A Hadamard Fractional Boundary Value Problem on an Infinite Interval at Resonance

Abstract

This paper addresses the existence of solutions to a Hadamard fractional differential equation of arbitrary order on an infinite interval, subject to integral boundary conditions that incorporate both Riemann–Stieltjes integrals and Hadamard fractional derivatives. Due to the presence of nontrivial solutions in the associated homogeneous boundary value problem, the problem is classified as resonant. The Mawhin continuation theorem is utilized to derive the main findings.

1. Introduction

In this paper, we examine the following fractional differential equation:

$${}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)=\mathfrak{f}\left(\theta ,\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{\alpha -n+2}\mathfrak{u}\left(\theta \right),\dots ,{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\right),\theta \in (1,\infty ),$$

(1)

subject to the following nonlocal boundary conditions:

$$\left\{\begin{array}{c}{\mathfrak{u}}^{\left(j\right)}\left(1\right)=0,j=0,\dots ,n-3;{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)={\int }_1^A{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)d\mathcal{P}\left(\sigma \right),\hfill \\ {}^{H}D_{1+}^{\alpha -1}\mathfrak{u}(\infty )={\int }_1^B{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)d\mathcal{Q}\left(\sigma \right),\hfill \end{array}\right.$$

(2)

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

The problem under consideration is resonant in the sense that the associated homogeneous boundary value problem admits nontrivial solutions (refer to assumption $\left(I1\right)$ and Lemma 2 below). By imposing various conditions on the nonlinearity $\mathfrak{f}$, we establish the existence of solutions to the boundary value problem (1), (2). The proofs of our main results are based on the coincidence degree theory developed by J. Mawhin, specifically the Mawhin continuation theorem (see Theorem IV.13 from [1,2]). In the subsequent discussion, we briefly review several works that are related to the present study. In [3], the authors studied the Hadamard fractional Equation (1) on the interval of $(1,e)$ at resonance, supplemented with the following integral boundary conditions:

$$\left\{\begin{array}{c}{\mathfrak{u}}^{\left(j\right)}\left(1\right)=0,j=0,\dots ,n-3,{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)={\int }_1^e{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)d\mathcal{P}\left(\sigma \right),\hfill \\ {}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(e\right)={\int }_1^e{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)d\mathcal{Q}\left(\sigma \right).\hfill \end{array}\right.$$

(3)

They proved the existence of solutions for the problem (1), (3) on $(1,e)$ by using the coincidence degree theory. In [4], the authors analyzed a Hadamard fractional differential equation at resonance given by

$$\left\{\begin{array}{c}\left({}^{H}D_{1+}^{\gamma }\mathfrak{u}\right)\left(\theta \right)+\mathfrak{f}(\theta ,\mathfrak{u}\left(\theta \right))=0,\theta \in (1,e),\hfill \\ \mathfrak{u}\left(1\right)=0,\mathfrak{u}\left(e\right)={\int }_1^e\mathfrak{u}\left(\sigma \right)dA\left(\sigma \right),\hfill \end{array}\right.$$

(4)

where $\gamma \in (1,2)$ and $\mathfrak{f}:(1,e)\times \mathbb{R}\to \mathbb{R}$ satisfies Carathéodory conditions. The boundary conditions include a Riemann–Stieltjes integral. Using the Mawhin continuation theorem, they established the existence of solutions for problem (4). In [5], the authors investigated the existence of solutions to a Riemann–Liouville fractional boundary value problem at resonance on an infinite interval. Utilizing the coincidence degree theory, they consider the following problem:

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

where $\alpha \in (2,3]$, $0<{\eta }_1<{\eta }_2<\cdots <{\eta }_n<\infty$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_m<\infty$, $0<{\eta }_1<{\eta }_2<\cdots <{\eta }_n<\infty$, and the coefficients are ${\alpha }_i,{\beta }_k\in \mathbb{R}$ for $i=1,\dots ,m$, $k=1,\dots ,n$. The function $\mathfrak{f}:(0,\infty )\times {\mathbb{R}}^3\to \mathbb{R}$ is assumed to be a Carathéodory function, and $D_{0+}^{\kappa }$ denotes the Riemann–Liouville fractional derivative of order $\kappa$ for $\kappa \in \{\alpha ,\alpha -1,\alpha -2\}$. In [6], the authors examined the multipoint boundary value problem of the following fractional differential equation:

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha }\mathfrak{u}\left(\theta \right)=\mathfrak{f}(\theta ,\mathfrak{u}\left(\theta \right),D_{0+}^{\alpha -1}\mathfrak{u}\left(\theta \right)),\theta \in (0,\infty ),\hfill \\ I_{0+}^{2-\alpha }\mathfrak{u}\left(0\right)=0,\underset{\theta \to \infty }{lim}{D}_{0+}^{\alpha -1}\mathfrak{u}\left(\theta \right)=\sum _{j=1}^{m-2}{\beta }_j{D}_{0+}^{\alpha -1}\mathfrak{u}\left({\xi }_j\right),\hfill \end{array}\right.$$

(5)

where $\alpha \in (1,2]$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_{m-2}<\infty$, ${\beta }_j>0$ for $j=1,\dots ,m-2$, $D_{0+}^{\alpha }$ is the Riemann–Liouville fractional derivative, $I_{0+}^{\alpha }$ is the Riemann–Liouville fractional integral, and ${\sum }_{j=1}^{m-2}{\beta }_j=1$. By applying the Mawhin coincidence degree theory, they showed the existence of solutions for problem (5). In [7], the authors explored the Riemann–Liouville fractional differential equation with a p-Laplacian operator on the half-line, subject to integral boundary conditions at resonance as follows:

$$\left\{\begin{array}{c}{\left({\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}\left(\theta \right)\right)\right)}^{\prime }+v\left(t\right)w(\theta ,\mathfrak{u}\left(\theta \right),D_{0+}^{\alpha }\mathfrak{u}\left(\theta \right))=0,\theta \in (0,\infty ),\hfill \\ {\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}\left(0\right)\right)={\int }_0^{\infty }v\left(\theta \right){\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}\left(\theta \right)\right)d\theta ,\hfill \\ {\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}(\infty )\right)={\int }_0^{\infty }v\left(\theta \right){\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}\left(\theta \right)\right)d\theta ,\hfill \end{array}\right.$$

(6)

where $\alpha \in (0,1]$, $w:[0,\infty )\times {\mathbb{R}}^2\to \mathbb{R}$ is a v-Carathéodory function; $v\in L^1[0,\infty )$; $v\left(\theta \right)>0$ with ${\int }_0^{\infty }v\left(\theta \right)d\theta =1$; and ${\phi }_p\left(r\right)={\left|r\right|}^{p-2}r$, where $p>1$. By using the Ge and Ren extension of the Mawhin coincidence degree theory, they proved the existence of solutions to problem (6). In [8], the authors studied the multipoint boundary value problem of the following Riemann–Liouville fractional p-Laplacian equation at resonance:

$$\left\{\begin{array}{c}{\left({\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}\left(\theta \right)\right)\right)}^{\prime }+\mathfrak{f}(\theta ,\mathfrak{u}\left(\theta \right),D_{0+}^{\alpha -1}\mathfrak{u}\left(\theta \right),D_{0+}^{\alpha }\mathfrak{u}\left(\theta \right))=0,\theta \in (0,\infty ),\hfill \\ \mathfrak{u}\left(0\right)={\mathfrak{u}}^{\prime }\left(0\right)=0,{\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}(\infty )\right)=\sum _{j=1}^n{\alpha }_j{\phi }_p\left(D_{0+}^{\alpha }\mathfrak{u}\left({\xi }_j\right)\right),\hfill \end{array}\right.$$

(7)

where $\alpha \in (1,2]$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_n<\infty$, ${\alpha }_j>0$, ${\sum }_{j=1}^n{\alpha }_j=1$, and $\mathfrak{f}$ is a $L^1$-Carathéodory function. By using an extension of Mawhin continuation theorem, they showed the existence of solutions for problem (7). In [9], the authors examined the system of Riemann-Liouville fractional differential equations supplemented with the following boundary conditions:

$$\left\{\begin{array}{c}{D}_{0+}^{\alpha }{\mathfrak{u}}_1\left(\theta \right)={\mathfrak{f}}_1\left(\theta ,{\mathfrak{u}}_1\left(\theta \right),D_{0+}^{\beta -1}{\mathfrak{u}}_2\left(\theta \right)\right),\theta \in (0,\infty ),\hfill \\ D_{0+}^{\beta }{\mathfrak{u}}_2\left(\theta \right)={\mathfrak{f}}_2\left(\theta ,{\mathfrak{u}}_2\left(\theta \right),D_{0+}^{\alpha -1}{\mathfrak{u}}_1\left(\theta \right)\right),\theta \in (0,\infty ),\hfill \\ {\mathfrak{u}}_1\left(0\right)=0,\underset{\theta \to \infty }{lim}{D}_{0+}^{\alpha -1}{\mathfrak{u}}_1\left(\theta \right)=\sum _{j=1}^{\infty }{\gamma }_j{\mathfrak{u}}_1\left({\eta }_j\right),\hfill \\ {\mathfrak{u}}_2\left(0\right)=0,\underset{\theta \to \infty }{lim}{D}_{0+}^{\beta -1}{\mathfrak{u}}_2\left(\theta \right)=\sum _{j=1}^{\infty }{\sigma }_j{\mathfrak{u}}_2\left({\xi }_j\right),\hfill \end{array}\right.$$

(8)

where $\alpha ,\beta \in (1,2]$, $0<{\eta }_1<{\eta }_2\cdots <{\eta }_j<\cdots$, $0<{\xi }_1<{\xi }_2<\cdots <{\xi }_j<\cdots$, ${lim}_{j\to \infty }{\eta }_j=\infty$, ${lim}_{j\to \infty }{\xi }_j=\infty$, ${\sum }_{j=1}^{\infty }\left|{\gamma }_j\right|{\eta }_j^{\alpha }<\infty$, ${\sum }_{j=1}^{\infty }\left|{\sigma }_j\right|{\xi }_j^{\beta }<\infty$, and ${\mathfrak{f}}_1,{\mathfrak{f}}_2$ satisfy Carathéodory conditions. By use of the coincidence degree theory of Mawhin, they proved the existence of solutions of problem (8). We also mention recent papers [10,11,12,13] where the authors investigated various boundary value problems for Riemann–Liouville fractional differential equations on finite intervals at resonance. Different boundary value problems for Caputo fractional differential equations and systems on finite intervals at resonance were studied in [14,15,16,17].

Hadamard fractional calculus finds applications across diverse fields, including rheology and probability (see [18,19] and the references therein). For instance, extensions of the Lomnitz logarithmic creep law are used to model geophysical processes in fluid-like materials such as igneous rocks. For further recent developments concerning Hadamard fractional differential equations and systems with a variety of applications, we refer the reader to [20,21,22,23]. The novelty of the present problem (1), (2) lies in the treatment of a resonant boundary value problem for a Hadamard fractional differential equation of arbitrary order also containing varied fractional derivatives on an infinite interval. Moreover, the general boundary conditions considered here include Riemann–Stieltjes integrals and encompass a broad class of conditions such as finite or infinite multipoint boundary conditions, classical integral conditions, and combinations thereof.

The organization of our paper is outlined as follows. Section 2 provides several supporting lemmas used in the proofs of the main results. Section 3 is dedicated to presenting the existence theorems for problem (1), (2). Section 4 is devoted to two examples that demonstrates the applicability of the theoretical findings. Finally, Section 5 presents the conclusions of the paper.

2. Auxiliary Results

In this section, we provide the supporting results for our main theorems presented in Section 3.

We introduce the following expressions:

$$\begin{array}{c}\mathfrak{a}={\int }_1^A\left({\int }_1^s(ln{\displaystyle \frac{\tau }{s}}){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(s\right)={\int }_1^A{\displaystyle \frac{s^2-2s^{2}lns-1}{4s^2}}d\mathcal{P}\left(s\right),\hfill \\ \mathfrak{b}={\int }_1^A\left({\int }_1^s(ln{\displaystyle \frac{s}{\tau }}){\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{P}\left(s\right)={\int }_1^A{\displaystyle \frac{lns+s^{2}lns+1-s^2}{4s^2}}d\mathcal{P}\left(s\right),\hfill \\ \mathfrak{c}={\int }_1^B\left({\int }_s^{\infty }{\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(s\right)={\int }_1^B{\displaystyle \frac{1}{2s^2}}d\mathcal{Q}\left(s\right),\hfill \\ \mathfrak{d}={\int }_1^B\left({\int }_s^{\infty }{\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{Q}\left(s\right)={\int }_1^B{\displaystyle \frac{2lns+1}{4s^2}}d\mathcal{Q}\left(s\right),\hfill \\ \Delta =\mathfrak{ad}-\mathfrak{bc}.\hfill \end{array}$$

(9)

Firstly, we present our main assumptions:

(I1)

The functions $\mathcal{P}:[1,A]\to \mathbb{R}$ and $\mathcal{Q}:[1,B]\to \mathbb{R}$ are bounded variation functions satisfying the resonant conditions of ${\int }_1^{A}d\mathcal{P}\left(\sigma \right)=1$ ($\iff \mathcal{P}\left(A\right)-\mathcal{P}\left(1\right)=1$), ${\int }_1^{B}d\mathcal{Q}\left(\sigma \right)=1$ ($\iff \mathcal{Q}\left(B\right)-\mathcal{Q}\left(1\right)=1$), and ${\int }_1^{A}ln\sigma d\mathcal{P}\left(\sigma \right)=0$. In addition, $\Delta \ne 0$ (given in (9)).

(I2)

The function $\mathfrak{f}:(1,\infty )\times {\mathbb{R}}^n\to \mathbb{R}$ is a Carathéodory function that satisfies the following conditions

(a)

$\mathfrak{f}(·,y_1,y_2,\dots ,y_n)$ is Lebesgue measurable for each $(y_1,y_2,\dots ,y_n)\in {\mathbb{R}}^n$;

(b)

$\mathfrak{f}(\theta ,·,·,\dots ,·)$ is continuous (in the last n variables) for a.e. $\theta \in (1,\infty )$;

(c)

There exist functions, i.e., $p_j\in L^1(1,\infty )$, $j=1,\dots ,n+1$, and $p_j\left(\theta \right)\ge 0$, a.e. $\theta \in (1,\infty ),j=1,\dots ,n+1$ such that

$$\begin{array}{c}|\mathfrak{f}(\theta ,y_1,y_2,\dots ,y_n)|\le p_1\left(\theta \right){\displaystyle \frac{|y_1|}{1+{(ln\theta )}^{\alpha -1}}}+p_2\left(\theta \right){\displaystyle \frac{|y_2|}{1+{(ln\theta )}^{n-2}}}+\cdots +\\ +p_{n-1}\left(\theta \right){\displaystyle \frac{|y_{n-1}|}{1+ln\theta }}+p_n\left(\theta \right)\left|y_n\right|+p_{n+1}\left(\theta \right)\end{array}$$

(10)

for all $y_j\in \mathbb{R},j=1,\dots ,n$ and a.e. $\theta \in (1,\infty ).$

 Remark 1. 

If assumption $\left(I1\right)$ is satisfied, then the boundary value problem (1), (2) is at resonance because the corresponding fractional differential equation, i.e.,

$${}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)=0,\theta \in (1,\infty ),$$

subject to the boundary conditions (2), admits nontrivial solutions of the form $\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2}$ for $\theta \in [1,\infty )$ for arbitrary $a_1,a_2\in \mathbb{R}$ (see Lemma 2 below).

We define the Banach space of

$$\begin{array}{c}\mathcal{U}=\left\{\mathfrak{u};\mathfrak{u},{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u},\dots ,{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\in C[1,\infty ),\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|\mathfrak{u}\right(\theta \left)\right|}{1+{(ln\theta )}^{\alpha -1}}}<\infty ,\right.\\ \left.\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}<\infty ,\dots ,\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|}{1+ln\theta }}<\infty ,\underset{\theta \ge 1}{sup}\left|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\right|<\infty \right\},\end{array}$$

with the following norm:

$${\parallel \mathfrak{u}\parallel }_{\mathcal{U}}=max\left\{{\parallel \mathfrak{u}\parallel }_{\alpha -1}{,\parallel }^H{D}_{1+}^{\alpha -n+1}{\mathfrak{u}\parallel }_{n-2}{,\dots ,\parallel }^H{D}_{1+}^{\alpha -2}{\mathfrak{u}\parallel }_1,{{\parallel }^H{D}_{1+}^{\alpha -1}\mathfrak{u}\parallel }_0\right\},$$

where

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\alpha -1}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|\mathfrak{u}\right(\theta \left)\right|}{1+{(ln\theta )}^{\alpha -1}}},{{\parallel }^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\parallel }_{n-2}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}},\dots ,\hfill \\ {\parallel }^H{D}_{1+}^{\alpha -2}{\mathfrak{u}\parallel }_1=\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|}{1+ln\theta }}{,\parallel }^H{D}_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0=\underset{\theta \ge 1}{sup}{|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|,\mathrm{for}\mathrm{u}\in \mathcal{U}.\hfill \end{array}$$

We also define the Banach space of $\mathcal{V}=L^1(1,\infty )$ with a norm of ${\parallel \mathfrak{v}\parallel }_{\mathcal{V}}={\int }_1^{\infty }\left|\mathfrak{v}\left(\theta \right)\right|d\theta$.

We now define the linear operator as $\mathcal{A}:D\left(\mathcal{A}\right)\subset \mathcal{U}\to \mathcal{V}$, given by

$$\mathcal{A}\mathfrak{u}\left(\theta \right){=}^H{D}_{1+}^{\alpha }\mathfrak{u}\left(\theta \right),\mathfrak{u}\in D\left(\mathcal{A}\right),\theta \in (1,\infty ),$$

where

$$D\left(\mathcal{A}\right)=\{\mathfrak{u}\in \mathcal{U};{}^{H}D_{1+}^{\alpha }\mathfrak{u}\in \mathcal{V},and\mathfrak{u}\mathrm{verifies}\mathrm{the}\mathrm{conditions}\left(2\right)\},$$

and the nonlinear operator as $\mathcal{B}:\mathcal{U}\to \mathcal{V}$, given by

$$\mathcal{B}\mathfrak{u}\left(\theta \right)=\mathfrak{f}(\theta ,\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right),\dots ,{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)),\theta \in (1,\infty ),\mathfrak{u}\in \mathcal{U}.$$

With the aid of the above operators ($\mathcal{A}$ and $\mathcal{B}$), the problem (1), (2) can be written equivalently as the following operator equation:

$$\mathcal{A}\mathfrak{u}=\mathcal{B}\mathfrak{u},\mathfrak{u}\in D\left(\mathcal{A}\right).$$

(11)

 Lemma 1 

([24]). Let $\mathcal{K}\subset \mathcal{U}$ be a bounded set satisfying the following conditions:

(a) The functions from $\mathcal{K}$ are equicontinuous on any compact interval of $[1,\infty )$;

(b) The functions from $\mathcal{K}$ are equiconvergent at infinity.

Therefore, set $\mathcal{K}$ is relatively compact in $\mathcal{U}$.

 Lemma 2. 

We assume that $\left(I1\right)$ holds. Then, the operator $\mathcal{A}:D\left(\mathcal{A}\right)\subset \mathcal{U}\to \mathcal{V}$ exhibits the following properties:

$$Ker\left(\mathcal{A}\right)=\{\mathfrak{u}\in D\left(\mathcal{A}\right);\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2},\forall \theta \in [1,\infty ),a_1,a_2\in \mathbb{R}\}\subset \mathcal{U},$$

(12)

$$Im\mathcal{A}=\{\mathfrak{v}\in \mathcal{V};\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0,\forall \theta \in [1,\infty )\}\subset \mathcal{V},$$

(13)

where

$$\begin{array}{c}\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right),\hfill \\ \left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)={\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right),\hfill \end{array}$$

(14)

for all $\mathfrak{v}\in \mathcal{V}$, $\theta \in [1,\infty )$.

 Proof. 

We consider $\mathfrak{u}\in Ker\mathcal{A}$; in other words, $\mathfrak{u}\in D\left(\mathcal{A}\right)$ and $\mathcal{A}\mathfrak{u}=0$. Then, ${}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)=0$ for all $\theta \in [1,\infty )$. Hence, according to Corollary 2.4 from [25], we find

$$\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+\cdots +a_n{(ln\theta )}^{\alpha -n},\theta \in [1,\infty ),$$

(15)

with $a_k\in \mathbb{R},k=1,\dots ,n$. By using the conditions of ${\mathfrak{u}}^{\left(k\right)}\left(1\right)=0$, $k=0,\dots ,n-3$, we infer that $a_k=0$ for $k=3,\dots ,n$. Therefore, according to (15), we obtain

$$\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2},\theta \in [1,\infty ).$$

Because ${}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)=a_1\Gamma \left(\alpha \right)$ and ${}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)=a_1\Gamma \left(\alpha \right)ln\theta +a_2\Gamma (\alpha -1)$, according to $\left(I1\right)$, the following conditions are verified:

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)={\int }_1^A{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)d\mathcal{P}\left(\sigma \right)\hfill \\ \iff a_2\Gamma (\alpha -1)={\int }_1^A(a_1\Gamma \left(\alpha \right)ln\sigma +a_2\Gamma (\alpha -1))d\mathcal{P}\left(\sigma \right)\hfill \\ \iff a_1\Gamma \left(\alpha \right){\int }_1^{A}ln\sigma d\mathcal{P}\left(\sigma \right)+a_2\Gamma (\alpha -1)\left({\int }_1^{A}d\mathcal{P}\left(\sigma \right)-1\right)=0,\hfill \\ {}^{H}D_{1+}^{\alpha -1}\mathfrak{u}(\infty )={\int }_1^B{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)d\mathcal{Q}\left(\sigma \right)\iff a_1\Gamma \left(\alpha \right)={\int }_1^B{a}_1\Gamma \left(\alpha \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff a_1\Gamma \left(\alpha \right)\left({\int }_1^{B}d\mathcal{Q}\left(\sigma \right)-1\right)=0.\hfill \end{array}$$

Therefore, we infer

$$Ker\mathcal{A}\subset \{\mathfrak{u}\in D\left(\mathcal{A}\right),\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2},\theta \in [1,\infty ),a_1,a_2\in \mathbb{R}\}.$$

For the reverse inclusion, let $\mathfrak{u}\in D\left(\mathcal{A}\right)$, $\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2},\theta \in [1,\infty )$ with $a_1,a_2\in \mathbb{R}$. We easily remark that ${}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)=0$ for all $\theta \in [1,\infty )$, and $\mathfrak{u}$ satisfies the conditions of (2), that is, $\mathfrak{u}\in Ker\mathcal{A}$. So,

$$\{\mathfrak{u}\in D\left(\mathcal{A}\right);\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2},\theta \in [1,\infty ),a_1,a_2\in \mathbb{R}\}\subset Ker\mathcal{A}.$$

Therefore, relation (12) holds. We see that $dimKer\mathcal{A}=2$.

Now, we prove relation (13). We consider $\mathfrak{v}\in Im\mathcal{A}$. Hence, there exists a function $\mathfrak{u}\in D\left(\mathcal{A}\right)$ such that $\mathcal{A}\mathfrak{u}=\mathfrak{v}$ or

$${}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)=\mathfrak{v}\left(\theta \right),\theta \in (1,\infty ).$$

(16)

By applying the integral operator ${}^{H}I_{1+}^{\alpha }$ to Equation (16), we find

$$\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(\theta \right)+a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2}+\cdots +a_n{(ln\theta )}^{\alpha -1},$$

(17)

with $a_j\in \mathbb{R}$, $j=1,\dots ,n$. We use now the conditions of ${\mathfrak{u}}^{\left(j\right)}\left(1\right)=0$, $j=0,\dots ,n-3$, and we obtain $a_3=a_4=\cdots =a_n=0$. Then, according to (17), we obtain

$$\mathfrak{u}\left(\theta \right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma +a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2},\theta \in [1,\infty ).$$

Then,

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^1\mathfrak{v}\left(\theta \right)+a_1\Gamma \left(\alpha \right)={\int }_1^{\theta }{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma +a_1\Gamma \left(\alpha \right),\hfill \\ {}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^2\mathfrak{v}\left(\theta \right)+a_1\Gamma \left(\alpha \right)ln\theta +a_2\Gamma (\alpha -1)\hfill \\ ={\int }_1^{\theta }\left(ln{\displaystyle \frac{\theta }{\sigma }}\right){\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma +a_1\Gamma \left(\alpha \right)ln\theta +a_2\Gamma (\alpha -1).\hfill \end{array}$$

So, the conditions of ${}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)={\int }_1^A{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)d\mathcal{P}\left(\sigma \right)$ and ${}^{H}D_{1+}^{\alpha -1}\mathfrak{u}(\infty )={\int }_1^B{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)d\mathcal{Q}\left(\sigma \right)$ give us

$$\left\{\begin{array}{c}{a}_2\Gamma (\alpha -1)={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau +a_1\Gamma \left(\alpha \right)ln\sigma +a_2\Gamma (\alpha -1)\right)d\mathcal{P}\left(\sigma \right),\hfill \\ {\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma +a_1\Gamma \left(\alpha \right)={\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau +a_1\Gamma \left(\alpha \right)\right)d\mathcal{Q}\left(\sigma \right),\hfill \end{array}\right.$$

or

$$\left\{\begin{array}{c}-a_1\Gamma \left(\alpha \right){\int }_1^{A}ln\sigma d\mathcal{P}\left(\sigma \right)-a_2\Gamma (\alpha -1)\left({\int }_1^{A}d\mathcal{P}\left(\sigma \right)-1\right)\hfill \\ ={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right),\hfill \\ a_1\Gamma \left(\alpha \right)\left(1-{\int }_1^{B}d\mathcal{Q}\left(\sigma \right)\right)={\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)-{\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma .\hfill \end{array}\right.$$

According to the resonant conditions from $\left(I1\right)$, we find

$$\left\{\begin{array}{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)=0,\hfill \\ {\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)-{\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma =0.\hfill \end{array}\right.$$

The second relation from the above system is equivalent to

$$\begin{array}{c}{\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)={\int }_1^B\left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff {\int }_1^B\left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau -{\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0\hfill \\ \iff {\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0.\hfill \end{array}$$

Thus, $\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$, that is,

$$Im\mathcal{A}\subset \{\mathfrak{v}\in \mathcal{V};\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0,\forall \theta \in [1,\infty )\}.$$

(18)

For the reverse inclusion, let $\mathfrak{v}\in \mathcal{V}$ with $\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$. We consider $\mathfrak{u}\left(\theta \right){}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(\theta \right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}$, $\theta \in [1,\infty )$. It is evident that $\mathfrak{u}\in D\left(\mathcal{A}\right)$. Indeed, we have $\mathfrak{u},{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u}={}^{H}I_{1+}^{n-1}\mathfrak{u},\dots ,{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}={}^{H}I_{1+}^1\mathfrak{u}\in C[1,\infty )$, and ${}^{H}D_{1+}^{\alpha }\mathfrak{u}=\mathfrak{v}\in \mathcal{V}$, with ${\parallel \mathfrak{u}\parallel }_{\mathcal{U}}<\infty$. In addition, $\mathfrak{u}$ verifies the conditions of (2), that is, ${\mathfrak{u}}^{\left(i\right)}\left(0\right)=0,i=0,\dots ,n-3$. Because

$${}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^2\mathfrak{v}\left(\theta \right)={\int }_1^{\theta }\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }},{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^1\mathfrak{v}\left(\theta \right)={\int }_1^{\theta }\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }},$$

the conditions of

$${}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)={\int }_1^A{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)d\mathcal{P}\left(\sigma \right)\iff 0={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)\mathfrak{v}\left(\tau \right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{P}\left(\sigma \right)$$

and

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}(\infty )={\int }_1^B{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)d\mathcal{Q}\left(\sigma \right)\iff {\int }_1^{\infty }\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}={\int }_1^B\left({\int }_1^{\sigma }v\left(\tau \right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff \left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma \right)\left({\int }_1^{B}d\mathcal{Q}\left(\sigma \right)\right)={\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff {\int }_1^B\left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)={\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff {\int }_1^B\left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau -{\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0\hfill \\ \iff {\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0,\hfill \end{array}$$

are verified. We also find $\mathcal{A}\mathfrak{u}\left(\theta \right)=\mathfrak{v}\left(\theta \right)$ or ${}^{H}D_{1+}^{\alpha }\left({}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(\theta \right)\right)=\mathfrak{v}\left(\theta \right)$ for all $\theta \in (1,\infty )$. Hence, we deduce

$$\{\mathfrak{v}\in \mathcal{V};\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0,\forall \theta \in [1,\infty )\}\subset Im\mathcal{A}.$$

(19)

Then, according to (18) and (19), we derive relation (13). □

 Remark 2. 

The relation $\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=0$ can be equivalently written as

$$\begin{array}{c}{\int }_1^A\left({\int }_{\tau }^{A}ln{\displaystyle \frac{\sigma }{\tau }}d\mathcal{P}\left(\sigma \right)\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau =0\hfill \\ \stackrel{\sigma \leftrightarrow \tau }{\iff }{\int }_1^A\left({\int }_{\sigma }^{A}ln{\displaystyle \frac{\tau }{\sigma }}d\mathcal{P}\left(\tau \right)\right){\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma =0,\hfill \end{array}$$

(20)

 and the relation $\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0$ can be equivalently written as

$$\begin{array}{c}{\int }_1^B\left({\int }_{\sigma }^B{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)+{\int }_1^B\left({\int }_B^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0\hfill \\ \iff {\int }_1^B\left({\int }_1^{\tau }d\mathcal{Q}\left(\sigma \right)\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau +{\int }_B^{\infty }\left({\int }_1^{B}d\mathcal{Q}\left(\sigma \right)\right){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau =0\hfill \\ \iff {\int }_1^B(\mathcal{Q}\left(\tau \right)-\mathcal{Q}\left(1\right)){\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau +(\mathcal{Q}\left(B\right)-\mathcal{Q}\left(1\right)){\int }_B^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau =0\hfill \\ \stackrel{\tau \stackrel{not}{=}\sigma }{\iff }{\int }_1^B(\mathcal{Q}\left(\sigma \right)-\mathcal{Q}\left(1\right)){\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma +{\int }_B^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\sigma \right)}{\sigma }}d\sigma =0.\hfill \end{array}$$

(21)

We now define the following linear operators:

$$\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)={\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\theta }^2}},\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)={\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{1}{{\theta }^2}},\theta \in [1,\infty ),\mathfrak{v}\in \mathcal{V}.$$

(22)

 Lemma 3. 

Assume that $\left(I1\right)$ holds. Then, the operator $\mathcal{A}:D\left(\mathcal{A}\right)\subset \mathcal{U}\to \mathcal{V}$ is a Fredholm operator of index zero. In addition, the operators $\mathcal{E}:\mathcal{U}\to \mathcal{U}$ and $\mathcal{F}:\mathcal{V}\to \mathcal{V}$ given by

$$\begin{array}{c}\mathcal{E}\mathfrak{u}\left(\theta \right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right){(ln\theta )}^{\alpha -1}+{\displaystyle \frac{1}{\Gamma (\alpha -1)}}{}^{H}D_{+}^{\alpha -2}\mathfrak{u}\left(1\right){(ln\theta )}^{\alpha -2},\theta \in [1,\infty ),\mathfrak{u}\in \mathcal{U},\hfill \\ \mathcal{F}\mathfrak{v}\left(\theta \right)={\mathcal{G}}_1\mathfrak{v}\left(\theta \right)+\left({\mathcal{G}}_2\mathfrak{v}\left(\theta \right)\right)ln\theta ,\theta \in [1,\infty ),\mathfrak{v}\in \mathcal{V},\hfill \end{array}$$

(23)

 are linear projectors and verify the following relations:

$$Im\mathcal{E}=Ker\mathcal{A},Ker\mathcal{F}=Im\mathcal{A},\mathcal{U}=Ker\mathcal{E}\oplus Ker\mathcal{A},\mathcal{V}=Im\mathcal{A}\oplus Im\mathcal{F}.$$

(24)

Proof. 

According to the definition of operator $\mathcal{E}$, we deduce that it is a continuous and linear projector operator and $Im\mathcal{E}=Ker\mathcal{A}$ (see also [3], Lemma 6). Furthermore, for $\mathfrak{u}\in \mathcal{U}$, we have $\mathfrak{u}=(\mathfrak{u}-\mathcal{E}\mathfrak{u})+\mathcal{E}\mathfrak{u}$, that is, $\mathcal{U}=Ker\mathcal{E}+Ker\mathcal{A}$. We can easily show that $Ker\mathcal{E}\cap Ker\mathcal{A}=\left\{0\right\}$, so $\mathcal{U}=Ker\mathcal{E}\oplus Ker\mathcal{A}$. We also observe that operator $\mathcal{F}$ is a continuous and linear operator and $dimIm\mathcal{F}=2$.

According to the definitions of ${\mathcal{G}}_1$ and ${\mathcal{G}}_2$ (see (22)), we deduce that for any $\mathfrak{v}\in \mathcal{V}$ and $\theta \in [1,\infty )$,

$$\begin{array}{c}{\mathcal{G}}_1\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)={\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\left(\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)\right)-\mathfrak{b}{\mathcal{C}}_2\left(\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)\right)){\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{d}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ -\mathfrak{b}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{d}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ -\mathfrak{b}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[{\displaystyle \frac{\mathfrak{d}}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\tau }{\sigma }}\right){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ -{\displaystyle \frac{\mathfrak{b}}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v})[\mathfrak{d}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\tau }{\sigma }}\right){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ -\mathfrak{b}{\int }_1^B{\displaystyle \frac{1}{2{\sigma }^2}}d\mathcal{Q}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}={\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\theta }^2}}=\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right),\hfill \end{array}$$

$$\begin{array}{c}{\mathcal{G}}_2(\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)ln\theta )\left(\theta \right)={\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2(\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)ln\theta )-\mathfrak{c}{\mathcal{C}}_1(\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)ln\theta )){\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{a}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\tau \right)ln\tau }{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -\mathfrak{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\tau \right)ln\tau }{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{a}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -\mathfrak{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[{\displaystyle \frac{\mathfrak{a}}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -{\displaystyle \frac{\mathfrak{c}}{\Delta }}(a{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v})[\mathfrak{a}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -\mathfrak{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{1}{{\theta }^2}}=\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right),\hfill \end{array}$$

$$\begin{array}{c}{\mathcal{G}}_1\left(\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)\ln \theta \right)\left(\theta \right)={\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\left(\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)\ln \theta \right)-\mathfrak{b}{\mathcal{C}}_2(\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\theta \right)ln\theta )){\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{d}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\tau \right)\ln \tau }{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ -\mathfrak{b}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\left({\mathcal{G}}_2\mathfrak{v}\right)\left(\tau \right)\ln \tau }{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}\left[\mathfrak{d}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{\ln \tau }{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)\right.\hfill \\ -\left.\mathfrak{b}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{\ln \tau }{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)\right]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v})[\mathfrak{d}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\tau }{\sigma }}\right){\displaystyle \frac{\ln \tau }{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ -\mathfrak{b}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\ln \tau }{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v})(\mathfrak{d}\mathfrak{b}-\mathfrak{b}\mathfrak{d})=0,\hfill \end{array}$$

$$\begin{array}{c}{\mathcal{G}}_2\left(\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)\right)\left(\theta \right)={\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\left(\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)\right)-\mathfrak{c}{\mathcal{C}}_1\left(\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\theta \right)\right)){\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{a}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -\mathfrak{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\left({\mathcal{G}}_1\mathfrak{v}\right)\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}[\mathfrak{a}{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -\mathfrak{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v})[{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ -\mathfrak{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)]{\displaystyle \frac{1}{{\theta }^2}}\hfill \\ ={\displaystyle \frac{1}{\Delta }}·{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_1\mathfrak{v})(\mathfrak{a}\mathfrak{c}-\mathfrak{c}\mathfrak{a})=0.\hfill \end{array}$$

Then, based on the above relations, we deduce that ${\mathcal{F}}^2\mathfrak{v}\left(\theta \right)=\mathcal{F}\left(\mathcal{F}\mathfrak{v}\right)\left(\theta \right)=\mathcal{F}\mathfrak{v}\left(\theta \right)$ for $\theta \in [1,\infty )$ and $\mathfrak{v}\in \mathcal{V}$, that is, $\mathcal{F}$ is a projector operator.

By using Lemma 2, we find $Im\mathcal{A}\subset Ker\mathcal{F}$. In what follows, we show that $Ker\mathcal{F}\subset Im\mathcal{A}$. To prove this, we consider $\mathfrak{v}\in Ker\mathcal{F}$ arbitrary but fixed for the moment. So, $\mathcal{F}v=0$ or ${\mathcal{G}}_1\mathfrak{v}\left(\theta \right)+\left({\mathcal{G}}_2\mathfrak{v}\left(\theta \right)\right)ln\theta =0$ for all $\theta \in [1,\infty )$. Then, we infer

$${\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathfrak{v}-\mathfrak{b}{\mathcal{C}}_2\mathfrak{v}){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathfrak{v}-\mathfrak{c}{\mathcal{C}}_1\mathfrak{v}){\displaystyle \frac{1}{{\theta }^2}}ln\theta =0,\forall \theta \in [1,\infty ).$$

Hence, we deduce the following homogeneous linear system in the unknowns of ${\mathcal{C}}_1\mathfrak{v}$ and ${\mathcal{C}}_2\mathfrak{v}$:

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

(25)

The determinant of the system (25) is $\Delta \ne 0$, and we conclude that this system has a zero solution (unique solution) of ${\mathcal{C}}_1\mathfrak{v}=0$ and ${\mathcal{C}}_2\mathfrak{v}=0$. This means that $\mathfrak{v}\in Im\mathcal{A}$, so $Ker\mathcal{F}\subset Im\mathcal{A}$. Then, we obtain $Ker\mathcal{F}=Im\mathcal{A}$.

With the aid of this last relation ($Ker\mathcal{F}=Im\mathcal{A}$), we find that $\mathcal{V}=Im\mathcal{A}\oplus Im\mathcal{F}$. Clearly, given a function $\mathfrak{v}\in \mathcal{V}$, we can formulate $\mathfrak{v}=(\mathfrak{v}-\mathcal{F}\mathfrak{v})+\mathcal{F}\mathfrak{v}$, so $\mathfrak{v}-\mathcal{F}\mathfrak{v}\in Ker\mathcal{F}=Im\mathcal{A}$. Hence, $\mathcal{F}\mathfrak{v}\in Im\mathcal{F}$. Hence, we derive $\mathfrak{v}\in Im\mathcal{A}+Im\mathcal{F}$. Furthermore, for any function $\mathfrak{v}\in Im\mathcal{A}\cap Im\mathcal{F}$, we find ${\mathcal{C}}_1\mathfrak{v}={\mathcal{C}}_2\mathfrak{v}=0$, and there exists $\mathfrak{w}\in D\left(\mathcal{F}\right)$ such that $\mathcal{F}\mathfrak{w}=\mathfrak{v}$. Hence, there exist $c_1,c_2\in \mathbb{R}$ such that $\mathfrak{v}\left(\theta \right)={\displaystyle \frac{c_1}{{\theta }^2}}+{\displaystyle \frac{c_2}{{\theta }^2}}ln\theta$, $\theta \in [1,\infty )$. So, we deduce the system as

$$\left\{\begin{array}{c}{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)\left({\displaystyle \frac{c_1}{{\tau }^2}}+{\displaystyle \frac{c_2}{{\tau }^2}}ln\tau \right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{P}\left(\sigma \right)=0,\hfill \\ {\int }_1^B\left({\int }_{\sigma }^{\infty }\left({\displaystyle \frac{c_1}{{\tau }^2}}+{\displaystyle \frac{c_2}{{\tau }^2}}ln\tau \right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{Q}\left(\sigma \right)=0,\hfill \end{array}\right.$$

or

$$\left\{\begin{array}{c}{c}_1{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{P}\left(\sigma \right)+c_2{\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{P}\left(\sigma \right)=0,\hfill \\ c_1{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{{\tau }^3}}d\tau \right)d\mathcal{Q}\left(\sigma \right)+c_2{\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{1}{{\tau }^3}}ln\tau d\tau \right)d\mathcal{Q}\left(\sigma \right)=0.\hfill \end{array}\right.$$

Therefore we find the following system:

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

which has the unique solution $c_1=c_2=0$. Hence, we deduce that $\mathfrak{v}=0$, that is, $Im\mathcal{A}\cap Im\mathcal{F}=\left\{0\right\}$. So, $\mathcal{V}=Im\mathcal{A}\oplus Im\mathcal{F}$, and all the relations from (24) hold.

Then, according to (24), we deduce that $dimKer\mathcal{A}=2$ and $codimIm\mathcal{A}=2$. Because $Im\mathcal{A}$ is closed, we infer that $\mathcal{A}$ is a Fredholm operator with the index $Ind\mathcal{A}=0$. □

 Lemma 4. 

Suppose that $\left(I1\right)$ holds. Then, the operator ${\mathcal{L}}_{\mathcal{E}}:Im\mathcal{A}\subset \mathcal{V}\to D\left(\mathcal{A}\right)\cap Ker\mathcal{E}\subset \mathcal{U}$, given by

$$\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(\theta \right)={}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(\theta \right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }},\mathfrak{v}\in Im\mathcal{A},$$

(26)

 represents the inverse of the operator ${\mathcal{A}|}_{D\left(\mathcal{A}\right)\cap Ker\mathcal{E}}$. In addition, we have $\parallel {\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_{\mathcal{U}}\le {\parallel \mathfrak{v}\parallel }_{\mathcal{V}}$ for all $\mathfrak{v}\in Im\mathcal{A}$.

Proof. 

We show that ${\mathcal{L}}_{\mathcal{E}}$ is a well-defined operator on $Im\mathcal{A}$. For $\mathfrak{v}\in Im\mathcal{A}$ (that is, $\mathfrak{v}\in \mathcal{V}$, $\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$), we first prove that ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\in D\left(\mathcal{A}\right)$.

Indeed, we have ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\in \mathcal{U}$ because

• ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\in C[1,\infty )$ and

  $$\begin{array}{c}\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}{1+{(ln\theta )}^{\alpha -1}}}\le \underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}·{\displaystyle \frac{1}{1+{(ln\theta )}^{\alpha -1}}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ =\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\displaystyle \frac{{(ln\theta -ln\sigma )}^{\alpha -1}}{1+{(ln\theta )}^{\alpha -1}}}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }{\displaystyle \frac{{(ln\theta )}^{\alpha -1}}{1+{(ln\theta )}^{\alpha -1}}}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ \le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma \le {\int }_1^{\infty }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma ={\parallel \mathfrak{v}\parallel }_{\mathcal{V}}<\infty ;\hfill \end{array}$$
• ${}^{H}D_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}={}^{H}D_{1+}^{\alpha -n+1} {}^{H}I_{1+}^{\alpha }\mathfrak{v}={}^{H}I_{1+}^{n-1}\mathfrak{v}\in C[1,\infty )$, and

  $$\begin{array}{c}\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{I}_{1+}^{n-1}\mathfrak{v}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}\hfill \\ \le \underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{1+{(ln\theta )}^{n-2}}}{\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{n-2}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ =\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_1^{\theta }{\displaystyle \frac{{(ln\theta -ln\sigma )}^{n-2}}{1+{(ln\theta )}^{n-2}}}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ \le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }{\displaystyle \frac{{(ln\theta )}^{n-2}}{1+{(ln\theta )}^{n-2}}}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma \hfill \\ \le {\int }_1^{\infty }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma ={\parallel v\parallel }_{\mathcal{V}}<\infty ;\hfill \end{array}$$
  $⋮$
• ${}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}={}^{H}D_{1+}^{\alpha -2} {}^{H}I_{1+}^{\alpha }\mathfrak{v}={}^{H}I_{1+}^2\mathfrak{v}\in C[1,\infty ),$ and

  $$\begin{array}{c}\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}{1+ln\theta }}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}I_{1+}^2\mathfrak{v}\left(\theta \right)|}{1+ln\theta }}\hfill \\ \le \underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{1+ln\theta }}·{\displaystyle \frac{1}{\Gamma \left(2\right)}}{\int }_1^{\theta }\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ =\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{\Gamma \left(2\right)}}{\int }_1^{\theta }{\displaystyle \frac{ln\theta -ln\sigma }{1+ln\theta }}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ \le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }{\displaystyle \frac{ln\theta }{1+ln\theta }}\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma \hfill \\ \le {\int }_1^{\infty }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma ={\parallel \mathfrak{v}\parallel }_{\mathcal{V}}<\infty ;\hfill \end{array}$$
• ${}^{H}D_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}={}^{H}D_{1+}^{\alpha -1}{}^{H}I_{1+}^{\alpha }\mathfrak{v}={}^{H}I_{1+}^1\mathfrak{v}\in C[1,\infty )$, and

  $$\begin{array}{c}\underset{\theta \ge 1}{sup}{|}^H{D}_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|=\underset{\theta \ge 1}{sup}{|}^H{I}_{1+}^1\mathfrak{v}\left(\theta \right)|\le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^0\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ =\underset{\theta \ge 1}{sup}{\int }_1^{\theta }\left|\mathfrak{v}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\le \underset{\theta \ge 1}{sup}{\int }_1^{\theta }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma \le {\int }_1^{\infty }\left|\mathfrak{v}\left(\sigma \right)\right|d\sigma ={\parallel v\parallel }_{\mathcal{V}}<\infty .\hfill \end{array}$$

So, ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\in \mathcal{U}$. In addition, we have ${}^{H}D_{1+}^{\alpha }{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}={}^{H}D_{1+}^{\alpha }{}^{H}I_{1+}^{\alpha }\mathfrak{v}=\mathfrak{v}\in \mathcal{V}$.

Additionally, ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}$ satisfies the conditions of (2). First, we find the following relations:

$${\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)}^{\left(i\right)}\left(1\right)=D^i {}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(1\right)=0,i=0,\dots ,n-3,$$

where $D^i$ is the classical derivative of order i. Indeed, for these relations, we obtain

• For $i=0$, we have

  $${\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)}^{\left(0\right)}\left(1\right)=\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)={}^{H}D_{1+}^{\alpha }\mathfrak{v}\left(1\right)=0;$$
• For $i=1$, we find

  $$\begin{array}{c}{\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)}^{\prime }\left(1\right)={\displaystyle \frac{d}{d\theta }}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)=\left(\theta {\displaystyle \frac{d}{d\theta }}\right)\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)=\delta \left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)\hfill \\ ={}^{H}D_{1+}^1\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)={}^{H}D_{1+}^1{}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(1\right)={}^{H}I_{1+}^{\alpha -1}\mathfrak{v}\left(1\right)=0;\hfill \end{array}$$
• For $i=2$, we have

  $$\begin{array}{c}{\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)}^{″}\left(1\right)={\displaystyle \frac{d^2}{d{\theta }^2}}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right){|}_{\theta =1}={\displaystyle \frac{d}{d\theta }}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right){{|}_{\theta =1}+{\displaystyle \frac{d^2}{d{\theta }^2}}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}_{\theta =1}\hfill \\ =\left(\theta {\displaystyle \frac{d}{d\theta }}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)+{\theta }^2{\displaystyle \frac{d^2}{d{\theta }^2}}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)\right){|}_{\theta =1}=\theta \left({\displaystyle \frac{d}{d\theta }}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)+\theta {\displaystyle \frac{d^2}{d{\theta }^2}}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)\right){|}_{\theta =1}\hfill \\ =\left(\theta {\displaystyle \frac{d}{d\theta }}\right)\left(\theta {\displaystyle \frac{d}{d\theta }}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(\theta \right){|}_{\theta =1}={\left(\theta {\displaystyle \frac{d}{d\theta }}\right)}^2\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)={\delta }^2\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)\hfill \\ ={}^{H}D_{1+}^2\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right){=}^H{D}_{1+}^2{}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(1\right)={}^{H}I_{1+}^{\alpha -2}\mathfrak{v}\left(1\right)=0;\hfill \end{array}$$
  $⋮$
• For $i=n-3$, we find

  $$\begin{array}{c}{\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)}^{(n-3)}\left(1\right)={\displaystyle \frac{d^{n-3}}{d{\theta }^{n-3}}}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right){|}_{\theta =1}=\cdots ={\left(\theta {\displaystyle \frac{d}{d\theta }}\right)}^{n-3}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)={\delta }^{n-3}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)\hfill \\ ={}^{H}D_{1+}^{n-3}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)={}^{H}D_{1+}^{n-3} {}^{H}I_{1+}^{\alpha }\mathfrak{v}\left(1\right)={}^{H}I_{1+}^{\alpha -n+3}\mathfrak{v}\left(1\right)=0.\hfill \end{array}$$

In fact, we can observe (as we proved before) that $u^{\left(i\right)}\left(1\right)=0$ for $i=0,\dots ,n-3$ if and only if ${}^{H}D_{1+}^{i}u\left(1\right)=\left({\delta }^{i}u\right)\left(1\right)={\left(\theta {\displaystyle \frac{d}{d\theta }}\right)}^{i}u\left(\theta \right){|}_{\theta =1}=0$ for $i=0,\dots ,n-3$.

In addition, we obtain

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -2}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(\theta \right)={}^{H}D_{1+}^{\alpha -2}\left({}^{H}I_{1+}^{\alpha }\mathfrak{v}\right)\left(\theta \right)={}^{H}I_{1+}^2\mathfrak{v}\left(\theta \right)={\int }_1^{\theta }\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }},\hfill \\ {}^{H}D_{1+}^{\alpha -1}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(\theta \right)={}^{H}D_{1+}^{\alpha -1}\left({}^{H}I_{1+}^{\alpha }\mathfrak{v}\right)\left(\theta \right)={}^{H}D_{1+}^1\mathfrak{v}\left(\theta \right)={\int }_1^{\theta }\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}.\hfill \end{array}$$

So, we deduce

$${}^{H}D_{1+}^{\alpha -2}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(1\right)=0,{}^{H}D_{1+}^{\alpha -1}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)(\infty )={\int }_1^{\infty }\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}.$$

Then, the last two conditions from (2) for ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}$, namely

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(1\right)={\int }_1^A{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\sigma \right)d\mathcal{P}\left(\sigma \right),and{}^{H}D_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}(\infty )={\int }_1^B{}^{H}D_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\sigma \right)d\mathcal{Q}\left(\sigma \right),\hfill \end{array}$$

can be written as

$$\begin{array}{c}0={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)\mathfrak{v}\left(\tau \right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{P}\left(\sigma \right),\hfill \end{array}$$

and

$$\begin{array}{c}{\int }_1^{\infty }\mathfrak{v}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}={\int }_1^B\left({\int }_1^{\sigma }\mathfrak{v}\left(\tau \right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff {\int }_1^B\left({\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)={\int }_1^B\left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ \iff {\int }_1^B\left({\int }_1^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau -{\int }_1^{\sigma }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0\hfill \\ \iff {\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathfrak{v}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)=0,\hfill \end{array}$$

respectively. The last relations are valid because $\left({\mathcal{C}}_1\mathfrak{v}\right)\left(\theta \right)=\left({\mathcal{C}}_2\mathfrak{v}\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$, ($\mathfrak{v}\in Im\mathcal{A}$).

By a simple argument, we obtain ${\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\in Ker\mathcal{E}$. Hence, we deduce that ${\mathcal{L}}_{\mathcal{E}}$ is a well-defined operator on $Im\mathcal{A}$.

We now show that ${\mathcal{L}}_{\mathcal{E}}$ represents the inverse of the operator ${\mathcal{A}|}_{D\left(\mathcal{A}\right)\cap Ker\mathcal{E}}$, that is,

 (a) 

$\mathcal{A}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(\theta \right)=\mathfrak{v}\left(\theta \right),\forall \mathfrak{v}\in Im\mathcal{A},\theta \in (1,\infty )$;

 (b) 

${\mathcal{L}}_{\mathcal{E}}\left(\mathcal{A}\mathfrak{u}\right)\left(\theta \right)=\mathfrak{u}\left(\theta \right),\forall \mathfrak{u}\in D\left(\mathcal{A}\right)\cap Ker\mathcal{E},\theta \in (1,\infty )$.

Relation $\left(a\right)$ is evident because

$$\mathcal{A}\left({\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\right)\left(\theta \right){=}^H{D}_{1+}^{\alpha }{(}^H{I}_{1+}^{\alpha }\mathfrak{v})\left(\theta \right)=\mathfrak{v}\left(\theta \right),\theta \in (1,\infty ),\mathfrak{v}\in Im\mathcal{A}.$$

To prove relation $\left(b\right)$, let $\mathfrak{u}\in D\left(\mathcal{A}\right)\cap Ker\mathcal{E}$. Then, according to Theorem 2.3 from [25], we have

$${\mathcal{L}}_{\mathcal{E}}\left(\mathcal{A}\mathfrak{u}\right)\left(\theta \right)={}^{H}I_{1+}^{\alpha }{(}^H{D}_{1+}^{\alpha }\mathfrak{u})\left(\theta \right)=\mathfrak{u}\left(\theta \right)+b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2}+\cdots +b_n{(ln\theta )}^{\alpha -n},$$

(27)

where $b_i\in \mathbb{R},i=1,\dots ,n$. Since $\mathfrak{u}\in D\left(\mathcal{A}\right)$, we obtain ${\mathfrak{u}}^{\left(j\right)}\left(1\right)=0$ for $j=0,\dots ,n-3$ and $b_j=0$ for $j=3,\dots ,n$. So, relation (27) gives us

$${\mathcal{L}}_{\mathcal{E}}\left(\mathcal{A}\mathfrak{u}\right)\left(\theta \right)=\mathfrak{u}\left(\theta \right)+b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2},\theta \in [1,\infty ),b_1,b_2\in \mathbb{R}.$$

(28)

Because $\mathfrak{u}\in Ker\mathcal{E}$, we derive $\mathcal{E}\mathfrak{u}=0$. So, according to (28), we find

$$\begin{array}{c}\mathcal{E}\left({\mathcal{L}}_{\mathcal{E}}\left(\mathcal{A}\mathfrak{u}\right)\right)\left(\theta \right)=\mathcal{E}\mathfrak{u}\left(\theta \right)+\mathcal{E}(b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2})\hfill \\ =\mathcal{E}(b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2})=b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2},\forall \theta \in [1,\infty ).\hfill \end{array}$$

(29)

In addition, we have $\mathcal{E}\left({\mathcal{L}}_{\mathcal{E}}\left(\mathcal{A}\mathfrak{u}\right)\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$ due to ${\mathcal{L}}_{\mathcal{E}}\left(\mathfrak{w}\right)\in Ker\mathcal{E}$ for all $\mathfrak{w}\in Im\mathcal{A}$. Hence, according to (29), we obtain $b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2}=0$ for all $\theta \in [1,\infty )$. Therefore, relation (28) gives us ${\mathcal{L}}_{\mathcal{E}}\left(\mathcal{A}\mathfrak{u}\right)\left(\theta \right)=\mathfrak{u}\left(\theta \right)$ for all $\theta \in [1,\infty )$.

Then, according to $a)$ and $b)$, we deduce that ${\mathcal{L}}_{\mathcal{E}}={\left(\mathcal{A}{|}_{D\left(\mathcal{A}\right)\cap Ker\mathcal{E}}\right)}^{-1}.$

Next, we prove that $\parallel {\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_{\mathcal{U}}\le {\parallel \mathfrak{v}\parallel }_{\mathcal{V}}$ for all $\mathfrak{v}\in Im\mathcal{A}$. For any $\mathfrak{v}\in Im\mathcal{A}$, we saw before that

• $\parallel {\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_{\alpha -1}={sup}_{\theta \ge 1}{\displaystyle \frac{|{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}{1+{(ln\theta )}^{\alpha -1}}}\le {\parallel \mathfrak{v}\parallel }_{\mathcal{V}},$
• ${\parallel }^H{D}_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_{n-2}={sup}_{\theta \ge 1}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}\le {\parallel \mathfrak{v}\parallel }_{\mathcal{V}},$
  $⋮$
• ${\parallel }^H{D}_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_1={sup}_{\theta \ge 1}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E}}\mathfrak{v}\left(\theta \right)|}{1+ln\theta }}\le {\parallel \mathfrak{v}\parallel }_{\mathcal{V}},$
• ${\parallel }^H{D}_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_0={sup}_{\theta \ge 1}{|}^H{D}_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\left(\theta \right)|\le \parallel \mathfrak{v}\parallel }_{\mathcal{V}}.$

Therefore, $\parallel {\mathcal{L}}_{\mathcal{E}}{\mathfrak{v}\parallel }_{\mathcal{U}}\le {\parallel \mathfrak{v}\parallel }_{\mathcal{V}}$ for any $\mathfrak{v}\in Im\mathcal{A}$.

□

 Lemma 5. 

Assume that $\left(I1\right)$ and $\left(I2\right)$ hold. Let $\Omega \subset \mathcal{U}$ be an open bounded set that satisfies the condition of $D\left(\mathcal{A}\right)\cap \overline{\Omega }\ne \varnothing$. Then, $\mathcal{B}$ is $\mathcal{A}$-compact on $\overline{\Omega }$.

Proof. 

We use the characterization of the $\mathcal{A}$-compact operator from [2], and we prove that

 (a) 

$\mathcal{F}\mathcal{B}:\overline{\Omega }\to \mathcal{V}$ is a continuous operator;

 (b) 

$\mathcal{F}\mathcal{B}\left(\overline{\Omega }\right)$ is a bounded set;

 (c) 

${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}={\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}:\overline{\Omega }\to \mathcal{U}$ is a continuous operator;

 (d) 

${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\left(\overline{\Omega }\right)$ is a relatively compact set in $\mathcal{U}$.

 (a) 

We obtain

$$\begin{array}{c}\mathcal{F}\mathcal{B}\left(\mathfrak{u}\right)\left(\theta \right)={\mathcal{G}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)+\left({\mathcal{G}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)\right)ln\theta \hfill \\ ={\displaystyle \frac{1}{\Delta }}\left(\mathfrak{d}{\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)-\mathfrak{b}{\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)\right){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}\left(\mathfrak{a}{\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)-\mathfrak{c}{\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)\right){\displaystyle \frac{1}{{\theta }^2}}ln\theta ,\hfill \end{array}$$

where

$$\begin{array}{c}{\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}\right)d\mathcal{P}\left(\sigma \right)\hfill \\ ={\int }_1^A\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)\mathfrak{f}\left(\tau ,\mathfrak{u}\left(\tau \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\tau \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\tau \right)\right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{P}\left(\sigma \right),\hfill \\ {\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)={\int }_1^B\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ ={\int }_1^B\left({\int }_{\sigma }^{\infty }\mathfrak{f}\left(\tau ,\mathfrak{u}\left(\tau \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\tau \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\tau \right)\right){\displaystyle \frac{d\tau }{\tau }}\right)d\mathcal{Q}\left(\sigma \right),\hfill \end{array}$$

for all $\theta \in [1,\infty )$ and $\mathfrak{u}\in \overline{\Omega }$. The $\mathfrak{f}$ function is continuous in the last n variables; then, by way of the Lebesgue dominated convergence theorem, we infer that the ${\mathcal{C}}_1\mathcal{B}$ and ${\mathcal{C}}_2\mathcal{B}$ operators are continuous. We conclude that $\mathcal{F}\mathcal{B}$ is a continuous operator.

 (b) 

First, we observe that there exists $C>0$ such that ${\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\le C$ for all $\mathfrak{u}\in \overline{\Omega }$. Then, according to $\left(I2\right)$ and (20), we obtain

$$\begin{array}{c}|{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)|=\left|{\int }_1^A\left({\int }_{\tau }^A\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)d\mathcal{P}\left(\sigma \right)\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right|\hfill \\ \le {\int }_1^A\left|{\int }_{\tau }^A\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)d\mathcal{P}\left(\sigma \right)\right|·\left|\mathfrak{f}\left(\tau ,\mathfrak{u}\left(\tau \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\tau \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\tau \right)\right)\right|{\displaystyle \frac{d\tau }{\tau }}\hfill \\ \le {\int }_1^A\left|{\int }_{\tau }^A\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)d\mathcal{P}\left(\sigma \right)\right|·\left(p_1\left(\tau \right){\displaystyle \frac{\left|\mathfrak{u}\right(\tau \left)\right|}{1+{(ln\tau )}^{\alpha -1}}}+p_2\left(\tau \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\tau \right)|}{1+{(ln\tau )}^{n-2}}}\right.\hfill \\ +\cdots +\left.p_{n-1}\left(\tau \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\tau \right)|}{1+ln\tau }}+p_n\left(\tau \right){|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\tau \right)|+p_{n+1}\left(\tau \right)\right){\displaystyle \frac{d\tau }{\tau }}\hfill \\ \le {\int }_1^A\left|{\int }_{\tau }^A\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)d\mathcal{P}\left(\sigma \right)\right|\left(\underset{\tau \ge 1}{max}{\displaystyle \frac{\left|\mathfrak{u}\right(\tau )}{1+{(ln\tau )}^{\alpha -1}}}{p}_1\left(\tau \right)+\underset{\tau \ge 1}{max}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\tau \right)|}{1+{(ln\tau )}^{n-2}}}{p}_2\left(\tau \right)\right.\hfill \\ +\cdots +\left.\underset{\tau \ge 1}{max}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\tau \right)|}{1+ln\tau }}{p}_{n-1}\left(\tau \right)+\underset{\tau \ge 1}{max}{|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\tau \right)|p_n\left(\tau \right)+p_{n+1}\left(\tau \right)\right){\displaystyle \frac{d\tau }{\tau }}.\hfill \end{array}$$

Because

$$\left|{\int }_{\tau }^A\left(ln{\displaystyle \frac{\sigma }{\tau }}\right)d\mathcal{P}\left(\sigma \right)\right|\le \left(\underset{\sigma \in [\tau ,A]}{sup}\left|ln{\displaystyle \frac{\sigma }{\tau }}\right|\right)\underset{\tau }{\stackrel{A}{V}}\left(\mathcal{P}\right)=\left(ln{\displaystyle \frac{A}{\tau }}\right)\underset{1}{\stackrel{A}{V}}\left(\mathcal{P}\right)\le (lnA)\underset{1}{\stackrel{A}{V}}\left(\mathcal{P}\right),$$

where $\underset{1}{\stackrel{A}{V}}\left(\mathcal{P}\right)$ is the total variation of function $\mathcal{P}$ on the interval of $[1,A]$, the above inequalities give us

$$\begin{array}{c}|{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)|\le (lnA)\underset{1}{\stackrel{A}{V}}\left(\mathcal{P}\right){\int }_1^{\infty }\left({\parallel \mathfrak{u}\parallel }_{\mathcal{U}}(p_1\left(\tau \right)+\cdots +p_n\left(\tau \right))+p_{n+1}\left(\tau \right)\right)d\tau \hfill \\ \le (lnA)\underset{1}{\stackrel{A}{V}}\left(\mathcal{P}\right)\left({\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}\right):=M_1,\hfill \end{array}$$

where ${\Xi }_0={\sum }_{i=1}^n\parallel p_i{\parallel }_{\mathcal{V}}={\sum }_{i=1}^n{\int }_1^{\infty }\left|p_i\left(\tau \right)\right|d\tau$ and $\parallel p_{n+1}{\parallel }_{\mathcal{V}}={\int }_1^{\infty }\left|p_{n+1}\left(\tau \right)\right|d\tau .$

Next, according to $\left(I2\right)$ and (21), we find

$$\begin{array}{c}|{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)|=\left|{\int }_1^B(\mathcal{Q}\left(\sigma \right)-\mathcal{Q}\left(1\right)){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\sigma \right)}{\sigma }}d\sigma +{\int }_B^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\sigma \right)}{\sigma }}d\sigma \right|\hfill \\ \le 2\underset{\sigma \in [1,B]}{sup}\left|\mathcal{Q}\left(\sigma \right)\right|{\int }_1^B{\displaystyle \frac{\left|\mathcal{B}\mathfrak{u}\right(\sigma \left)\right|}{\sigma }}d\sigma +{\int }_B^{\infty }{\displaystyle \frac{\left|\mathcal{B}\mathfrak{u}\right(\sigma \left)\right|}{\sigma }}d\sigma \le M_0{\int }_1^{\infty }{\displaystyle \frac{\left|\mathcal{B}\mathfrak{u}\right(\sigma \left)\right|}{\sigma }}d\sigma \hfill \\ \le M_0{\int }_1^{\infty }\left|\mathfrak{f}\left(\sigma ,\mathfrak{u}\left(\sigma \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)\right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ \le M_0{\int }_1^{\infty }\left(p_1\left(\sigma \right){\displaystyle \frac{\left|\mathfrak{u}\right(\sigma \left)\right|}{1+{(ln\sigma )}^{\alpha -1}}}+p_2\left(\sigma \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right)|}{1+{(ln\sigma )}^{n-2}}}\right.\hfill \\ \left.+\cdots +p_{n-1}\left(\sigma \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)|}{1+ln\sigma }}+p_n\left(\sigma \right){|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)|+p_{n+1}\left(\sigma \right)\right)d\sigma \hfill \\ \le M_0{\int }_1^{\infty }\left(p_1\left(\sigma \right)\underset{\sigma \ge 1}{max}{\displaystyle \frac{\left|\mathfrak{u}\right(\sigma \left)\right|}{1+{(ln\sigma )}^{\alpha -1}}}+p_2\left(\sigma \right)\underset{\sigma \ge 1}{max}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right)|}{1+{(ln\sigma )}^{n-2}}}\right.\hfill \\ +\left.\dots +p_{n-1}\left(\sigma \right)\underset{\sigma \ge 1}{max}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)|}{1+ln\sigma }}+p_n\left(\sigma \right)\underset{\sigma \ge 1}{max}{|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)|+p_{n+1}\left(\sigma \right)\right)d\sigma \hfill \\ \le M_0{\int }_1^{\infty }\left({\parallel \mathfrak{u}\parallel }_{\mathcal{U}}(p_1\left(\sigma \right)+p_2\left(\sigma \right)+\dots +p_n\left(\sigma \right))+p_{n+1}\left(\sigma \right)\right)d\sigma \hfill \\ \le M_0\left({\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}\right):=M_2,\hfill \end{array}$$

where $M_0=max\left\{2{sup}_{\sigma \in [1,B]}\left|\mathcal{Q}\left(\sigma \right)\right|,1\right\}$.

Hence, based on the above inequalities, we deduce

$$\begin{array}{c}\left|\mathcal{F}\mathcal{B}\mathfrak{u}\left(\theta \right)\right|\le {\displaystyle \frac{1}{|\Delta |}}\left(\right|\mathfrak{d}|M_1+\left|\mathfrak{b}\right|M_2){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{|\Delta |}}\left(\right|\mathfrak{a}|M_2+\left|\mathfrak{c}\right|M_1){\displaystyle \frac{1}{{\theta }^2}}ln\theta ,\hfill \\ \forall \theta \in [1,\infty ),\mathfrak{u}\in \overline{\Omega }.\hfill \end{array}$$

(30)

Therefore, we conclude

$$\begin{array}{c}{\parallel \mathcal{F}\mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}={\int }_1^{\infty }\left|\mathcal{F}\mathcal{B}\mathfrak{u}\left(\theta \right)\right|d\theta \le {\int }_1^{\infty }{\displaystyle \frac{1}{|\Delta |}}\left(\right|\mathfrak{d}|M_1+\left|\mathfrak{b}\right|M_2){\displaystyle \frac{1}{{\theta }^2}}d\theta \hfill \\ +{\int }_1^{\infty }{\displaystyle \frac{1}{|\Delta |}}\left(\right|\mathfrak{a}|M_2+\left|\mathfrak{c}\right|M_1){\displaystyle \frac{1}{{\theta }^2}}ln\theta d\theta \hfill \\ ={\displaystyle \frac{1}{|\Delta |}}\left[\right(\left|\mathfrak{d}\right|+\left|\mathfrak{c}\right|)M_1+\left(\right|\mathfrak{b}|+\left|\mathfrak{a}\right|)M_2]:=M_3,\forall \mathfrak{u}\in \overline{\Omega }.\hfill \end{array}$$

So, we infer that set $\mathcal{F}\mathcal{B}\left(\overline{\Omega }\right)$ is bounded.

 (c) 

We remark that $(I-\mathcal{F})\mathcal{B}$ is a continuous operator on $\overline{\Omega }$. In addition, we find

$$\begin{array}{c}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\left(\mathfrak{u}\right)={\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}\left[\mathcal{B}\mathfrak{u}\left(\sigma \right)-{\mathcal{G}}_1\mathcal{B}\mathfrak{u}\left(\sigma \right)-\left({\mathcal{G}}_2\mathcal{B}\mathfrak{u}\left(\sigma \right)\right)ln\sigma \right]{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{\alpha -1}\left[\mathcal{B}\mathfrak{u}\left(\sigma \right)-{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)-\mathfrak{b}{\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)){\displaystyle \frac{1}{{\sigma }^2}}\right.\hfill \\ -\left.{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)-\mathfrak{c}{\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)){\displaystyle \frac{1}{{\sigma }^2}}ln\sigma \right]{\displaystyle \frac{d\sigma }{\sigma }}.\hfill \end{array}$$

With the aid of the Lebesgue dominated convergence theorem for ${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\left(\mathfrak{u}\right)$ and its fractional derivatives, ${}^{H}D_{1+}^{\alpha -1}\left({\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\right)\left(\mathfrak{u}\right)={}^{H}I_{1+}^1(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)$, ${}^{H}D_{1+}^{\alpha -2}\left({\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\right)\left(\mathfrak{u}\right)=$${}^{H}I_{1+}^2(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)$, …, and ${}^{H}D_{1+}^{\alpha -n+1}\left({\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\right)\left(\mathfrak{u}\right)={}^{H}I_{1+}^{n-1}(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)$, we conclude that ${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}$ is a continuous operator in space $\mathcal{U}$.

 (d) 

First, we prove that ${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}:\overline{\Omega }\to \mathcal{U}$ is bounded. In fact, for any $\mathfrak{u}\in \overline{\Omega }$, we have

$$\begin{array}{c}{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}={\int }_1^{\infty }\left|\mathfrak{f}\left(\sigma ,\mathfrak{u}\left(\sigma \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)\right)\right|d\sigma \hfill \\ \le {\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}:=M_4.\hfill \end{array}$$

Then by use of Lemma 4, we find

$$\parallel {\mathcal{L}}_{\mathcal{E}}{(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)\parallel }_{\mathcal{U}}\le {\parallel (I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)\parallel }_{\mathcal{V}}\le {\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}+{\parallel \mathcal{F}\mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\le M_4+M_3.$$

Then the functions ${\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right)$ and $\mathfrak{u}\in \overline{\Omega }$ are uniformly bounded in $\mathcal{U}$. Hence, ${\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\left(\overline{\Omega }\right)$ is a bounded set in $\mathcal{U}$.

Next, we prove that $\{{\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right),\mathfrak{u}\in \overline{\Omega }\}$ are equicontinuous on any subcompact interval of $[1,\infty )$. Indeed, for $\mathfrak{u}\in \overline{\Omega }$, according to $\left(I2\right)$, we find

$$\begin{array}{c}\left|\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|=\left|\mathfrak{f}(\sigma ,\mathfrak{u}\left(\sigma \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right))\right|\hfill \\ \le p_1\left(\sigma \right){\displaystyle \frac{\left|\mathfrak{u}\right(\sigma \left)\right|}{1+{(ln\sigma )}^{\alpha -1}}}+p_2\left(\sigma \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right)|}{1+{(ln\sigma )}^{n-2}}}+\dots +p_{n-1}\left(\sigma \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)|}{1+ln\sigma }}\hfill \\ +p_n\left(\sigma \right){|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)|+p_{n+1}\left(\sigma \right),\mathrm{a}.\mathrm{e}.\sigma \in (1,\infty ).\hfill \end{array}$$

Then, using similar inequalities as those obtained in part $\left(b\right)$, we deduce

$$\begin{array}{c}\left|\mathcal{F}\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|=|{\mathcal{G}}_1\mathcal{B}\mathfrak{u}\left(\sigma \right)+\left({\mathcal{G}}_2\mathcal{B}\mathfrak{u}\left(\sigma \right)\right)ln\sigma |\hfill \\ \le {\displaystyle \frac{1}{|\Delta |}}\left|(\mathfrak{d}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\sigma \right)-\mathfrak{b}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\sigma \right)){\displaystyle \frac{1}{{\sigma }^2}}\right|+{\displaystyle \frac{1}{|\Delta |}}\left|(\mathfrak{a}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\sigma \right)-\mathfrak{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\sigma \right)){\displaystyle \frac{ln\sigma }{{\sigma }^2}}\right|\hfill \\ \le {\displaystyle \frac{1}{|\Delta |}}\left[\left(\right|\mathfrak{d}|·|{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\sigma \right)|+|\mathfrak{b}|·|{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\sigma \right)\left|\right)+\left(\right|\mathfrak{a}|·|{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\sigma \right)|+|\mathfrak{c}|·|{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\sigma \right)\left|\right)ln\sigma \right]{\displaystyle \frac{1}{{\sigma }^2}}\hfill \\ \le {\displaystyle \frac{1}{|\Delta |}}\left[\right(\left|\mathfrak{d}\right|M_1+\left|\mathfrak{b}\right|M_2)+(\left|\mathfrak{a}\right|M_2+\left|\mathfrak{c}\right|M_1)ln\sigma ]{\displaystyle \frac{1}{{\sigma }^2}},\forall \sigma \in [1,\infty ),\hfill \end{array}$$

and

$$\begin{array}{c}\left|\right(I-\mathcal{F}\left)\mathcal{B}\mathfrak{u}\right(\sigma \left)\right|=\left|\mathcal{B}\mathfrak{u}\right(\sigma )-\mathcal{F}\mathcal{B}\mathfrak{u}(\sigma \left)\right|\le \left|\mathcal{B}\mathfrak{u}\right(\sigma \left)\right|+\left|\mathcal{F}\mathcal{B}\mathfrak{u}\right(\sigma \left)\right|\hfill \\ \le [p_1\left(\sigma \right){\displaystyle \frac{\left|\mathfrak{u}\right(\sigma \left)\right|}{1+{(ln\sigma )}^{\alpha -1}}}+p_2\left(\sigma \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\sigma \right)|}{1+{(ln\sigma )}^{n-2}}}+\cdots +p_{n-1}\left(\sigma \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\sigma \right)|}{1+ln\sigma }}\hfill \\ +p_n\left(\sigma \right){|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)|+p_{n+1}\left(\sigma \right)]\hfill \\ +{\displaystyle \frac{1}{|\Delta |}}\left[\right(\left|\mathfrak{d}\right|M_1+\left|\mathfrak{b}\right|M_2)+(\left|\mathfrak{a}\right|M_2+\left|\mathfrak{c}\right|M_1)ln\sigma ]{\displaystyle \frac{1}{{\sigma }^2}}.\hfill \end{array}$$

Now, let R be a finite positive constant from $(1,\infty )$. Then, for any ${\theta }_1,{\theta }_2\in [1,R]$ (we assume, without loss of generality, that ${\theta }_1<{\theta }_2$), we obtain

$$\begin{array}{c}\left|{\displaystyle \frac{{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+{(ln{\theta }_1)}^{\alpha -1}}}-{\displaystyle \frac{{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+{(ln{\theta }_2)}^{\alpha -1}}}\right|\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left|{\displaystyle \frac{1}{1+{(ln{\theta }_1)}^{\alpha -1}}}{\int }_1^{{\theta }_1}{\left(ln{\displaystyle \frac{{\theta }_1}{\sigma }}\right)}^{\alpha -1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right.\hfill \\ -\left.{\displaystyle \frac{1}{1+{(ln{\theta }_2)}^{\alpha -1}}}{\int }_1^{{\theta }_2}{\left(ln{\displaystyle \frac{{\theta }_2}{\sigma }}\right)}^{\alpha -1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left|{\int }_1^{{\theta }_2}{\displaystyle \frac{1}{1+{(ln{\theta }_2)}^{\alpha -1}}}{\left(ln{\displaystyle \frac{{\theta }_2}{\sigma }}\right)}^{\alpha -1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right.\hfill \\ -\left.{\int }_1^{{\theta }_1}{\displaystyle \frac{1}{1+{(ln{\theta }_2)}^{\alpha -1}}}{\left(ln{\displaystyle \frac{{\theta }_2}{\sigma }}\right)}^{\alpha -1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ +{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left|{\int }_1^{{\theta }_1}{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}-{\int }_1^{{\theta }_1}{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left|{\int }_{{\theta }_1}^{{\theta }_2}{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ +{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left|{\int }_1^{{\theta }_1}\left({\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}-{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}\right)(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_{{\theta }_1}^{{\theta }_2}\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}+{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{{\theta }_1}\left|{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}-{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}\right|\hfill \\ \times \left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|d\sigma \to 0,\mathrm{as}{\theta }_1\to {\theta }_2.\hfill \end{array}$$

In a similar manner, we find

$$\begin{array}{c}\left|{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+{(ln{\theta }_1)}^{n-2}}}-{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+{(ln{\theta }_2)}^{n-2}}}\right|\hfill \\ =\left|{\displaystyle \frac{{}^{H}I_{1+}^{n-1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+{(ln{\theta }_1)}^{n-2}}}-{\displaystyle \frac{{}^{H}I_{1+}^{n-1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+{(ln{\theta }_2)}^{n-2}}}\right|\to 0,\mathrm{as}{\theta }_1\to {\theta }_2,\hfill \\ ⋮\hfill \\ \left|{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+ln{\theta }_1}}-{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+ln{\theta }_2}}\right|\hfill \\ =\left|{\displaystyle \frac{{}^{H}I_{1+}^2(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+ln{\theta }_1}}-{\displaystyle \frac{{}^{H}I_{1+}^2(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+ln{\theta }_2}}\right|\to 0,\mathrm{as}{\theta }_1\to {\theta }_2,\hfill \\ {|}^H{D}_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right){-}^H{D}_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)|\hfill \\ =|{}^{H}I_{1+}^1(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_1\right)-{}^{H}I_{1+}^1(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_2\right)|\hfill \\ \le {\int }_{{\theta }_1}^{{\theta }_2}\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|d\sigma \to 0,\mathrm{as}{\theta }_1\to {\theta }_2.\hfill \end{array}$$

Hence, we infer that $\{{\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\left(\mathfrak{u}\right),\mathfrak{u}\in \overline{\Omega }\}$ (or ${\mathcal{L}}_{\mathcal{E}}(I-\mathcal{F})\mathcal{B}\left(\overline{\Omega }\right)$) is a family of equicontinuous functions on $[1,R]$.

Finally, we show that ${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\left(\overline{\Omega }\right)$ is equiconvergent at infinity. For any $\mathfrak{u}\in \overline{\Omega }$, we have

$${\int }_1^{\infty }\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\theta \right)\right|d\theta \le {\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}+{\parallel \mathcal{F}\mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\le M_4+M_3.$$

Hence, for $\epsilon >0$, there exists a constant $T_1>1$ such that

$${\int }_{T_1}^{\infty }\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\theta \right)\right|d\theta <\epsilon .$$

(31)

We also have

$$\underset{\theta \to \infty }{lim}{\displaystyle \frac{{\left(ln\frac{\theta }{T_1}\right)}^{\alpha -1}}{1+{(ln\theta )}^{\alpha -1}}}=1,\underset{\theta \to \infty }{lim}{\displaystyle \frac{{\left(ln\frac{\theta }{T_1}\right)}^{n-2}}{1+{(ln\theta )}^{n-2}}}=1,\dots ,\underset{\theta \to \infty }{lim}{\displaystyle \frac{ln\frac{\theta }{T_1}}{1+ln\theta }}=1.$$

So, for $\epsilon >0$ above, there exists a constant $T_2>T_1>1$ such that for any ${\theta }_1,{\theta }_2\ge T_2$ and $1\le \sigma \le T_1$, we obtain

$$\begin{array}{c}\left|{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}-{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}\right|\hfill \\ \le \left(1-{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{T_1}\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}\right)+\left(1-{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{T_1}\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}\right)<\epsilon ,\hfill \\ \left|{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{n-2}}{1+{(ln{\theta }_1)}^{n-2}}}-{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{n-2}}{1+{(ln{\theta }_2)}^{n-2}}}\right|\hfill \\ \le \left(1-{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{T_1}\right)}^{n-2}}{1+{(ln{\theta }_1)}^{n-2}}}\right)+\left(1-{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{T_1}\right)}^{n-2}}{1+{(ln{\theta }_2)}^{n-2}}}\right)<\epsilon ,\hfill \\ ⋮\hfill \\ \left|{\displaystyle \frac{ln\frac{{\theta }_1}{\sigma }}{1+ln{\theta }_1}}-{\displaystyle \frac{ln\frac{{\theta }_2}{\sigma }}{1+ln{\theta }_2}}\right|\le \left(1-{\displaystyle \frac{ln\frac{{\theta }_1}{T_1}}{1+ln{\theta }_1}}\right)+\left(1-{\displaystyle \frac{ln\frac{{\theta }_2}{T_1}}{1+ln{\theta }_2}}\right)<\epsilon .\hfill \end{array}$$

(32)

Thus, for any ${\theta }_1,{\theta }_2\ge T_2$, according to (31) and (32), we deduce

$$\begin{array}{c}\left|{\displaystyle \frac{{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+{(ln{\theta }_1)}^{\alpha -1}}}-{\displaystyle \frac{{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+{(ln{\theta }_2)}^{\alpha -1}}}\right|\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left|{\int }_1^{{\theta }_1}{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}-{\int }_1^{{\theta }_2}{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{T_1}\left|{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}-{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}\right|·\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ +{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_{T_1}^{{\theta }_1}{\displaystyle \frac{{\left(ln\frac{{\theta }_1}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_1)}^{\alpha -1}}}\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ +{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_{T_1}^{{\theta }_2}{\displaystyle \frac{{\left(ln\frac{{\theta }_2}{\sigma }\right)}^{\alpha -1}}{1+{(ln{\theta }_2)}^{\alpha -1}}}\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ \le {\displaystyle \frac{\epsilon }{\Gamma \left(\alpha \right)}}{\int }_1^{T_1}\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}+{\displaystyle \frac{2}{\Gamma \left(\alpha \right)}}{\int }_{T_1}^{\infty }\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ <{\displaystyle \frac{\epsilon }{\Gamma \left(\alpha \right)}}(M_4+M_3)+{\displaystyle \frac{2\epsilon }{\Gamma \left(\alpha \right)}}\le \epsilon (M_4+M_3+2).\hfill \end{array}$$

In a similar manner, we obtain

$$\begin{array}{c}\left|{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+{(ln{\theta }_1)}^{n-2}}}-{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -n+1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+{(ln{\theta }_2)}^{n-2}}}\right|\hfill \\ =\left|{\displaystyle \frac{{}^{H}I_{1+}^{n-1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+{(ln{\theta }_1)}^{n-2}}}-{\displaystyle \frac{{}^{H}I_{1+}^{n-1}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+{(ln{\theta }_2)}^{n-2}}}\right|\hfill \\ \le {\displaystyle \frac{\epsilon }{\Gamma (n-1)}}(M_4+M_3+2)\le \epsilon (M_4+M_3+2),\hfill \\ ⋮\hfill \\ \left|{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+ln{\theta }_1}}-{\displaystyle \frac{{}^{H}D_{1+}^{\alpha -2}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+ln{\theta }_2}}\right|\hfill \\ =\left|{\displaystyle \frac{{}^{H}I_{1+}^2(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_1\right)}{1+ln{\theta }_1}}-{\displaystyle \frac{{}^{H}I_{1+}^2(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_2\right)}{1+ln{\theta }_2}}\right|\hfill \\ \le {\displaystyle \frac{\epsilon }{\Gamma \left(2\right)}}(M_4+M_3+2)=\epsilon (M_4+M_3+2),\hfill \\ |{}^{H}D_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_1\right)-{}^{H}D_{1+}^{\alpha -1}{\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\mathfrak{u}\left({\theta }_2\right)|\hfill \\ =|{}^{H}I_{1+}^1(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_1\right)-{}^{H}I_{1+}^1(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left({\theta }_2\right)|\hfill \\ =\left|{\int }_{{\theta }_1}^{{\theta }_2}(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\le {\int }_{T_1}^{\infty }\left|(I-\mathcal{F})\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|d\sigma <\epsilon .\hfill \end{array}$$

Thus, we deduce that ${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\left(\overline{\Omega }\right)$ is equiconvergent at infinity.

Therefore, by using Lemma 1, we conclude that ${\mathcal{L}}_{\mathcal{E},\mathcal{F}}\mathcal{B}\left(\overline{\Omega }\right)$ is relatively compact in $\mathcal{U}$. Hence, according to $\left(a\right)-\left(d\right)$, we find that operator $\mathcal{B}$ is $\mathcal{A}$-compact on $\overline{\Omega }$. □

3. Main Results

In this section, we state our main theorems concerning the existence of solutions to problem (1), (2).

 Theorem 1. 

Assume that $\left(I1\right)$ and $\left(I2\right)$ hold. We also suppose that

 (I3 )

There exist positive constants $E>0$ and $F>1$ such that for all $\mathfrak{u}\in D\left(\mathcal{A}\right)\setminus Ker\mathcal{A}$, if one of the following conditions is satisfied, then either ${\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\ne 0$ or ${\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\ne 0$:

${\left(i\right)|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|>E$ for any $\theta \in [1,F]$;

${\left(ii\right)|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|>E$ for any $\theta \in [1,\infty )$.

 (I4) 

There exists a positive constant r such that for any $b_1,b_2\in \mathbb{R}$ satisfying $|b_1|>r$ or $|b_2|>r$ and $\mathfrak{u}\left(\theta \right)=b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2}$, then either

$$b_1{\mathcal{C}}_1\mathcal{B}\left(\mathfrak{u}\left(\theta \right)\right)+b_2{\mathcal{C}}_2\mathcal{B}\left(\mathfrak{u}\left(\theta \right)\right)<0,$$

(33)

$$or{b}_1{\mathcal{C}}_1\mathcal{B}\left(\mathfrak{u}\left(\theta \right)\right)+b_2{\mathcal{C}}_2\mathcal{B}\left(\mathfrak{u}\left(\theta \right)\right)>0.$$

(34)

 (I5) 

$${\Xi }_0(2+{\Xi }_{1}lnF)<1,$$

(35)

where ${\Xi }_0={\sum }_{i=1}^n{\parallel p_i\parallel }_{\mathcal{V}}$ and ${\Xi }_1=max\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},1\}$.

Then, the boundary value problem (1), (2) has at least one solution in space $\mathcal{U}$.

The proof of Theorem 1 is based on the following three lemmas.

 Lemma 6. 

We assume that $\left(I1\right)$, $\left(I2\right)$, $\left(I3\right)$, and $\left(I5\right)$ hold. Then, the set

$${\Lambda }_1=\{\mathfrak{u}\in D\left(\mathcal{A}\right)\setminus Ker\mathcal{A};\mathcal{A}\mathfrak{u}=\nu \mathcal{B}\mathfrak{u},\nu \in (0,1)\}\subset \mathcal{U}$$

 is bounded in $\mathcal{U}$.

Proof. 

For $\mathfrak{u}\in {\Lambda }_1$, $\mathcal{B}\mathfrak{u}\in Im\mathcal{A}$, which implies ${\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)=0$ and ${\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$. Hence, according to assumption $\left(I3\right)$, we find that there exist ${\theta }_1\in [1,F]$ and ${\theta }_2\in [1,\infty )$ such that ${|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_1\right)|\le E$ and ${|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_2\right)|\le E$.

By using Lemma 4 and Remark 1 from [3], with $\alpha$ replaced by $\alpha -1$ and $\alpha$ replaced by $\alpha -2$, we deduce

$$\begin{array}{c}|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|=\left|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_2\right)+{\int }_{{\theta }_2}^{\theta }{(}^H{D}_{1+}^{\alpha }\mathfrak{u})\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ \le \left|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_2\right)\right|+{\int }_{{\theta }_2}^{\theta }\left|{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ \le E+{\int }_1^{\infty }\left|\mathcal{B}\mathfrak{u}\left(\sigma \right)\right|d\sigma \le E+{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}},\forall \theta \in [1,\infty ),\hfill \end{array}$$

and

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|=\left|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_1\right)+{\int }_{{\theta }_1}^{\theta }{(}^H{D}_{1+}^{\alpha -1}\mathfrak{u})\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ \le |{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_1\right)|+{\int }_{{\theta }_1}^{\theta }\left|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ {\le E+\parallel }^H{D}_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0\left|{\int }_{{\theta }_1}^{\theta }{\displaystyle \frac{d\sigma }{\sigma }}\right|\le E+{\parallel {}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\parallel }_0\left|ln{\displaystyle \frac{\theta }{{\theta }_1}}\right|,\forall \theta \in [1,\infty ).\hfill \end{array}$$

So, we obtain

$$\begin{array}{c}|{}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\left(1\right)|\le E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}},\hfill \\ |{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)|\le E+\parallel {}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0\left|ln{\displaystyle \frac{1}{{\theta }_1}}\right|\hfill \\ =E+\parallel {}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_{0}ln{\theta }_1\le E+lnF{\parallel {}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\parallel }_0\hfill \\ \le {E+lnF(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}{)=E(1+lnF)+lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}.\hfill \end{array}$$

Hence, according to the definition of $\mathcal{E}$, we infer

$$\begin{array}{c}{\parallel \mathcal{E}\mathfrak{u}\parallel }_{\alpha -1}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|\mathcal{E}\mathfrak{u}\right(\theta \left)\right|}{1+{(ln\theta )}^{\alpha -1}}}\hfill \\ =\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{1+{(ln\theta )}^{\alpha -1}}}\left|{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right){(ln\theta )}^{\alpha -1}+{\displaystyle \frac{1}{\Gamma (\alpha -1)}}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right){(ln\theta )}^{\alpha -2}\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)|\underset{\theta \ge 1}{sup}{\displaystyle \frac{{(ln\theta )}^{\alpha -1}}{1+{(ln\theta )}^{\alpha -1}}}+{\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)\right|\underset{\theta \ge 1}{sup}{\displaystyle \frac{{(ln\theta )}^{\alpha -2}}{1+{(ln\theta )}^{\alpha -1}}}\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right].\hfill \end{array}$$

Because ${}^{H}D_{1+}^{\alpha -1}\mathcal{E}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)$ and ${}^{H}D_{1+}^{\alpha -2}\mathcal{E}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)ln\theta +{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)$, we obtain

$$\begin{array}{c}\parallel {}^{H}D_{1+}^{\alpha -1}{\mathcal{E}\mathfrak{u}\parallel }_0=\underset{\theta \ge 1}{sup}|{}^{H}D_{1+}^{\alpha -1}\mathcal{E}\mathfrak{u}\left(\theta \right)|=\underset{\theta \ge 1}{sup}|{}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\left(1\right)|\le E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}},\hfill \\ \parallel {}^{H}D_{1+}^{\alpha -2}{\mathcal{E}\mathfrak{u}\parallel }_1=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -2}\mathcal{E}\mathfrak{u}\left(\theta \right)|}{1+ln\theta }}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)ln\theta +{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)|}{1+ln\theta }}\hfill \\ \le |{}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\left(1\right)|+|}^H{D}_{1+}^{\alpha -2}{\mathfrak{u}\left(1\right)|\le E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}+E(1+lnF)+lnF{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(2\right)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma \left(1\right)}}{[E(1+lnF)+lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}].\hfill \end{array}$$

Next, we find

$${}^{H}D_{1+}^{\alpha -3}\mathcal{E}\mathfrak{u}\left(\theta \right)={\displaystyle \frac{1}{\Gamma \left(3\right)}}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right){(ln\theta )}^2+{\displaystyle \frac{1}{\Gamma \left(2\right)}}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)ln\theta ,\forall \theta \ge 1,$$

then

$$\begin{array}{c}\parallel {}^{H}D_{1+}^{\alpha -3}{\mathcal{E}\mathfrak{u}\parallel }_2=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -3}\mathcal{E}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^2}}\hfill \\ =\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|\frac{1}{\Gamma \left(3\right)}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right){(ln\theta )}^2+\frac{1}{\Gamma \left(2\right)}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)ln\theta \right|}{1+{(ln\theta )}^2}}\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(3\right)}}|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)|+{\displaystyle \frac{1}{\Gamma \left(2\right)}}\left|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(3\right)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma \left(2\right)}}{[E(1+lnF)+lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}].\hfill \end{array}$$

We go on in this way until the fractional derivative of order $\alpha -n+1$ of $\mathcal{E}\mathfrak{u}$, and we obtain

$${}^{H}D_{1+}^{\alpha -n+1}\mathcal{E}\mathfrak{u}\left(\theta \right)={\displaystyle \frac{1}{\Gamma (n-1)}}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right){(ln\theta )}^{n-2}+{\displaystyle \frac{1}{\Gamma (n-2)}}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right){(ln\theta )}^{n-3},\forall \theta \ge 1,$$

so

$$\begin{array}{c}\parallel {}^{H}D_{1+}^{\alpha -n+1}{\mathcal{E}\mathfrak{u}\parallel }_{n-2}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -n+1}\mathcal{E}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}\hfill \\ =\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{1+{(ln\theta )}^{n-2}}}\left[{\displaystyle \frac{1}{\Gamma (n-1)}}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right){(ln\theta )}^{n-2}+{\displaystyle \frac{1}{\Gamma (n-2)}}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right){(ln\theta )}^{n-3}\right]\hfill \\ \le {\displaystyle \frac{1}{\Gamma (n-1)}}|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(1\right)|+{\displaystyle \frac{1}{\Gamma (n-2)}}\left|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma (n-1)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma (n-2)}}{[E(1+lnF)+lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}].\hfill \end{array}$$

By use of the above inequalities, we derive

$$\begin{array}{c}{\parallel \mathcal{E}\mathfrak{u}\parallel }_{\mathcal{U}}=max\left\{{\parallel \mathcal{E}\mathfrak{u}\parallel }_{\alpha -1},\parallel {}^{H}D_{1+}^{\alpha -n+1}{\mathcal{E}\mathfrak{u}\parallel }_{n-2},\dots ,{\parallel {}^{H}D_{1+}^{\alpha -3}\mathcal{E}\mathfrak{u}\parallel }_2,\right.\hfill \\ \left.\parallel {}^{H}D_{1+}^{\alpha -2}{\mathcal{E}\mathfrak{u}\parallel }_1,{\parallel {}^{H}D_{1+}^{\alpha -1}\mathcal{E}\mathfrak{u}\parallel }_0\right\}\hfill \\ =max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right]\right.,\hfill \\ {\displaystyle \frac{1}{\Gamma (n-1)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma (n-2)}}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right],\dots \hfill \\ {\displaystyle \frac{1}{\Gamma \left(3\right)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma \left(2\right)}}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right],\hfill \\ \left.{\displaystyle \frac{1}{\Gamma \left(2\right)}}{(E+\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}})+{\displaystyle \frac{1}{\Gamma \left(1\right)}}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right],{\displaystyle \frac{1}{\Gamma \left(1\right)}}\left(E+{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right)\right\}.\hfill \end{array}$$

So, we obtain

$$\begin{array}{c}{\parallel \mathcal{E}\mathfrak{u}\parallel }_{\mathcal{U}}\le max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}},{\displaystyle \frac{1}{\Gamma (n-1)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(3\right)}},{\displaystyle \frac{1}{\Gamma \left(2\right)}},{\displaystyle \frac{1}{\Gamma \left(1\right)}}\right\}\left(E+{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right)\hfill \\ +max\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},{\displaystyle \frac{1}{\Gamma (n-2)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(2\right)}},{\displaystyle \frac{1}{\Gamma \left(1\right)}}\right\}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right]\hfill \\ =max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}},1\right\}\left(E+{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right)+max\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},1\right\}\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right]\hfill \\ =E+{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}+{\Xi }_1\left[E(1+lnF)+{lnF\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\right].\hfill \end{array}$$

Therefore, we conclude

$${\parallel \mathcal{E}\mathfrak{u}\parallel }_{\mathcal{U}}\le E[1+{\Xi }_1(1+lnF)]+(1+{\Xi }_{1}lnF){\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}.$$

(36)

For $\mathfrak{u}\in {\Lambda }_1$, we have $(I-\mathcal{E})\mathfrak{u}\in D\left(\mathcal{A}\right)\cap Ker\mathcal{E}$ and $\mathcal{A}\mathcal{E}\mathfrak{u}=0$. Then, by virtue of Lemma 4, we find

$${\parallel (I-\mathcal{E})\mathfrak{u}\parallel }_{\mathcal{U}}=\parallel {\mathcal{L}}_{\mathcal{E}}{\mathcal{A}(I-\mathcal{E})\mathfrak{u}\parallel }_{\mathcal{U}}=\parallel {\mathcal{L}}_{\mathcal{E}}{\mathcal{A}\mathfrak{u}\parallel }_{\mathcal{U}}\le {\parallel \mathcal{A}\mathfrak{u}\parallel }_{\mathcal{V}}\le {\nu \parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\le {\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}.$$

(37)

Therefore, according to (36) and (37), we deduce

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}={\parallel \mathcal{E}\mathfrak{u}+(I-\mathcal{E})\mathfrak{u}\parallel }_{\mathcal{U}}\le {\parallel \mathcal{E}\mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel (I-\mathcal{E})\mathfrak{u}\parallel }_{\mathcal{U}}\hfill \\ \le E[1+{\Xi }_1(1+lnF)]+(1+{\Xi }_{1}lnF){\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}+{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\hfill \\ =E[1+{\Xi }_1(1+lnF)]+(2+{\Xi }_{1}lnF){\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}.\hfill \end{array}$$

(38)

According to $\left(I2\right)$, we obtain

$$\begin{array}{c}{\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}={\int }_1^{\infty }\left|\mathcal{B}\mathfrak{u}\left(\theta \right)\right|d\theta ={\int }_1^{\infty }\left|\mathfrak{f}\left(\theta ,\mathfrak{u}\left(\theta \right){,}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right),\dots {,}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\right)\right|d\theta \hfill \\ \le {\int }_1^{\infty }(p_1\left(\theta \right){\displaystyle \frac{\left|\mathfrak{u}\right(\theta \left)\right|}{1+{(ln\theta )}^{\alpha -1}}}+p_2\left(\theta \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}+\cdots +p_{n-1}\left(\theta \right){\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|}{1+ln\theta }}\hfill \\ +p_n\left(\theta \right){|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|+p_{n+1}\left(\theta \right))d\theta \hfill \\ \le {\parallel \mathfrak{u}\parallel }_{\alpha -1}\parallel p_1{\parallel }_{\mathcal{V}}{+\parallel }^H{D}_{1+}^{\alpha -n+1}{\mathfrak{u}\parallel }_{n-2}\parallel p_2{\parallel }_{\mathcal{V}}{+\cdots + \parallel }^H{D}_{1+}^{\alpha -2}{\mathfrak{u}\parallel }_1{\parallel p_{n-1}\parallel }_{\mathcal{V}}\hfill \\ {+\parallel }^H{D}_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0\parallel p_n{\parallel }_{\mathcal{V}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}\hfill \\ \le {\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\left(\parallel p_1{\parallel }_{\mathcal{V}}+\parallel p_2{\parallel }_{\mathcal{V}}+\cdots + \parallel p_{n-1}{\parallel }_{\mathcal{V}}+{\parallel p_n\parallel }_{\mathcal{V}}\right)+{\parallel p_{n+1}\parallel }_{\mathcal{V}}\hfill \\ ={\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}.\hfill \end{array}$$

(39)

According to (38), (39), and (35), we find

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\le E[1+{\Xi }_1(1+lnF)]+(2+{\Xi }_{1}lnF)\left({\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}\right)\hfill \\ =E[1+{\Xi }_1(1+lnF)]+(2+{\Xi }_{1}lnF){\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+(2+{\Xi }_{1}lnF){\parallel p_{n+1}\parallel }_{\mathcal{V}},\hfill \end{array}$$

then

$${\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\le {\displaystyle \frac{E[1+{\Xi }_1(1+lnF)]+(2+{\Xi }_{1}lnF){\parallel p_{n+1}\parallel }_{\mathcal{V}}}{1-(2+{\Xi }_{1}lnF){\Xi }_0}}.$$

Thus, we deduce that ${\Lambda }_1$ is a bounded set in space $\mathcal{U}$. □

 Lemma 7. 

Suppose that $\left(I1\right)$, $\left(I2\right)$, and $\left(I4\right)$ hold. Then, the set

$${\Lambda }_2=\{\mathfrak{u}\in Ker\mathcal{A};\mathcal{B}\mathfrak{u}\in Im\mathcal{A}\}\subset \mathcal{U},$$

 is bounded in $\mathcal{U}$.

Proof. 

Let $\mathfrak{u}\in {\Lambda }_2$; then, $\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2}$, with $a_1,a_2\in \mathbb{R}$ and ${\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)={\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)=0$ for all $\theta \in [1,\infty )$. From $\left(I4\right)$, we deduce that $|a_1|\le r$ and $|a_2|\le r$. Hence, we have

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\alpha -1}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|\mathfrak{u}\right(\theta \left)\right|}{1+{(ln\theta )}^{\alpha -1}}}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2}|}{1+{(ln\theta )}^{\alpha -1}}}\le |a_1|+|a_2|\le 2r,\hfill \\ {\parallel }^H{D}_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0=\underset{\theta \ge 1}{sup}{|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|=\underset{\theta \ge 1}{sup}|a_1\Gamma \left(\alpha \right)|= |a_1\Gamma \left(\alpha \right)|\le r\Gamma \left(\alpha \right),\hfill \\ {\parallel }^H{D}_{1+}^{\alpha -2}{\mathfrak{u}\parallel }_1=\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|}{1+ln\theta }}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|a_1\Gamma \left(\alpha \right)ln\theta +a_2\Gamma (\alpha -1)|}{1+ln\theta }}\hfill \\ \le |a_1\Gamma \left(\alpha \right)|+|a_2\Gamma (\alpha -1)|\le r(\Gamma \left(\alpha \right)+\Gamma (\alpha -1)),\hfill \\ {\parallel }^H{D}_{1+}^{\alpha -3}{\mathfrak{u}\parallel }_2=\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -3}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^2}}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|a_1\frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}{(ln\theta )}^2+a_2\frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}ln\theta \right|}{1+{(ln\theta )}^2}}\hfill \\ \le \left|a_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}}\right|+\left|a_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}}\right|\le r\left({\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}}+{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}}\right)=r\left({\displaystyle \frac{\Gamma \left(\alpha \right)}{2}}+\Gamma (\alpha -1)\right).\hfill \end{array}$$

We proceed with these estimations up to the fractional derivative of order $\alpha -n+1$ of function $\mathfrak{u}$, and we infer

$$\begin{array}{c}{\parallel }^H{D}_{1+}^{\alpha -n+1}{\mathfrak{u}\parallel }_{n-2}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{{|}^H{D}_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{\left|a_1\frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}{(ln\theta )}^{n-2}+a_2\frac{\Gamma (\alpha -1)}{\Gamma (n-2)}{(ln\theta )}^{n-3}\right|}{1+{(ln\theta )}^{n-2}}}\hfill \\ \le \left|a_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}}\right|+\left|a_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}\right|\le r\left({\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}}+{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}\right).\hfill \end{array}$$

With the help of the above inequalities, we deduce

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}=max\{2r,r\left({\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}}+{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}\right),\dots ,r\left({\displaystyle \frac{\Gamma \left(\alpha \right)}{2}}+\Gamma (\alpha -1)\right),\\ r(\Gamma (\alpha )+\Gamma (\alpha -1\left)\right),r\Gamma \left(\alpha \right)\}<\infty .\end{array}$$

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

 Lemma 8. 

Suppose that $\left(I1\right)$, $\left(I2\right)$, and $\left(I4\right)$ hold. Then the set

$${\Lambda }_3=\{\mathfrak{u}\in Ker\mathcal{A},\varsigma \zeta \mathcal{H}\mathfrak{u}+(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u}=0,\zeta \in [0,1]\}\subset \mathcal{U}$$

 is bounded in $\mathcal{U}$, where $\varsigma =-1$ if relation (33) holds and $\varsigma =1$ if relation (34) holds. The operator $\mathcal{H}:Ker\mathcal{A}\to Im\mathcal{F}$ is a linear isomorphism given by

$$\mathcal{H}\left(\mathfrak{u}\right)\left(\theta \right)={\displaystyle \frac{1}{\Delta }}\left[(\mathfrak{d}{b}_1-\mathfrak{b}{b}_2){\displaystyle \frac{1}{{\theta }^2}}+(\mathfrak{a}{b}_2-\mathfrak{c}{b}_1){\displaystyle \frac{ln\theta }{{\theta }^2}}\right],\theta \in [1,\infty ),$$

 for $\mathfrak{u}\in Ker\mathcal{A}$, $\mathfrak{u}\left(\theta \right)=b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2},\theta \ge 1$, with $b_1,b_2\in \mathbb{R}.$

Proof. 

We assume that assumption (34) holds, that is, $\varsigma =1$. For $\mathfrak{u}\in {\Lambda }_3$, we can write $\mathfrak{u}$ in the form of $\mathfrak{u}\left(\theta \right)=b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2}$, $\theta \ge 1$, and $\zeta \mathcal{H}\mathfrak{u}=-(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u}$ for $\zeta \in [0,1]$. Using the same argument as in the proof of Lemma 7, we need only to prove that $|b_1|\le r$ and $|b_2|\le r$. We have three cases:

$\left(1\right)$ For $\zeta =0$, $\mathcal{F}\mathcal{B}\mathfrak{u}=0$, that is, ${\mathcal{G}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)+\left({\mathcal{G}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)\right)ln\theta =0$ for all $\theta \in [1,\infty ),$ or

$${\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}-\mathfrak{b}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}-\mathfrak{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}){\displaystyle \frac{ln\theta }{{\theta }^2}}=0,\forall \theta \in [1,\infty ).$$

Then, we find

$$\left\{\begin{array}{c}\mathfrak{d}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}-\mathfrak{b}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}=0,\hfill \\ -\mathfrak{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}+\mathfrak{a}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}=0.\hfill \end{array}\right.$$

Since the determinant of the above system is $\Delta \ne 0$, we deduce that ${\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)={\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)=0$. Hence, according to $\left(I4\right)$, we obtain $|b_1|\le r$ and $|b_2|\le r$.

$\left(2\right)$ For $\zeta =1$, $\mathcal{H}\mathfrak{u}=0$, that is.

$${\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{b}_1-\mathfrak{b}{b}_2){\displaystyle \frac{1}{{\theta }^2}}+(\mathfrak{a}{b}_2-\mathfrak{c}{b}_1){\displaystyle \frac{ln\theta }{{\theta }^2}}=0.$$

From this relation, it follows that

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

Since $\Delta \ne 0$, we obtain $b_1=b_2=0$.

$\left(3\right)$ For $\zeta \in (0,1)$, based on $\zeta \mathcal{H}\mathfrak{u}=-(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u}$, we have

$$\begin{array}{c}\zeta \left[{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{b}_1-\mathfrak{b}{b}_2){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{b}_2-\mathfrak{c}{b}_1){\displaystyle \frac{ln\theta }{{\theta }^2}}\right]\hfill \\ =-(1-\zeta )\left[{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}-\mathfrak{b}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}-\mathfrak{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}){\displaystyle \frac{ln\theta }{{\theta }^2}}\right],\hfill \end{array}$$

from which we deduce

$$\left\{\begin{array}{c}\zeta \mathfrak{d}{b}_1-\zeta \mathfrak{b}{b}_2=-(1-\zeta )\mathfrak{d}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}+(1-\zeta )\mathfrak{b}{\mathcal{C}}_2\mathcal{B}\mathfrak{u},\hfill \\ \zeta \mathfrak{a}{b}_2-\zeta \mathfrak{c}{b}_1=-(1-\zeta )\mathfrak{a}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}+(1-\zeta )\mathfrak{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}.\hfill \end{array}\right.$$

Because $\Delta \ne 0$, we find

$$\left\{\begin{array}{c}\zeta b_1=-(1-\zeta ){\mathcal{C}}_1\mathcal{B}\mathfrak{u},\hfill \\ \zeta b_2=-(1-\zeta ){\mathcal{C}}_2\mathcal{B}\mathfrak{u}.\hfill \end{array}\right.$$

Now, we show that $|b_1|\le r$ and $|b_2|\le r$. If this assertion were not to hold, then according to (34), we would obtain

$$\begin{array}{c}0\le \zeta (b_1^2+b_2^2)=\left(\zeta b_1\right)b_1+\left(\zeta b_2\right)b_2=-b_1(1-\zeta ){\mathcal{C}}_1\mathcal{B}\mathfrak{u}-b_2(1-\zeta ){\mathcal{C}}_2\mathcal{B}\mathfrak{u}\hfill \\ =-(1-\zeta )(b_1{\mathcal{C}}_1\mathcal{B}\mathfrak{u}+b_2{\mathcal{C}}_2\mathcal{B}\mathfrak{u})<0.\hfill \end{array}$$

This is a contradiction. Consequently, we derive $|b_1|\le r$ and $|b_2|\le r$. We infer that ${\Lambda }_3$ is a bounded set in $\mathcal{U}$. If relation (33) holds (for $\zeta =-1$), then by way of similar arguments, we find that ${\Lambda }_3$ is also bounded. □

We now present the proof of Theorem 1.

 Proof of Theorem 1. 

Let ${\Lambda }_0\subset \mathcal{U}$ be a bounded open set such that ${\cup }_{i=1}^3{\overline{\Lambda }}_i\subset {\Lambda }_0$. In Lemma 3, we saw that $\mathcal{A}$ is a Fredholm operator of index 0, and in Lemma 5, we found that $\mathcal{B}$ is an $\mathcal{A}$-compact operator on ${\overline{\Lambda }}_0$. Furthermore, according to Lemmas 6 and 7, we have

 (i) 

$\mathcal{A}\mathfrak{u}\ne \nu \mathcal{B}\mathfrak{u}$ for any $\mathfrak{u}\in (D\left(\mathcal{A}\right)\setminus Ker\mathcal{A})\cap \partial {\Lambda }_0,\nu \in (0,1)$;

 (ii) 

$\mathcal{B}\mathfrak{u}\notin Im\mathcal{A}$ for any $\mathfrak{u}\in Ker\mathcal{A}\cap \partial {\Lambda }_0.$

Next, we prove that condition $\left(iii\right)$ from the Mawhin continuation theorem (see Theorem IV.13 from [1] or Theorem 1 from [3]) is also satisfied. For this, we define the operator $\mathcal{S}(\mathfrak{u},\zeta )=\varsigma \zeta \mathcal{H}\mathfrak{u}+(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u}$, where $\varsigma$ is given in Lemma 8. So, by way of Lemma 8, we find $\mathcal{S}(\mathfrak{u},\zeta )\ne 0$ for all $\mathfrak{u}\in Ker\mathcal{A}\cap \partial {\Lambda }_0$ and $\zeta \in [0,1]$. Hence, by the homotopy of degree, we conclude

$$\begin{array}{c}deg\{\mathcal{F}\mathcal{B}{|}_{Ker\mathcal{A}},{\Lambda }_0\cap Ker\mathcal{A},0\}=deg\{\mathcal{S}(·,0),{\Lambda }_0\cap Ker\mathcal{A},0\}\hfill \\ =deg\{\mathcal{S}(·,1),{\Lambda }_0\cap Ker\mathcal{A},0\}=deg\{\varsigma \mathcal{H},{\Lambda }_0\cap Ker\mathcal{A},0\}\ne 0.\hfill \end{array}$$

Then, all assumptions of Theorem IV.13 from [1] ($\left(i\right)$, $\left(ii\right)$, and $\left(iii\right)$) are satisfied. Therefore, we deduce that the operator Equation (11)—specifically, $\mathcal{A}\mathfrak{u}=\mathcal{B}\mathfrak{u}$—admits at least one solution $\mathfrak{u}\in D\left(\mathcal{A}\right)\cap {\overline{\Lambda }}_0$. This implies that problem (1), (2) has at least one solution in space $\mathcal{U}$. □

 Theorem 2. 

Suppose that $\left(I1\right)$ and $\left(I2\right)$ hold. We also suppose that

 (I6)

There exists a positive constant G such that for each $\mathfrak{u}\in D\left(\mathcal{A}\right)$ satisfying ${|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|>G$ for all $\theta \in [1,\infty )$, we have either

$$sgn{\{}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)>0,\forall \theta \in [1,\infty ),$$

(40)

$$\mathrm{or}sgn{\{}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)<0,\forall \theta \in [1,\infty ).$$

(41)

 (I7)

There exist positive constants $L>0$ and $M>1$ such that for every $\mathfrak{u}\in D\left(\mathcal{A}\right)$ satisfying ${|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|>L$ for all $\theta \in [1,M]$, we have either

$$sgn{\{}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)\}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)>0,\forall \theta \in [1,M],$$

(42)

$$\mathrm{or}sgn{\{}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)\}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)>0,\forall \theta \in [1,M].$$

(43)

 (I8)

$${\Xi }_0{\Xi }_2<1,$$

(44)

where ${\Xi }_0={\sum }_{i=1}^n{\parallel p_i\parallel }_{\mathcal{V}}$, ${\Xi }_2=max\{{\displaystyle \frac{3}{\Gamma \left(\alpha \right)}},{\displaystyle \frac{3}{2}}\}+2{\Xi }_{1}lnM$, with ${\Xi }_1=max\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},1\}$.

Then, the boundary value problem (1), (2) has at least one solution in space $\mathcal{U}$.

The proof of Theorem 2 is based on the following three lemmas.

 Lemma 9. 

Assume that $\left(I1\right)$, $\left(I2\right)$, $\left(I6\right)$, $\left(I7\right)$, and $\left(I8\right)$ hold. Then, the set ${\Lambda }_1$ (defined in Lemma 6) is bounded in $\mathcal{U}$.

Proof. 

We consider $\mathfrak{u}\in {\Lambda }_1$; then, we have $\mathcal{B}\mathfrak{u}\in Im\mathcal{A}=Ker\mathcal{A}$. By virtue of $\left(I6\right)$ and $\left(I7\right)$, there exist constants ${\theta }_1\in [1,\infty )$ and ${\theta }_2\in [1,M]$ such that ${|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_1\right)|\le G$ and ${|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_2\right)|\le L$. This, together with Lemma 4 and Remark 1 from [3], gives us

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_1\right)+{\int }_{{\theta }_1}^{\theta }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }},\hfill \\ {}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_2\right)+{\int }_{{\theta }_2}^{\theta }{}^{H}D_1^{\alpha -1}\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ ={}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_2\right)+(ln\theta -ln{\theta }_2){}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_1\right)+{\int }_{{\theta }_2}^{\theta }\left({\int }_{{\theta }_1}^{\sigma }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\tau \right){\displaystyle \frac{d\tau }{\tau }}\right){\displaystyle \frac{d\sigma }{\sigma }}.\hfill \end{array}$$

Then, we obtain

$$\parallel {}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0=\underset{\theta \ge 1}{sup}|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|\le G+\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}},$$

(45)

and

$$\begin{array}{c}\parallel {}^{H}D_{1+}^{\alpha -2}{\mathfrak{u}\parallel }_1=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)|}{1+ln\theta }}\le \underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_2\right)|}{1+ln\theta }}\hfill \\ +\underset{\theta \ge 1}{sup}{\displaystyle \frac{|ln\theta -ln{\theta }_2|}{1+ln\theta }}\left|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_1\right)\right|+\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{1+ln\theta }}\left|{\int }_{{\theta }_2}^{\theta }\left({\int }_{{\theta }_1}^{\sigma }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\tau \right){\displaystyle \frac{d\tau }{\tau }}\right){\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ \le L+G+\left({\int }_1^{\infty }|{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\tau \right)d\tau \right)\underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{1+ln\theta }}\left|{\int }_{{\theta }_2}^{\theta }{\displaystyle \frac{d\sigma }{\sigma }}\right|\hfill \\ =L+G+\underset{\theta \ge 1}{sup}{\displaystyle \frac{|ln\theta -ln{\theta }_2|}{1+ln\theta }}\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}=L+G+{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}.\hfill \end{array}$$

(46)

In addition, for $\mathfrak{u}\in {\Lambda }_1\subset D\left(\mathcal{A}\right)$, we find

$$\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^{\alpha }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{(ln\theta )}^{\alpha -1}+c_2{(ln\theta )}^{\alpha -2}+\dots +c_n{(ln\theta )}^{\alpha -n},\forall \theta \ge 1,$$

with $c_i\in \mathbb{R}$, $i=1,\dots ,n$. Because $\mathfrak{u}$ satisfies the boundary conditions of (2), we obtain

$$\mathfrak{u}\left(\theta \right)={}^{H}I_{1+}^{\alpha }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{(ln\theta )}^{\alpha -1}+c_2{(ln\theta )}^{\alpha -2},\theta \ge 1.$$

Therefore, we find

$$\begin{array}{c}{\displaystyle \frac{\mathfrak{u}\left(\theta \right)}{1+{(ln\theta )}^{\alpha -1}}}={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\int }_1^{\theta }{\displaystyle \frac{{(ln\frac{\theta }{\sigma })}^{\alpha -1}}{1+{(ln\theta )}^{\alpha -1}}}{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right)d\sigma +c_1{\displaystyle \frac{{(ln\theta )}^{\alpha -1}}{1+{(ln\theta )}^{\alpha -1}}}+c_2{\displaystyle \frac{{(ln\theta )}^{\alpha -2}}{1+{(ln\theta )}^{\alpha -1}}},\hfill \end{array}$$

(47)

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -1}{}^{H}I_{1+}^{\alpha }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{}^{H}D_{1+}^{\alpha -1}\left({(ln\theta )}^{\alpha -1}\right)+c_2{}^{H}D_{1+}^{\alpha -1}\left({(ln\theta )}^{\alpha -2}\right)\hfill \\ ={}^{H}I_{1+}^1{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1\Gamma \left(\alpha \right)={\int }_1^{\theta }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+c_1\Gamma \left(\alpha \right),\hfill \end{array}$$

(48)

and

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -2}{}^{H}I_{1+}^{\alpha }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{}^{H}D_{1+}^{\alpha -2}\left({(ln\theta )}^{\alpha -1}\right)+c_2{}^{H}D_{1+}^{\alpha -2}\left({(ln\theta )}^{\alpha -2}\right)\hfill \\ ={}^{H}I_{1+}^2{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(2\right)}}ln\theta +c_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(1\right)}}\hfill \\ ={\int }_1^{\theta }ln{\displaystyle \frac{\theta }{\sigma }}{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+c_1\Gamma \left(\alpha \right)ln\theta +c_2\Gamma (\alpha -1)\hfill \\ =-{\int }_1^{\theta }ln\sigma {}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+ln\theta {\int }_1^{\theta }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+c_1\Gamma \left(\alpha \right)ln\theta +c_2\Gamma (\alpha -1)\hfill \\ =-{\int }_1^{\theta }ln\sigma {}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+ln\theta \left({}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)-c_1\Gamma \left(\alpha \right)\right)+c_1\Gamma \left(\alpha \right)ln\theta +c_2\Gamma (\alpha -1)\hfill \\ =-{\int }_1^{\theta }ln\sigma {}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+ln\theta {}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)+c_2\Gamma (\alpha -1).\hfill \end{array}$$

(49)

So, for $\theta ={\theta }_2$, relation (49) gives us

$${}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_2\right)=-{\int }_1^{{\theta }_2}ln\sigma {}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+ln{\theta }_2{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_2\right)+c_2\Gamma (\alpha -1).$$

(50)

Hence, from (48) and (50), we obtain the following relations for constants $c_1$ and $c_2$:

$$\begin{array}{c}{c}_1={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left({}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)-{\int }_1^{\theta }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right),\hfill \\ c_2={\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left({}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_2\right)-ln{\theta }_2{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_2\right)+{\int }_1^{{\theta }_2}ln\sigma {}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}\right).\hfill \end{array}$$

(51)

Now, by using the relations (45), (46), and (51), we derive

$$\begin{array}{c}|c_1| \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left(\parallel {}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0+{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}\right)\le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left(G+2\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}\right),\hfill \\ |c_2| \le {\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left(L+lnM\parallel {}^{H}D_{1+}^{\alpha -1}{\mathfrak{u}\parallel }_0+lnM{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}\right)\hfill \\ \le {\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left(L+GlnM+2lnM\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}\right).\hfill \end{array}$$

(52)

We substitute (52) in (47), and we obtain

$$\begin{array}{c}\left|{\displaystyle \frac{\mathfrak{u}\left(\theta \right)}{1+{(ln\theta )}^{\alpha -1}}}\right|\le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}} + |c_1| + |c_2|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}\left(G+2\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}\right)\hfill \\ +{\displaystyle \frac{1}{\Gamma (\alpha -1)}}\left(L+GlnM+2lnM\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}\right)\hfill \\ ={\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}[3+2(\alpha -1)lnM]+{\displaystyle \frac{G}{\Gamma \left(\alpha \right)}}+{\displaystyle \frac{L+GlnM}{\Gamma (\alpha -1)}},\forall \theta \in [1,\infty ).\hfill \end{array}$$

From this last inequality, we find

$${\parallel \mathfrak{u}\parallel }_{\alpha -1}\le {\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}[3+2(\alpha -1)lnM]{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma \left(\alpha \right)}}+{\displaystyle \frac{L+GlnM}{\Gamma (\alpha -1)}}.$$

(53)

We continue with the fractional derivative of order $\alpha -3$ of function $\mathfrak{u}$, and we obtain

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -3}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -3}{}^{H}D_{1+}^{\alpha }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{}^{H}D_{1+}^{\alpha -3}\left({(ln\theta )}^{\alpha -1}\right)+c_2{}^{H}D_{1+}^{\alpha -3}\left({(ln\theta )}^{\alpha -2}\right)\hfill \\ ={}^{H}I_{1+}^3{}^{H}D_{1+}^1\mathfrak{u}\left(\theta \right)+c_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}}{(ln\theta )}^2+c_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}}ln\theta \hfill \\ ={\displaystyle \frac{1}{\Gamma \left(3\right)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^2{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+c_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}}{(ln\theta )}^2+c_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}}ln\theta ,\hfill \end{array}$$

and

$$\begin{array}{c}\parallel {}^{H}D_{1+}^{\alpha -3}{\mathfrak{u}\parallel }_2=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -3}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^2}}\le \underset{\theta \ge 1}{sup}{\displaystyle \frac{1}{\Gamma \left(3\right)}}{\int }_1^{\theta }{\displaystyle \frac{{(ln\frac{\theta }{\sigma })}^2}{1+{(ln\theta )}^2}}\left|{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ + |c_1|{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}}\underset{\theta \ge 1}{sup}{\displaystyle \frac{{(ln\theta )}^2}{1+{(ln\theta )}^2}}+\left|c_2\right|{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}}\underset{\theta \ge 1}{sup}{\displaystyle \frac{ln\theta }{1+{(ln\theta )}^2}}\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(3\right)}}\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma \left(3\right)}}|c_1|+{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma \left(2\right)}}\left|c_2\right|\hfill \\ \le {\displaystyle \frac{1}{\Gamma \left(3\right)}}\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{1}{\Gamma \left(3\right)}}(G+2\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}{\parallel }_{\mathcal{V}})\hfill \\ +{\displaystyle \frac{1}{\Gamma \left(2\right)}}(L+GlnM+2lnM\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}{\parallel }_{\mathcal{V}}).\hfill \end{array}$$

(54)

We proceed with these estimations until the fractional derivative of order $\alpha -n+1$ of function $\mathfrak{u}$, and we infer

$$\begin{array}{c}{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right)={}^{H}D_{1+}^{\alpha -n+1}{}^{H}I_{1+}^{\alpha }{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{}^{H}D_{1+}^{\alpha -n+1}\left({(ln\theta )}^{\alpha -1}\right)\hfill \\ +c_2{}^{H}D_{1+}^{\alpha -n+1}\left({(ln\theta )}^{\alpha -2}\right)\hfill \\ ={}^{H}I_{1+}^{n-1}{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\theta \right)+c_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}}{(ln\theta )}^{n-2}+c_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}{(ln\theta )}^{n-3}\hfill \\ ={\displaystyle \frac{1}{\Gamma (n-1)}}{\int }_1^{\theta }{\left(ln{\displaystyle \frac{\theta }{\sigma }}\right)}^{n-2}{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right){\displaystyle \frac{d\sigma }{\sigma }}+c_1{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}}{(ln\theta )}^{n-2}+c_2{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}{(ln\theta )}^{n-3},\hfill \end{array}$$

and

$$\begin{array}{c}\parallel {}^{H}D_{1+}^{\alpha -n+1}{\mathfrak{u}\parallel }_{n-2}=\underset{\theta \ge 1}{sup}{\displaystyle \frac{|{}^{H}D_{1+}^{\alpha -n+1}\mathfrak{u}\left(\theta \right)|}{1+{(ln\theta )}^{n-2}}}\le {\displaystyle \frac{1}{\Gamma (n-1)}}\underset{\theta \ge 1}{sup}{\int }_1^{\theta }{\displaystyle \frac{{\left(ln\frac{\theta }{\sigma }\right)}^{n-2}}{1+{(ln\theta )}^{n-2}}}\left|{}^{H}D_{1+}^{\alpha }\mathfrak{u}\left(\sigma \right)\right|{\displaystyle \frac{d\sigma }{\sigma }}\hfill \\ + |c_1|{\displaystyle \frac{\Gamma \left(\alpha \right)}{\Gamma (n-1)}}\underset{\theta \ge 1}{sup}{\displaystyle \frac{{(ln\theta )}^{n-2}}{1+{(ln\theta )}^{n-2}}}+\left|c_2\right|{\displaystyle \frac{\Gamma (\alpha -1)}{\Gamma (n-2)}}\underset{\theta \ge 1}{sup}{\displaystyle \frac{{(ln\theta )}^{n-3}}{1+{(ln\theta )}^{n-2}}}\hfill \\ \le {\displaystyle \frac{1}{\Gamma (n-1)}}\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{1}{\Gamma (n-1)}}(G+2\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}{\parallel }_{\mathcal{V}})\hfill \\ +{\displaystyle \frac{1}{\Gamma (n-2)}}(L+GlnM+2lnM\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}{\parallel }_{\mathcal{V}}).\hfill \end{array}$$

(55)

Combining relations (45), (46) and (53)–(55), we deduce

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}=max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}[3+2(\alpha -1)lnM]{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma \left(\alpha \right)}}+{\displaystyle \frac{L+GlnM}{\Gamma (\alpha -1)}},\right.\hfill \\ \left[{\displaystyle \frac{1}{\Gamma (n-1)}}+{\displaystyle \frac{2}{\Gamma (n-1)}}+{\displaystyle \frac{2lnM}{\Gamma (n-2)}}\right]{{}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma (n-1)}}+{\displaystyle \frac{L+GlnM}{\Gamma (n-2)}},\dots ,\hfill \\ \left[{\displaystyle \frac{1}{\Gamma \left(3\right)}}+{\displaystyle \frac{2}{\Gamma \left(3\right)}}+{\displaystyle \frac{2lnM}{\Gamma \left(2\right)}}\right]{{}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma \left(3\right)}}+{\displaystyle \frac{L+GlnM}{\Gamma \left(2\right)}},\hfill \\ \left.\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}+L+G,{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+G\right\}\hfill \\ =max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}[3+2(\alpha -1)lnM]{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma \left(\alpha \right)}}+{\displaystyle \frac{L+GlnM}{\Gamma (\alpha -1)}},\right.\hfill \\ {\displaystyle \frac{1}{\Gamma (n-1)}}[3+2(n-2)lnM]{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma (n-1)}}+{\displaystyle \frac{L+GlnM}{\Gamma (n-2)}},\dots ,\hfill \\ {\displaystyle \frac{1}{\Gamma \left(3\right)}}[3+2·2lnM]{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+{\displaystyle \frac{G}{\Gamma \left(3\right)}}+{\displaystyle \frac{L+GlnM}{\Gamma \left(2\right)}},\hfill \\ \parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}+L+G,{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+G\}\hfill \\ \le max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}[3+2(\alpha -1)lnM],{\displaystyle \frac{1}{\Gamma (n-1)}}[3+2(n-2)lnM],\dots ,\right.\hfill \\ \left.{\displaystyle \frac{1}{\Gamma \left(3\right)}}[3+2·2lnM],1\right\}{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+Gmax\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}},{\displaystyle \frac{1}{\Gamma (n-1)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(3\right)}},{\displaystyle \frac{1}{\Gamma \left(2\right)}},{\displaystyle \frac{1}{\Gamma \left(1\right)}}\right\}\hfill \\ +Lmax\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},{\displaystyle \frac{1}{\Gamma (n-2)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(2\right)}},1\right\}+GlnMmax\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},{\displaystyle \frac{1}{\Gamma (n-2)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(2\right)}}\right\}.\hfill \end{array}$$

Because $max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}},{\displaystyle \frac{1}{\Gamma (n-1)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(2\right)}},{\displaystyle \frac{1}{\Gamma \left(1\right)}}\right\}=max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}},1\right\}=1$ and

$$\begin{array}{c}max\left\{{\displaystyle \frac{1}{\Gamma \left(\alpha \right)}}[3+2(\alpha -1)lnM],{\displaystyle \frac{1}{\Gamma (n-1)}}[3+2(n-2)lnM],\dots ,{\displaystyle \frac{1}{\Gamma \left(3\right)}}[3+2·2lnM],1\right\}\hfill \\ \le max\left\{{\displaystyle \frac{3}{\Gamma \left(\alpha \right)}},{\displaystyle \frac{3}{\Gamma (n-1)}},\dots ,{\displaystyle \frac{3}{\Gamma \left(3\right)}},1\right\}+2lnMmax\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},{\displaystyle \frac{1}{\Gamma (n-2)}},\dots ,{\displaystyle \frac{1}{\Gamma \left(2\right)}}\right\}\hfill \\ =max\left\{{\displaystyle \frac{3}{\Gamma \left(\alpha \right)}},{\displaystyle \frac{3}{2}}\right\}+2lnMmax\left\{{\displaystyle \frac{1}{\Gamma (\alpha -1)}},1\right\}={\Xi }_2,\hfill \end{array}$$

we conclude that

$${\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\le {\Xi }_2{\parallel {}^{H}D_{1+}^{\alpha }\mathfrak{u}\parallel }_{\mathcal{V}}+G+(L+GlnM){\Xi }_1.$$

(56)

On the other hand, because $\mathcal{A}\mathfrak{u}=\nu \mathcal{B}\mathfrak{u}$, by $\left(I2\right)$, we have

$$\parallel {}^{H}D_{1+}^{\alpha }{\mathfrak{u}\parallel }_{\mathcal{V}}\le {\parallel \mathcal{B}\mathfrak{u}\parallel }_{\mathcal{V}}\le {\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\parallel p_{n+1}\parallel }_{\mathcal{V}}.$$

(57)

Hence, according to (56) and (57), we deduce

$$\begin{array}{c}{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\le {\Xi }_2({\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+\parallel p_{n+1}{\parallel }_{\mathcal{V}})+G+(L+GlnM){\Xi }_1\hfill \\ ={\Xi }_2{\Xi }_0{\parallel \mathfrak{u}\parallel }_{\mathcal{U}}+{\Xi }_2{\parallel p_{n+1}\parallel }_{\mathcal{V}}+G+(L+GlnM){\Xi }_1.\hfill \end{array}$$

(58)

So, by using (58) and (44) (from $\left(I8\right)$), we obtain

$${\parallel \mathfrak{u}\parallel }_{\mathcal{U}}\le {\displaystyle \frac{{\Xi }_2{\parallel p_{n+1}\parallel }_{\mathcal{V}}+G+(L+GlnM){\Xi }_1}{1-{\Xi }_0{\Xi }_2}},$$

which means that ${\Lambda }_1$ is bounded. □

 Lemma 10. 

Assume that $\left(I1\right)$, $\left(I2\right)$, $\left(I6\right)$, and $\left(I7\right)$ hold. Then, the set ${\Lambda }_2$ (defined in Lemma 7) is bounded in $\mathcal{U}$.

Proof. 

For any $\mathfrak{u}\in {\Lambda }_2$, $\mathfrak{u}\left(\theta \right)=a_1{(ln\theta )}^{\alpha -1}+a_2{(ln\theta )}^{\alpha -2}$ with $a_1,a_2\in \mathbb{R}$ and ${\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)={\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)=0$. Using the same arguments as those used in the proof of Lemma 7, we prove that $|a_1|$ and $|a_2|$ are bounded and, therefore, ${\Lambda }_2$ is bounded.

According to $\left(I6\right)$ and $\left(I7\right)$, we deduce that there exist the constants ${\theta }_3\in [1,\infty )$ and ${\theta }_4\in [1,M]$ such that $|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_3\right)|\le G$ and $|{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_4\right)|\le G$, that is,

$$\begin{array}{c}|{}^{H}D_{1+}^{\alpha -1}\mathfrak{u}\left({\theta }_3\right)|=|a_1|\Gamma \left(\alpha \right)\le G,\hfill \\ |{}^{H}D_{1+}^{\alpha -2}\mathfrak{u}\left({\theta }_4\right)|=|a_1\Gamma \left(\alpha \right)ln{\theta }_4+a_2\Gamma (\alpha -1)|\le L.\hfill \end{array}$$

Therefore, we obtain

$$|a_1|\le {\displaystyle \frac{G}{\Gamma \left(\alpha \right)}},\left|a_2\right|\le {\displaystyle \frac{L+|a_1|\Gamma \left(\alpha \right)lnM}{\Gamma (\alpha -1)}}\le {\displaystyle \frac{L+GlnM}{\Gamma (\alpha -1)}}.$$

□

 Lemma 11. 

Assume that $\left(I1\right)$, $\left(I2\right)$, $\left(I6\right)$, and $\left(I7\right)$ hold. Then, the set

$${\Lambda }_4=\{\mathfrak{u}\in Ker\mathcal{A};\varsigma \zeta \tilde{\mathcal{H}}\mathfrak{u}+(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u}=0,\zeta \in [0,1]\}\subset \mathcal{U}$$

 is bounded in $\mathcal{U}$, where

$$\varsigma =\left\{\begin{array}{cc}1,& if\left(40\right)and\left(42\right)hold-Case\left(I\right),\hfill \\ -1,& if\left(41\right)and\left(43\right)hold-Case\left(II\right),\hfill \\ 1,& if\left(41\right)and\left(42\right)hold-Case\left(III\right),\hfill \\ -1,& if\left(40\right)and\left(43\right)hold-Case\left(IV\right),\hfill \end{array}\right.$$

and $\tilde{\mathcal{H}}:Ker\mathcal{A}\to Im\mathcal{F}$ is the linear isomorphism operator defined by

$$\tilde{\mathcal{H}}\left(\mathfrak{u}\left(\theta \right)\right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{b}_2-\mathfrak{b}{b}_1){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}(\mathfrak{a}{b}_1-\mathfrak{c}{b}_2){\displaystyle \frac{ln\theta }{{\theta }^2}},inthecases\left(I\right),\left(II\right),\hfill \\ {\displaystyle \frac{1}{\Delta }}(\mathfrak{d}{b}_2+\mathfrak{b}{b}_1){\displaystyle \frac{1}{{\theta }^2}}+{\displaystyle \frac{1}{\Delta }}(-\mathfrak{a}{b}_1-\mathfrak{c}{b}_2){\displaystyle \frac{ln\theta }{{\theta }^2}},inthecases\left(III\right),\left(IV\right),\hfill \end{array}\right.$$

for $\mathfrak{u}\in Ker\mathcal{A}$, $\mathfrak{u}\left(\theta \right)=b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2}$, $\theta \ge 1$, with $b_1,b_2\in \mathbb{R}.$

Proof. 

We prove this lemma, without loss of generality, in the case of $\left(I\right)$—specifically, (40) and (42) hold, and $\varsigma =1$. Let $\mathfrak{u}\in {\Lambda }_4$, $\mathfrak{u}\left(\theta \right)=b_1{(ln\theta )}^{\alpha -1}+b_2{(ln\theta )}^{\alpha -2}$ with $b_1,b_2\in \mathbb{R}$ and

$$\zeta \tilde{\mathcal{H}}\mathfrak{u}\left(\theta \right)=-(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u}\left(\theta \right),\zeta \in [0,1].$$

(59)

As in Lemma 8, we can show that $|b_1|$ and $|b_2|$ are bounded when $\zeta =0$ or $\zeta =1$. Now, we prove that $|b_1|$ and $|b_2|$ are bounded in the case of $\zeta \in (0,1)$. According to relation (59), we find

$$\left\{\begin{array}{c}\zeta (\mathfrak{d}{b}_2-\mathfrak{b}{b}_1)=-(1-\zeta )(\mathfrak{d}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}-\mathfrak{b}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}),\hfill \\ \zeta (\mathfrak{a}{b}_1-\mathfrak{c}{b}_2)=-(1-\zeta )(\mathfrak{a}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}-\mathfrak{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}).\hfill \end{array}\right.$$

Since $\Delta \ne 0$, from the above system, we obtain

$$\zeta b_1=-(1-\zeta ){\mathcal{C}}_2\mathcal{B}\mathfrak{u},$$

(60)

$$\zeta b_2=-(1-\zeta ){\mathcal{C}}_1\mathcal{B}\mathfrak{u}.$$

(61)

According to (40) and (60), we deduce that $|b_1|\Gamma \left(\alpha \right)\le G$, so $|b_1| \le {\displaystyle \frac{G}{\Gamma \left(\alpha \right)}}$. We use the method of reduction to the absurd. If we suppose that $|b_1|\Gamma \left(\alpha \right)>G$, then ${|}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)|=|b_1|\Gamma \left(\alpha \right)>G$. Now, using (40), we obtain $sgn{\{}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)>0$ for all $\theta \in [1,\infty )$ or $sgn\left\{b_1\right\}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)>0$ for all $\theta \in [1,\infty )$. Then, by using (60), we conclude

$$0\le \zeta b_{1}sgn\left\{b_1\right\}=\zeta b_{1}sgn{\{}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\}=-(1-\zeta ){\mathcal{C}}_2\mathcal{B}\mathfrak{u}sgn{\{}^H{D}_{1+}^{\alpha -1}\mathfrak{u}\left(\theta \right)\}<0,$$

which is a contradiction.

Next, according to (42) and (61), we deduce that $|b_2|\Gamma (\alpha -1)\le L+GlnM$; then, $|b_2| \le {\displaystyle \frac{1}{\Gamma (\alpha -1)}}(L+GlnM).$ We also use the method of reduction to the absurd. We suppose that $|b_2|\Gamma (\alpha -1)>L+GlnM$. So, we find

$$\begin{array}{c}{|}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(M\right)|=|b_1\Gamma \left(\alpha \right)lnM+b_2\Gamma (\alpha -1)|\ge -|b_1|\Gamma \left(\alpha \right)lnM+|b_2|\Gamma (\alpha -1)\hfill \\ >L+GlnM-|b_1|\Gamma \left(\alpha \right)lnM\ge L+GlnM-GlnM=L.\hfill \end{array}$$

So, according to (42), we have $sgn{\{}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(\theta \right)\}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)>0$ for all $\theta \in [1,M]$. For $\theta =1$, we obtain $sgn{\{}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)\}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(1\right)>0$. Hence, by using (61), we deduce

$$0\le \zeta b_{2}sgn\left\{b_2\right\}=\zeta b_{2}sgn{\{}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)\}=-(1-\zeta ){\mathcal{C}}_1\mathcal{B}\mathfrak{u}sgn{\{}^H{D}_{1+}^{\alpha -2}\mathfrak{u}\left(1\right)\}<0,$$

which is a contradiction.

So, we find that set ${\Lambda }_4$ is bounded. □

We now present the proof of Theorem 2.

Proof of Theorem 2. 

Let ${\Lambda }_5\subset \mathcal{U}$ be a bounded open set such that ${\overline{\Lambda }}_1\cup {\overline{\Lambda }}_2\cup {\overline{\Lambda }}_4\subset {\Lambda }_5$. According to Lemmas 3 and 5, operator $\mathcal{A}$ is a Fredholm operator of index 0 and operator $\mathcal{B}$ is $\mathcal{A}$-compact on ${\overline{\Lambda }}_5$. According to Lemmas 9 and 10, we see that conditions $\left(i\right)$ and $\left(ii\right)$ of the Mawhin continuation theorem are satisfied. In what follows, we prove that condition $\left(iii\right)$ is also satisfied. For this, we define the following operator:

$$\mathcal{T}(\mathfrak{u},\zeta )=\varsigma \zeta \tilde{\mathcal{H}}\left(\mathfrak{u}\right)+(1-\zeta )\mathcal{F}\mathcal{B}\mathfrak{u},$$

where $\varsigma$ is given in Lemma 11. By use of Lemma 11, we have $\mathcal{T}(\mathfrak{u},\zeta )\ne 0$ for all $\mathfrak{u}\in Ker\mathcal{A}\cap \partial {\Lambda }_5$ and $\zeta \in [0,1]$. Then, according to the homotopy of degree, we obtain

$$\begin{array}{c}deg\{\mathcal{F}\mathcal{B}{|}_{Ker\mathcal{A}},{\Lambda }_5\cap Ker\mathcal{A},0\}=deg\{\mathcal{T}(·,0),{\Lambda }_5\cap Ker\mathcal{A},0\}\hfill \\ =deg\{\mathcal{T}(·,1),{\Lambda }_5\cap Ker\mathcal{A},0\}=deg\{\varsigma \tilde{\mathcal{H}},{\Lambda }_5\cap Ker\mathcal{A},0\}\ne 0.\hfill \end{array}$$

Therefore, by applying Theorem IV.13 from [1], we deduce that the equation $\mathcal{A}\mathfrak{u}=\mathcal{B}\mathfrak{u}$, that is, Equation (11), has at least one solution in $D\left(\mathcal{A}\right)\cap {\overline{\Lambda }}_5$. This signifies that problem (1), (2) has at least one solution in space $\mathcal{U}$. □

4. Examples

 Example 1. 

Let $\alpha ={\displaystyle \frac{11}{3}}$, $n=4$, $A=B=20$,

$$\mathcal{P}\left(\sigma \right)=\left\{\begin{array}{c}{a}_0(\sigma -1)+{\displaystyle \frac{1}{6}},\sigma \in [1,10),\hfill \\ {\displaystyle \frac{7}{6}},\sigma \in [10,20],\hfill \end{array}\right.\mathcal{Q}\left(\sigma \right)=\left\{\begin{array}{c}1,\sigma \in [1,10),\hfill \\ {\displaystyle \frac{1}{100}}{(\sigma -10)}^2+1,\sigma \in [10,20],\hfill \end{array}\right.$$

where $a_0={\displaystyle \frac{ln10}{9-ln10}}$.

We consider the following fractional differential equation:

$${}^{H}D_{1+}^{11/3}\mathfrak{u}\left(\theta \right)=\mathfrak{f}(\theta ,\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\theta \right)),\theta \in (1,\infty ),$$

(62)

subject to the following boundary conditions:

$$\left\{\begin{array}{c}\mathfrak{u}\left(1\right)={\mathfrak{u}}^{\prime }\left(1\right)=0,{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\theta \right)=a_0{\int }_1^{10}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\sigma \right)d\sigma +(1-9a_0){}^{H}D_{1+}^{5/3}\mathfrak{u}\left(10\right),\hfill \\ {}^{H}D_{1+}^{8/3}\mathfrak{u}(\infty )={\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10){}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\sigma \right)d\left(\sigma \right).\hfill \end{array}\right.$$

(63)

After some computations, we find $\mathfrak{a}\approx 0.17788008$, $\mathfrak{b}\approx -0.13237075$, $\mathfrak{c}\approx 0.00193147$, $\mathfrak{d}\approx 0.00628112$, and $\Delta \approx 0.00137296\ne 0$. In addition, we have ${\int }_1^{20}d\mathcal{P}\left(\sigma \right)=1$, ${\int }_1^{20}d\mathcal{Q}\left(\sigma \right)=1$, and ${\int }_1^{20}ln\sigma d\mathcal{P}\left(\sigma \right)=0$. So, assumption $\left(I1\right)$ is satisfied.

We consider the following function for $x_i\in \mathbb{R}$, $i=1,\dots ,4$:

$$\mathfrak{f}(\theta ,x_1,x_2,x_3,x_4)=\left\{\begin{array}{c}{e}^{-4\theta +3}\left({\displaystyle \frac{{\theta }^2+1}{2\theta }}+{\displaystyle \frac{1}{3}}arctan{x}_1-{\displaystyle \frac{x_2}{4(x_2^2+1)}}+{\displaystyle \frac{x_3}{30}}\right),\theta \in [1,10],\hfill \\ {\displaystyle \frac{c_0{x}_4}{5{\theta }^2}},\theta \in (10,\infty ),\hfill \end{array}\right.$$

(64)

where

$$c_0={\displaystyle \frac{-25}{2(2ln2-1)}}{\int }_1^{10}\left[a_0\left(ln{\displaystyle \frac{10}{\tau }}-10+\tau \right)+ln{\displaystyle \frac{10}{\tau }}\right]e^{-4\tau +3}{\displaystyle \frac{d\tau }{\tau }}\approx 0.4106.$$

We obtain

$$\begin{array}{c}|\mathfrak{f}(\theta ,x_1,x_2,x_3,x_4)|\le p_1\left(\theta \right){\displaystyle \frac{|x_1|}{1+{(ln\theta )}^{8/3}}}+p_2\left(\theta \right){\displaystyle \frac{|x_2|}{1+{(ln\theta )}^2}}\hfill \\ +p_3\left(\theta \right){\displaystyle \frac{|x_3|}{1+ln\theta }}+p_4\left(\theta \right)\left|x_4\right|+p_5\left(\theta \right),\theta \in [1,\infty ),x_i\in \mathbb{R},i=1,\dots ,4,\hfill \end{array}$$

with

$$\begin{array}{c}{p}_1\left(\theta \right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{3}}{e}^{-4\theta +3}(1+{(ln\theta )}^{8/3}),\theta \in [1,10],\hfill \\ 0,\theta \in (10,\infty ),\hfill \end{array}\right.\hfill \\ p_2\left(\theta \right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{4}}{e}^{-4\theta +3}(1+{(ln\theta )}^2),\theta \in [1,10],\hfill \\ 0,\theta \in (10,\infty ),\hfill \end{array}\right.\hfill \\ p_3(\theta )=\{\begin{array}{c}{\displaystyle \frac{1}{30}}{e}^{-4\theta +3}(1+ln\theta ),\theta \in [1,10],\hfill \\ 0,\theta \in (10,\infty ),\hfill \end{array}\hfill \\ p_4(\theta )=\{\begin{array}{c}0,\theta \in [1,10],\hfill \\ {\displaystyle \frac{c_0}{5{\theta }^2}},\theta \in (10,\infty ),\hfill \end{array}\hfill \\ p_5(\theta )=\{\begin{array}{c}{e}^{-4\theta +3}{\displaystyle \frac{{\theta }^2+1}{2\theta }},\theta \in [1,10],\hfill \\ 0,\theta \in (10,\infty ).\hfill \end{array}\hfill \end{array}$$

We have $p_i\in L^1(1,\infty )$ for $i=1,\dots ,5$ and $\parallel p_1{\parallel }_{\mathcal{V}} \approx 0.03200099$, $\parallel p_2{\parallel }_{\mathcal{V}} \approx 0.02468634$, $\parallel p_3{\parallel }_{\mathcal{V}} \approx 0.00369825$, $\parallel p_4{\parallel }_{\mathcal{V}} \approx 0.00821168$, and $\parallel p_5{\parallel }_{\mathcal{V}} \approx 0.09543632.$ We also find ${\Xi }_0\approx 0.06859726$, and ${\Xi }_1=1$. So, assumption $\left(I2\right)$ is also satisfied.

We now verify assumption $\left(I3\right)$. We consider $E=200$ and $F=10$. Let $\mathfrak{u}\in D\left(\mathcal{A}\right)\setminus Ker\mathcal{A}$.

If ${|}^H{D}_{1+}^{5/3}\mathfrak{u}\left(\theta \right)|>E$, then we obtain

$$\begin{array}{c}{\Theta }_1={\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)={\int }_1^{20}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ =a_0{\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma +(1-9a_0)\left({\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)\hfill \\ =a_0{\int }_1^{10}[{\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}\left({\displaystyle \frac{{\tau }^2+1}{2\tau }}+{\displaystyle \frac{1}{3}}arctan\mathfrak{u}\left(\tau \right)-{\displaystyle \frac{{}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\tau \right)}{4({\left({}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\tau \right)\right)}^2+1)}}\right.\hfill \\ +\left.\left.{\displaystyle \frac{1}{30}}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)\right)d\tau \right]d\sigma +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}\left({\displaystyle \frac{{\tau }^2+1}{2\tau }}\right.\hfill \\ +\left.{\displaystyle \frac{1}{3}}arctan\mathfrak{u}\left(\tau \right)-{\displaystyle \frac{{}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\tau \right)}{4({\left({}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\tau \right)\right)}^2+1)}}+{\displaystyle \frac{1}{30}}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)\right)d\tau \hfill \\ =a_0{\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \right)d\sigma +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +a_0{\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \right)d\sigma \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau ,\hfill \end{array}$$

where

$$h\left(\tau \right)={\displaystyle \frac{{\tau }^2+1}{2\tau }}+{\displaystyle \frac{1}{3}}arctan\mathfrak{u}\left(\tau \right)-{\displaystyle \frac{{}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\tau \right)}{4({\left({}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\tau \right)\right)}^2+1)}},\tau \in [1,10].$$

Therefore, we find

$$\begin{array}{c}{\Theta }_1={\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)=a_0{\int }_1^{10}\left({\int }_{\tau }^{10}\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\sigma \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +a_0{\int }_1^{10}\left({\int }_{\tau }^{10}\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\sigma \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \hfill \\ =a_0{\int }_1^{10}{\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)\left({\int }_{\tau }^{10}ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +a_0{\int }_1^{10}{\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)\left({\int }_{\tau }^{10}ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \hfill \end{array}$$

$$\begin{array}{c}=a_0{\int }_1^{10}{\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right)h\left(\tau \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +a_0{\int }_1^{10}{\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \hfill \\ ={\int }_1^{10}V\left(\tau \right)h\left(\tau \right)d\tau +{\displaystyle \frac{1}{30}}{\int }_1^{10}V\left(\tau \right){}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau ,\hfill \end{array}$$

where

$$V\left(\tau \right)={\displaystyle \frac{a_0}{\tau }}{e}^{-4\tau +3}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right)+(1-9a_0)\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3},\tau \in [1,10].$$

We see that $V\left(\tau \right)\le 0$ for all $\tau \in [1,10]$ and $h\left(\tau \right)\in [h_1\left(\tau \right),h_2\left(\tau \right)]$, where $h_1\left(\tau \right)={\displaystyle \frac{{\tau }^2+1}{2\tau }}-{\displaystyle \frac{\pi }{6}}-{\displaystyle \frac{1}{8}}$ and $h_2\left(\tau \right)={\displaystyle \frac{{\tau }^2+1}{2\tau }}+{\displaystyle \frac{\pi }{6}}+{\displaystyle \frac{1}{8}}$ for $\tau \in [1,10]$. We have ${min}_{\sigma \in [1,10]}{h}_1\left(\sigma \right)\approx 0.3514>0$ and $V\left(\tau \right)h_2\left(\tau \right)\le V\left(\tau \right)h\left(\tau \right)\le V\left(\tau \right)h_1\left(\tau \right)$ for all $\tau \in [1,10]$.

If ${}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\theta \right)>E$, then ${\Theta }_1<E{\int }_1^{10}{\displaystyle \frac{1}{30}}V\left(\tau \right)d\tau +{\int }_1^{10}V\left(\tau \right)h_1\left(\tau \right)d\tau \approx -0.0898<0$.

If ${}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\theta \right)<-E$, then ${\Theta }_1>-E{\int }_1^{10}{\displaystyle \frac{1}{30}}V\left(\tau \right)d\tau +{\int }_1^{10}V\left(\tau \right)h_2\left(\tau \right)d\tau \approx 0.06283>0$.

So, ${\Theta }_1\ne 0$.

If $|{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\theta \right)|>E$, we obtain

$$\begin{array}{c}{\Theta }_2={\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)={\int }_1^{20}\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)={\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma \hfill \\ ={\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\tau \right)d\tau \right)d\sigma .\hfill \end{array}$$

If ${}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\theta \right)>E$, then ${\Theta }_2>E{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -1)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}d\tau \right)d\sigma >0$.

If ${}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\theta \right)<-E$, then ${\Theta }_2<-E{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -1)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}d\tau \right)d\sigma <0$.

So, ${\Theta }_2\ne 0$.

We deduce that assumption $\left(I3\right)$ is satisfied.

In what follows, we verify assumption $\left(I4\right)$. We consider $r=50$ and $\mathfrak{u}\left(\theta \right)=b_1{(ln\theta )}^{8/3}+b_2{(ln\theta )}^{5/3}$ for $\theta \in [1,\infty )$, with $b_1,b_2\in \mathbb{R}$ and $|b_1|>r$ or $|b_2|>r$. We obtain

$$\begin{array}{c}{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\theta \right)=b_1\Gamma \left({\displaystyle \frac{11}{3}}\right),{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\theta \right)=b_1\Gamma \left({\displaystyle \frac{11}{3}}\right)ln\theta +b_2\Gamma \left({\displaystyle \frac{8}{3}}\right),\hfill \\ {}^{H}D_{1+}^{2/3}\mathfrak{u}\left(\theta \right)={\displaystyle \frac{b_1}{2}}\Gamma \left({\displaystyle \frac{11}{3}}\right){(ln\theta )}^2+b_2\Gamma \left({\displaystyle \frac{8}{3}}\right)ln\theta .\hfill \end{array}$$

Then, we find

$$\begin{array}{c}{\Theta }_0=b_1{\mathcal{C}}_1\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)+b_2{\mathcal{C}}_2\left(\mathcal{B}\mathfrak{u}\right)\left(\theta \right)\hfill \\ =b_1{\int }_1^{20}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)+b_2{\int }_1^{20}\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ =b_1\left[a_0{\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right]\hfill \\ +b_2{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma \hfill \\ =b_1\left\{a_0{\int }_1^{10}\left[{\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}\left(h\left(\tau \right)+{\displaystyle \frac{1}{30}}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)\right)d\tau \right]d\sigma \right.\hfill \\ +\left.(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}\left(h\left(\tau \right)+{\displaystyle \frac{1}{30}}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)\right)d\tau \right\}\hfill \\ +b_2{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\tau \right)d\tau \right)d\sigma \hfill \\ =b_1\left[a_0{\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \right)d\sigma \right.\hfill \\ +a_0{\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \right)d\sigma \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +\left.(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \right]\hfill \\ +b_2{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\tau \right)d\tau \right)d\sigma .\hfill \end{array}$$

So, we obtain

$$\begin{array}{c}{\Theta }_0=b_1\left\{a_0{\int }_1^{10}\left({\int }_{\tau }^{10}ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \right.\hfill \\ +a_0{\int }_1^{10}\left({\int }_{\tau }^{\infty }ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +\left.(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}{}^{H}D_{1+}^{5/3}\mathfrak{u}\left(\tau \right)d\tau \right\}\hfill \\ +b_2{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}{}^{H}D_{1+}^{8/3}\mathfrak{u}\left(\tau \right)d\tau \right)d\sigma \hfill \\ =b_1\left\{a_0{\int }_1^{10}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \right.\hfill \\ +a_0{\int }_1^{10}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}\left(b_1\Gamma \left({\displaystyle \frac{11}{3}}\right)ln\tau +b_2\Gamma \left({\displaystyle \frac{8}{3}}\right)\right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +\left.(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}\left(b_1\Gamma \left({\displaystyle \frac{11}{3}}\right)ln\tau +b_2\Gamma \left({\displaystyle \frac{8}{3}}\right)\right)d\tau \right\}\hfill \\ +b_2{\int }_{10}^{20}{\displaystyle \frac{1}{50}}(\sigma -10)\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{c_0}{5{\tau }^3}}{b}_1\Gamma \left({\displaystyle \frac{11}{3}}\right)d\tau \right)d\sigma \hfill \\ =C_1{b}_1^2+C_2{b}_1=b_1(C_1{b}_1+C_2),\hfill \end{array}$$

where

$$\begin{array}{c}{C}_1=a_0{\int }_1^{10}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}\Gamma \left({\displaystyle \frac{11}{3}}\right)ln\tau d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{30\tau }}{e}^{-4\tau +3}\Gamma \left({\displaystyle \frac{11}{3}}\right)ln\tau d\tau ,\hfill \\ C_2=a_0{\int }_1^{10}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}h\left(\tau \right)d\tau .\hfill \end{array}$$

We obtain $C_1\approx -0.00052405<0$ and $C_2\in [C_2^{\prime },C_2^{″}]$, where

$$\begin{array}{c}{C}_2^{\prime }=a_0{\int }_1^{10}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}{h}_2\left(\tau \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}{h}_2\left(\tau \right)d\tau \approx -0.02176398,\hfill \\ C_2^{″}=a_0{\int }_1^{10}\left(10ln{\displaystyle \frac{10}{\tau }}-10+\tau \right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}{h}_1\left(\tau \right)d\tau \hfill \\ +(1-9a_0){\int }_1^{10}\left(ln{\displaystyle \frac{10}{\tau }}\right){\displaystyle \frac{1}{\tau }}{e}^{-4\tau +3}{h}_1\left(\tau \right)d\tau \approx -0.00530448.\hfill \end{array}$$

So, we find $-{\displaystyle \frac{C_2}{C_1}}\in [-{\displaystyle \frac{C_2^{″}}{C_1}},-{\displaystyle \frac{C_2^{\prime }}{C_1}}],$ with $-{\displaystyle \frac{C_2^{″}}{C_1}}\approx -10.12211324$ and $-{\displaystyle \frac{C_2^{\prime }}{C_1}}\approx -41.53048712$. Then, if $|b_1|>50$, we deduce that ${\Theta }_0<0$. So, assumption $\left(I4\right)$ is satisfied.

In addition, ${\Xi }_0(2+{\Xi }_{1}lnF)\approx 0.29514557<1$, that is, $\left(I5\right)$ is also satisfied. Therefore, according to Theorem 1, we conclude that problem (62), (63), with the function $\mathfrak{f}$ given by (64), has at least one solution.

 Example 2. 

Let $\alpha ={\displaystyle \frac{9}{4}}$, $n=3$, $A=B=10$, and

$$\mathcal{P}\left(\sigma \right)=\left\{\begin{array}{c}{\sigma }^2+b_0(\sigma -1),\sigma \in [1,5],\hfill \\ 2,\sigma \in [5,10],\hfill \end{array}\right.\mathcal{Q}\left(\sigma \right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{4}},\sigma \in [1,5),\hfill \\ {\displaystyle \frac{2}{15}}(\sigma -10)+{\displaystyle \frac{5}{4}},\sigma \in [5,10],\hfill \end{array}\right.$$

where $b_0={\displaystyle \frac{2ln5-12}{4-ln5}}\approx -3.6732$.

We consider the following fractional differential equation:

$${}^{H}D_{1+}^{9/4}\mathfrak{u}\left(\theta \right)=\mathfrak{f}\left(\theta ,\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\theta \right),{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\theta \right)\right),\theta \in (1,\infty ),$$

(65)

supplemented with the following boundary conditions:

$$\left\{\begin{array}{c}\mathfrak{u}\left(1\right)=0,{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(1\right)={\int }_1^5(2\sigma +b_0){}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\sigma \right)d\sigma -(23+4b_0){}^{H}D_{1+}^{1/4}\mathfrak{u}\left(5\right),\hfill \\ {}^{H}D_{1+}^{5/4}\mathfrak{u}(\infty )={\displaystyle \frac{1}{3}}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(5\right)+{\displaystyle \frac{2}{15}}{\int }_5^{10}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\sigma \right)d\sigma .\hfill \end{array}\right.$$

(66)

After some computations, we obtain $\mathfrak{a}\approx 0.26300051$, $\mathfrak{b}\approx -0.18818034$, $\mathfrak{c}\approx 0.01333333$, $\mathfrak{d}\approx 0.03017152$, and $\Delta \approx 0.01044419\ne 0$. We also find ${\int }_1^{10}d\mathcal{P}\left(\sigma \right)=1$, ${\int }_1^{10}d\mathcal{Q}\left(\sigma \right)=1$, and ${\int }_1^{10}ln\sigma d\mathcal{P}\left(\sigma \right)=0$. So, assumption $\left(I1\right)$ is satisfied.

We consider the following function:

$$\begin{array}{c}\mathfrak{f}(\theta ,x_1,x_2,x_3)={\displaystyle \frac{1}{10}}{e}^{-2\theta +1}sin\left({\displaystyle \frac{x_1}{1+{(ln\theta )}^{5/4}}}\right)+{\displaystyle \frac{1}{20}}g\left(t\right)e^{-3\theta +2}{x}_2\hfill \\ +{\displaystyle \frac{1}{17}}(1-g\left(\theta \right))e^{-\theta +3}{x}_3+{\displaystyle \frac{1}{12{\theta }^2}},\theta \in [1,\infty ),x_i\in \mathbb{R},i=1,\dots ,3,\hfill \end{array}$$

(67)

where

$$g\left(\theta \right)=\left\{\begin{array}{c}1,\theta \in [1,5],\hfill \\ 0,\theta \in (5,\infty ).\hfill \end{array}\right.$$

We obtain

$$|\mathfrak{f}(\theta ,x_1,x_2,x_3)|\le p_1\left(\theta \right){\displaystyle \frac{|x_1|}{1+{(ln\theta )}^{5/4}}}+p_2\left(\theta \right){\displaystyle \frac{|x_2|}{1+ln\theta }}+p_3\left(\theta \right)\left|x_3\right|+p_4\left(\theta \right),$$

for all $\theta \in [1,\infty )$ and $x_i\in \mathbb{R},i=1,\dots ,3$, with

$$\begin{array}{c}{p}_1\left(\theta \right)={\displaystyle \frac{1}{10}}{e}^{-2\theta +1},p_2\left(\theta \right)=\left\{\begin{array}{c}{\displaystyle \frac{1}{20}}{e}^{-3\theta +2}(1+ln\theta ),\theta \in [1,5],\hfill \\ 0,\theta \in (5,\infty ),\hfill \end{array}\right.\hfill \\ p_3\left(\theta \right)=\left\{\begin{array}{c}0,\theta \in [1,5],\hfill \\ {\displaystyle \frac{1}{17}}{e}^{-\theta +3},\theta \in (5,\infty ),\hfill \end{array}\right.p_4\left(\theta \right)={\displaystyle \frac{1}{12{\theta }^2}}.\hfill \end{array}$$

We have $p_i\in L^1(1,\infty )$ for $i=1,\dots ,4$ and $\parallel p_1{\parallel }_{\mathcal{V}}\approx 0.01839397$, $\parallel p_2{\parallel }_{\mathcal{V}}\approx 0.00773814$, and $\parallel p_3{\parallel }_{\mathcal{V}}\approx 0.00796089$. In addition, we find ${\Xi }_0\approx 0.03409301$, ${\Xi }_1\approx 1.10326265$, ${\Xi }_2\approx 6.19909584$, and ${\Xi }_0{\Xi }_2\approx 0.21134587<1$. So, assumptions $\left(I2\right)$ and $\left(I8\right)$ are satisfied.

We now verify assumption $\left(I7\right)$. We consider $L=25$ and $M=5$. Let $\mathfrak{u}\in D\left(\mathcal{A}\right)$ satisfy ${|}^H{D}_{1+}^{1/4}\mathfrak{u}\left(\theta \right)|>25$ for all $\theta \in [1,5]$. Then, we obtain

$$\begin{array}{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)={\int }_1^{10}\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{P}\left(\sigma \right)\hfill \\ ={\int }_1^5(2\sigma +b_0)\left({\int }_1^{\sigma }\left(ln{\displaystyle \frac{\sigma }{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma -(23+4b_0)\left({\int }_1^5\left(ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)\hfill \\ ={\int }_1^5\left({\int }_{\tau }^5(2\sigma +b_0)ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau -(23+4b_0){\int }_1^5\left(ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \hfill \\ ={\int }_1^5\left({\int }_{\tau }^5(2\sigma +b_0)ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right){\displaystyle \frac{1}{\tau }}\left({\displaystyle \frac{1}{10}}{e}^{-2\tau +1}sin\left({\displaystyle \frac{\mathfrak{u}\left(\tau \right)}{1+{(ln\tau )}^{5/4}}}\right)+{\displaystyle \frac{1}{12{\tau }^2}}\right.\hfill \\ +\left.{\displaystyle \frac{1}{20}}{e}^{-3\tau +2}{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\tau \right)\right)d\tau \hfill \\ -(23+4b_0){\int }_1^5\left(ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{1}{\tau }}\left({\displaystyle \frac{1}{10}}{e}^{-2\tau +1}sin\left({\displaystyle \frac{\mathfrak{u}\left(\tau \right)}{1+{(ln\tau )}^{5/4}}}\right)+{\displaystyle \frac{1}{12{\tau }^2}}\right.\hfill \\ \left.+{\displaystyle \frac{1}{20}}{e}^{-3\tau +2}{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\tau \right)\right)d\tau \hfill \\ ={\int }_1^5\left({\int }_{\tau }^5(2\sigma +b_0)ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right){\displaystyle \frac{1}{\tau }}k\left(\tau \right)d\tau \hfill \\ +{\int }_1^5\left({\int }_{\tau }^5(2\sigma +b_0)ln{\displaystyle \frac{\sigma }{\tau }}d\sigma \right){\displaystyle \frac{1}{20\tau }}{e}^{-3\tau +2}{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\tau \right)d\tau \hfill \\ -(23+4b_0){\int }_1^5\left(ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{1}{\tau }}k\left(\tau \right)d\tau -(23+4b_0){\int }_1^5\left(ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{1}{20\tau }}{e}^{-3\tau +2}{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\tau \right)d\tau ,\hfill \end{array}$$

where

$$k\left(\tau \right)={\displaystyle \frac{1}{10}}{e}^{-2\tau +1}sin\left({\displaystyle \frac{\mathfrak{u}\left(\tau \right)}{1+{(ln\tau )}^{5/4}}}\right)+{\displaystyle \frac{1}{12{\tau }^2}},\tau \in [1,5].$$

Then, we find

$$\begin{array}{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)={\int }_1^5\left(U\left(\tau \right)-(23+4b_0)ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{1}{\tau }}k\left(\tau \right)d\tau \hfill \\ +{\int }_1^5\left(U\left(\tau \right)-(23+4b_0)ln{\displaystyle \frac{5}{\tau }}\right){\displaystyle \frac{1}{20\tau }}{e}^{-3\tau +2}{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\tau \right)d\tau ,\hfill \end{array}$$

where $U\left(\tau \right)=(25+5b_0)ln{\displaystyle \frac{5}{\tau }}-{\displaystyle \frac{25}{2}}-5b_0+{\displaystyle \frac{{\tau }^2}{2}}+b_0\tau$ for $\tau \in [1,5].$

We have $k_1\left(\tau \right)\le k\left(\tau \right)\le k_2\left(\tau \right)$ for all $\tau \in [1,5]$, where $k_1\left(\tau \right)=-{\displaystyle \frac{1}{10}}{e}^{-2\tau +1}+{\displaystyle \frac{1}{12{\tau }^2}}$ and $k_2\left(\tau \right)={\displaystyle \frac{1}{10}}{e}^{-2\tau +1}+{\displaystyle \frac{1}{12{\tau }^2}}$ for all $\tau \in [1,5]$. In addition, we find $U\left(\tau \right)\le 0$ for all $\tau \in [1,5]$ and ${inf}_{\tau \in [1,5]}{k}_1\left(\tau \right)\approx 0.0033>0$, so $k_1\left(\tau \right)>0$ for all $\tau \in [1,5]$. Then we have the following inequalities:

$$k_2\left(\tau \right)U\left(\tau \right)\le k\left(\tau \right)U\left(\tau \right)\le k_1\left(\tau \right)U\left(\tau \right),\forall \tau \in [1,5].$$

If ${}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\theta \right)>L$, then

$${\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)\le {\int }_1^{5}U\left(\tau \right){\displaystyle \frac{1}{\tau }}{k}_1\left(\tau \right)d\tau +L{\int }_1^{5}U\left(\tau \right){\displaystyle \frac{1}{20\tau }}{e}^{-3\tau +2}d\tau <0,\forall \theta \in [1,5],$$

so $sgn\left\{{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\theta \right)\right\}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)<0$ for all $\theta \in [1,5]$.

If ${}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\theta \right)<-L$, then

$$\begin{array}{c}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)\ge {\int }_1^{5}U\left(\tau \right){\displaystyle \frac{1}{\tau }}{k}_2\left(\tau \right)d\tau -L{\int }_1^{5}U\left(\tau \right){\displaystyle \frac{1}{20\tau }}{e}^{-3\tau +2}d\tau \hfill \\ \approx -0.02701926+0.00135717L>0,\hfill \end{array}$$

therefore, $sgn\left\{{}^{H}D_{1+}^{1/4}\mathfrak{u}\left(\theta \right)\right\}{\mathcal{C}}_1\mathcal{B}\mathfrak{u}\left(\theta \right)<0$ for all $\theta \in [1,5]$.

We deduce that assumption $\left(I7\right)$ is satisfied.

Finally, we verify assumption $\left(I6\right)$. We consider $G=5$. Let $\mathfrak{u}\in D\left(\mathcal{A}\right)$, which satisfies $|{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\theta \right)|>G$ for all $\theta \in [1,\infty )$. Then, we obtain

$$\begin{array}{c}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)={\int }_1^{10}\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\mathcal{Q}\left(\sigma \right)\hfill \\ ={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +{\displaystyle \frac{2}{15}}{\int }_5^{10}\left({\int }_{\sigma }^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma \hfill \\ ={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +{\displaystyle \frac{2}{15}}\left[{\int }_5^{10}\left({\int }_{\sigma }^{10}{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma +{\int }_5^{10}\left({\int }_{10}^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)d\sigma \right]\hfill \\ ={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +{\displaystyle \frac{2}{15}}\left[{\int }_5^{10}\left({\int }_5^{\tau }d\sigma \right){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +\left({\int }_{10}^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right)\left({\int }_5^{10}d\sigma \right)\right]\hfill \\ ={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +{\displaystyle \frac{2}{15}}\left[{\int }_5^{10}(\tau -5){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +5{\int }_{10}^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau \right]\hfill \\ ={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +{\displaystyle \frac{2}{15}}{\int }_5^{10}(\tau -5){\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{\mathcal{B}\mathfrak{u}\left(\tau \right)}{\tau }}d\tau .\hfill \end{array}$$

So, we find

$$\begin{array}{c}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{\tau }}\left[k\left(\tau \right)+{\displaystyle \frac{1}{17}}{e}^{-\tau +3}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\tau \right)\right]d\tau \hfill \\ +{\displaystyle \frac{2}{15}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}\left[k\left(\tau \right)+{\displaystyle \frac{1}{17}}{e}^{-\tau +3}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\tau \right)\right]d\tau \hfill \\ +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{\tau }}\left[k\left(\tau \right)+{\displaystyle \frac{1}{17}}{e}^{-\tau +3}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\tau \right)\right]d\tau \hfill \\ ={\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{\tau }}k\left(\tau \right)d\tau +{\displaystyle \frac{2}{15}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}k\left(\tau \right)d\tau +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{\tau }}k\left(\tau \right)d\tau \hfill \\ +{\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{17\tau }}{e}^{-\tau +3}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\tau \right)d\tau +{\displaystyle \frac{2}{255}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}{e}^{-\tau +3}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\tau \right)d\tau \hfill \\ +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{17\tau }}{e}^{-\tau +3}{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\tau \right)d\tau .\hfill \end{array}$$

If ${}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\theta \right)>G$, then we obtain

$$\begin{array}{c}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)\ge {\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{\tau }}{k}_1\left(\tau \right)d\tau +{\displaystyle \frac{2}{15}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}{k}_1\left(\tau \right)d\tau +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{\tau }}{k}_1\left(\tau \right)d\tau \hfill \\ +G\left({\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{17\tau }}{e}^{-\tau +3}d\tau +{\displaystyle \frac{2}{255}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}{e}^{-\tau +3}d\tau +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{17\tau }}{e}^{-\tau +3}d\tau \right)>0,\hfill \end{array}$$

so $sgn\left\{{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\theta \right)\right\}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)>0$ for all $\theta \in [1,\infty )$.

If ${}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\theta \right)<-G$, then we find

$$\begin{array}{c}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)\le {\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{\tau }}{k}_2\left(\tau \right)d\tau +{\displaystyle \frac{2}{15}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}{k}_2\left(\tau \right)d\tau +{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{\tau }}{k}_2\left(\tau \right)d\tau \hfill \\ -G\left({\displaystyle \frac{1}{3}}{\int }_5^{\infty }{\displaystyle \frac{1}{17\tau }}{e}^{-\tau +3}d\tau +{\displaystyle \frac{2}{255}}{\int }_5^{10}{\displaystyle \frac{\tau -5}{\tau }}{e}^{-\tau +3}d\tau \right.\hfill \\ +\left.{\displaystyle \frac{2}{3}}{\int }_{10}^{\infty }{\displaystyle \frac{1}{17\tau }}{e}^{-\tau +3}d\tau \right)\approx 0.00111156-0.00060861G<0,\hfill \end{array}$$

so $sgn\left\{{}^{H}D_{1+}^{5/4}\mathfrak{u}\left(\theta \right)\right\}{\mathcal{C}}_2\mathcal{B}\mathfrak{u}\left(\theta \right)>0$ for all $\theta \in [1,\infty )$.

We conclude that assumption $\left(I6\right)$ is satisfied.

Therefore, according to Theorem 2, we deduce that the boundary value problem (65), (66), with the function $\mathfrak{f}$ given by (67), has at least one solution.

5. Conclusions

In this paper, we studied the Hadamard fractional differential Equation (1) defined on the infinite interval of $(1,\infty )$, which involves fractional derivatives of various orders, supplemented with the integral boundary conditions (2). These nonlocal conditions encompass Hadamard fractional derivatives and Riemann–Stieltjes integrals. Problem (1), (2) is a resonant problem, because the corresponding homogeneous boundary value problem admits nontrivial solutions. We first expressed our problem as an operator equation in suitable Banach spaces containing two operators whose key properties were analyzed in detail. Subsequently, we established two existence theorems for the solutions of (1), (2) by employing the coincidence degree theory developed by J. Mawhin—more concretely, the Mawhin continuation theorem.