Existence and Uniqueness of Positive Solutions for the Fractional Differential Equation Involving the ρ(τ)-Laplacian Operator and Nonlocal Integral Condition

Abstract

This paper aims to investigate the Caputo fractional differential equation involving the $\rho \left(\tau \right)$ Laplacian operator and nonlocal multi-point of Riemann–Liouville’s fractional integral. We also prove the uniqueness of the positive solutions for Boyd and Wong’s nonlinear contraction via the Guo–Krasnoselskii fixed-point theorem in Banach spaces. Finally, we illustrate the theoretical results and show that by solving the nonlocal problems, it is possible to obtain accurate approximations of the solutions. An example is also provided to illustrate the applications of our theorem.

1. Introduction

In the analysis of many occurrences in science and engineering, the concept of fractional differential equations has become a potent and well-organized mathematical instrument. Diverse fields, including control systems, elasticity, signal analysis, biomathematics, biomedicine, social systems, bioengineering, management, financial systems, traffic flow, turbulence, complex systems, pollution control, image processing, etc., use fractional differential equation research. More information can be found in [1,2,3,4,5,6,7,8,9].

Mathematicians study and prove the existence and numerical and approximate solutions of fractional differential equations in different ways. The existence and uniqueness of the solution to a nonlinear fractional differential equation with nonlinear integral boundary conditions on time scales have been established by Kumar and Malik [10]. They applied the fixed-point theorems of Banach, Schaefer, Leray Schauder’s nonlinear alternative, and Krasnoselskii to arrive at these conclusions. A class of nonlinear fractional differential equations with Atangana–Baleanu derivatives and fractional boundary conditions has recently been studied by Saha et al. [11]. To establish their key findings, the authors made use of three common fixed-point theorems. While the Banach contraction principle ensures uniqueness, the Schauder alternative guarantees the existence of solutions.

The uniqueness of the solutions to a two-term nonlinear fractional integrodifferential equation with nonlocal boundary conditions and variable coefficients was investigated by Li [12] using the Mittag–Leffler function, Babenko’s method, and the contractive principle of Banach. Georgiev et al. [13] also looked at whether impulsive differential equations with two-point integral boundary conditions have at least one solution and at least two non-negative solutions. For the sum of two operators on Banach spaces, they used the most recent fixed-point theorems, and Guo cited [14] to obtain the correct answers using iterative techniques.

Additionally, some mathematicians have focused on the $\rho \left(\tau \right)$-Laplacian operator, which has been utilized for simulating electrorheological fluids in elastic mechanics, image restoration, magnetostatic issues, etc. (see [15,16,17,18,19,20,21]). In 2017, Qiao and Zhou [22] defined the concept of singular fractional differential equations with boundary value conditions:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\delta }\nu \left(\tau \right)+q\left(\tau \right)h(\tau ,\nu \left(\tau \right))=0,0<\tau <1;\hfill & \\ {\displaystyle \nu \left(0\right)={\nu }^{\prime }\left(0\right)=\dots ={\nu }^{(k-2)}\left(0\right)=0,{}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\rho }\nu \left(1\right)=\sum _{i=1}^{\infty }{\eta }_i\nu \left({\lambda }_i\right),}\hfill \end{array}\right.\end{array}$$

where $\delta \in (2,\infty )$, $k-1<\delta <k$, $\rho \in [1,\delta -1]$ is a fixed number, ${\lambda }_i\in (0,1)$, ${\eta }_i\ge 0$ for $i=1,2,3\dots$, ${}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\delta }$ and ${}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\rho }$ denote the Riemann-Liouville fractional derivatives of orders $\delta$ and $\rho$, and ${\displaystyle \frac{\Gamma \left(\delta \right)}{\Gamma (\delta -\rho )}-\sum _{i=1}^{\infty }{\eta }_i{\lambda }_i^{\delta -1}>0}$.

Recently, Borisut and Kumam [23] presented a solution for the Caputo–Hadamard fractional under the following conditions:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}^C{\mathcal{D}}_{0^{+}}^{\delta }\nu \left(\tau \right)=h(\tau ,\nu \left(\tau \right)),\tau \in [0,\mathcal{S}];\hfill & \\ {\displaystyle {\nu }^{(\ell )}\left(0\right)={\Im }_{\ell },\nu \left(\mathcal{S}\right)=\sum _{i=1}^j{\eta }_i{}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{{\rho }_i}\nu \left({\lambda }_i\right),}\hfill \end{array}\right.\end{array}$$

where $j,k\in \mathbb{N}$, ${\Im }_{\ell },{\eta }_i\in \mathbb{R}$, and $k-1<\delta <k$, with $k\ge 2$, $\ell =0,1,\dots ,k-2$, and the order $i=1,2,\dots ,j$. By ${\displaystyle \sum _{i=1}^j{\eta }_i{\lambda }_i^{{\rho }_i+k-1}\frac{\Gamma \left(k\right)}{\Gamma (k+{\rho }_i)}\ne {\mathcal{S}}^{k-1}}$, ${}^C{\mathcal{D}}_{0^{+}}^{\delta }$ denotes the Caputo fractional derivative, and the function $h:[0,\mathcal{S}]\times \mathcal{C}\left([0,\mathcal{S}],\mathcal{E}\right)\to \mathcal{E}$, ${}_{\mathcal{RL}}{\mathcal{I}}^{{\rho }_i}$ is the Riemann–Liouville fractional derivative of order ${\rho }_i>0$ and ${\lambda }_i\in (0,\mathcal{S})$.

In the same year, Wang et al. [24] investigated the positive solution of the Caputo–Hadamard fractional derivative defined by:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}^{{}_C\mathcal{H}}{\mathcal{D}}^{L_2}{\psi }_{\rho \left(\tau \right)}\left({{}^{{}_C\mathcal{H}}\mathcal{D}}^{L_1}w\left(\tau \right)\right)+h(w\left(\tau \right),{\mathcal{I}}_{\lambda }^{\Lambda ,\phi }w\left(\tau \right))=0,\tau \in [1,e];\hfill & \\ w^{\prime }\left(1\right)=\eta w^{\prime }\left(e\right),w\left(1\right)=w^{″}\left(1\right)=0,{{}^{{}_C\mathcal{H}}\mathcal{D}}^{L_1}w\left(1\right)=0,\hfill \end{array}\right.\end{array}$$

for all $\Lambda \in \mathbb{R}$, where $L_1\in (2,3],L_2\in (0,1],\frac{e}{2}<\eta \le \frac{2e}{4-e}$. Note that ${}^{{}_C\mathcal{H}}\mathcal{D}$ is the Caputo–Hadamard fractional derivative, ${\psi }_{\rho \left(\tau \right)}(·)$ is the $\rho \left(\tau \right)$-Laplacian operator with $\rho \left(\tau \right)\in {\mathcal{C}}^1[1,e]$ such that $\rho \left(\tau \right)>1$, and ${\mathcal{I}}_{\lambda }^{\Lambda ,\phi }$ is the generalized Erdelyi–Kober fractional integral order $\phi ,\lambda >0$.

Inspired by the above papers [22,23,24], we derive and prove the uniqueness of the positive solutions for Boyd and Wong’s nonlinear contraction using the Guo–Krasnoselskii fixed-point theorem in Banach spaces, provided that the following condition in Equation (1) holds:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}^C{\mathcal{D}}_{0^{+}}^{\delta }{\psi }_{\rho \left(\tau \right)}\left({{}^C\mathcal{D}}_{0^{+}}^{\varrho }\nu \left(\tau \right)\right)+h(\tau ,\nu \left(\tau \right))=0,\tau \in [0,1]\hfill & \\ {\displaystyle \nu \left(0\right)=\ell ,\nu \left(1\right)=\sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{b_i}\nu \left({\lambda }_i\right),}\hfill \end{array}\right.\end{array}$$

(1)

for $j,i\in \mathbb{N},\ell \in \mathbb{R}$, where $\delta ,\varrho ,b_i\in (0,1)$, $b_i+\varrho >1,{\eta }_i\ge 0,\rho \left(\tau \right)>2,0<{\lambda }_1<{\lambda }_2<\cdots <{\lambda }_j<1$. The function $h:[0,1]\times \mathcal{C}\left(\right[0,1],\mathcal{E})\to \mathcal{E}$ is continuous and $\mathcal{E}$ represents Banach spaces. The fractional integral ${{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{b_i}$ is the Riemann–Liouville-type order $b_i$ and ${}^C{\mathcal{D}}_{0^{+}}^{\delta }$ and ${}^C{\mathcal{D}}_{0^{+}}^{\varrho }$ are the Caputo-type fractional derivatives of orders $\delta$ and $\varrho$.

In this work, we investigate the existence and uniqueness of solutions to the problem using Guo–Krasnoselskii’s fixed-point theorem and Boyd–Wong’s theorem. Solving fractional differential equations involving the $\rho \left(\tau \right)$-Laplacian operator using the Guo–Krasnoselskii fixed-point theorem and Boyd-Wong’s theorem has not yet been studied by any author. In this paper, we are interested in using the above theorems to solve the problem. Finally, we illustrate the theoretical results and show that by solving the nonlocal problems, it is possible to obtain accurate approximations of the solutions. As an example that demonstrates the technique, we present the following lemma, which plays an essential role in deriving our main theorem.

2. Preliminaries

In this section, we present some definitions, theorems, and lemmas that are important for proving the solutions of nonlinear fractional differential equations.

Definition 1 

([25,26]). Let $\delta ,\lambda >0$, and ν be positive functions of real numbers. The fractional integral

$${{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\nu \left(\tau \right)=\frac{1}{\Gamma \left(\delta \right)}{\int }_0^{\tau }{(\tau -\Im )}^{\delta -1}\nu (\Im )d\Im$$

is the Riemann–Liouville fractional integral of order δ of ν if it exists. The fractional derivative ${}^C{\mathcal{D}}_{0^{+}}^{\delta }\nu \left(\tau \right)$ is the Caputo-type fractional derivative given by

$${}^C{\mathcal{D}}_{0^{+}}^{\delta }\nu \left(\tau \right)=\frac{1}{\Gamma (k-\delta )}{\int }_0^{\tau }{(\tau -\Im )}^{k-\delta -1}{\nu }^{\left(k\right)}(\Im )d\Im ,$$

which is valid pointwise on $(0,\infty )$, where $k=\left[\delta \right]+1$, $k-1<\delta <k$, and Γ represents the gamma function. If $\delta =k$, then ${}^C{\mathcal{D}}_{0^{+}}^{\delta }\nu \left(\tau \right)={\nu }^{\left(k\right)}\left(\tau \right)$.

Lemma 1 

([25,26]). Let $k-1<\delta <k$. If $\nu \in {\mathcal{C}}^k\left([m,n]\right)$, then

$${{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }{(}^C{\mathcal{D}}_{0^{+}}^{\delta }\nu )\left(\tau \right)=\nu \left(\tau \right)+{\tilde{a}}_0+{\tilde{a}}_1\tau +{\tilde{a}}_2{\tau }^2+\cdots +{\tilde{a}}_{k-1}{\tau }^{k-1}$$

where $i=0,1,2,\cdots ,k-1$, for some ${\tilde{a}}_i\in \mathbb{R}$ and k is the smallest integer greater than or equal to δ.

Proposition 1 

([27]). If $\rho ,\delta >0$, then:

(i) 

$h\left(\tau \right)=\ell \ne 0,\ell$ is a constant, then ${}^C{\mathcal{D}}_{0^{+}}^{\delta }\ell =0$ and ${}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\delta }\ell =\frac{{\tau }^{-\delta }\ell }{\Gamma (1-\delta )}$.

(ii) 

${}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\delta }{\tau }^{\delta -1}=0$.

(iii) 

${}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }{\tau }^{\rho }=\frac{\Gamma (\rho +1)}{\Gamma (\delta +\rho +1)}{\tau }^{\delta +\rho }$, for $\rho >1$.

(iv) 

${}_{\mathcal{RL}}{\mathcal{D}}_{0^{+}}^{\delta }{\tau }^{\rho }=\frac{\Gamma (\rho +1)}{\Gamma (1+\rho -\delta )}{\tau }^{\rho -\delta }$, for $\rho >-1+\delta$.

(v) 

${}^C{\mathcal{D}}_{0^{+}}^{\delta }{\tau }^{\rho }=\frac{\Gamma (\rho +1)}{\Gamma (1+\rho -\delta )}{\tau }^{\rho -\delta }$, for $\rho >0$.

Lemma 2 

([27]). $h\left(\tau \right)\in {\mathcal{L}}^1[a,b]$ for all $\zeta ,\eta \in \mathbb{R}$, then ${\mathcal{I}}_{0^{+}}^{\zeta }{\mathcal{I}}_{0^{+}}^{\eta }h\left(\tau \right)={\mathcal{I}}_{0^{+}}^{\zeta +\eta }h\left(\tau \right)$.

Lemma 3 

([24]). Let ${\psi }_{\rho \left(\tau \right)}^{-1}$ be the inverse operator of the $\rho \left(\tau \right)$-Laplacian operator for $(\tau ,\nu )\in [0,1]\times \mathbb{R}$. The equation ${\psi }_{\rho \left(\tau \right)}\left(\nu \right)=|\nu {|}^{\rho \left(\tau \right)-2}\nu$ is given by

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{\psi }_{\rho \left(\tau \right)}^{-1}\left(\nu \right)=|\nu {|}^{\frac{2-\rho \left(\tau \right)}{\rho \left(\tau \right)-1}}\nu ,\nu \in \mathbb{R}\setminus \left\{0\right\},\hfill & \\ {\psi }_{\rho \left(\tau \right)}^{-1}\left(0\right)=0,\nu =0.\hfill \end{array}\right.\end{array}$$

Now, we present the theorem of Bernoulli’s inequality in Theorem 1.

Theorem 1 

([28]). Consider the following conditions on ζ:

($\mathcal{B}1$)

If $0<\zeta <1$, then ${(1+w)}^{\zeta }\le 1+\zeta w$, $\forall w\ge -1$.

($\mathcal{B}2$)

If $\zeta \ge 1$, then ${(1+w)}^{\zeta }\ge 1+\zeta w$, $\forall w\ge -1$.

($\mathcal{B}3$)

If $\zeta \le 0$, then ${(1+w)}^{\zeta }\ge 1+\zeta w$, $\forall w\ge -1$.

Definition 2 

([29]). Let $\mathcal{J}:\mathcal{E}⟶\mathcal{E}$ with $\mathcal{E}$ be a Banach space. $\mathcal{J}$ is said to be a nonlinear contraction if there exists $\mathsf{\Phi }:{\mathbb{R}}^{+}⟶{\mathbb{R}}^{+}$ such that Φ is a continuous non-decreasing function, $\mathsf{\Phi }\left(0\right)=0$, and $\mathsf{\Phi }\left(\iota \right)<\iota$,

$$\parallel \mathcal{J}w-\mathcal{J}z\parallel \le \Phi (\parallel w-z\parallel ),$$

$\forall \iota >0$ and for all $w,z\in \mathcal{E}.$

Now, we present Theorems 2 and 3, which are the theorems of Guo–Krasnoselskii and Boyd and Wong. The theorems are used to prove the existence and uniqueness of the fixed-point solution.

Theorem 2 

([29], Guo–Krasnoselskii fixed-point theorem). Suppose that $\mathcal{E}$ is a Banach space. Let $\mathcal{Q}\subset \mathcal{E}$ be a cone, and assume that ${\Gamma }_1,{\Gamma }_2\subseteq \mathcal{E}$ are open. $\mathcal{J}:\mathcal{Q}\cap ({\overline{\Gamma }}_2\setminus {\Gamma }_1)\to \mathcal{Q}$ is a completely continuous operator and $0\in {\Gamma }_1,{\overline{\Gamma }}_1\subset {\Gamma }_2$ such that:

 (G1) 

For $\nu \in \mathcal{Q}\cap \partial {\Gamma }_1$, $\parallel \mathcal{J}\nu \parallel \ge \parallel \nu \parallel$, and for $\nu \in \mathcal{Q}\cap \partial {\Gamma }_2$, $\parallel \mathcal{J}\nu \parallel \le \parallel \nu \parallel$, or

 (G2) 

For $\nu \in \mathcal{Q}\cap \partial {\Gamma }_1$, $\parallel \mathcal{J}\nu \parallel \le \parallel \nu \parallel$, and for $\nu \in \mathcal{Q}\cap \partial {\Gamma }_2$, $\parallel \mathcal{J}\nu \parallel \ge \parallel \nu \parallel$.

Then, $\mathcal{J}$ has a fixed point in $\mathcal{E}$ in $\mathcal{Q}\cap ({\overline{\Gamma }}_2\setminus {\Gamma }_1).$

Theorem 3 

([29], Boyd and Wong’s fixed-point theorem). Let $\mathcal{J}$ be a Banach space and let $\mathcal{J}:\mathcal{E}\to \mathcal{E}$ be a nonlinear contraction. Then, $\mathcal{J}$ has a unique fixed point in $\mathcal{E}$.

Lemma 4. 

If $g\left(\tau \right)\in \mathcal{C}[0,1]$, the unique solution of the following linear fractional differential equation involving the $\rho \left(\tau \right)$-Laplacian operator

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}^C{\mathcal{D}}_{0^{+}}^{\delta }{\psi }_{\rho \left(\tau \right)}\left({{}^C\mathcal{D}}_{0^{+}}^{\varrho }\nu \left(\tau \right)\right)=-g\left(\tau \right),\tau \in [0,1]\hfill & \\ {\displaystyle \nu \left(0\right)=\ell ,\nu \left(1\right)=\sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{b_i}\nu \left({\lambda }_i\right),}\hfill \end{array}\right.\end{array}$$

is equivalent to $\nu \left(\tau \right)={\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho \left(\tau \right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g(\Im )\right)d\Im$ where

$$\begin{array}{c}\hfill \mathcal{A}(\tau ,\Im )=\left\{\begin{array}{c}\frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}[-\mathcal{Z}{(\tau -\Im )}^{\varrho -1}\hfill \\ +\mu (\Im ){(1-\Im )}^{\varrho -1}],\hfill & 0<\Im <\tau <1\hfill \\ \frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[\mu (\Im ){(1-\Im )}^{\varrho -1}\right],\hfill & 0<\tau <\Im <1,\hfill \end{array}\right.\end{array}$$

(2)

with ${\displaystyle \mu (\Im )=\left[1-\Gamma \left(\varrho \right)\sum _{i=1}^j\frac{{\eta }_i{({\lambda }_i-\Im )}^{b_i+\varrho -1}}{\Gamma (\varrho +b_i)}{(1-\Im )}^{1-\varrho }\right]}$, ${\displaystyle \mu \left(0\right)=\left[1-\Gamma \left(\varrho \right)\sum _{i=1}^j\frac{{\eta }_i{\lambda }_i^{b_i+\varrho -1}}{\Gamma (\varrho +b_i)}\right]}$ and ${\displaystyle \mathcal{Z}=1-\sum _{i=1}^j\frac{{\eta }_i{\lambda }_i^{b_i}}{\Gamma (b_i+1)}}$.

Proof. 

By utilizing Lemma 1, we can see that $\psi {(}^C{\mathcal{D}}_{0^{+}}^{\varrho }\nu \left(\tau \right))=-{{}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}}^{\delta }g\left(\tau \right)+{\tilde{a}}_0$. From the first condition in Definition 1 and $\nu \left(0\right)=\ell$ in Lemma 3, along with ${\tilde{a}}_0=0$, we have

$$\begin{array}{c}\hfill \nu \left(\tau \right)={-}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\varrho }{\psi }_{\rho \left(\tau \right)}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }g\left(\tau \right)\right)+{\tilde{a}}_0^{\ast }.\end{array}$$

(3)

By substituting $\tau ={\lambda }_i$ and applying ${\displaystyle \sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{b_i}}$ to (3), we obtain

$$\begin{array}{ccc}\hfill {\displaystyle \sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{b_i}\nu \left({\lambda }_i\right)}& =& {\displaystyle -\sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\varrho +b_i}{\psi }_{\rho \left({\lambda }_i\right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g\left({\lambda }_i\right)\right)+{\tilde{a}}_0^{\ast }\sum _{i=1}^j\frac{{\eta }_i{\lambda }_i^{b_i}}{\Gamma (b_i+1)}.}\hfill \end{array}$$

From the second condition, we obtain ${\displaystyle \nu \left(1\right)=\sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{b_i}\nu \left({\lambda }_i\right)}$. Also, we deduce that

$$\begin{array}{ccc}\hfill \nu \left(1\right)& =& -{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\varrho }{\psi }_{\rho \left(1\right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g\left(1\right)\right)+{\tilde{a}}_0^{\ast }\hfill \\ \hfill {\tilde{a}}_0^{\ast }& =& {\displaystyle \frac{1}{\mathcal{Z}}\left(-\sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\varrho +b_i}{\psi }_{\rho \left({\lambda }_i\right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g\left({\lambda }_i\right)\right)+{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\varrho }{\psi }_{\rho \left(1\right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g\left(1\right)\right)\right),}\hfill \end{array}$$

and upon substituting ${\tilde{a}}_0^{\ast }$ into (3), we can see that

$$\begin{array}{ccc}\hfill \nu \left(\tau \right)& =& -{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\varrho }{\psi }_{\rho \left(\tau \right)}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }g\left(\tau \right)\right)\hfill \\ & & {\displaystyle +\frac{1}{\mathcal{Z}}\left(-\sum _{i=1}^j{\eta }_i{{}_{\mathcal{RL}}I}_{0^{+}}^{\varrho +b_i}{\psi }_{\rho \left({\lambda }_i\right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g\left({\lambda }_i\right)\right)+{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\varrho }{\psi }_{\rho \left(1\right)}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }g\left(1\right)\right)\right).}\hfill \end{array}$$

Hence, $\nu \left(\tau \right)={\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho \left(\tau \right)}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }g(\Im )\right)d\Im .$ □

Lemma 5. 

Let $\mu \left(0\right)>0$, $\mathcal{Z}>0$ and then $\mu (\Im )$ is non-decreasing on $[0,1]$, where $b_i+\varrho >1,{\eta }_i\ge 0$, and $\mu (\Im )\ge 0$ for $\Im \in [0,1]$.

Proof. 

It easy to see that

$$\begin{array}{ccc}\hfill {\mu }^{\prime }(\Im )& =& {\displaystyle \Gamma \left(\varrho \right)\sum _{i=1}^j{\eta }_i(b_i+\varrho -1){({\lambda }_i-\Im )}^{b_i+\varrho -2}{(1-\Im )}^{1-\varrho }}\hfill \\ & \ge & {\displaystyle \Gamma \left(\varrho \right)\sum _{i=1}^j\frac{{\eta }_i{({\lambda }_i-\Im )}^{b_i+\varrho -1}}{\Gamma (\varrho +b_i)}(1-\varrho ){(1-\Im )}^{-\varrho }\ge 0,}\hfill \end{array}$$

when ${\displaystyle \underset{i⟶\infty }{lim}{\lambda }_i\le \Im \le 1.\mu (\Im )=1,{\mu }^{\prime }(\Im )=0.}$ This implies that $\mu (\Im )$ is non-decreasing on $[0,1]$ and $\mu (\Im )\ge \mu \left(0\right)>0$ for $\Im \in [0,1]$. □

Lemma 6. 

By utilizing Equation (2), $\mathcal{A}(\tau ,\Im )$ satisfies three conditions:

$\left[\mathcal{A}\right(1\left)\right]$

$\mathcal{A}(\tau ,\Im )\ge 0,\frac{\partial }{\partial \tau }\mathcal{A}(\tau ,\Im )\ge 0$ for$0<\tau ,\Im <1$.

$\left[\mathcal{A}\right(2\left)\right]$

${\displaystyle \underset{\tau \in [0,1]}{max}\mathcal{A}(\tau ,\Im )=\mathcal{A}(1,\Im )=\frac{{(1-\Im )}^{\varrho -1}}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}+\mu (\Im )\right]}$.

$\left[\mathcal{A}\right(3\left)\right]$

$\mathcal{A}(\tau ,\Im )\ge {\tau }^{\varrho +1}\mathcal{A}(1,\Im )$, for$0\le \tau ,\Im <1.$

Proof. 

For $0<\Im <\tau <1$, notice that ${\displaystyle \mathcal{Z}=1-\sum _{i=1}^j\frac{{\eta }_i{\lambda }_i^{b_i}}{\Gamma (b_i+1)}>0}$, and from Lemma 5, we can see that $\mathcal{A}(\tau ,\Im )\ge (-\mathcal{Z}+\mu \left(0\right)){(1-\Im )}^{\varrho -1}$. Hence, ${\displaystyle \sum _{i=1}^j{\eta }_i{\lambda }_i^{b_i}\left(\frac{1}{\Gamma (b_i+1)}-\frac{{\lambda }_i^{\varrho -1}\Lambda \left(\varrho \right)}{\Gamma (\varrho +b_i)}\right)>0}$, and for $0<\tau <\Im <1,\mathcal{A}(\tau ,\Im )\ge 0$, we obtain

$$\begin{array}{c}\hfill \mathcal{A}(\tau ,\Im )=\frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}{(\tau -\Im )}^{\varrho -1}+\mu (\Im ){(1-\Im )}^{\varrho -1}\right]\ge 0.\end{array}$$

Also, we can show that

$$\begin{array}{c}\hfill \frac{\partial }{\partial \tau }\mathcal{A}(\tau ,\Im )=\left\{\begin{array}{cc}-\frac{(\varrho -1){(\tau -\Im )}^{\varrho -2}}{\Gamma \left(\varrho \right)},\hfill & 0<\Im <\tau <1\hfill \\ 0,\hfill & 0<\tau <\Im <1.\hfill \end{array}\right.\end{array}$$

From the above equation, we can see that $\frac{\partial }{\partial \tau }\mathcal{A}(\tau ,\Im )\ge 0$, and $\frac{\partial }{\partial \tau }\mathcal{A}(\tau ,\Im )$ is a continuous function on $[0,1]\times [0,1]$. From condition $\mathcal{A}\left(1\right)$, we know that $\mathcal{A}(\tau ,\Im )$ is non-decreasing with respect to τ. Hence, $\mathcal{A}(1,\Im )=\frac{{(1-\Im )}^{\varrho -1}}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}+\mu (\Im )\right]$. For $0\le \Im \le \tau <1,$ we obtain

$$\begin{array}{ccc}\hfill \mathcal{A}(\tau ,\Im )& =& \frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}{(\tau -\Im )}^{\varrho -1}+\mu (\Im ){(1-\Im )}^{\varrho -1}\right]\hfill \\ & \ge & \frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}{(1-\Im )}^{\varrho -1}+\mu (\Im ){(1-\Im )}^{\varrho -1}\right]\hfill \\ & \ge & \frac{{(1-\Im )}^{\varrho -1}{\tau }^{\varrho +1}}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}+\mu (\Im )\right]\hfill \\ & =& {\tau }^{\varrho +1}\mathcal{A}(1,\Im ).\hfill \end{array}$$

Also, we obtain

$$\begin{array}{ccc}\hfill \mathcal{A}(\tau ,\Im )& =& \frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[\mu (\Im ){(1-\Im )}^{\varrho -1}\right]\hfill \\ & \ge & \frac{1}{\mathcal{Z}\Gamma \left(\varrho \right)}\left[-\mathcal{Z}{(\tau -\Im )}^{\varrho -1}+\mu (\Im ){(1-\Im )}^{\varrho -1}\right]\hfill \\ & \ge & {\tau }^{\varrho +1}\mathcal{A}(1,\Im ),\mathrm{for}0\le \tau \le \Im <1.\hfill \end{array}$$

 □

Let $\mathcal{E}$ be a Banach space with $\parallel ·\parallel$, where $\mathcal{E}=\mathcal{C}[0,1]$, and the norm is defined as ${\displaystyle \parallel \nu \parallel =\underset{\tau \in [0,1]}{max}\left|\nu \left(\tau \right)\right|}$. Let $\mathcal{Q}=\{\nu \in \mathcal{E}:\nu \left(\tau \right)\ge t^{\varrho +1}\parallel \nu \parallel ,\tau \in [0,1]\}$ be set, then $\mathcal{Q}$ is a cone in $\mathcal{E}$. The function $\mathcal{B}\left(d\right)=\{\nu \in \mathcal{Q}:\parallel \nu \parallel <d\}$, and $\partial \mathcal{B}\left(d\right)=\{\nu \in \mathcal{Q}:\parallel \nu \parallel =d\}$ for $d>0$. The operator $\mathcal{J}:\mathcal{Q}\setminus \varnothing \to \mathcal{E}$ defined by

$$\begin{array}{c}\hfill \left(\mathcal{J}\nu \right)\left(\tau \right)={\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho \left(\tau \right)}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im .\end{array}$$

(4)

Then, the problem in Equation (1) has a solution if and only if $\mathcal{J}$ has a fixed point in $\mathcal{E}$.

3. Main Results

In this section, we consider the nonlocal integral boundary value problem of a nonlinear fractional differential Equation (1). Here, we establish the existence and uniqueness of positive solutions through the application of fixed points in the sense of Guo–Krasnoselskii and Boyd and Wong’s non-linear contraction.

Now, we consider and prove Lemma 7, which is the tool used to prove the main results.

Lemma 7. 

Suppose that the condition below satisfies:

• $\left(\mathcal{P}1\right)$ h is a continuous function with $h\in \mathcal{C}((0,1)\times {\mathbb{R}}^{+},{\mathbb{R}}^{+})$.
• $\left(\mathcal{P}2\right)$For any positive number $d_1<d_2$, there exists a continuous function $b_{d_1,d_2}\left(\tau \right):(0,1)\to [0,+\infty )$ and $\rho \left(\tau \right)>2$ such that $0{<}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }{b}_{d_1,d_2}\left(\tau \right)\le 1,0<\tau <1,{\tau }^{\varrho +1}{d}_1\le \nu \le d_2$ and $0<d_1<d_2,$$h(\tau ,\nu \left(\tau \right))\le b_{d_1,d_2}\left(\tau \right)$. Then, $\mathcal{J}:\overline{\mathcal{B}\left(d_2\right)}\setminus \mathcal{B}\left(d_1\right)\to \mathcal{Q}$ is completely continuous.

Proof. 

For any $\nu \in \overline{\mathcal{B}\left(d_2\right)}\setminus \mathcal{B}\left(d_1\right)$, we can deduce from (4) in Lemma 6 that

$$\begin{array}{ccc}\hfill \left(\mathcal{J}\nu \right)\left(\tau \right)& =& {\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im \hfill \\ & \le & {\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im ,\hfill \end{array}$$

for $\tau \in [0,1]$. Also, we obtain

$$\begin{array}{ccc}\hfill \left(\mathcal{J}\nu \right)\left(\tau \right)& =& {\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im \hfill \\ & \ge & {\tau }^{\varrho +1}{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im ,\tau \in [0,1].\hfill \end{array}$$

So, $\left(\mathcal{J}\nu \right)\left(\tau \right)\ge {\tau }^{\varrho +1}\parallel \mathcal{J}\nu \parallel ,\tau \in [0,1]$, indicating that $\mathcal{J}:\overline{\mathcal{B}\left(d_2\right)}\setminus \mathcal{B}\left(d_1\right)\to \mathcal{Q}$. Given the continuity of the function $\mathcal{A}(\tau ,\Im )$, as well as the fulfillment of the conditions $\left(\mathcal{P}1\right)$ and $\left(\mathcal{P}2\right)$, we can see that $\mathcal{J}$ is the continuous function in $\overline{\mathcal{B}\left(d_2\right)}\setminus \mathcal{B}\left(d_1\right)$. Next, we show that $\mathcal{J}$ is relatively compact. For any $\nu \in \overline{\mathcal{B}\left(d_2\right)}\setminus \mathcal{B}\left(d_1\right)$, we have

$$\begin{array}{ccc}\hfill \left|\mathcal{J}\nu \right(\tau \left)\right|& =& |{\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\le {\int }_0^1\mathcal{A}(1,\Im )d\Im ,\mathrm{for}\tau \in [0,1].\hfill \end{array}$$

This means that $\mathcal{J}(\overline{\Gamma \left(d_2\right)}\setminus \Gamma \left(d_1\right))$ is uniformly bounded. $\mathcal{A}(\tau ,\Im )$ is continuous and uniformly continuous on $[0,1]\times [0,1]$ for any $\iota >0$. There exists $\phi >0$ such that $|\mathcal{A}({\varpi }_2,\Im )-\mathcal{A}({\varpi }_1,\Im )|<\iota$. If $|{\varpi }_2-{\varpi }_1|<\phi$ and $({\varpi }_2,\Im ),({\varpi }_1,\Im )\in [0,1]\times [0,1]$, then for any $\nu \in \overline{\Gamma \left(d_2\right)}\setminus \Gamma \left(d_1\right)$ and ${\varpi }_1,{\varpi }_2\in [0,1]$ such that $|{\varpi }_2-{\varpi }_1|<\phi$, we have

$$\begin{array}{ccc}\hfill |\mathcal{J}\nu \left({\varpi }_2\right)-\mathcal{J}\nu \left({\varpi }_1\right)|& =& |{\int }_0^1\left(\mathcal{A}({\varpi }_2,\Im )-\mathcal{A}({\varpi }_1,\Im )\right){\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \le & {\int }_0^1|\mathcal{A}({\varpi }_2,\Im )-\mathcal{A}({\varpi }_1,\Im )|{\psi }_{\rho (\Im )}^{-1}\left({}_{\mathcal{RL}}{\mathcal{I}}_{0^{+}}^{\delta }{b}_{d_1,d_2}(\Im )\right)d\Im \le \iota .\hfill \end{array}$$

We can see that the functions in $\mathcal{J}(\overline{\Gamma \left(d_2\right)}\setminus \Gamma \left(d_1\right))$ are equicontinuous. So, $\mathcal{J}(\overline{\Gamma \left(d_2\right)}\setminus \mathit{\Gamma }\left(d_1\right))$ is relatively compact according to the theorem of Arzela–Ascoli. Thus, $\mathcal{J}:\overline{\mathit{\Gamma }\left(d_2\right)}\setminus \mathit{\Gamma }\left(d_1\right)\to \mathcal{Q}$ is completely continuous. □

Before proving the solution of Problem 1 using the theorem of Guo–Krasnoselskii, we introduce the height function equations to control the growth of the nonlinear term $h(\tau ,\nu )$:

$$\kappa (\tau ,d)=max\{h(\tau ,\nu ):{\tau }^{\varrho +1}d\le \nu \le d\},0<\tau <1,d>0,$$

$$\theta (\tau ,d)=min\{h(\tau ,\nu ):{\tau }^{\varrho +1}d\le \nu \le d\},0<\tau <1,d>0.$$

Theorem 4. 

Suppose that there exists a positive number $a^{\ast }<b^{\ast }$ and the conditions $\left(\mathcal{P}1\right)$ and $\left(\mathcal{P}2\right)$ hold such that:

$\left(\mathcal{P}3\right)$

$a^{\ast }\le |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\theta (s,a^{\ast })\right)d\Im |$

$<+\infty \mathit{and}|{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\kappa (s,b^{\ast })\right)d\Im |\le b^{\ast };$

$\left(\mathcal{P}4\right)$

$|{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\kappa (s,a^{\ast })\right)d\Im |\le a^{\ast }$

and $b^{\ast }\le |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\theta (s,b^{\ast })\right)d\Im |<+\infty .$

This implies that at least one of these conditions is true. Then, Problem (1) has at least one positive solution ${\nu }^{\ast }\in \mathcal{Q}$, which is non-decreasing such that $a^{\ast }\le \parallel {\nu }^{\ast }\parallel \le b^{\ast }$.

Proof. 

If $\nu \in \partial \mathcal{B}\left(a^{\ast }\right)$, then $\parallel \nu \parallel =a^{\ast }$ and ${\tau }^{\varrho +1}{a}^{\ast }\le \nu \left(\tau \right)\le a^{\ast }$ for $0\le \tau \le 1$. From the equation for $\theta$, we can see that

$$h(\tau ,\nu \left(\tau \right))\ge \theta (\tau ,a^{\ast }),\mathrm{for}\tau \in (0,1).$$

(5)

From Equation (5) and the condition of $\mathcal{A}(\tau ,\Im )$ in Lemma 6, we obtain

$$\begin{array}{ccc}\hfill \parallel \mathcal{J}\nu \parallel & =& \underset{\tau \in [0,1]}{max}|{\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \ge & \underset{\tau \in [0,1]}{max}{\tau }^{\varrho +1}|{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \ge & |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\theta (\Im ,a^{\ast })\right)d\Im |\hfill \\ & \ge & a^{\ast }=\parallel \nu \parallel .\hfill \end{array}$$

If $\nu \in \partial \mathcal{B}\left(b^{\ast }\right)$, then $\parallel \nu \parallel =b^{\ast }$ and ${\tau }^{\varrho +1}{b}^{\ast }\le \nu \left(\tau \right)\le b^{\ast }$ for $\tau \in [0,1]$. From the equation for $\kappa$, we obtain

$$h(\tau ,\nu \left(\tau \right))\le \kappa (\tau ,b^{\ast }),\mathrm{for}\tau \in (0,1).$$

(6)

From (6), we obtain

$$\begin{array}{ccc}\hfill \parallel \mathcal{J}\nu \parallel & =& \underset{\tau \in [0,1]}{max}|{\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \le & |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \le & |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\kappa (\Im ,b^{\ast })\right)d\Im |\hfill \\ & \le & b^{\ast }=\parallel \nu \parallel .\hfill \end{array}$$

If $\nu \in \partial \mathcal{B}\left(a^{\ast }\right)$, then $\parallel \nu \parallel =a^{\ast }$ and ${\tau }^{\varrho +1}{a}^{\ast }\le \nu \left(\tau \right)\le a^{\ast }$ for $\tau \in [0,1]$. From the equation for $\kappa$, we obtain

$$\begin{array}{c}\hfill h(\tau ,\nu \left(\tau \right))\le \kappa (\tau ,a^{\ast }),\mathrm{for}\tau \in (0,1).\end{array}$$

(7)

From (7), we know that

$$\begin{array}{ccc}\hfill \parallel \mathcal{J}\nu \parallel & =& \underset{\tau \in [0,1]}{max}|{\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \le & |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \le & |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\kappa (\Im ,a^{\ast })\right)d\Im |\hfill \\ & \le & a^{\ast }=\parallel \nu \parallel .\hfill \end{array}$$

If $\nu \in \partial \mathcal{B}\left(b^{\ast }\right)$, then $\parallel \nu \parallel =b^{\ast }$ and ${\tau }^{\varrho +1}{b}^{\ast }\le \nu \left(\tau \right)\le b^{\ast }$ for $\tau \in [0,1]$. From the equation for $\theta (\tau ,b^{\ast })$, we obtain

$$\begin{array}{c}\hfill h(\tau ,\nu \left(\tau \right))\ge \theta (\tau ,b^{\ast }),\mathrm{for}\tau \in (0,1).\end{array}$$

(8)

From the inequality in (8), we can see that

$$\begin{array}{ccc}\hfill \parallel \mathcal{J}\nu \parallel & =& \underset{\tau \in [0,1]}{max}|{\int }_0^1\mathcal{A}(\tau ,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \ge & \underset{\tau \in [0,1]}{max}{t}^{\varrho +1}|{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im |\hfill \\ & \ge & |{\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\theta (\Im ,b^{\ast })\right)d\Im |\hfill \\ & \ge & b^{\ast }=\parallel \nu \parallel .\hfill \end{array}$$

So, $\mathcal{J}$ has a fixed point ${\nu }^{\ast }\in \overline{\mathcal{B}\left(b^{\ast }\right)}\setminus \mathcal{B}\left(a^{\ast }\right)$. From the Guo–Krasnoselskii theorem, we deduce that $\mathcal{J}$ is a solution of (1), where $a^{\ast }\le \parallel {\nu }^{\ast }\parallel \le b^{\ast }$. For $\tau \in (0,1]$, ${\nu }^{\ast }\left(\tau \right)\ge {\tau }^{\varrho +1}\parallel {\nu }^{\ast }\parallel \ge a^{\ast }{\tau }^{\varrho +1}>0,$ then ${\nu }^{\ast }$ is the positive solution of Problem (1). The equation ${\left({\nu }^{\ast }\right)}^{\prime }\left(\tau \right)={\left(\mathcal{J}{\nu }^{\ast }\right)}^{\prime }\left(\tau \right)={\int }_0^1\frac{\partial \mathcal{A}(\tau ,\Im )}{\partial \tau }{\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)d\Im \ge 0,$ which shows that ${\nu }^{\ast }$ is a non-decreasing positive solution. □

Continuing from Theorem 4, we prove the unique positive solution, which is the solution of Problem (1), by using the tool from Definition 2 and applying Theorem 3.

Theorem 5. 

Let $h\in \mathcal{C}((0,1)\times {\mathbb{R}}^{+},{\mathbb{R}}^{+})$. For $\tau \in [0,1],\nu ,\varsigma \ge 0$. Assume that $|h(\tau ,\nu )-h(\tau ,\varsigma )|\le \frac{y\left(\tau \right)|\nu -\varsigma |}{\mathcal{M}+|\nu -\varsigma |},$ with the continuous function $y\left(\tau \right):[0,1]⟶{\mathbb{R}}^{+}$ and the constant $\mathcal{M}:={\int }_0^1\frac{\mathcal{A}(1,\Im )}{\left(\rho \right(\Im )-1)}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }y\left(1\right)d\Im \ne 0$. Then, there exists a unique positive solution on $[0,1]$ in Problem (1).

Proof. 

Let $\mathcal{G}:{\mathbb{R}}^{+}\to {\mathbb{R}}^{+}$ be a continuous non-decreasing function defined by $\mathcal{G}\left(\iota \right)=\frac{\mathcal{M}\iota }{\mathcal{M}+\iota },\forall \iota >0,$ such that $\mathcal{G}\left(0\right)=0$ and $\mathcal{G}\left(\iota \right)>\iota ,\forall \iota >0$. For any $\nu ,\varsigma \in \mathcal{E}$ and for each $\tau \in [0,1]$, we know that

$$\begin{array}{ccc}\hfill \left|\mathcal{J}\nu \right(\tau )-\mathcal{J}\varsigma (\tau \left)\right|& =& |{\int }_0^1\mathcal{A}(\tau ,\Im )\left[{\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))\right)}^{\frac{1}{\rho (\Im )-1}}-{\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\varsigma (\Im ))\right)}^{\frac{1}{\rho (\Im )-1}}\right]d\Im |\hfill \\ & \le & {\int }_0^1\mathcal{A}(\tau ,\Im )\left|\frac{1}{\rho (\Im )-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\nu (\Im ))-{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }h(\Im ,\varsigma (\Im ))\right)\right|d\Im \hfill \\ & \le & {\int }_0^1\left(\frac{\mathcal{A}(\tau ,\Im )}{\rho (\Im )-1}\right)\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\right)|h(\Im ,\nu (\Im ))-h(\Im ,\varsigma (\Im ))|d\Im \hfill \end{array}$$

for the constant $\mathcal{M}$, we obtain

$$\begin{array}{ccc}\hfill \left|\mathcal{J}\nu \right(\tau )-\mathcal{J}\varsigma (\tau \left)\right|& =& {\int }_0^1\left(\frac{\mathcal{A}(\tau ,\Im )}{\left(\rho \right(\Im )-1)}\right)\left(\frac{|\nu -\varsigma |}{(\mathcal{M}+|\nu -\varsigma \left|\right)}\right)\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\right)y(\Im )d\Im \hfill \\ & \le & \frac{\mathcal{G}(\parallel \nu -\varsigma \parallel )}{\mathcal{M}}{\int }_0^1\frac{\mathcal{A}(1,\Im )}{\left(\rho \right(\Im )-1)}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }y\left(1\right)d\Im \hfill \\ & \le & \mathcal{G}(\parallel \nu -\varsigma \parallel ).\hfill \end{array}$$

This implies that $\parallel \mathcal{J}\nu -\mathcal{J}\varsigma \parallel \le \mathcal{G}(\parallel \nu -\varsigma \parallel )$. Therefore, $\mathcal{J}$ is a nonlinear contraction. Hence, in Problem (1), $\mathcal{J}$ has a unique fixed point and $\mathcal{J}$ is a unique positive solution. □

Next, we confirm that the main results shown in Example 1 are true.

Example 1. 

Firstly, we present the following problem:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}^C{\mathcal{D}}_{0^{+}}^{\frac{9}{10}}{\psi }_{\tau +2}\left({}^C{\mathcal{D}}_{0^{+}}^{\frac{4}{5}}\nu \left(\tau \right)\right)+\frac{1}{10+\tau }\left(3-\frac{1}{\nu +1}+\sqrt{\tau }\right)=0,\tau \in [0,1]\hfill & \\ \nu \left(0\right)=\sqrt{3},\hfill & \\ \nu \left(1\right)=\frac{1}{5}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{1}{4}}\nu \left(\frac{1}{5}\right)+\frac{1}{7}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{1}{3}}\nu \left(\frac{3}{10}\right)+\frac{4}{25}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{1}{2}}\nu \left(\frac{2}{5}\right),\hfill \end{array}\right.\end{array}$$

(9)

where $\delta =\frac{9}{10},\varrho =\frac{4}{5},\rho \left(\tau \right)=\tau +2,{\rho }_1=\frac{1}{4},{\rho }_2=\frac{1}{3},{\rho }_3=\frac{1}{2},h(\tau ,\nu )=\frac{1}{10+\tau }\left(3-\frac{1}{\nu +1}+\sqrt{\tau }\right),{\eta }_1=\frac{1}{5},{\eta }_2=\frac{1}{7},{\eta }_3=\frac{4}{25},\ell =\sqrt{3},{\lambda }_1=\frac{1}{5},{\lambda }_2=\frac{3}{10},{\lambda }_3=\frac{2}{5}$. It is evident that $h\in \mathcal{C}((0,1)\times {\mathbb{R}}^{+},{\mathbb{R}}^{+})$. For positive numbers $d_1<d_2$, we can see that $\left(\mathcal{P}1\right)–\left(\mathcal{P}3\right)$ hold and $\left(\mathcal{P}2\right)$ holds for $b_{d_1,d_2}\left(\tau \right)=\frac{1}{10}\left(4-\frac{1}{d_2+1}\right)$, such that $h(\tau ,\nu )\le b_{d_1,d_2}\left(\tau \right)$ and $0<{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }{b}_{d_1,d_2}\left(\tau \right)\le 1.$

$$\kappa (\tau ,d)=max\{\frac{1}{10+\tau }\left(3-\frac{1}{\nu +1}+\sqrt{\tau }\right):{\tau }^{\frac{9}{5}}d\le \nu \le d\}\le \frac{1}{10}\left(4-\frac{1}{d+1}\right)$$

$$\theta (\tau ,d)=min\{\frac{1}{10+\tau }\left(3-\frac{1}{\nu +1}+\sqrt{\tau }\right):{\tau }^{\frac{9}{5}}d\le \nu \le d\}\le \frac{1}{10}\left(4-\frac{1}{t^{\frac{9}{5}}d+1}\right).$$

It follows that ${\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\rho (\Im )}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\delta }\kappa (\Im ,b)\right)d\Im \le b$. Consequently, we obtain

$${\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\Im +2}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{9}{10}}\kappa (\Im ,1)\right)d\Im \le 0.7250<1,$$

and

$${\int }_0^1\mathcal{A}(1,\Im ){\psi }_{\Im +2}^{-1}\left({{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{9}{10}}\theta (\Im ,0.1)\right)d\Im \ge 0.1032>0.1.$$

From Theorem 4, we know that (9) has at least one non-decreasing positive solution ${\nu }^{\ast }$ and $0.1\le \parallel {\nu }^{\ast }\parallel \le 1.$ We choose $y\left(\tau \right)=\frac{1}{10+\tau }$. Then, we find that $\mathcal{M}=0.0475$. Clearly, $|h(\tau ,\nu )-h(\tau ,\varsigma )|=\frac{1}{(10+\tau )}\left|\frac{\nu -\varsigma }{\nu \varsigma +1+\nu +\varsigma }\right|\le \frac{1}{(10+\tau )}\frac{|\nu -\varsigma |}{(0.0475+|\nu -\varsigma \left|\right)}.$ Therefore, Problem (9) has a unique positive solution on $[0,1]$ by utilizing Theorem 5.

In the next example, we consider the boundary value problem that is similar to Example 1 but with a specific condition at $\tau =0$.

Example 2. 

Consider the following boundary value problem:

$$\begin{array}{c}\hfill \left\{\begin{array}{cc}{}^C{\mathcal{D}}_{0^{+}}^{\frac{17}{10}}\nu \left(\tau \right)+\frac{1}{10+\tau }\left(3-\frac{1}{\nu +1}+\sqrt{\tau }\right)=0,\tau \in [0,1]\hfill & \\ \nu \left(0\right)=\sqrt{3},\hfill & \\ \nu \left(1\right)=\frac{1}{5}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{1}{4}}\nu \left(\frac{1}{5}\right)+\frac{1}{7}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{1}{3}}\nu \left(\frac{3}{10}\right)+\frac{4}{25}{{}_{\mathcal{RL}}\mathcal{I}}_{0^{+}}^{\frac{1}{2}}\nu \left(\frac{2}{5}\right).\hfill \end{array}\right.\end{array}$$

Therefore, this problem has a unique positive solution on $[0,1]$.

4. Conclusions and Remarks

In this work, we consider the nonlocal integral boundary value problem of a nonlinear fractional differential Equation (1). In the main results, we use the important conditions of the fixed-point theorem in the sense of Guo–Krasnoselskii and Boyd and Wong’s nonlinear contraction. Our results show that Problem (1) has a unique fixed point and a positive solution. Moreover, we confirm these results in Example 1. From Examples 1 and 2, it can be seen that the $\rho \left(\tau \right)$-Laplacian operator is an extension of the fractional differential equation, which represents the general case of Qiao and Zhou [22] and Borisut and Kumam [23], making it more useful. In the future, changing boundary value problems that arise in the laboratory and real life will be studied and further developed.