Existence and exact asymptotic behaviour of positive solutions for fractional boundary value problem with P-Laplacian operator

ABSTRACT

This paper deals with existence, uniqueness and global behaviour of a positive solution for the fractional boundary value problem $D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma }$ in $(0,1)$ with the condition $\underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)=\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\mathrm{and}\underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,$ where $\beta ,\alpha \in (1,2]$, ${\mathrm{\Phi }}_p\left(t\right)=t|t{|}^{p-2}$, p>1, $\sigma \in (1-p,p-1)$, the differential operator is taken in the Riemann–Liouville sense and $\psi ,a:(0,1)⟶\mathbb{R}$ are non-negative and continuous functions that may are singular at x=0 or x=1 and satisfies some appropriate conditions.

KEYWORDS:

• Fractional differential equation
• Dirichlet problem
• positive solution
• Schauder fixed point theorem

1. Introduction

In this paper, we consider the following nonlinear boundary value problem of fractional differential equation with p-Laplacian (1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) where $\alpha ,\beta \in (1,2]$ and for $t\in \mathbb{R}$, ${\mathrm{\Phi }}_p\left(t\right):=t|t{|}^{p-2}$, $(p>1)$ and $\sigma \in (1-p,p-1)$. The operator $D^{\alpha }$ (and $D^{\beta }$) is the Riemann–Liouville derivative of order α (of order β). The functions ψ and a are positive and continuous in $(0,1)$ that may are singular at x=0 or x=1 and satisfying some conditions detailed below. Then, we will adress the question of existence, uniqueness and exact asymptotic behaviour of positive solutions to problem (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ).

We believe that there are few papers dedicated to the study of fractional differential equations with p-Laplacian operator, see for instance [1–11]. These types of problems arise in applied field such as turbulent flow of a gas in a porous medium, biophysics, plasma physics and material science.

This work is motivated by recent advances in the study of p-Laplacian fractional differential equations involving singular or sublinear nonlinearities with different boundary conditions. Namely, in [12], Liu considered the fractional differential equation (2) $D^{\beta }\left({\mathrm{\Phi }}_p\right(\psi \left(x\right)D^{\alpha }u\left(x\right)\left)\right)=f(x,u(x\left)\right),x\in (0,1),$(2) where $0<\alpha ,\beta \le 1$, $\psi \in C\left(\right(0,1\left)\right)$ and f is a non-negative function on $(0,1]\times \mathbb{R}$ allowed to be singular at t=0. The author proved the existence of positive solution with fractional non-local integral boundary conditions.

Recently, in [13], Mâagli et al. considered the following problem (3) $\begin{array}{rl}& D^{\alpha }u\left(x\right)=-a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0^{+}}{lim}{D}^{\alpha -1}u\left(x\right)=0,u\left(1\right)=0,\end{array}$(3) where $1<\alpha \le 2$, $-1<\sigma <1$ and the function a is required to satisfy some assumptions related to $\mathcal{K}$, the set of all Karamata functions L defined on $(0,\eta ]$, by $L\left(t\right):=c\exp \left({\int }_t^{\eta }\frac{z\left(s\right)}{s}\mathrm{d}s\right),$ for some $\eta >1,$ where c>0 and z is a continuous function on $[0,\eta ],$ with $z\left(0\right)=0.$ To describe the result of [13] in more details, we need some notations.

• For two non-negative functions f and g defined on a set S, the notation $f\left(x\right)\approx g\left(x\right)$, $x\in S$ means that there exists c>0 such that $\left(1/c\right)f\left(x\right)\le g\left(x\right)\le cf\left(x\right)$, for all $x\in S$.

• ${\mathcal{C}}_{2-\alpha }\left(\right[0,1\left]\right)=\left\{f\right|t\to t^{2-\alpha }f\left(t\right)\in \mathcal{C}\left(\right[0,1\left]\right)\}$.

In [13], Mâagli et al. studied problem (3(3) $\begin{array}{rl}& D^{\alpha }u\left(x\right)=-a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0^{+}}{lim}{D}^{\alpha -1}u\left(x\right)=0,u\left(1\right)=0,\end{array}$(3) ) where a verifies the assumption ${\mathbf{(}\mathbf{H}}_0)$:

${\mathbf{(}\mathbf{H}}_0):$ $a\in C\left(\right(0,1\left)\right)$ satisfying for each $x\in$ $(0,1),$ (4) $a\left(x\right)\approx x^{-\lambda }(1-x{)}^{-\mu }L(x\left)\tilde{L}\right(1-x),$(4) where $\lambda +(2-\alpha )\sigma \le 1$, $\mu \le \alpha$ and $L,\tilde{L}\in \mathcal{K}$ such that (5) ${\int }_0^{\eta }\frac{L\left(t\right)}{t^{\lambda +(2-\alpha )\sigma }}\mathrm{d}t<\mathrm{\infty }\mathrm{and}{\int }_0^{\eta }\frac{\tilde{L}\left(t\right)}{t^{\mu -\alpha +1}}\mathrm{d}t<\mathrm{\infty }.$(5) Based on the Schauder fixed-point theorem, the authors showed in [13] the following result.

Theorem 1.1

Assume that a satisfies $\left(H_0\right)$. Then problem (3(3) $\begin{array}{rl}& D^{\alpha }u\left(x\right)=-a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0^{+}}{lim}{D}^{\alpha -1}u\left(x\right)=0,u\left(1\right)=0,\end{array}$(3) ) has a unique positive solution $u\in C_{2-\alpha }\left(\right[0,1\left]\right)$ satisfying for $x\in (0,1)$, (6) $u\left(x\right)\approx x^{\alpha -2}(1-x{)}^{min(1,(\left(\alpha -\mu \right)/\left(1-\sigma \right)\left)\right)}{\mathrm{\Pi }}_{\tilde{L},\mu ,\sigma ,\alpha }(1-x),$(6) where ${\mathrm{\Pi }}_{\tilde{L},\mu ,\sigma ,\alpha }$ defined on $(0,1)$ by (7) $\begin{array}{rl}& {\mathrm{\Pi }}_{\tilde{L},\mu ,\sigma ,\alpha }\left(t\right)\\ & :=\left\{\begin{array}{ll}1,& \mathrm{if}\mu <\sigma +\alpha -1,\\ {\left({\int }_t^{\eta }\frac{\tilde{L}\left(s\right)}{s}\mathrm{d}s\right)}^{1/\left(1-\sigma \right)},& \mathrm{if}\mu =\sigma +\alpha -1,\\ {\left(\tilde{L}\left(t\right)\right)}^{1/\left(1-\sigma \right)},& \mathrm{if}\sigma +\alpha -1<\mu <\alpha ,\\ {\left({\int }_0^t\frac{\tilde{L}\left(s\right)}{s}\mathrm{d}s\right)}^{1/\left(1-\sigma \right)},& \mathrm{if}\mu =\alpha .\end{array}\right.\end{array}$(7)

The main goal of this paper is to improve and extend the above results on the boundary behaviour of solutions to problem (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ). More precisely, we consider the following assumption: ${\mathbf{(}\mathbf{H}}_1):$ a and ψ are non-negative functions in $C\left(\right(0,1\left)\right)$ satisfying for each $x\in$ $(0,1),$ (8) $\begin{array}{rl}a\left(x\right)& \approx x^{-\lambda }(1-x{)}^{-\mu }{L}_1(x\left)L_2\right(1-x)\mathrm{and}\\ \psi \left(x\right)& \approx (1-x{)}^{(p-1)\alpha +\beta -\mu },\end{array}$(8) where $\lambda +(2-\alpha )\sigma \le 1$, $\beta -1<\mu <\beta$, $3<\beta +p$ and $L_1,L_2\in \mathcal{K}$ such that (9) ${\int }_0^{\eta }\frac{L_1\left(t\right)}{t^{\lambda +(2-\alpha )\sigma }}\mathrm{d}t<\mathrm{\infty }\mathrm{and}{\int }_0^{\eta }\frac{\left(L_2\right(t){)}^{\frac{1}{p-1}}}{t}\mathrm{d}t<\mathrm{\infty }.$(9)

Theorem 1.2

Assume ${\mathbf{(}\mathbf{H}}_1)$, then problem (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ) has a unique positive solution $u\in C_{2-\alpha }\left(\right[0,1\left]\right)$ satisfying for $x\in (0,1)$, (10) $u\left(x\right)\approx x^{\alpha -2}{\left({\int }_0^{1-x}\frac{\left(L_2\right(t\left){)}^1/\right(p-1)}{t}\mathrm{d}t\right)}^{\left(p-1\right)/\left(p-1-\sigma \right)}.$(10)

The rest of the paper is as follows. In Section 2, we state some already known results on functions in $\mathcal{K}$ and some definitions and lemmas from fractional calculus theory. In Section 3, we present some necessary conditions to existence result and we prove our main result.

2. Preliminaries: Karamata properties and fractional calculus

In what follows, we are quoting without proof some fundamental properties of functions belonging to the class $\mathcal{K}$ collected from [14, 15]. We recall that a function L defined on $(0,\eta ]$ belongs to the class $\mathcal{K}$, if (11) $L\left(t\right):=c\exp \left({\int }_t^{\eta }\frac{z\left(s\right)}{s}\mathrm{d}s\right),$(11) for some $\eta >1,$ c>0 and z is a continuous function on $[0,\eta ]$ with $z\left(0\right)=0.$

Proposition 2.1

1. A function L is in $\mathcal{K}$ if and only if L is a positive function in ${\mathcal{C}}^1\left(\right(0,\eta \left]\right)$ such that (12) $\underset{t\to 0^{+}}{lim}\frac{t{L}^{{}^{\mathrm{\prime }}}\left(t\right)}{L\left(t\right)}=0.$(12)

2. Let $L_1,L_2\in \mathcal{K}$ and $p\in \mathbb{R}$. Then we have (13) $L_1+L_2\in \mathcal{K},L_1{L}_2\in \mathcal{K}\mathrm{and}{L}_1^p\in \mathcal{K}.$(13)

3. Let $L\in \mathcal{K}$ and $\epsilon >0.$ Then we have (14) $\underset{t\to 0^{+}}{lim}{t}^{\epsilon }L\left(t\right)=0.$(14)

As a standard example of functions belonging to the class $\mathcal{K}$ (see [15]), we give

Example 2.2

Let $m\in {\mathbb{N}}^{\ast }$ and $\eta >0$. Let $({\mu }_1,{\mu }_2,\dots ,{\mu }_m)\in {\mathbb{R}}^m$ and w be a sufficiently large positif real number such that the function (15) $L\left(t\right)=\prod _{1\le i\le m}{\left({\log }_i\left(\frac{w}{t}\right)\right)}^{{\mu }_i}$(15) is defined and positive on $(0,\eta ]$, where ${\log }_{i}t=\log \circ \cdots .\circ \log t(i\times )$. Then we have $L\in \mathcal{K}$.

Lemma 2.3

1. Let L be a function in $\mathcal{K}$. Then we have (16) $\underset{t\to 0^{+}}{lim}\frac{L\left(t\right)}{{\displaystyle {\int }_t^{\eta }\frac{L\left(s\right)}{s}\mathrm{d}s}}=0.$(16) In particular, (17) $t⟼{\int }_t^{\eta }\frac{L\left(s\right)}{s}\mathrm{d}s\in \mathcal{K}.$(17)

2. If ${\displaystyle {\int }_0^{\eta }\frac{L\left(s\right)}{s}\mathrm{d}s}$ converges, then (18) $\underset{t\to 0^{+}}{lim}\frac{L\left(t\right)}{{\displaystyle {\int }_0^t\frac{L\left(s\right)}{s}\mathrm{d}s}}=0.$(18) In particular, (19) $t⟼{\int }_0^t\frac{L\left(s\right)}{s}\mathrm{d}s\in \mathcal{K}.$(19)

Applying Karamata's theorem, we get the following.

Lemma 2.4

Let $\gamma \in \mathbb{R}$ and L be a function in $\mathcal{K}$ defined on $(0,\eta ]$ for some $\eta >1$. We have

1. If $\gamma >-1$, then ${\int }_0^{\eta }{s}^{\gamma }L\left(s\right)\mathrm{d}s$ converges and (20) ${\int }_0^t{s}^{\gamma }L\left(s\right)\mathrm{d}s\underset{t\to 0^{+}}{\sim }\frac{t^{1+\gamma }L\left(t\right)}{1+\gamma }.$(20)

2. If $\gamma <-1$, then ${\int }_0^{\eta }{s}^{\gamma }L\left(s\right)\mathrm{d}s$ diverges and (21) ${\int }_t^{\eta }{s}^{\gamma }L\left(s\right)\mathrm{d}s\underset{t\to 0^{+}}{\sim }-\frac{t^{1+\gamma }L\left(t\right)}{1+\gamma }.$(21)

Next, we give some definitions and fundamental facts of fractional calculus theory, which can be found in [16, 17].

Definition 2.5

Let $\gamma >0$, the Riemann–Liouville fractional integral of order γ of a measurable function $f:(0,\mathrm{\infty })⟶\mathbb{R}$ is given by (22) $I^{\gamma }f\left(x\right)=\frac{1}{\mathrm{\Gamma }\left(\gamma \right)}{\int }_0^x(x-t{)}^{\gamma -1}f(t)\mathrm{d}t,$(22) provided that the right-hand side is pointwise defined on $(0,\mathrm{\infty })$.

Here Γ is the Euler Gamma function.

Definition 2.6

The Riemann–Liouville derivative of order $\gamma >0$ of a measurable function

$f:(0,\mathrm{\infty })⟶\mathbb{R}$ is given by $\begin{array}{rl}{D}^{\gamma }f\left(x\right)& =\frac{1}{\mathrm{\Gamma }(n-\gamma )}(\frac{\mathrm{d}}{\mathrm{d}x}{)}^n{\int }_0^x(x-t{)}^{n-\gamma -1}f(t)\mathrm{d}t\\ & =(\frac{\mathrm{d}}{\mathrm{d}x}{)}^n{I}^{n-\gamma }f\left(x\right),\end{array}$ provided that the right-hand side is pointwise defined on $(0,\mathrm{\infty })$.

Here $n=\left[\gamma \right]+1$, where $\left[\gamma \right]$ means the integer part of the number γ.

Lemma 2.7

Let $\alpha ,\gamma >0$, n be the smallest integer greater than or equal to α and $f\in L^1\left(\right(0,1\left]\right)$. Then, we have

1. $I^{\alpha }{I}^{\gamma }f\left(x\right)=I^{\alpha +\gamma }f\left(x\right)$ for $x\in [0,1]$ and $\alpha +\gamma \ge 1.$

2. $D^{\alpha }{I}^{\alpha }f\left(x\right)=f\left(x\right)$ for a.e. $x\in [0,1]$.

3. $D^{\alpha }f\left(x\right)=0$ if and only if $f\left(x\right)=\sum _{j=1}^n{c}_j{x}^{\alpha -j}$, where $(c_1,\dots ,c_n)\in {\mathbb{R}}^n$.

Since our approach is based on potential theory, we recall in the following some basic tools. For $\gamma \in (1,2]$, we denote by $G_{\gamma }(x,t)$ the Green's function for the boundary value problem (3(3) $\begin{array}{rl}& D^{\alpha }u\left(x\right)=-a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0^{+}}{lim}{D}^{\alpha -1}u\left(x\right)=0,u\left(1\right)=0,\end{array}$(3) ). From [13], we have (23) $G_{\gamma }(x,t)=\frac{1}{\mathrm{\Gamma }\left(\gamma \right)}\left[x^{\gamma -2}\right(1-t{)}^{\gamma -1}-\left(\right(x-t{)}^{+}{)}^{\gamma -1}],$(23) where $x^{+}=max\{x,0\}$.

Proposition 2.8

see [13]

Let $1<\gamma \le 2$ and f be a non-negative measurable function on $(0,1)$. Then we have

1. For $x,t\in (0,1)$, (24) $G_{\gamma }(x,t)\approx x^{\gamma -2}(1-t{)}^{\gamma -2}(1-max(x,t)).$(24)

2. For $x\in (0,1)$, $G_{\gamma }f\left(x\right):={\int }_0^1{G}_{\gamma }(x,t)f\left(t\right)\mathrm{d}t<\mathrm{\infty }$ if and only if ${\int }_0^1(1-t{)}^{\gamma -1}f(t)\mathrm{d}t<\mathrm{\infty }$.

3. If the map $t\to (1-t{)}^{\gamma -1}f(t)$ is continuous and integrable on $(0,1)$, then $G_{\gamma }f$ is the unique solution in $C_{2-\gamma }\left(\right[0,1\left]\right)$ of the boundary value problem (25) $\begin{array}{rl}& D^{\gamma }u\left(x\right)=-f\left(x\right),x\in (0,1),\\ & \underset{x\to 0^{+}}{lim}{D}^{\gamma -1}u\left(x\right)=0,u\left(1\right)=0.\end{array}$(25)

Lemma 2.9

[13]

Let ${\tilde{L}}_1,{\tilde{L}}_2\in \mathcal{K}$ and let for $x\in (0,1)$ (26) $b\left(t\right)=t^{-{\lambda }_1}(1-t{)}^{-{\mu }_1}{\tilde{L}}_1(t\left){\tilde{L}}_2\right(1-t),$(26) with ${\lambda }_1\le 1$ and ${\mu }_1\le \beta$. Assume that (27) ${\int }_0^{\eta }{t}^{-{\lambda }_1}{\tilde{L}}_1\left(t\right)\mathrm{d}t<\mathrm{\infty }\mathrm{and}{\int }_0^{\eta }{t}^{\beta -1-{\mu }_1}{\tilde{L}}_2\left(t\right)\mathrm{d}t<\mathrm{\infty }.$(27) Then we have for $x\in (0,1)$, (28) $x^{2-\beta }{G}_{\beta }b\left(x\right)\approx (1-x{)}^{min(1,\beta -{\mu }_1}){\mathrm{\Pi }}_{{\tilde{L}}_2,{\mu }_1,0,\beta }(1-x).$(28)

Remark 2.1

We need to verify condition ${\int }_0^{\eta }{t}^{-{\lambda }_1}{\tilde{L}}_1\left(t\right)\mathrm{d}t<\mathrm{\infty }$ and ${\int }_0^{\eta }{t}^{\beta -1-{\mu }_1}{\tilde{L}}_2\left(t\right)\mathrm{d}t<\mathrm{\infty }$ in Lemma 6, only if ${\lambda }_1=1$ and ${\mu }_1=\beta$. This is due to Lemma 2.

3. Proof of main result

We begin this section by announcing and proving some propositions and lemmas, that will play a crucial role in the proof of our main result.

Let p>1 and let $q=p/\left(p-1\right)$. Then ${\mathrm{\Phi }}_q\left(t\right)=t|t{|}^{\left(2-p\right)/\left(p-1\right)}$ is the function inverse of ${\mathrm{\Phi }}_p$. We say easily that ${\mathrm{\Phi }}_q$ is increasing monotone and multiplicative on $(0,\mathrm{\infty })$.

Lemma 3.1

Let $\alpha ,\beta \in (1,2]$. Let f be a non-negative function in $\mathcal{B}\left(\right(0,1\left)\right)$ such that $x⟼(1-x{)}^{\beta -1}f(x)$ and $x⟼(1-x{)}^{\alpha -1}{\mathrm{\Phi }}_q(\left(1/\psi \right)G_{\beta }f\left)\right(x)$ are continuous and integrable on $(0,1)$, then the boundary value problem (29) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)=f\left(x\right),x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right)\left(x\right)=0,\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(29) has a unique positive solution in ${\mathcal{C}}_{2-\alpha }\left(\right[0,1\left]\right)$ given by (30) $\begin{array}{rl}u\left(x\right)& =G_{\alpha }({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }f\right))\left(x\right)\\ & :={\int }_0^1{G}_{\alpha }(x,t){\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{\int }_0^1{G}_{\beta }(t,s)f\left(s\right)\mathrm{d}s\right)\mathrm{d}t.\end{array}$(30)

Proof.

Since $x⟼(1-x{)}^{\beta -1}f(x)$ is continuous and integrable on $(0,1)$, we deduce by Proposition 2 $\left(iii\right)$ that for $x\in (0,1)$, we have (31) $\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right)\left(x\right)=-G_{\beta }f\left(x\right).$(31) Thus (32) $D^{\alpha }u\left(x\right)=-{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(x\right)}{G}_{\beta }f\left(x\right)\right).$(32) In addition, using the fact that $x⟼(1-x{)}^{\alpha -1}{\mathrm{\Phi }}_q(\left(1/\psi \left(x\right)\right)G_{\beta }f\left(x\right))$ is continuous and integrable on $(0,1)$, we conclude again by Proposition 2 $\left(iii\right)$ that problem (29(29) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)=f\left(x\right),x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right)\left(x\right)=0,\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(29) ) has a unique solution u in ${\mathcal{C}}_{2-\alpha }\left(\right[0,1\left]\right)$ given by (33) $u\left(x\right)=G_{\alpha }({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }f\right))\left(x\right),0<x<1.$(33)

Here, below we provide a crucial property concerning continuity.

Lemma 3.2

Let ϕ and ψ be tow non-negative functions in $\mathcal{B}\left(\right(0,1\left)\right)$ such that (34) ${\int }_0^1(1-t{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\varphi \left(t\right)\right)\mathrm{d}t<\mathrm{\infty }.$(34) Then the family (35) $\mathcal{F}=\{Sf:x⟼x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }f\right)\right)(x);|f|\le \varphi \}$(35) is relatively compact in $C\left(\right[0,1\left]\right)$.

Proof.

Let $f\in \mathcal{B}\left(\right(0,1\left)\right)$ such that $|f\left(x\right)|\le \varphi \left(x\right)$ for all $x\in (0,1)$.

Let $x\in (0,1)$, we have $\begin{array}{rl}Sf\left(x\right)& =x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }f\right)\right)\left(x\right)\\ & =x^{2-\alpha }{\int }_0^1{G}_{\alpha }(x,t){\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }f\left(t\right)\right)\mathrm{d}t.\end{array}$ Using Proposition 2 $\left(i\right)$, we obtain that (36) $|Sf\left(x\right)|\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1(1-t{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\varphi \left(t\right)\right)\mathrm{d}t<\mathrm{\infty }.$(36) Thus $\mathcal{F}$ is uniformly bounded. Now, let us prove that $\mathcal{F}$ is equicontinuous in $[0,1]$.

Let $x,y\in (0,1)$, then we have $\begin{array}{rl}& |Sf\left(x\right)-Sf\left(y\right)|=|x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }f\right)\right)\left(x\right)\\ & -y^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }f\right)\right)\left(y\right)|\\ & =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}|{\int }_0^1\left(y^{2-\alpha }\left(\right(y-t{)}^{+}{)}^{\alpha -1}-x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}\right)\\ & {\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }f\left(t\right)\right)\mathrm{d}t|\\ & \le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1|y^{2-\alpha }\left(\right(y-t{)}^{+}{)}^{\alpha -1}-x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}|\\ & {\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\varphi \left(t\right)\right)\mathrm{d}t.\end{array}$ For every $t\in (0,1)$, we have (37) $\begin{array}{rl}& |y^{2-\alpha }\left(\right(y-t{)}^{+}{)}^{\alpha -1}-x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}|⟶0\\ & \mathrm{as}|x-y|⟶0,\end{array}$(37) and (38) $\begin{array}{rl}& |y^{2-\alpha }\left(\right(y-t{)}^{+}{)}^{\alpha -1}-x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}|\\ & \le 2(1-t{)}^{\alpha -1}.\end{array}$(38) Then we obtain by Lesbegue's theorem that (39) $|Sf\left(x\right)-Sf\left(y\right)|⟶0\mathrm{as}|x-y|⟶0.$(39) Now, let $x\in (0,1)$, we have $\begin{array}{rl}& |Sf\left(x\right)-\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1(1-t{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }f\left(t\right)\right)\mathrm{d}t|\\ & =\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}|{\int }_0^1{x}^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }f\left(t\right)\right)\mathrm{d}t|\\ & \le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1{x}^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\varphi \left(t\right)\right)\mathrm{d}t.\end{array}$ Using the fact that for $t\in (0,1)$, (40) $x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}⟶0\mathrm{as}x⟶0$(40) and (41) $0\le x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}\le (1-t{)}^{\alpha -1},$(41) we get again by Lesbegue's theorem that (42) $\begin{array}{rl}& |Sf\left(x\right)-\frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1(1-t{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }f\left(t\right)\right)\mathrm{d}t|\\ & ⟶0\mathrm{as}x⟶0.\end{array}$(42) Furthermore, for $x\in (0,1)$, we have (43) $\begin{array}{rl}& |Sf\left(x\right)|\le \frac{1}{\mathrm{\Gamma }\left(\alpha \right)}{\int }_0^1\left(\right(1-t{)}^{\alpha -1}-x^{2-\alpha }\left(\right(x-t{)}^{+}{)}^{\alpha -1}){\mathrm{\Phi }}_q\\ & \left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\varphi \left(t\right)\right)\mathrm{d}t.\end{array}$(43) By similar arguments as above, we deduce that (44) $|Sf\left(x\right)|⟶0\mathrm{as}x⟶1.$(44) Finally, we conclude that the familly $\mathcal{F}$ is equicontinuous in $[0,1]$. Hence, by Ascoli's theorem, we deduce that $\mathcal{F}$ is relatively compact in $\mathcal{C}\left(\right[0,1\left]\right)$.

Proposition 3.3

Assume ${\mathbf{(}\mathbf{H}}_1\mathbf{)}$ and suppose that there exist a non-negative function θ in $C\left(\right[0,1\left]\right)$ such that (45) ${\int }_0^1(1-t{)}^{\beta -1}w(t)\mathrm{d}t<\mathrm{\infty }$(45) and (46) $x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\approx \theta (1-x),$(46) where $w\left(t\right)=a\left(t\right)t^{(\alpha -2)\sigma }{\theta }^{\sigma }(1-t)$ for $t\in (0,1)$.

Then problem  (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ) has a unique solution $u\in C_{2-\alpha }\left(\right[0,1\left]\right)$ satisfying for each $x\in (0,1)$ (47) $u\left(x\right)\approx x^{\alpha -2}\theta (1-x).$(47)

Proof.

Let $m\ge 1$ and θ be a non-negative function satisfying for each $x\in (0,1)$ (48) $\frac{1}{m}\theta (1-x)\le x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\le m\theta (1-x).$(48) Put $c_0:=m^{\left(p-1\right)/\left(p-1-|\sigma |\right)}$. We consider the closed convex set given by (49) $Y:=\{v\in \mathcal{C}([0,1]);\frac{1}{c_0}\theta (1-x)\le v(x)\le c_0\theta (1-x\left)\right\}.$(49) Using Lemma 3.2 and Proposition 2 (ii), we easily see that the function $x\mapsto x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\right(\left(1/\psi \right)G_{\beta }w\left)\right)\left(x\right)$ belongs to $\mathcal{C}\left(\right[0,1\left]\right)$ and satisfies (48(48) $\frac{1}{m}\theta (1-x)\le x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\le m\theta (1-x).$(48) ). So Y is not empty. In order to use the Schauder's fixed point theorem, we denote $\tilde{a}\left(x\right)=x^{(\alpha -2)\sigma }a\left(x\right)$ and we define the operator T on Y by (50) $Tv\left(x\right)=x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }\left(\tilde{a}{v}^{\sigma }\right)\right)\right)\left(x\right).$(50) We need to check that the operator T has a fixed point v in Y. For this choice of $c_0$, we will prove that T map Y into itself. Indeed, let $v\in Y$, by using (48(48) $\frac{1}{m}\theta (1-x)\le x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\le m\theta (1-x).$(48) ), we have $\begin{array}{rl}Tv\left(x\right)& \le {\mathrm{\Phi }}_q\left(c_0^{|\sigma |}\right)x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\\ & \le c_0^{|\sigma |/p-1}m\theta (1-x)\\ & =c_0\theta (1-x)\end{array}$ and $\begin{array}{rl}Tv\left(x\right)& \ge {\mathrm{\Phi }}_q\left(c_0^{-|\sigma |}\right)x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\\ & \ge \frac{c_0^{-|\sigma |/p-1}}{m}\theta (1-x)\\ & =\frac{1}{c_0}\theta (1-x).\end{array}$ Furthermore, using (48(48) $\frac{1}{m}\theta (1-x)\le x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\le m\theta (1-x).$(48) ), we have for each $x\in (0,1)$ (51) $G_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)<\mathrm{\infty }.$(51) This implies by Proposition 2 (ii) that (52) ${\int }_0^1(1-t{)}^{\alpha -1}{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }w\left(t\right)\right)\mathrm{d}t<\mathrm{\infty }.$(52) Hence, it follows from Lemma 3.2 that the family TY is relatively compact in $\mathcal{C}\left(\right[0,1\left]\right)$. So Y is invariant under T.

Next, we shall prove the continuity of T. Let $\left\{v_k\right\}$ be a sequence in Y which converges uniformly to v in Y.

For $x\in (0,1)$, we have $\begin{array}{rl}& |T{v}_k\left(x\right)-Tv\left(x\right)|=x^{2-\alpha }|G_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }\left(\tilde{a}{v}_k^{\sigma }\right)\right)\right)\left(x\right)\\ & -G_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }\left(\tilde{a}{v}^{\sigma }\right)\right)\right)\left(x\right)|\\ & \le x^{2-\alpha }{\int }_0^1{G}_{\alpha }(x,t)|{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\left(\tilde{a}{v}_k^{\sigma }\right)\left(t\right)\right)\\ & -{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\left(\tilde{a}{v}^{\sigma }\right)\left(t\right)\right)|\mathrm{d}t.\end{array}$ For $t\in (0,1)$, we have (53) $\begin{array}{rl}& |G_{\beta }\left(\tilde{a}{v}_k^{\sigma }\right)\left(t\right)-G_{\beta }\left(\tilde{a}{v}^{\sigma }\right)\left(t\right)|\\ & \le {\int }_0^1{G}_{\beta }(t,s)|\left(\tilde{a}{v}_k^{\sigma }\right)\left(s\right)-\left(\tilde{a}{v}^{\sigma }\right)\left(s\right)|\mathrm{d}s\end{array}$(53) and for every $s\in (0,1)$, (54) $|\left(\tilde{a}{v}_k^{\sigma }\right)\left(s\right)-\left(\tilde{a}{v}^{\sigma }\right)\left(s\right)|\le 2c_0^{|\sigma |}w\left(s\right).$(54) Using Proposition 2 (ii) and (45(45) ${\int }_0^1(1-t{)}^{\beta -1}w(t)\mathrm{d}t<\mathrm{\infty }$(45) ), we obtain by Lesbegue's theorem that (55) $|G_{\beta }\left(\tilde{a}{v}_k^{\sigma }\right)\left(t\right)-G_{\beta }\left(\tilde{a}{v}^{\sigma }\right)\left(t\right)|⟶0\mathrm{as}k⟶\mathrm{\infty }.$(55) Since ${\mathrm{\Phi }}_q$ is continuous, we deduce that (56) $\begin{array}{rl}& |{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\left(\tilde{a}{v}_k^{\sigma }\right)\left(t\right)\right)-{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\left(\tilde{a}{v}^{\sigma }\right)\left(t\right)\right)|\\ & ⟶0\mathrm{as}k⟶\mathrm{\infty }.\end{array}$(56) We have (57) $\begin{array}{rl}& |{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }\left(\tilde{a}{v}_k^{\sigma }\right)\left(t\right)\right)-{\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }(\tilde{a}{v}^{\sigma }\right)\left(t\right))|\\ & \le 2{\mathrm{\Phi }}_q\left(c_0^{|\sigma |}\right){\mathrm{\Phi }}_q\left(\frac{1}{\psi \left(t\right)}{G}_{\beta }w\left(t\right)\right).\end{array}$(57) Using (48(48) $\frac{1}{m}\theta (1-x)\le x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\le m\theta (1-x).$(48) ), we obtain by Lesbegue's theorem that for $x\in [0,1]$ (58) $T{v}_k\left(x\right)⟶Tv\left(x\right),\mathrm{as}k⟶\mathrm{\infty }.$(58) Since TY is a relatively compact in $\mathcal{C}\left(\right[0,1\left]\right)$, we have the uniform convergence, namely (59) $\parallel T{v}_k-Tv{\parallel }_{\mathrm{\infty }}⟶0,\mathrm{as}k⟶\mathrm{\infty }.$(59) Thus, we have proved that T is a compact mapping from Y into itself. It follows by the Schauder fixed point theorem that there exists $v\in Y$ such that Tv=v. Put (60) $u\left(x\right)=x^{\alpha -2}v\left(x\right),$(60) then $u\in {\mathcal{C}}_{2-\alpha }\left(\right[0,1\left]\right)$ and u satisfies the equation (61) $u\left(x\right)=G_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }\left(a{u}^{\sigma }\right)\right)\right)\left(x\right).$(61) Then due to Lemma 3.1, u is a positive continuous solution of problem (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ).

Finally, let us prove that u is the unique positive continuous solution satisfying (47(47) $u\left(x\right)\approx x^{\alpha -2}\theta (1-x).$(47) ). To this aim, we assume that $\left(1.1\right)$ has two positive solutions u and v satisfying (47(47) $u\left(x\right)\approx x^{\alpha -2}\theta (1-x).$(47) ). Then, there exists a constant m>1 such that (62) $\frac{1}{m}v\le u\le mv.$(62) This implies that the set (63) $J:=\left\{t\in \left(1,\mathrm{\infty }\right):\frac{1}{t}v\le u\le tv\right\}$(63) is not empty. Now, put $c:=infJ,$ then we aim to show that $c=1.$ Suppose that c>1. Then by simple calculus, we obtain that (64) $\begin{array}{rl}& D^{\beta }\left(\psi \left(x\right)\left({\mathrm{\Phi }}_p\right(D^{\alpha }\left(c^{\left(|\sigma |/\right(p-1\left)\right)}v\right))-{\mathrm{\Phi }}_p(D^{\alpha }u\left)\right)\right)\\ & =a(c^{|\sigma |}{v}^{\sigma }-u^{\sigma })\ge 0,\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \left(x\right)\left({\mathrm{\Phi }}_p\right(D^{\alpha }\left(c^{\left(|\sigma |/\right(p-1\left)\right)}v\right))-{\mathrm{\Phi }}_p(D^{\alpha }u\left)\right)\right)\\ & \left(x\right)=0,\\ & \underset{x\to 1}{lim}\psi \left(x\right)\left({\mathrm{\Phi }}_p\left(D^{\alpha }\right(c^{\left(|\sigma |/\right(p-1\left)\right)}v\left)\right(x\left)\right)-{\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)\right)\\ & =0.\end{array}$(64) We conclude by Proposition 2 (iii) that (65) $\begin{array}{rl}& \psi \left({\mathrm{\Phi }}_p\left(D^{\alpha }\right(c^{|\sigma |/p-1}v\left)\right)-{\mathrm{\Phi }}_p\left(D^{\alpha }u\right)\right)\\ & =-G_{\beta }\left(a\right(c^{|\sigma |}{v}^{\sigma }-u^{\sigma }\left)\right)\le 0.\end{array}$(65) Then we have (66) ${\mathrm{\Phi }}_p\left(D^{\alpha }\right(c^{|\sigma |/p-1}v\left)\right)\le {\mathrm{\Phi }}_p\left(D^{\alpha }u\right).$(66) Which implies with the fact that ${\mathrm{\Phi }}_p$ is increasing monotone that (67) $\begin{array}{rl}& D^{\alpha }(c^{|\sigma |/p-1}v-u)\le 0,\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}(c^{|\sigma |/p-1}v-u)\left(x\right)=0,\\ & c^{|\sigma |/p-1}v\left(1\right)-u\left(1\right)=0.\end{array}$(67) Using again Proposition 2 (iii), we conclude that (68) $c^{|\sigma |/p-1}v-u\ge 0.$(68) By symmetry, we obtain that $v\le c^{|\sigma |/p-1}u$. So $c^{|\sigma |/p-1}\in J$. Since $\left(|\sigma |/\right(p-1\left)\right)<1$ and c>1, we have $c^{|\sigma |/p-1}<c$. This yields to a contradiction with the fact that $c=infJ$. Hence c=1 and consequently u=v.

3.1. Proof of Theorem  2

Suppose that $\beta -1<\mu <\beta$ and $L_2$ satisfy (69) ${\int }_0^{\eta }\frac{\left(L_2\right(t\left)\right)\left(1/\right(p-1\left)\right)}{t}\mathrm{d}t<\mathrm{\infty }.$(69) Let θ be the function defined on $[0,1]$ by (70) $\theta \left(x\right)={\left({\int }_0^x\frac{\left(L_2\right(t\left)\right)\left(1/\right(p-1\left)\right)}{t}\mathrm{d}t\right)}^{\left(p-1\right)/\left(p-1-\sigma \right)}.$(70) Then, for $x\in (0,1)$ $\begin{array}{rl}w\left(x\right)& :=a\left(x\right)x^{(\alpha -2)\sigma }{\theta }^{\sigma }(1-x)\\ & \approx x^{-\lambda +(\alpha -2)\sigma }(1-x{)}^{-\mu }{L}_1(x\left)L_2\right(1-x)\\ & {\left({\int }_0^{1-x}\frac{\left(L_2\right(t\left)\right)\left(1/\right(p-1\left)\right)}{t}\mathrm{d}t\right)}^{\sigma (p-1)/\left(p-1-\sigma \right)}.\end{array}$ We conclude by Lemma 2.9 that for $x\in (0,1)$ (71) $\begin{array}{rl}{G}_{\beta }w\left(x\right)& \approx x^{\beta -2}(1-x{)}^{\beta -\mu }{L}_2(1-x)\\ & {\left({\int }_0^{1-x}\frac{\left(L_2\right(t\left)\right)\left(1/\right(p-1\left)\right)}{t}\mathrm{d}t\right)}^{\sigma (p-1)/\left(p-1-\sigma \right)}.\end{array}$(71) This implies by Proposition 2 (ii) that ${\displaystyle {\int }_0^1(1-t{)}^{\beta -1}w(t)\mathrm{d}t<\mathrm{\infty }}$ and for $x\in (0,1)$, we have (72) $\begin{array}{rl}& {\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\left(x\right)\approx x^{-\left(\right(2-\beta \left)/\right(p-1\left)\right)}(1-x{)}^{-\alpha }\\ & \left(L_2\right(1-x){)}^{1/\left(p-1\right)}\\ & {\left({\int }_0^{1-x}\frac{\left(L_2\right(t\left)\right)\left(1/\right(p-1\left)\right)}{t}\mathrm{d}t\right)}^{\sigma /\left(p-1-\sigma \right)}.\end{array}$(72) Using again ${\mathbf{(}\mathbf{H}}_1\mathbf{)}$ and by Lemma 2.9, we deduce that for $x\in (0,1)$ (73) $x^{2-\alpha }{G}_{\alpha }\left({\mathrm{\Phi }}_q\left(\frac{1}{\psi }{G}_{\beta }w\right)\right)\left(x\right)\approx \theta (1-x).$(73) Hence it follows from Proposition 3 that problem (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ) has a unique positive solution u in $C_{2-\alpha }\left(\right[0,1\left]\right)$ satisfying for $x\in (0,1)$, (74) $u\left(x\right)\approx x^{\alpha -2}\theta (1-x).$(74) As an application of our main result, we give the following example.

3.2. Example 1

Let $\beta ,\alpha \in (1,2]$, p>1 and $\sigma \in (1-p,p-1)$. Let a and ψ be two positive continuous functions on $(0,1)$ such that (75) $a_1\left(x\right)\approx (1-x{)}^{-\mu }{\left(\log \left(\frac{3}{1-x}\right)\right)}^{-p}$(75) and (76) $\psi \left(x\right)\approx (1-x{)}^{(p-1)\alpha +\beta -\mu },$(76) where $\beta -1<\mu <\beta$ and $\beta +p>3$. Therefore by Theorem 2, problem (1(1) $\begin{array}{rl}& D^{\beta }\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left)\right)=a\left(x\right)u^{\sigma },x\in (0,1),\\ & \underset{x\to 0}{lim}{D}^{\beta -1}\left(\psi \right(x\left){\mathrm{\Phi }}_p\right(D^{\alpha }u\left(x\right)\left)\right)\\ & =\underset{x\to 1}{lim}\psi \left(x\right){\mathrm{\Phi }}_p\left(D^{\alpha }u\right(x\left)\right)=0\\ & \underset{x\to 0}{lim}{D}^{\alpha -1}u\left(x\right)=u\left(1\right)=0,\end{array}$(1) ) has a unique positive solution $u\in C_{2-\alpha }\left(\right[0,1\left]\right)$ satisfying for each $x\in (0,1)$ (77) $u\left(x\right)\approx x^{\alpha -2}{\left(\log \left(\frac{3}{1-x}\right)\right)}^{-1/\left(p-1-\sigma \right)}.$(77)

Acknowledgments

The authors would like to thank the anonymous reviewers for their valuable comments and suggestions to improve the manuscript.

Disclosure statement

No potential conflict of interest was reported by the authors.

ORCID

Adnane Hamiaz  http://orcid.org/0000-0001-5605-8728

Additional information

Funding

This work was supported by Taibah University.