alpha -1}{\mathcal{T}}_{\mathcal{H}}v{=}^H{I}_{1+}^{1}v\in C[1,e]$, and ${}^H{D}_{1+}^{\alpha }{\mathcal{T}}_{\mathcal{H}}v=v\in \mathcal{Y}$. Moreover, ${\mathcal{T}}_{\mathcal{H}}v$ satisfies the conditions (2), due to: ${\left({\mathcal{T}}_{\mathcal{H}}v\right)}^{\left(i\right)}\left(1\right)=D^i{}^H{I}_{1+}^{\alpha }v\left(1\right)=0$ for $i=0,\dots ,n-3$; in addition, we find ${}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(t\right){=}^H{I}_{1+}^{1}v\left(t\right)={\int }_1^{t}v\left(s\right){\displaystyle \frac{ds}{s}}$. So the conditions ${}^H{D}_{1+}^{\alpha -1}u\left(1\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dP\left(s\right)$ and ${}^H{D}_{1+}^{\alpha -1}u\left(e\right)={\int }_1^e{}^H{D}_{1+}^{\alpha -1}u\left(s\right)dQ\left(s\right)$ are equivalent to

$$\left\{\begin{array}{c}{\displaystyle {\int }_1^{1}v\left(s\right)\frac{ds}{s}={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dP\left(s\right),}\hfill \\ {\displaystyle {\int }_1^{e}v\left(s\right)\frac{ds}{s}={\int }_1^e\left({\displaystyle {\int }_1^{s}v\left(\tau \right)\frac{d\tau }{\tau }}\right)dQ\left(s\right),}\hfill \end{array}\right.$$

or

$$\left\{\begin{array}{c}{\displaystyle {\int }_1^e(P\left(e\right)-P\left(s\right))v\left(s\right)\frac{ds}{s}=0,}\hfill \\ {\displaystyle {\int }_1^e(Q\left(1\right)-Q\left(s\right))v\left(s\right)\frac{ds}{s}=0.}\hfill \end{array}\right.$$

(26)

The relationships from the system (26) are verified, because ${\mathcal{S}}_{1}v={\mathcal{S}}_{2}v=0$, ( $v\in Im\mathcal{L}$).

Then we show that ${\mathcal{T}}_{\mathcal{H}}v\in Ker\mathcal{H}$, i.e., $\mathcal{H}\left({\mathcal{T}}_{\mathcal{H}}v\right)=0$ or

$${\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(1\right){(lnt)}^{\alpha -2}=0.}$$

Indeed, we have

$$\begin{array}{c}{\displaystyle {}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(t\right){=}^H{I}_{1+}^{1}v\left(t\right)={\int }_1^{t}v\left(s\right)\frac{ds}{s},}\hfill \\ {\displaystyle {}^H{D}_{1+}^{\alpha -2}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(t\right){=}^H{I}_{1+}^{2}v\left(t\right)={\int }_1^{t}ln\frac{t}{s}v\left(s\right)\frac{ds}{s}.}\hfill \end{array}$$

Therefore, in the point 1 we obtain ${}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(1\right)=0$ and ${}^H{D}_{1+}^{\alpha -2}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(1\right)=0$, and then $\mathcal{H}\left({\mathcal{T}}_{\mathcal{H}}v\right)=0$. We deduce that ${\mathcal{T}}_{\mathcal{H}}v\in D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$. Therefore ${\mathcal{T}}_{\mathcal{H}}$ is well defined on $Im\mathcal{L}$.

We show next that ${\mathcal{T}}_{\mathcal{H}}$ is the inverse of operator ${\mathcal{L}|}_{D\left(\mathcal{L}\right)\cap Ker\mathcal{H}}$, i.e.,

(i)

$\mathcal{L}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(t\right)=v\left(t\right),\forall v\in Im\mathcal{L}$, $t\in (1,e)$;

(ii)

${\mathcal{T}}_{\mathcal{H}}\left(\mathcal{L}u\right)\left(t\right)=u\left(t\right),\forall u\in D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$, $t\in [1,e]$.

The above Relationship (i) follows immediately, in view of

$$\mathcal{L}\left({\mathcal{T}}_{\mathcal{H}}v\right)\left(t\right){=}^H{D}_{1+}^{\alpha }{(}^H{I}_{1+}^{\alpha }v)\left(t\right)=v\left(t\right),\forall v\in Im\mathcal{L},t\in (1,e).$$

To show Relationship (ii), we consider $u\in D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$. Then

$${\mathcal{T}}_{\mathcal{H}}\left(\mathcal{L}u\right)\left(t\right){=}^H{I}_{1+}^{\alpha }{(}^H{D}_{1+}^{\alpha }u)\left(t\right)=u\left(u\right)+a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2}+\cdots +a_n{(lnt)}^{\alpha -n},$$

(27)

with $a_i\in \mathbb{R}$, $i=1,\dots ,n$. Because $u\in D\left(\mathcal{L}\right)$, we have $u^{\left(i\right)}\left(1\right)=0$ for $i=0,\dots ,n-3$, and so $a_i=0$ for $i=3,\dots ,n$. Then (27) gives us

$${\mathcal{T}}_{\mathcal{H}}\left(\mathcal{L}u\right)\left(t\right)=u\left(t\right)+a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},t\in [1,e],a_1,a_2\in \mathbb{R}.$$

(28)

Due to $u\in Ker\mathcal{H}$, we find $\mathcal{H}u=0$. Therefore, by (28) we obtain

$$\begin{array}{c}\mathcal{H}\left({\mathcal{T}}_{\mathcal{H}}\left(\mathcal{L}u\right)\right)\left(t\right)=\mathcal{H}u\left(t\right)+\mathcal{H}(a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2})\hfill \\ =\mathcal{H}(a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2})=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2},\forall t\in [1,e].\hfill \end{array}$$

(29)

On the other hand, $\mathcal{H}\left({\mathcal{T}}_{\mathcal{H}}\left(\mathcal{L}u\right)\right)\left(t\right)=0$ for all $t\in [1,e]$, because ${\mathcal{T}}_{\mathcal{H}}\left(w\right)\in Ker\mathcal{H}$ for all $w\in Im\mathcal{L}$. Therefore, by (29) we deduce that $a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2}=0$ for all $t\in [1,e]$. Hence (28) gives us ${\mathcal{T}}_{\mathcal{H}}\left(\mathcal{L}u\right)\left(t\right)=u\left(t\right)$ for all $t\in [1,e]$.

Hence, by (i) and (ii), we conclude that ${\mathcal{T}}_{\mathcal{H}}$ is the inverse of ${\mathcal{L}|}_{D\left(\mathcal{L}\right)\cap Ker\mathcal{H}}$.

In addition, by using Lemma 3, we find for any $v\in Im\mathcal{L}$

$$\begin{array}{c}{\displaystyle |{\mathcal{T}}_{\mathcal{H}}v\left(t\right)|\le \frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}|v\left(s\right)|\frac{ds}{s}\le \frac{1}{\Gamma \left(\alpha \right)}{\int }_1^e|v\left(s\right)|\frac{ds}{s}}\hfill \\ {\displaystyle \le \frac{1}{\Gamma \left(\alpha \right)}{\int }_1^e|v\left(s\right)|ds=\frac{1}{\Gamma \left(\alpha \right)}{\parallel v\parallel }_{\mathcal{Y}}\le {\parallel v\parallel }_{\mathcal{Y}},\forall t\in [1,e],}\hfill \\ {|}^H{D}_{1+}^{\alpha -1}{\mathcal{T}}_{\mathcal{H}}{v\left(t\right)|=|}^H{D}_{1+}^{\alpha -1}{}^H{I}_{1+}^{\alpha }{v\left(t\right)|=|}^H{I}_{1+}^{1}v\left(t\right)|=\left|{\displaystyle {\int }_1^{t}v\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle \le {\int }_1^e|v\left(s\right)|ds={\parallel v\parallel }_{\mathcal{Y}},\forall t\in [1,e],}\hfill \\ {|}^H{D}_{1+}^{\alpha -2}{\mathcal{T}}_{\mathcal{H}}v\left(t\right)|=\left|{}^H{D}_{1+}^{\alpha -2}{}^H{I}_{1+}^{\alpha }v\left(t\right)\right|=\left|{}^H{I}_{1+}^{2}v\left(t\right)\right|=\left|{\displaystyle \frac{1}{\Gamma \left(2\right)}{\int }_1^{t}ln\frac{t}{s}v\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle \le {\int }_1^e|v\left(s\right)|ds={\parallel v\parallel }_{\mathcal{Y}},\forall t\in [1,e],}\hfill \\ ⋮\hfill \\ {|}^H{D}_{1+}^{\alpha -n+1}{\mathcal{T}}_{\mathcal{H}}v\left(t\right)|=\left|{}^H{D}_{1+}^{\alpha -n+1}{}^H{I}_{1+}^{\alpha }v\left(t\right)\right|={|}^H{I}_{1+}^{n-1}v\left(t\right)|=\left|{\displaystyle \frac{1}{\Gamma (n-1)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{n-2}v\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle \le {\int }_1^e|v\left(s\right)|ds={\parallel v\parallel }_{\mathcal{Y}},\forall t\in [1,e].}\hfill \end{array}$$

Therefore $\parallel {\mathcal{T}}_{\mathcal{H}}{v\parallel }_{\mathcal{X}}\le {\parallel v\parallel }_{\mathcal{Y}}$ for any $v\in Im\mathcal{L}$.  □

Lemma 8.

We suppose that $\left(A1\right)$, $\left(A2\right)$ hold. Let $\Omega \subset \mathcal{X}$ be an open bounded set satisfying the condition $D\left(\mathcal{L}\right)\cap \overline{\Omega }\ne \varnothing$. Then the operator $\mathcal{N}$ is $\mathcal{L}$-compact on $\overline{\Omega }$.

Proof. 

Using the characterization of the $\mathcal{L}$-compact operator from [2], we will show the following conditions:

(a)

the operator $\mathcal{K}\mathcal{N}:\overline{\Omega }\to \mathcal{Y}$ is continuous;

(b)

the set $\mathcal{K}\mathcal{N}\left(\overline{\Omega }\right)$ is bounded;

(c)

the operator ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}={\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}:\overline{\Omega }\to \mathcal{X}$ is continuous;

(d)

the set ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}\left(\overline{\Omega }\right)$ is relatively compact in $\mathcal{X}$.

For condition (a), we have

$$\begin{array}{c}\mathcal{K}\mathcal{N}\left(u\right)\left(t\right)={\mathcal{E}}_1\mathcal{N}u\left(t\right)+\left({\mathcal{E}}_2\mathcal{N}u\left(t\right)\right)lnt\hfill \\ {\displaystyle =\frac{1}{\Delta }(d{\mathcal{S}}_1\left(\mathcal{N}u\right)\left(t\right)-b{\mathcal{S}}_2\left(\mathcal{N}u\right)\left(t\right))+\frac{1}{\Delta }(a{\mathcal{S}}_2\left(\mathcal{N}u\right)\left(t\right)-c{\mathcal{S}}_1\left(\mathcal{N}u\right)\left(t\right))lnt,}\hfill \end{array}$$

where

$$\begin{array}{c}{\displaystyle {\mathcal{S}}_1\left(\mathcal{N}u\right)\left(t\right)={\int }_1^e(P\left(e\right)-P\left(s\right))\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle ={\int }_1^e(P\left(e\right)-P\left(s\right))f\left(s,u\left(s\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(s\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(s\right)\right)\frac{ds}{s},}\hfill \\ {\displaystyle {\mathcal{S}}_2\left(\mathcal{N}u\right)\left(t\right)={\int }_1^e(Q\left(1\right)-Q\left(s\right))\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle ={\int }_1^e(Q\left(1\right)-Q\left(s\right))f\left(s,u\left(s\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(s\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(s\right)\right)\frac{ds}{s},}\hfill \end{array}$$

for all $t\in [1,e]$ and $u\in \overline{\Omega }$. Because f is continuous in the last n variables, by the Lebesgue-dominated convergence theorem, we deduce that the operators ${\mathcal{S}}_1\mathcal{N}$ and ${\mathcal{S}}_2\mathcal{N}$ are continuous, and so $\mathcal{K}\mathcal{N}$ is also a continuous operator.

For condition (b), we remark first that there exists $C>0$ such that ${\parallel u\parallel }_{\mathcal{X}}\le C$ for all $u\in \overline{\Omega }$. Then, by $\left(A1\right)$, we deduce that there exists ${\omega }_C\in L^{\infty }(1,e)$, ${\omega }_C\left(t\right)\ge 0,a.e.t\in (1,e)$ such that $|f(t,u_1,\dots ,u_n)|\le {\omega }_C\left(t\right)$ for all $u_i\in \mathbb{R}$, $|u_i|\le C$, for all $i=1,\dots ,n$, and a.e. $t\in (1,e)$. Therefore, for any $u\in \overline{\Omega }$ and $t\in [1,e]$, we obtain

$$\begin{array}{c}{\displaystyle |{\mathcal{S}}_1\mathcal{N}u\left(t\right)|\le {\int }_1^e\left|(P\left(e\right)-P\left(s\right))f(s,u\left(s\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(s\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(s\right))\frac{1}{s}\right|ds}\hfill \\ {\displaystyle \le {\int }_1^e|P\left(e\right)-P\left(s\right)|{\omega }_C\left(s\right)\frac{ds}{s}\le {\mathrm{ess}\sup }_{t\in (1,e)}{\omega }_C\left(t\right){\int }_1^e|P\left(e\right)-P\left(s\right)|ds}\hfill \\ \le 2(e-1){\mathrm{ess}\sup }_{t\in (1,e)}{\omega }_C\left(t\right)\underset{t\in [1,e]}{sup}|P\left(t\right)|=:M_1,\hfill \\ |{\mathcal{S}}_2\mathcal{N}u\left(t\right)\left|{\displaystyle \le {\int }_1^e|(Q\left(1\right)-Q\left(s\right))f(s,u\left(s\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(s\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(s\right))\frac{1}{s}}\right|ds\hfill \\ {\displaystyle \le {\int }_1^e|Q\left(1\right)-Q\left(s\right)|{\omega }_C\left(s\right)\frac{ds}{s}\le {\mathrm{ess}\sup }_{t\in (1,e)}{\omega }_C\left(t\right){\int }_1^e|Q\left(1\right)-Q\left(s\right)|ds}\hfill \\ \le 2(e-1){\mathrm{ess}\sup }_{t\in (1,e)}{\omega }_C\left(t\right)\underset{t\in [1,e]}{sup}|Q\left(t\right)|=:M_2.\hfill \end{array}$$

Hence, we find

$${\displaystyle |\mathcal{K}\mathcal{N}u\left(t\right)|\le \frac{1}{\Delta }\left[(|c|+|d|)M_1+(|a|+|b|)M_2\right]=:M_3,\forall u\in \overline{\Omega },t\in [1,e].}$$

We deduce that

$${\parallel \mathcal{K}\mathcal{N}u\parallel }_{\mathcal{Y}}=\left|{\displaystyle {\int }_1^e\mathcal{K}\mathcal{N}u\left(t\right)dt}\right|\le (e-1)M_3,\forall u\in \overline{\Omega },$$

which means that the set $\mathcal{K}\mathcal{N}\left(\overline{\Omega }\right)$ is bounded.

For condition (c), we observe that the operator $(I-\mathcal{K})\mathcal{N}$ is continuous on $\overline{\Omega }$. Next, we have

$$\begin{array}{c}{\displaystyle {\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}\left(u\right)={\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right)=\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}\left[\mathcal{N}u\left(s\right)-{\mathcal{E}}_1\mathcal{N}u\left(s\right)-\left({\mathcal{E}}_2\mathcal{N}u\left(s\right)\right)lns\right]\frac{ds}{s}}\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^t{\left({\displaystyle ln\frac{t}{s}}\right)}^{\alpha -1}\left[{\displaystyle \mathcal{N}u\left(s\right)-\frac{1}{\Delta }(d{\mathcal{S}}_1\left(\mathcal{N}u\right)-b{\mathcal{S}}_2\left(\mathcal{N}u\right))}\right.}\hfill \\ {\displaystyle \left.{\displaystyle -\frac{1}{\Delta }(a{\mathcal{S}}_2\left(\mathcal{N}u\right)-c{\mathcal{S}}_1\left(\mathcal{N}u\right))lns}\right]\frac{ds}{s}.}\hfill \end{array}$$

Using the Lebesgue dominated convergence theorem for ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}\left(u\right)$ and its fractional derivatives ${}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)\left(u\right){=}^H{I}_{1+}^1(I-\mathcal{K})\mathcal{N}\left(u\right)$, ${}^H{D}_{1+}^{\alpha -2}{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right){=}^H{I}_{1+}^2(I-\mathcal{K})\mathcal{N}\left(u\right),\dots$, and ${}^H{D}_{1+}^{\alpha -n+1}{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right){=}^H{I}_{1+}^{n-1}(I-\mathcal{K})\mathcal{N}\left(u\right)$, we conclude that the operator ${\mathcal{T}}_{\mathcal{H},\mathcal{K}}\mathcal{N}$ is continuous in the space $\mathcal{X}$.

For Condition (d), similarly to how we proved Condition (b), we deduce first that there exists $M_4>0$ such that

$$|(I-\mathcal{K})\mathcal{N}u\left(t\right)|\le M_4,\forall u\in \overline{\Omega },a.e.t\in (1,e).$$

Then, by using Lemma 7, we obtain

$${\displaystyle \parallel {\mathcal{T}}_{\mathcal{H}}{(I-\mathcal{K})\mathcal{N}\left(u\right)\parallel }_{\mathcal{X}}\le {\parallel (I-\mathcal{K})\mathcal{N}\left(u\right)\parallel }_{\mathcal{Y}}={\int }_1^e|(I-\mathcal{K})\mathcal{N}u\left(t\right)|dt\le (e-1)M_4,\forall u\in \overline{\Omega }.}$$

So, the functions ${\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right),u\in \overline{\Omega }$ are uniformly bounded in $\mathcal{X}$. Therefore, the set ${\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(\overline{\Omega }\right)$ is bounded in $\mathcal{X}$.

We will prove now that $\{{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right),u\in \overline{\Omega }\}$ are uniformly equicontinuous, and thus equicontinuous in $\mathcal{X}$. For this, let $u\in \overline{\Omega }$ and $t_1,t_2\in [1,e]$, $t_1<t_2$. Then we obtain

$$\begin{array}{c}|{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right)\left(t_1\right)-{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right)\left(t_2\right)|\hfill \\ =\left|{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{\int }_1^{t_1}{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{\alpha -1}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}-\frac{1}{\Gamma \left(\alpha \right)}{\int }_1^{t_2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha -1}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\hfill \\ =\left|{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{\int }_1^{t_1}\left[{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{\alpha -1}-{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha -1}\right](I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right.\hfill \\ -\left.{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{\int }_{t_1}^{t_2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha -1}(I-\mathcal{K})\mathcal{N}\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle \le \frac{M_4}{\Gamma \left(\alpha \right)}{\int }_1^{t_1}\left[{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha -1}-{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{\alpha -1}\right]\frac{ds}{s}+\frac{M_4}{\Gamma \left(\alpha \right)}{\int }_{t_1}^{t_2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha -1}\frac{ds}{s}}\hfill \\ {\displaystyle =\frac{M_4}{\Gamma \left(\alpha \right)}\left[{\displaystyle -\frac{1}{\alpha }{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha }{|}_{s=1}^{s=t_1}+\frac{1}{\alpha }{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{\alpha }{|}_{s=1}^{s=t_1}}\right.\left.{\displaystyle -\frac{1}{\alpha }{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{\alpha }{|}_{s=t_1}^{s=t_2}}\right]}\hfill \\ {\displaystyle =\frac{M_4}{\Gamma (\alpha +1)}\left[{(ln{t}_2)}^{\alpha }-{(ln{t}_1)}^{\alpha }\right].}\hfill \end{array}$$

Because the function $g_0\left(t\right)={(lnt)}^{\alpha },t\in [1,e]$ is uniformly continuous on $[1,e]$, we deduce that the family of functions $\{{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right),u\in \overline{\Omega }\}$ is uniformly equicontinuous on $[1,e]$, and so, equicontinuous on $[1,e]$ in $C[1,e]$.

In what follows, we find

$$\begin{array}{c}\left|{}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}u\right)\left(t_1\right){-}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}u\right)\left(t_2\right)\right|\hfill \\ {=|}^H{I}_{1+}^1(I-\mathcal{K})\mathcal{N}u\left(t_1\right){-}^H{I}_{1+}^1(I-\mathcal{K})\mathcal{N}u\left(t_2\right)|\hfill \\ =\left|{\displaystyle {\int }_1^{t_1}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}-{\int }_1^{t_2}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\hfill \\ =\left|{\displaystyle {\int }_{t_1}^{t_2}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\le M_4\left|{\displaystyle {\int }_{t_1}^{t_2}\frac{ds}{s}}\right|=M_4(ln{t}_2-ln{t}_1).\hfill \end{array}$$

Due to the function $g_1\left(t\right)=lnt,t\in [1,e]$ is uniformly continuous on $[1,e]$, we obtain that ${\{}^H{D}_{1+}^{\alpha -1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)\left(u\right),u\in \overline{\Omega }\}$ is uniformly equicontinuous on $[1,e]$, and so equicontinuous on $[1,e]$ in $C[1,e]$.

Then we have

$$\begin{array}{c}\left|{}^H{D}_{1+}^{\alpha -2}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)u\left(t_1\right){-}^H{D}_{1+}^{\alpha -2}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)u\left(t_2\right)\right|\hfill \\ {=|}^H{I}_{1+}^2(I-\mathcal{K})\mathcal{N}u\left(t_1\right){-}^H{I}_{1+}^2(I-\mathcal{K})\mathcal{N}u\left(t_2\right)|\hfill \\ =\left|{\displaystyle \frac{1}{\Gamma \left(2\right)}{\int }_1^{t_1}ln\frac{t_1}{s}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}-\frac{1}{\Gamma \left(2\right)}{\int }_1^{t_2}ln\frac{t_2}{s}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\hfill \\ =\left|{\displaystyle {\int }_1^{t_1}\left({\displaystyle ln\frac{t_1}{s}-ln\frac{t_2}{s}}\right)(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}-{\int }_{t_1}^{t_2}ln\frac{t_2}{s}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\hfill \\ \le M_4\left[{\displaystyle {\int }_1^{t_1}\left({\displaystyle ln\frac{t_2}{s}-ln\frac{t_1}{s}}\right)\frac{ds}{s}+{\int }_{t_1}^{t_2}ln\frac{t_2}{s}\frac{ds}{s}}\right]\hfill \\ =M_4\left[{\displaystyle -\frac{1}{2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^2{|}_{s=1}^{s=t_1}+\frac{1}{2}{\left({\displaystyle ln\frac{t_1}{s}}\right)}^2{|}_{s=1}^{s=t_1}-\frac{1}{2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^2{|}_{s=t_1}^{s=t_2}}\right]\hfill \\ {\displaystyle =\frac{M_4}{2}\left[{(ln{t}_2)}^2-{(ln{t}_1)}^2\right].}\hfill \end{array}$$

In view of the function $g_2\left(t\right)={(lnt)}^2,t\in [1,e]$ is uniformly continuous on $[1,e]$, we deduce that ${\{}^H{D}_{1+}^{\alpha -2}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)\left(u\right),u\in \overline{\Omega }\}$ is uniformly equicontinuous on $[1,e]$ in $C[1,e]$, and so equicontinuous on $[1,e]$ in $C[1,e]$.

We continue in the same manner until the case

$$\begin{array}{c}\left|{}^H{D}_{1+}^{\alpha -n+1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)u\left(t_1\right){-}^H{D}_{1+}^{\alpha -n+1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)u\left(t_2\right)\right|\hfill \\ {=|}^H{I}_{1+}^{n-1}(I-\mathcal{K})\mathcal{N}u\left(t_1\right){-}^H{I}_{1+}^{n-1}(I-\mathcal{K})\mathcal{N}u\left(t_2\right)|\hfill \\ =\left|{\displaystyle \frac{1}{\Gamma (n-1)}{\int }_1^{t_1}{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{n-2}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}-\frac{1}{\Gamma (n-1)}{\int }_1^{t_2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-2}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle =\frac{1}{\Gamma (n-1)}\left|{\displaystyle {\int }_1^{t_1}\left[{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{n-2}-{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-2}\right](I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right.}\hfill \\ \left.{\displaystyle -{\int }_{t_1}^{t_2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-2}(I-\mathcal{K})\mathcal{N}u\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle \le \frac{M_4}{\Gamma (n-1)}\left\{{\displaystyle {\int }_1^{t_1}\left[{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-2}-{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{n-2}\right]\frac{ds}{s}+{\int }_{t_1}^{t_2}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-2}\frac{ds}{s}}\right\}}\hfill \\ {\displaystyle =\frac{M_4}{\Gamma (n-1)}\left[{\displaystyle -\frac{1}{n-1}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-1}{|}_{s=1}^{s=t_1}+\frac{1}{n-1}{\left({\displaystyle ln\frac{t_1}{s}}\right)}^{n-1}{|}_{s=1}^{s=t_1}-\frac{1}{n-1}{\left({\displaystyle ln\frac{t_2}{s}}\right)}^{n-1}{|}_{s=t_1}^{s=t_2}}\right]}\hfill \\ {\displaystyle =\frac{M_4}{\Gamma \left(n\right)}\left[{(ln{t}_2)}^{n-1}-{(ln{t}_1)}^{n-1}\right]=\frac{M_4}{(n-1)!}\left[{(ln{t}_2)}^{n-1}-{(ln{t}_1)}^{n-1}\right].}\hfill \end{array}$$

Because the function $g_{n-1}\left(t\right)={(lnt)}^{n-1},t\in [1,e]$ is uniformly continuous on $[1,e]$, we obtain that ${\{}^H{D}_{1+}^{\alpha -n+1}\left({\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\right)\left(u\right),u\in \overline{\Omega }\}$ is uniformly equicontinuous on $[1,e]$ in $C[1,e]$, and so equicontinuous on $[1,e]$ in $C[1,e]$.

By the above statements, we conclude that the family of functions $\{{\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(u\right),$$u\in \overline{\Omega }\}$ (or ${\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(\overline{\Omega }\right)$) is equicontinuous on $[1,e]$ in the space $\mathcal{X}$. By the Ascoli-Arzela theorem, we deduce that the set ${\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}\left(\overline{\Omega }\right)$ is relatively compact in $\mathcal{X}$, and so ${\mathcal{T}}_{\mathcal{H}}(I-\mathcal{K})\mathcal{N}:\overline{\Omega }\to \mathcal{X}$ is a compact operator.

Therefore, we conclude that operator $\mathcal{N}$ is $\mathcal{L}$-compact on $\overline{\Omega }$.  □

Now, we present the following assumptions that will be used in our main result.

(A3)

There exist the nonnegative functions $p_i\in \mathcal{Y},i=1,\dots ,n+1$, $p_i\left(t\right)\ge 0$ a.e. $t\in (1,e),i=1,\dots ,n+1$, such that

$${\displaystyle |f(t,x_1,x_2,\dots ,x_n)|\le {\displaystyle \sum _{i=1}^n}{p}_i\left(t\right)|x_i|+p_{n+1}\left(t\right),}$$

for all $x_i\in \mathbb{R}$, $i=1,\dots ,n$, and a.e. $t\in (1,e)$, and ${\sum }_{i=1}^n\parallel p_i{\parallel }_{\mathcal{Y}}={\sum }_{i=1}^n{\int }_1^e|p_i\left(t\right)|dt<{\displaystyle \frac{1}{{\Xi }_1+1}}$, where ${\Xi }_1=max\left\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)}},2\right\}$.

(A4)

There exist the constants ${\sigma }_1,{\sigma }_2>0$ such that if for all $t\in [1,e]$ and $u\in D\left(\mathcal{L}\right)$, if ${|}^H{D}_{1+}^{\alpha -1}u\left(t\right)|>{\sigma }_1$ or ${|}^H{D}_{1+}^{\alpha -2}u\left(t\right)|>{\sigma }_2$, then either ${\mathcal{S}}_1\left(\mathcal{N}u\left(t\right)\right)\ne 0$ or ${\mathcal{S}}_2\left(\mathcal{N}u\left(t\right)\right)\ne 0$.

(A5)

There exists a constant $r>0$ such that for all $b_1,b_2\in \mathbb{R}$, and $u\left(t\right)=b_1{(lnt)}^{\alpha -1}+b_2{(lnt)}^{\alpha -2}\in Ker\mathcal{L}$, if $|b_1|>r$ or $|b_2>r$, then either

$$b_1{\mathcal{S}}_1\mathcal{N}u\left(t\right)+b_2{\mathcal{S}}_2\mathcal{N}u\left(t\right)>0,\mathrm{or}$$

(30)

$$b_1{\mathcal{S}}_1\mathcal{N}u\left(t\right)+b_2{\mathcal{S}}_2\mathcal{N}u\left(t\right)<0.$$

(31)

Before we present and prove the main existence result for Problem (1),(2), we will prove three additional lemmas.

Lemma 9.

We assume that $\left(A1\right)–\left(A4\right)$ hold. Then the set

$${\Lambda }_1=\{u\in D\left(\mathcal{L}\right)\setminus Ker\mathcal{L},\mathcal{L}u=\nu \mathcal{N}u,\nu \in (0,1)\}\subset \mathcal{X}$$

is bounded in $\mathcal{X}$.

Proof. 

Let $u\in {\Lambda }_1$; so $\mathcal{L}u=\nu \mathcal{N}u$ with $u\in D\left(\mathcal{L}\right)\setminus Ker\mathcal{L}$ and $\nu \in (0,1)$. Therefore $\mathcal{N}u\left(t\right)={\displaystyle \frac{1}{\nu }}\mathcal{L}u\left(t\right)={{\displaystyle \frac{1}{\nu }}}^H{D}_{1+}^{\alpha }u\left(t\right)$. Then we obtain $\nu \mathcal{N}u\in Im\mathcal{L}=Ker\mathcal{K}$, i.e., $\mathcal{K}\left(\nu \mathcal{N}u\right)=0$ or $\nu \mathcal{K}\left(\mathcal{N}u\right)=0$. So $\mathcal{K}\left(\mathcal{N}u\right)=0$, i.e., $\mathcal{N}u\in Ker\mathcal{K}=Im\mathcal{L}$. We deduce that ${\mathcal{S}}_1\left(\mathcal{N}u\right)\left(t\right)=0$ and ${\mathcal{S}}_2\left(\mathcal{N}u\right)\left(t\right)=0$ for all $t\in [1,e]$. Hence, by assumption $\left(A4\right)$, we find that that there exist $t_1,t_2\in [1,e]$ such that ${|}^H{D}_{1+}^{\alpha -1}u\left(t_1\right)|\le {\sigma }_1$ and ${|}^H{D}_{1+}^{\alpha -2}u\left(t_2\right)|\le {\sigma }_2$.

Using Formula (9) with $\alpha$ replaced by $\alpha -1$, and $\alpha$ replaced by $\alpha -2$, we deduce

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -1}u\left(t\right)|=\left|{\displaystyle {}^H{D}_{1+}^{\alpha -1}u\left(t_1\right)+{\int }_{t_1}^t{(}^H{D}_{1+}^{\alpha }u)\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle {\le |}^H{D}_{1+}^{\alpha -1}u\left(t_1\right)|+{\int }_{t_1}^t\left|{}^H{D}_{1+}^{\alpha }u\left(s\right)\right|\frac{ds}{s}}\hfill \\ {\displaystyle \le {\sigma }_1+{\int }_{t_1}^t{|}^H{D}_{1+}^{\alpha }u\left(s\right)|ds\le {\sigma }_1+{\int }_1^e|\mathcal{N}u\left(s\right)|ds={\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e],}\hfill \end{array}$$

and

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -2}u\left(t\right)|=\left|{\displaystyle {}^H{D}_{1+}^{\alpha -2}u\left(t_2\right)+{\int }_{t_2}^t{(}^H{D}_{1+}^{\alpha -1}u)\left(s\right)\frac{ds}{s}}\right|\hfill \\ {\displaystyle {\le |}^H{D}_{1+}^{\alpha -2}u\left(t_2\right)|+{\int }_{t_2}^t\left|{}^H{D}_{1+}^{\alpha -1}u\left(s\right)\right|\frac{ds}{s}}\hfill \\ \le {\sigma }_2{+\parallel }^H{D}_{1+}^{\alpha -1}{u\parallel }_{\infty }\left({\displaystyle {\int }_{t_2}^t\frac{ds}{s}}\right)\le {\sigma }_2+{{\parallel }^H{D}_{1+}^{\alpha -1}u\parallel }_{\infty }\left({\displaystyle {\int }_1^e\frac{1}{s}ds}\right)\hfill \\ ={\sigma }_2{+\parallel }^H{D}_{1+}^{\alpha -1}{u\parallel }_{\infty }\le {\sigma }_2+{\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e].\hfill \end{array}$$

So, we obtain

$${|}^H{D}_{1+}^{\alpha -1}u\left(1\right)|\le {\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}{,|}^H{D}_{1+}^{\alpha -2}u\left(1\right)|\le {\sigma }_1+{\sigma }_2+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}.$$

Then by the definition of operator $\mathcal{H}$ and Lemma 1, we find

$$\begin{array}{c}{}^H{D}_{1+}^{\alpha -1}\mathcal{H}u\left(t\right){=}^H{D}_{1+}^{\alpha -1}\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{\alpha -2}}\right)\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){}^H{D}_{1+}^{\alpha -1}\left({(lnt)}^{\alpha -1}\right)+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){}^H{D}_{1+}^{\alpha -1}\left({(lnt)}^{\alpha -2}\right)}\hfill \\ {=}^H{D}_{1+}^{\alpha -1}u\left(1\right)(=a_1\Gamma \left(\alpha \right)),\hfill \end{array}$$

and

$$\begin{array}{c}{}^H{D}_{1+}^{\alpha -2}\mathcal{H}u\left(t\right){=}^H{D}_{1+}^{\alpha -2}\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{\alpha -2}}\right)\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\left({}^H{D}_{1+}^{\alpha -2}{(lnt)}^{\alpha -1}\right)+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right)\left({}^H{D}_{1+}^{\alpha -2}{(lnt)}^{\alpha -2}\right)}\hfill \\ {=}^H{D}_{1+}^{\alpha -1}u\left(1\right)lnt{+}^H{D}_{1+}^{\alpha -2}u\left(1\right)(=a_1\Gamma \left(\alpha \right)lnt+a_2\Gamma (\alpha -1)).\hfill \end{array}$$

Therefore we deduce

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -1}{\mathcal{H}u\left(t\right)|=|}^H{D}_{1+}^{\alpha -1}u\left(1\right)|\le {\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e],\hfill \end{array}$$

and

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -2}{\mathcal{H}u\left(t\right)|\le |}^H{D}_{1+}^{\alpha -1}{u\left(1\right)|lnt+|}^H{D}_{1+}^{\alpha -2}u\left(1\right)|\hfill \\ {\displaystyle \le 2{\sigma }_1+{\sigma }_2+{2\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}=\frac{2}{\Gamma \left(2\right)}{\sigma }_1+\frac{1}{\Gamma \left(1\right)}{\sigma }_2+\frac{2}{\Gamma \left(2\right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e].}\hfill \end{array}$$

Next, we find

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -3}\mathcal{H}u\left(t\right){=}^H{D}_{1+}^{\alpha -3}\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{\alpha -2}}\right)\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){}^H{D}_{1+}^{\alpha -3}\left({(lnt)}^{\alpha -1}\right)+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){}^H{D}_{1+}^{\alpha -3}\left({(lnt)}^{\alpha -2}\right)}\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}{(lnt)}^2+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right)\frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}lnt}\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(3\right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^2+\frac{1}{\Gamma \left(2\right)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right)lnt,\forall t\in [1,e],}\hfill \end{array}$$

and so

$$\begin{array}{c}{\displaystyle {|}^H{D}_{1+}^{\alpha -3}\mathcal{H}u\left(t\right)|\le \frac{1}{\Gamma \left(3\right)}{|}^H{D}_{1+}^{\alpha -1}u\left(1\right)|+\frac{1}{\Gamma \left(2\right)}{|}^H{D}_{1+}^{\alpha -2}u\left(1\right)|}\hfill \\ {\displaystyle \le \frac{1}{\Gamma \left(3\right)}\left({\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}\right)+\frac{1}{\Gamma \left(2\right)}\left({\sigma }_1+{\sigma }_2+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}\right)}\hfill \\ {\displaystyle =\left({\displaystyle \frac{1}{\Gamma \left(3\right)}+\frac{1}{\Gamma \left(2\right)}}\right){\sigma }_1+\frac{1}{\Gamma \left(2\right)}{\sigma }_2+\left({\displaystyle \frac{1}{\Gamma \left(3\right)}+\frac{1}{\Gamma \left(2\right)}}\right){\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}}\hfill \\ {\displaystyle =\frac{3}{\Gamma \left(3\right)}{\sigma }_1+\frac{1}{\Gamma \left(2\right)}{\sigma }_2+\frac{3}{\Gamma \left(3\right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e].}\hfill \end{array}$$

We proceed in this way until the fractional derivative of order $\alpha -n+1$ of $\mathcal{H}u$, obtaining

$$\begin{array}{c}{}^H{D}_{1+}^{\alpha -n+1}\mathcal{H}u\left(t\right){=}^H{D}_{1+}^{\alpha -n+1}\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{\alpha -2}}\right)\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){}^H{D}_{1+}^{\alpha -n+1}\left({(lnt)}^{\alpha -1}\right)+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_1^{\alpha -2}u\left(1\right){}^H{D}_{1+}^{\alpha -n+1}\left({(lnt)}^{\alpha -2}\right)}\hfill \\ {\displaystyle =\frac{1}{\Gamma \left(\alpha \right)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}{(lnt)}^{n-2}+\frac{1}{\Gamma (\alpha -1)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right)\frac{\Gamma (\alpha -1)}{\Gamma (n-2)}{(lnt)}^{n-3}}\hfill \\ {\displaystyle =\frac{1}{\Gamma (n-1)}{}^H{D}_{1+}^{\alpha -1}u\left(1\right){(lnt)}^{n-2}+\frac{1}{\Gamma (n-2)}{}^H{D}_{1+}^{\alpha -2}u\left(1\right){(lnt)}^{n-3},\forall t\in [1,e],}\hfill \end{array}$$

and so

$$\begin{array}{c}{\displaystyle \left|{}^H{D}_{1+}^{\alpha -n+1}\mathcal{H}u\left(t\right)\right|\le \frac{1}{\Gamma (n-1)}\left|{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\right|{(lnt)}^{n-2}+\frac{1}{\Gamma (n-2)}\left|{}^H{D}_{1+}^{\alpha -2}u\left(1\right)\right|{(lnt)}^{n-3}}\hfill \\ {\displaystyle \le \frac{1}{\Gamma (n-1)}({\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}})+\frac{1}{\Gamma (n-2)}\left({\sigma }_1+{\sigma }_2+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}\right)}\hfill \\ {\displaystyle =\left({\displaystyle \frac{1}{\Gamma (n-1)}+\frac{1}{\Gamma (n-2)}}\right){\sigma }_1+\frac{1}{\Gamma (n-2)}{\sigma }_2+\left({\displaystyle \frac{1}{\Gamma (n-1)}+\frac{1}{\Gamma (n-2)}}\right){\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}}\hfill \\ {\displaystyle =\frac{n-1}{\Gamma (n-1)}{\sigma }_1+\frac{1}{\Gamma (n-2)}{\sigma }_2+\frac{n-1}{\Gamma (n-1)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e].}\hfill \end{array}$$

We also find

$$\begin{array}{c}{\displaystyle |\mathcal{H}u\left(t\right)|\le \frac{1}{\Gamma \left(\alpha \right)}\left|{}^H{D}_{1+}^{\alpha -1}u\left(1\right)\right|{(lnt)}^{\alpha -1}+\frac{1}{\Gamma (\alpha -1)}\left|{}^H{D}_{1+}^{\alpha -2}u\left(1\right)\right|{(lnt)}^{\alpha -2}}\hfill \\ {\displaystyle \le \frac{1}{\Gamma \left(\alpha \right)}({\sigma }_1+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}})+\frac{1}{\Gamma (\alpha -1)}({\sigma }_1+{\sigma }_2+{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}})}\hfill \\ {\displaystyle =\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}+\frac{1}{\Gamma (\alpha -1)}}\right){\sigma }_1+\frac{1}{\Gamma (\alpha -1)}{\sigma }_2+\left({\displaystyle \frac{1}{\Gamma \left(\alpha \right)}+\frac{1}{\Gamma (\alpha -1)}}\right){\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}}\hfill \\ {\displaystyle =\frac{\alpha }{\Gamma \left(\alpha \right)}{\sigma }_1+\frac{1}{\Gamma (\alpha -1)}{\sigma }_2+\frac{\alpha }{\Gamma \left(\alpha \right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\forall t\in [1,e].}\hfill \end{array}$$

From the above inequalities, we deduce

$$\begin{array}{c}{\parallel \mathcal{H}u\parallel }_{\mathcal{X}}=max\left\{{\parallel \mathcal{H}u\parallel }_{\infty }{,\parallel }^H{D}_{1+}^{\alpha -n+1}{\mathcal{H}u\parallel }_{\infty },\dots ,{{\parallel }^H{D}_{1+}^{\alpha -3}\mathcal{H}u\parallel }_{\infty },\right.\hfill \\ \left.{\parallel }^H{D}_{1+}^{\alpha -2}{\mathcal{H}u\parallel }_{\infty },{{\parallel }^H{D}_{1+}^{\alpha -1}\mathcal{H}u\parallel }_{\infty }\right\}\hfill \\ =max\left\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)}{\sigma }_1+\frac{1}{\Gamma (\alpha -1)}{\sigma }_2+\frac{\alpha }{\Gamma \left(\alpha \right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},}\right.\hfill \\ {\displaystyle \frac{n-1}{\Gamma (n-1)}{\sigma }_1+\frac{1}{\Gamma (n-2)}{\sigma }_2+\frac{n-1}{\Gamma (n-1)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\dots }\hfill \\ {\displaystyle \frac{3}{\Gamma \left(3\right)}{\sigma }_1+\frac{1}{\Gamma \left(2\right)}{\sigma }_2+\frac{3}{\Gamma \left(3\right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},}\hfill \\ \left.{\displaystyle \frac{2}{\Gamma \left(2\right)}{\sigma }_1+\frac{1}{\Gamma \left(1\right)}{\sigma }_2+\frac{2}{\Gamma \left(2\right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\frac{1}{\Gamma \left(1\right)}{\sigma }_1+\frac{1}{\Gamma \left(1\right)}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}}\right\}\hfill \\ \le max\left\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)},\frac{n-1}{\Gamma (n-1)},\cdots ,\frac{3}{\Gamma \left(3\right)},\frac{2}{\Gamma \left(2\right)},\frac{1}{\Gamma \left(1\right)}}\right\}{\sigma }_1\hfill \\ +max\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)},\frac{1}{\Gamma (n-2)},\cdots ,\frac{1}{\Gamma \left(2\right)},\frac{1}{\Gamma \left(1\right)}}\right\}{\sigma }_2\hfill \\ +max\left\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)},\frac{n-1}{\Gamma (n-1)},\cdots ,\frac{3}{\Gamma \left(3\right)},\frac{2}{\Gamma \left(2\right)},\frac{1}{\Gamma \left(1\right)}}\right\}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}\hfill \\ =max\left\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)},2}\right\}{\sigma }_1+max\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)},1}\right\}{\sigma }_2+max\left\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)},2}\right\}{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}\hfill \\ ={\Xi }_1{\sigma }_1+{\Xi }_2{\sigma }_2+{\Xi }_1{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}},\hfill \end{array}$$

where ${\Xi }_1=max\{{\displaystyle \frac{\alpha }{\Gamma \left(\alpha \right)}},2\}$, ${\Xi }_2=max\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},1\}$.

Therefore we conclude

$${\parallel \mathcal{H}u\parallel }_{\mathcal{X}}\le {\Xi }_1{\sigma }_1+{\Xi }_2{\sigma }_2+{\Xi }_1{\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}.$$

(32)

On the other hand, for $u\in {\Lambda }_1$, we find $(I-\mathcal{H})u\in D\left(\mathcal{L}\right)\cap Ker\mathcal{H}$ and $\mathcal{L}\mathcal{H}u=0$. Therefore, by Lemma 7 we obtain

$${\parallel (I-\mathcal{H})u\parallel }_{\mathcal{X}}=\parallel {\mathcal{T}}_{\mathcal{H}}{\mathcal{L}(I-\mathcal{H})u\parallel }_{\mathcal{X}}=\parallel {\mathcal{T}}_{\mathcal{H}}{\mathcal{L}u\parallel }_{\mathcal{X}}\le {\parallel \mathcal{L}u\parallel }_{\mathcal{Y}}={\nu \parallel \mathcal{N}u\parallel }_{\mathcal{Y}}\le {\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}.$$

(33)

Then, by (32) and (33), we deduce

$$\begin{array}{c}{\parallel u\parallel }_{\mathcal{X}}={\parallel \mathcal{H}u+(I-\mathcal{H})u\parallel }_{\mathcal{X}}\le {\parallel \mathcal{H}u\parallel }_{\mathcal{X}}+{\parallel (I-\mathcal{H})u\parallel }_{\mathcal{X}}\hfill \\ \le {\Xi }_1{\sigma }_1+{\Xi }_2{\sigma }_2+({\Xi }_1+1){\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}.\hfill \end{array}$$

(34)

By $\left(A3\right)$, we obtain

$$\begin{array}{c}{\displaystyle {\parallel \mathcal{N}u\parallel }_{\mathcal{Y}}={\int }_1^e|\mathcal{N}u\left(t\right)|dt={\int }_1^e\left|f(t,u\left(t\right){,}^H{D}_{1+}^{\alpha -n+1}u\left(t\right),\dots {,}^H{D}_{1+}^{\alpha -1}u\left(t\right))\right|dt}\hfill \\ {\displaystyle \le {\int }_1^e\left({\displaystyle p_1\left(t\right)|u\left(t\right)|+{\displaystyle \sum _{i=1}^{n-1}}{p}_{i+1}\left(t\right)\left|{}^H{D}_{1+}^{\alpha -n+i}u\left(t\right)\right|+p_{n+1}\left(t\right)}\right)dt}\hfill \\ {\displaystyle \le {\parallel u\parallel }_{\infty }\parallel p_1{\parallel }_{\mathcal{Y}}+{\displaystyle \sum _{i=1}^{n-1}}{\parallel }^H{D}_{1+}^{\alpha -n+i}{u\parallel }_{\infty }\parallel p_{i+1}{\parallel }_{\mathcal{Y}}+{\parallel p_{n+1}\parallel }_{\mathcal{Y}}}\hfill \\ \le \left({\displaystyle {\displaystyle \sum _{i=1}^n}{\parallel p_i\parallel }_{\mathcal{Y}}}\right){\parallel u\parallel }_{\mathcal{X}}+{\parallel p_{n+1}\parallel }_{\mathcal{Y}}.\hfill \end{array}$$

(35)

By (34), (35), and $\left(A3\right)$, we find

$${\parallel u\parallel }_{\mathcal{X}}\le {\Xi }_1{\sigma }_1+{\Xi }_2{\sigma }_2+({\Xi }_1+1)\left({\displaystyle {\displaystyle \sum _{i=1}^n}\parallel p_i{\parallel }_{\mathcal{Y}}{\parallel u\parallel }_{\mathcal{X}}+{\parallel p_{n+1}\parallel }_{\mathcal{Y}}}\right),$$

and then

$${\displaystyle {\parallel u\parallel }_{\mathcal{X}}\le \frac{{\Xi }_1{\sigma }_1+{\Xi }_2{\sigma }_2+({\Xi }_1+1){\parallel p_{n+1}\parallel }_{\mathcal{Y}}}{{\displaystyle 1-({\Xi }_1+1){\displaystyle \sum _{i=1}^n}{\parallel p_i\parallel }_{\mathcal{Y}}}}.}$$

Therefore, we deduce that the set ${\Lambda }_1$ is bounded in the space $\mathcal{X}$.  □

Lemma 10.

We assume that $\left(A1\right)$, $\left(A2\right)$ and $\left(A4\right)$ hold. Then the set

$${\Lambda }_2=\{u\in Ker\mathcal{L};\mathcal{N}u\in Im\mathcal{L}\}\subset \mathcal{X},$$

is bounded in $\mathcal{X}$.

Proof. 

Let $u\in {\Lambda }_2$, so $u\left(t\right)=a_1{(lnt)}^{\alpha -1}+a_2{(lnt)}^{\alpha -2}$, with $a_1,a_2\in \mathbb{R}$, and ${\mathcal{S}}_1\left(\mathcal{N}u\right)\left(t\right)={\mathcal{S}}_2\left(\mathcal{N}u\right)\left(t\right)=0$ for all $t\in [1,e]$. From $\left(A4\right)$ we deduce that there exist $t_3,t_4\in [1,e]$ such that ${|}^H{D}_{1+}^{\alpha -1}u\left(t_3\right)|\le {\sigma }_1$ and ${|}^H{D}_{1+}^{\alpha -2}u\left(t_4\right)|\le {\sigma }_2$. Then we obtain

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -1}u\left(t_3\right)|=|a_1\Gamma \left(\alpha \right)|\le {\sigma }_1,\hfill \\ {|}^H{D}_{1+}^{\alpha -2}u\left(t_4\right)|=|a_1\Gamma \left(\alpha \right)ln{t}_4+a_2\Gamma (\alpha -1)|\le {\sigma }_2.\hfill \end{array}$$

Therefore we find

$$\begin{array}{c}{\displaystyle |a_1|\le \frac{{\sigma }_1}{\Gamma \left(\alpha \right)},|a_2|\Gamma (\alpha -1)\le |a_1\Gamma \left(\alpha \right)ln{t}_4+a_2\Gamma (\alpha -1)|+|a_1\Gamma \left(\alpha \right)ln{t}_4|}\hfill \\ \le {\sigma }_2+a_1\Gamma \left(\alpha \right)ln{t}_4\le {\sigma }_2+{\sigma }_{1}ln{t}_4\Rightarrow \hfill \\ {\displaystyle |a_2|\le \frac{{\sigma }_2+{\sigma }_{1}ln{t}_4}{\Gamma (\alpha -1)}\le \frac{{\sigma }_1+{\sigma }_2}{\Gamma (\alpha -1)}.}\hfill \end{array}$$

Hence, we conclude

$$\begin{array}{c}{\displaystyle |u\left(t\right)|\le |a_1|+|a_2|\le \frac{{\sigma }_1}{\Gamma \left(\alpha \right)}+\frac{{\sigma }_1+{\sigma }_2}{\Gamma (\alpha -1)}=\frac{\alpha {\sigma }_1+(\alpha -1){\sigma }_2}{\Gamma \left(\alpha \right)},}\hfill \\ {|}^H{D}_{1+}^{\alpha -1}u\left(t\right)|=|a_1\Gamma \left(\alpha \right)|\le {\sigma }_1,\hfill \\ {\displaystyle {|}^H{D}_{1+}^{\alpha -2}u\left(t\right)|=|a_1\Gamma \left(\alpha \right)lnt+a_2\Gamma (\alpha -1)|\le 2{\sigma }_1+{\sigma }_2=\frac{2{\sigma }_1+{\sigma }_2}{\Gamma \left(2\right)},\forall t\in [1,e].}\hfill \end{array}$$

For the next fractional derivative, we have

$$\begin{array}{c}{}^H{D}_{1+}^{\alpha -3}u\left(t\right)=a_1{}^H{D}_{1+}^{\alpha -3}\left({(lnt)}^{\alpha -1}\right)+a_2{}^H{D}_{1+}^{\alpha -3}\left({(lnt)}^{\alpha -2}\right)\hfill \\ {\displaystyle =a_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}{(lnt)}^2+a_2\frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}(lnt),}\hfill \end{array}$$

and so

$${\displaystyle {|}^H{D}_{1+}^{\alpha -3}u\left(t\right)|\le |a_1|\frac{\Gamma \left(\alpha \right)}{2}+|a_2|\Gamma (\alpha -1)\le \frac{{\sigma }_1}{2}+{\sigma }_1+{\sigma }_2=\frac{3{\sigma }_1+2{\sigma }_2}{\Gamma \left(3\right)},\forall t\in [1,e].}$$

We continue these evaluations until the fractional derivative of order $\alpha -n+1$ of function u, and we find

$$\begin{array}{c}{}^H{D}_{1+}^{\alpha -n+1}u\left(t\right)=a_1{}^H{D}_{1+}^{\alpha -n+1}\left({(lnt)}^{\alpha -1}\right)+a_2{}^H{D}_{1+}^{\alpha -n+1}\left({(lnt)}^{\alpha -2}\right)\hfill \\ {\displaystyle =a_1\frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}{(lnt)}^{n-2}+a_2\frac{\Gamma (\alpha -1)}{\Gamma (n-2)}{(lnt)}^{n-3},}\hfill \end{array}$$

and then

$$\begin{array}{c}{\displaystyle {|}^H{D}_{1+}^{\alpha -n+1}u\left(t\right)|\le |a_1|\frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}+|a_2|\frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}\hfill \\ {\displaystyle \le \frac{{\sigma }_1}{\Gamma (n-1)}+\frac{{\sigma }_1+{\sigma }_2}{\Gamma (n-2)}=\frac{(n-1){\sigma }_1+(n-2){\sigma }_2}{\Gamma (n-1)},\forall t\in [1,e].}\hfill \end{array}$$

By the above inequalities, we deduce

$$\begin{array}{c}{\parallel u\parallel }_{\mathcal{X}}=max\left\{{\displaystyle \frac{\alpha {\sigma }_1+(\alpha -1){\sigma }_2}{\Gamma \left(\alpha \right)},\frac{(n-1){\sigma }_1+(n-2){\sigma }_2}{\Gamma (n-1)},\cdots ,}\right.\hfill \\ \left.{\displaystyle \frac{3{\sigma }_1+2{\sigma }_2}{\Gamma \left(3\right)},\frac{2{\sigma }_1+{\sigma }_2}{\Gamma \left(2\right)},{\sigma }_1}\right\}<\infty .\hfill \end{array}$$

We conclude that the set ${\Lambda }_2$ is bounded in the space $\mathcal{X}$.  □

Lemma 11.

We assume that $\left(A1\right)$, $\left(A2\right)$ and $\left(A5\right)$ hold. Then the set

$${\Lambda }_3=\{u\in Ker\mathcal{L};\zeta \nu \mathcal{G}u+(1-\nu )\mathcal{K}\mathcal{N}u=0,\nu \in [0,1]\}\subset \mathcal{X},$$

is bounded in $\mathcal{X}$, where $\zeta =1$ if Relationship (30) holds, and $\zeta =-1$ if Relationship (31) holds. Here $\mathcal{G}:Ker\mathcal{L}\to Im\mathcal{K}$ is the linear isomorphism defined by

$${\displaystyle \mathcal{G}\left(u\right)\left(t\right)=\frac{1}{\Delta }[(d{b}_1-b{b}_2)+(a{b}_2-c{b}_1)lnt],t\in [1,e],}$$

for $u\in Ker\mathcal{L},u\left(t\right)=b_1{(lnt)}^{\alpha -1}+b_2{(lnt)}^{\alpha -2},t\in [1,e]$, with $b_1,b_2\in \mathbb{R}$.

Proof. 

We suppose that Relationship (31) holds, i.e., $\zeta =-1$. Let $u\in {\Lambda }_3$; so there exist $b_1,b_2\in \mathbb{R}$, $\nu \in [0,1]$ such that $u\left(t\right)=b_1{(lnt)}^{\alpha -1}+b_2{(lnt)}^{\alpha -2}$, for all $t\in [1,e]$, and $-\nu \mathcal{G}u+(1-\nu )\mathcal{K}\mathcal{N}u=0$. By using Lemma 10, it is sufficient to prove that $|b_1|\le r$ and $|b_2|\le r$, with $r>0$. We have the following three cases:

(1)

For $\nu =0$, then $\mathcal{K}\mathcal{N}u=0$. So, we obtain ${\mathcal{E}}_1\left(\mathcal{N}u\right)\left(t\right)+\left({\mathcal{E}}_2\left(\mathcal{N}u\right)\left(t\right)\right)lnt=0$ for all $t\in [1,e]$. Then we find ${\mathcal{E}}_1\left(\mathcal{N}u\right)=0$ and ${\mathcal{E}}_2\left(\mathcal{N}u\right)=0$, or

$$\left\{\begin{array}{c}{\displaystyle \frac{1}{\Delta }(d{\mathcal{S}}_1\left(\mathcal{N}u\right)-b{\mathcal{S}}_2\left(\mathcal{N}u\right))=0,}\hfill \\ {\displaystyle \frac{1}{\Delta }(a{\mathcal{S}}_2\left(\mathcal{N}u\right)-c{\mathcal{S}}_1\left(\mathcal{N}u\right))=0.}\hfill \end{array}\right.$$

Because the determinant of the above system is $1/\Delta =1/(ad-bc)\ne 0$, we deduce that ${\mathcal{S}}_1\left(\mathcal{N}u\right)={\mathcal{S}}_2\left(\mathcal{N}u\right)=0$. Hence, by $\left(A5\right)$, we obtain $|b_1|\le r$ and $|b_2|\le r$.

(2)

For $\nu =1$, then $\mathcal{G}u=0$. So, we find the system

$$\left\{\begin{array}{c}d{b}_1-b{b}_2=0,\hfill \\ a{b}_2-c{b}_1=0,\hfill \end{array}\right.$$

which has the unique solution $b_1=b_2=0$. Then $|b_1|\le r$ and $|b_2|\le r$.

(3)

For $\nu \in (0,1)$, by the relationship $\nu \mathcal{G}u=(1-\nu )\mathcal{K}\mathcal{N}u$, we obtain

$$\left\{\begin{array}{c}{\displaystyle \nu \frac{1}{\Delta }(d{b}_1-b{b}_2)=\frac{1-\nu }{\Delta }(d{\mathcal{S}}_1\left(\mathcal{N}u\right)-b{\mathcal{S}}_2\left(\mathcal{N}u\right)),}\hfill \\ {\displaystyle \nu \frac{1}{\Delta }(a{b}_2-c{b}_1)=\frac{1-\nu }{\Delta }(a{\mathcal{S}}_2\left(\mathcal{N}u\right)-c{\mathcal{S}}_1\left(\mathcal{N}u\right)).}\hfill \end{array}\right.$$

The above system in the unknowns $b_1,b_2$ has the determinant $\Delta =ad-bc\ne 0$, and then we deduce

$$\left\{\begin{array}{c}\nu b_1=(1-\nu ){\mathcal{S}}_1\left(\mathcal{N}u\right),\hfill \\ \nu b_2=(1-\nu ){\mathcal{S}}_2\left(\mathcal{N}u\right).\hfill \end{array}\right.$$

(36)

Therefore if $|b_1|>r$ or $|b_2|>r$, we obtain using the system (36) and (31) that

$$\begin{array}{c}0<\nu (b_1^2+b_2^2)=\left(\nu b_1\right)b_1+\left(\nu b_2\right)b_2=b_1(1-\nu ){\mathcal{S}}_1\left(\mathcal{N}u\right)+b_2(1-\nu ){\mathcal{S}}_2\left(\mathcal{N}u\right)\hfill \\ =(1-\nu )(b_1{\mathcal{S}}_1\left(\mathcal{N}u\right)+b_2{\mathcal{S}}_2\left(\mathcal{N}u\right))<0,\hfill \end{array}$$

which is a contradiction. So $|b_1|\le r$ and $|b_2|\le r$.

We conclude that the set ${\Lambda }_3$ is bounded in $\mathcal{X}$. If Relationship (30) holds (for $\zeta =1$), by using similar arguments, we also deduce that ${\Lambda }_3$ is bounded.  □

Now, we give our main result related to the existence of solutions for Problem (1),(2).

Theorem 2.

We assume that assumptions $\left(A1\right)-\left(A5\right)$ hold. Then, the boundary value Problem (1),(2) have at least one solution in the space $\mathcal{X}$.

Proof. 

We consider a bounded open set ${\Lambda }_0$ in the space $\mathcal{X}$ such that ${\cup }_{i=1}^3{\overline{\Lambda }}_i\subset {\Lambda }_0$. In Lemma 6 we shown that $\mathcal{L}$ is a linear Fredholm operator of index 0, and in Lemma 8 we proved that the operator $\mathcal{N}$ is $\mathcal{L}$-compact on ${\overline{\Lambda }}_0$. In addition, from Lemmas 9 and 10, we have

(i) 

$\mathcal{L}u\ne \nu \mathcal{N}u$, for any $(u,\nu )\in [(D\left(\mathcal{L}\right)\setminus Ker\mathcal{L})\cap \partial {\Lambda }_0]\times (0,1)$;

(ii) 

$\mathcal{N}u\notin Im\mathcal{L}$, for any $u\in Ker\mathcal{L}\cap \partial {\Lambda }_0$,

that is, the first two conditions from Theorem 1 are satisfied.

We will show next that condition $\left(iii\right)$ from Theorem 1 is also satisfied. For this, we define the operator

$$\mathcal{V}(u,\nu )=\zeta \nu \mathcal{G}u+(1-\nu )\mathcal{K}\mathcal{N}u,$$

where $\zeta$ is defined in Lemma 11. Then by Lemma 11, we find $\mathcal{V}(u,\nu )\ne 0$ for all $u\in Ker\mathcal{L}\cap \partial {\Lambda }_0$ and $\nu \in [0,1]$. Hence, by the homotopy of degree, we deduce

$$\begin{array}{c}deg\{\mathcal{K}\mathcal{N}{|}_{Ker\mathcal{L}},{\Lambda }_0\cap Ker\mathcal{L},0\}=deg\{\mathcal{V}(·,0),{\Lambda }_0\cap Ker\mathcal{L},0\}\hfill \\ =deg\{\mathcal{V}(·,1),{\Lambda }_0\cap Ker\mathcal{L},0\}=deg\{\zeta \mathcal{G},{\Lambda }_0\cap Ker\mathcal{L},0\}\ne 0.\hfill \end{array}$$

Therefore, all assumptions $\left(i\right)$and $\left(ii\right)$and $\left(iii\right)$ from Theorem 1 are satisfied. Therefore, by Theorem 1, we conclude that the operator Equation (11), namely $\mathcal{L}u=\mathcal{N}u$, has at least one solution $u\in D\left(\mathcal{L}\right)\cap {\overline{\Lambda }}_0$. This means that our Problem (1),(2) have at least one solution in the space $\mathcal{X}$.  □

4. An Example

Let $\alpha ={\displaystyle \frac{10}{3}}$, $n=4$, the functions

$$\begin{array}{c}P\left(s\right)=\left\{\begin{array}{c}{\displaystyle \frac{45}{16}}{(s-1)}^3+1,s\in [1,{\displaystyle \frac{5}{3}}),\hfill \\ {\displaystyle \frac{11}{6}},s\in [{\displaystyle \frac{5}{3}},2),\hfill \\ 2,s\in [2,e],\hfill \end{array}\right.Q\left(s\right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{2}},s\in [1,2),\hfill \\ 3{(s-2)}^2+{\displaystyle \frac{1}{2}},s\in [2,{\displaystyle \frac{5}{2}}),\hfill \\ {\displaystyle \frac{3}{2}},s\in [{\displaystyle \frac{5}{2}},e],\hfill \end{array}\right.\hfill \end{array}$$

and the function $f:[1,e]\times {\mathbb{R}}^4\to \mathbb{R}$ given by

$$\begin{array}{c}{\displaystyle f(t,x_1,x_2,x_3,x_4)=\varsigma \left(t\right)\left({\displaystyle \frac{t^2+1}{5t}+\frac{1}{24}arctan{x}_1-\frac{x_2}{31(x_2^2+1)}+\frac{x_3}{35}}\right)+(1-\varsigma \left(t\right))\frac{c_0{x}_4}{35},}\hfill \end{array}$$

for all $t\in [1,e]$ and $x_i\in \mathbb{R},i=1,\dots ,4$, with $c_0=3/7\left({\int }_1^2(2-P\left(s\right)){\displaystyle \frac{ds}{s}}\right){\left({\int }_2^e(Q\left(s\right)-{\displaystyle \frac{1}{2}}){\displaystyle \frac{ds}{s}}\right)}^{-1}\approx 1.41399238,$ and $\varsigma \left(t\right)=\{1,t\in [1,2);0,t\in [2,e\left]\right\}$.

We consider the Hadamard fractional differential equation

$${}^H{D}_{1+}^{10/3}u\left(t\right)=f\left(t,u\left(t\right){,}^H{D}_{1+}^{1/3}u\left(t\right){,}^H{D}_{1+}^{4/3}u\left(t\right){,}^H{D}_{1+}^{7/3}u\left(t\right)\right),t\in (1,e),$$

(37)

subject to the boundary conditions

$$\left\{\begin{array}{c}u\left(1\right)=u^{\prime }\left(1\right)=0,\hfill \\ {\displaystyle {}^H{D}_{1+}^{7/3}u\left(1\right)=\frac{135}{16}{\int }_1^{5/3}{(s-1)}^2{}^H{D}_{1+}^{7/3}u\left(s\right)ds+\frac{1}{6}{}^H{D}_{1+}^{7/3}u\left(2\right),}\hfill \\ {\displaystyle {}^H{D}_{1+}^{7/3}u\left(e\right)=\frac{1}{4}{}^H{D}_{1+}^{7/3}u\left({\displaystyle \frac{5}{2}}\right)+6{\int }_2^{5/2}(s-2){}^H{D}_{1+}^{7/3}u\left(s\right)ds.}\hfill \end{array}\right.$$

(38)

The function f is a Carathéodory one, and the functions $P,Q:[1,e]\to \mathbb{R}$ have bounded variations, with $P\left(e\right)=P\left(1\right)+1=2$, (or ${\int }_1^{e}dP\left(s\right)=1$), and $Q\left(e\right)=Q\left(1\right)+1={\displaystyle \frac{3}{2}}$, (or ${\int }_1^{e}dQ\left(s\right)=1$). In addition, after some computations, we obtain $a\approx 0.45013184$, $b\approx 0.11060136$, $c\approx -0.13643188$, $d\approx -0.1256727$, and $\Delta \approx -0.04147973\ne 0$. So, assumptions $\left(A1\right)$ and $\left(A2\right)$ are satisfied.

Then, for the function f, we find the following inequality

$${\displaystyle |f(t,x_1,x_2,x_3,x_4)|\le {\displaystyle \sum _{i=1}^4}{p}_i\left(t\right)|x_i|+p_5\left(t\right),\forall t\in [1,e],x_i\in \mathbb{R},i=1,\dots ,4,}$$

where $p_1\left(t\right)={\displaystyle \frac{1}{24}}$, $p_2\left(t\right)={\displaystyle \frac{1}{31}}$, $p_3\left(t\right)={\displaystyle \frac{1}{35}}$, $p_4\left(t\right)={\displaystyle \frac{c_0}{35}}\approx 0.0403997$, $p_5\left(t\right)={\displaystyle \frac{t^2+1}{5t}}$, for all $t\in [1,e]$. We have here ${\Xi }_1=2$, and ${\sum }_{i=1}^4{\parallel p_i\parallel }_{\mathcal{Y}}\approx 0.2455355<{\displaystyle \frac{1}{{\Xi }_1+1}}={\displaystyle \frac{1}{3}}\approx 0.3333.$ Therefore assumption $\left(A3\right)$ is also satisfied.

Now, we verify assumption $\left(A4\right)$. We consider ${\sigma }_1=10$ (in fact it can be chosen arbitrarily, positive) and ${\sigma }_2=18$.

If $u\in D\left(\mathcal{L}\right)$ and ${|}^H{D}_{1+}^{7/3}u\left(t\right)|>{\sigma }_1=10$ for all $t\in [1,e]$, then we obtain

$$\begin{array}{c}{\displaystyle C_1:={\mathcal{S}}_2\mathcal{N}u\left(t\right)={\int }_1^e(Q\left(1\right)-Q\left(s\right))\mathcal{N}\left(s\right)\frac{ds}{s}={\int }_1^e\left(\frac{1}{2}-Q\left(s\right)\right)\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle =-{\int }_2^{5/2}3{(s-2)}^2\mathcal{N}u\left(s\right)\frac{ds}{s}-{\int }_{5/2}^e\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle =-3{\int }_2^{5/2}{(s-2)}^2\frac{c_0}{35}{}^H{D}_{1+}^{7/3}u\left(s\right)\frac{ds}{s}-{\int }_{5/2}^e\frac{c_0}{35}{}^H{D}_{1+}^{7/3}u\left(s\right)\frac{ds}{s}.}\hfill \end{array}$$

If ${}^H{D}_{1+}^{7/3}u\left(s\right)>{\sigma }_1=10$, then we have

$${\displaystyle C_1<-\frac{3c_0}{35}{\int }_2^{5/2}{(s-2)}^2\frac{ds}{s}-\frac{c_0}{35}{\int }_{5/2}^e\frac{ds}{s}\approx -0.05511818<0.}$$

If ${}^H{D}_{1+}^{7/3}u\left(s\right)<-{\sigma }_1=-10$, then we derive

$${\displaystyle C_1>\frac{3c_0}{35}{\int }_2^{5/2}{(s-2)}^2\frac{ds}{s}+\frac{c_0}{35}{\int }_{5/2}^e\frac{ds}{s}\approx 0.05511818>0.}$$

So $C_1\ne 0$.

If $u\in D\left(\mathcal{L}\right)$ and ${|}^H{D}_{1+}^{4/3}u\left(t\right)|>{\sigma }_2=18$ for all $t\in [1,e]$, then we find

$$\begin{array}{c}{\displaystyle C_2:={\mathcal{S}}_1\mathcal{N}u\left(t\right)={\int }_1^e(P\left(e\right)-P\left(s\right))\mathcal{N}u\left(s\right)\frac{ds}{s}={\int }_1^e(2-P\left(s\right))\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle ={\int }_1^{5/3}\left({\displaystyle 2-\frac{45}{16}{(s-1)}^3-1}\right)\mathcal{N}u\left(s\right)\frac{ds}{s}+{\int }_{5/3}^2\frac{1}{6}\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle ={\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\left[{\displaystyle \frac{s^2+1}{5s}+\frac{1}{24}arctanu\left(s\right)-\frac{1}{31}\frac{{}^H{D}_{1+}^{1/3}u\left(s\right)}{{{(}^H{D}_{1+}^{1/3}u\left(s\right))}^2+1}}\right]\frac{ds}{s}}\hfill \\ {\displaystyle +\frac{1}{6}{\int }_{5/3}^2\left[{\displaystyle \frac{s^2+1}{5s}+\frac{1}{24}arctanu\left(s\right)-\frac{1}{31}\frac{{}^H{D}_{1+}^{1/3}u\left(s\right)}{{{(}^H{D}_{1+}^{1/3}u\left(s\right))}^2+1}}\right]\frac{ds}{s}}\hfill \\ {\displaystyle +\frac{1}{35}{\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right){}^H{D}_{1+}^{4/3}u\left(s\right)\frac{ds}{s}+\frac{1}{210}{\int }_{5/3}^2{}^H{D}_{1+}^{4/3}u\left(s\right)\frac{ds}{s}.}\hfill \end{array}$$

We denote by

$${\displaystyle h\left(s\right)=\frac{s^2+1}{5s}+\frac{1}{24}arctanu\left(s\right)-\frac{1}{31}\frac{{}^H{D}_{1+}^{1/3}u\left(s\right)}{{{(}^H{D}_{1+}^{1/3}u\left(s\right))}^2+1},s\in [1,2].}$$

So, we obtain $h_1\left(s\right)\le h\left(s\right)\le h_2\left(s\right)$ for all $s\in [1,2]$, where

$${\displaystyle h_1\left(s\right)=\frac{s^2+1}{5s}-\frac{\pi }{48}-\frac{1}{62},h_2\left(s\right)=\frac{s^2+1}{5s}+\frac{\pi }{48}+\frac{1}{62},s\in [1,2].}$$

We see that ${min}_{s\in [1,2]}{h}_1\left(s\right)\approx 0.318421>0$. Because $1-{\displaystyle \frac{45}{16}}{(s-1)}^3>0,$ for all $s\in [1,5/3]$ (in fact, we have ${min}_{s\in [1,5/3]}(1-{\displaystyle \frac{45}{16}}{(s-1)}^3)\approx 0.166667>0$), we deduce

$$\begin{array}{c}{\displaystyle l_1\le {\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h\left(s\right)\frac{ds}{s}\le l_3,l_2\le \frac{1}{6}{\int }_{5/3}^{2}h\left(s\right)\frac{ds}{s}\le l_4,}\hfill \end{array}$$

where

$$\begin{array}{c}{\displaystyle l_1={\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h_1\left(s\right)\frac{ds}{s}\approx 0.13929499,}\hfill \\ {\displaystyle l_3={\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h_2\left(s\right)\frac{ds}{s}\approx 0.20777963,}\hfill \\ {\displaystyle l_2=\frac{1}{6}{\int }_{5/3}^2{h}_1\left(s\right)\frac{ds}{s}\approx 0.01196551,l_4=\frac{1}{6}{\int }_{5/3}^2{h}_2\left(s\right)\frac{ds}{s}\approx 0.01692338.}\hfill \end{array}$$

(a) If ${}^H{D}_{1+}^{4/3}u\left(s\right)>{\sigma }_2=18$, then we find

$$C_2\ge l_1+l_2+{\sigma }_2\left[{\displaystyle \frac{1}{35}{\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\frac{ds}{s}+\frac{1}{210}{\int }_{5/3}^2\frac{ds}{s}}\right]\approx 0.38275688>0.$$

(b) If ${}^H{D}_{1+}^{4/3}u\left(s\right)<-{\sigma }_2=-18$, then we have

$$C_2\le l_3+l_4-{\sigma }_2\left[{\displaystyle \frac{1}{35}{\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\frac{ds}{s}+\frac{1}{210}{\int }_{5/3}^2\frac{ds}{s}}\right]\approx -0.00679336<0.$$

So $C_2\ne 0$, and then assumption $\left(A4\right)$ is satisfied.

We finally verify assumption $\left(A5\right)$. We take $r=26$, and we consider $u\left(t\right)=b_1{(lnt)}^{7/3}+b_2{(lnt)}^{4/3}\in Ker\mathcal{L},$ with $b_1,b_2\in \mathbb{R}$. Then we obtain

$$\begin{array}{c}{C}_0:=b_1{\mathcal{S}}_1\mathcal{N}u\left(t\right)+b_2{\mathcal{S}}_2\mathcal{N}u\left(t\right)\hfill \\ {\displaystyle =b_1{\int }_1^e(P\left(e\right)-P\left(s\right))\mathcal{N}u\left(s\right)\frac{ds}{s}+b_2{\int }_1^e(Q\left(1\right)-Q\left(s\right))\mathcal{N}u\left(s\right)\frac{ds}{s}}\hfill \\ =b_1\left[{\displaystyle {\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\mathcal{N}u\left(s\right)\frac{ds}{s}+{\int }_{5/3}^2\frac{1}{6}\mathcal{N}u\left(s\right)\frac{ds}{s}}\right]\hfill \\ +b_2\left[{\displaystyle -{\int }_2^{5/2}3{(s-2)}^2\mathcal{N}u\left(s\right)\frac{ds}{s}-{\int }_{5/2}^e\mathcal{N}u\left(s\right)\frac{ds}{s}}\right]\hfill \\ =b_1\left\{{\displaystyle {\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\left({\displaystyle h\left(s\right)+\frac{1}{35}{}^H{D}_{1+}^{4/3}u\left(s\right)}\right)\frac{ds}{s}}\right.\hfill \\ \left.{\displaystyle +\frac{1}{6}{\int }_{5/3}^2\left({\displaystyle h\left(s\right)+\frac{1}{35}{}^H{D}_{1+}^{4/3}u\left(s\right)}\right)\frac{ds}{s}}\right\}\hfill \\ +b_2\left[{\displaystyle -{\int }_2^{5/2}3{(s-2)}^2\frac{c_0}{35}{}^H{D}_{1+}^{7/3}u\left(s\right)\frac{ds}{s}-{\int }_{5/2}^e\frac{c_0}{35}{}^H{D}_{1+}^{7/3}u\left(s\right)\frac{ds}{s}}\right].\hfill \end{array}$$

For the above function $u\left(t\right)$, we find

$$\begin{array}{c}{}^H{D}_{1+}^{7/3}u\left(t\right)=b_1\Gamma (10/3),{}^H{D}_{1+}^{4/3}u\left(t\right)=b_1\Gamma (10/3)lnt+b_2\Gamma (7/3),\hfill \\ {\displaystyle {}^H{D}_{1+}^{1/3}u\left(t\right)=b_1\frac{\Gamma (10/3)}{2}{(lnt)}^2+b_2\Gamma (7/3)lnt,t\in [1,e].}\hfill \end{array}$$

Then for $C_0$ we deduce

$$\begin{array}{c}{\displaystyle C_0=b_1{\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h\left(s\right)\frac{ds}{s}}\hfill \\ {\displaystyle +\frac{b_1}{35}{\int }_1^{5/3}\left(b_1\Gamma \left(\frac{10}{3}\right)lns+b_2\Gamma \left({\displaystyle \frac{7}{3}}\right)\right)\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\frac{ds}{s}}\hfill \\ {\displaystyle +\frac{b_1}{6}{\int }_{5/3}^{2}h\left(s\right)\frac{ds}{s}+\frac{b_1}{210}{\int }_{5/3}^2\left(b_1\Gamma \left({\displaystyle \frac{10}{3}}\right)lns+b_2\Gamma \left({\displaystyle \frac{7}{3}}\right)\right)\frac{ds}{s}}\hfill \\ {\displaystyle -\frac{3b_2{c}_0}{35}{\int }_2^{5/2}{(s-2)}^2{b}_1\Gamma \left({\displaystyle \frac{10}{3}}\right)\frac{ds}{s}-b_2{\int }_{5/2}^e\frac{c_0}{35}{b}_1\Gamma \left({\displaystyle \frac{10}{3}}\right)\frac{ds}{s}}\hfill \\ {\displaystyle =\frac{b_1^2}{35}\Gamma \left({\displaystyle \frac{10}{3}}\right){\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)lns\frac{ds}{s}+\frac{b_1^2}{210}\Gamma \left({\displaystyle \frac{10}{3}}\right){\int }_{5/3}^{2}lns\frac{ds}{s}}\hfill \\ {\displaystyle +b_1{\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h\left(s\right)\frac{ds}{s}+\frac{b_1}{6}{\int }_{5/3}^{2}h\left(s\right)\frac{ds}{s}}\hfill \\ +b_1{b}_2\left[{\displaystyle \frac{1}{35}{\int }_1^{5/3}\Gamma \left({\displaystyle \frac{7}{3}}\right)\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)\frac{ds}{s}+\frac{1}{210}{\int }_{5/3}^2\Gamma \left({\displaystyle \frac{7}{3}}\right)\frac{ds}{s}}\right.\hfill \\ \left.{\displaystyle -\frac{3c_0}{35}{\int }_2^{5/2}\Gamma \left({\displaystyle \frac{10}{3}}\right){(s-2)}^2\frac{ds}{s}-\frac{c_0}{35}\Gamma \left({\displaystyle \frac{10}{3}}\right){\int }_{5/2}^e\frac{ds}{s}}\right].\hfill \end{array}$$

So, we obtain $C_0=C_3{b}_1^2+C_4{b}_1,$ where

$$\begin{array}{c}{\displaystyle C_3=\frac{1}{35}\Gamma \left({\displaystyle \frac{10}{3}}\right){\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)lns\frac{ds}{s}+\frac{1}{210}\Gamma \left({\displaystyle \frac{10}{3}}\right){\int }_{5/3}^{2}lns\frac{ds}{s}\approx 0.00877909,}\hfill \\ {\displaystyle C_4={\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h\left(s\right)\frac{ds}{s}+\frac{1}{6}{\int }_{5/3}^{2}h\left(s\right)\frac{ds}{s}.}\hfill \end{array}$$

Hence $C_0=C_3{b}_1(b_1+C_4/C_3).$ For $C_4$ we find

$$\begin{array}{c}{C}_4\in \left[{\displaystyle {\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h_1\left(s\right)\frac{ds}{s}+\frac{1}{6}{\int }_{5/3}^2{h}_1\left(s\right)\frac{ds}{s};}\right.\hfill \\ \left.{\displaystyle {\int }_1^{5/3}\left({\displaystyle 1-\frac{45}{16}{(s-1)}^3}\right)h_2\left(s\right)\frac{ds}{s}+\frac{1}{6}{\int }_{5/3}^2{h}_2\left(s\right)\frac{ds}{s}}\right]\hfill \\ =[l_1+l_2;l_3+l_4]\approx [0.15126051;0.22470301].\hfill \end{array}$$

We obtain $C_4/C_3\in [(l_1+l_2)/C_3,(l_3+l_4)/C_3]\approx [17.22963609;$$25.59525372]$. Then, if $|b_1|>r=26$, we deduce that $C_0>0$. Therefore, assumption $\left(A5\right)$ is satisfied.

Therefore, by Theorem 2, we conclude that Problem (37),(38) have at least one solution.

5. Conclusions

In this paper, we investigated the Hadamard fractional differential Equation (1) containing fractional derivatives of various orders, subject to the nonlocal boundary conditions (2). These conditions are general ones encompassing Riemann–Stieltjes integrals and Hadamard fractional derivatives. Our Problem (1),(2) are resonant in the sense that the corresponding homogeneous boundary-value problem has nontrivial solutions. We equivalently wrote this problem as an operator equation in certain Banach spaces involving two operators for which we studied diverse properties. Then, based on the coincidence degree theory of J. Mawhin, more specifically the Mawhin continuation theorem, we proved the existence of solutions for (1) and (2).