Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition

Abstract

Here, we center on the solvability of a fractional-order quadratic functional integro-differential equation with a nonlocal fractional-order integro-differential condition in the class of continuous functions. The maximal and minimal solutions will be discussed. The continuous dependence of the solutions on a few parameters will be examined. Finally, the problems of conjugate orders and integer orders, and some other problems and remarks will be discussed and presented.

1. Introduction

The solvability of nonlocal problems of fractional orders has been discussed by many authors; see, for example, refs. [1,2,3,4,5,6,7,8,9].

Many existence and uniqueness results of differential equations and inclusions with nonlocal boundary conditions have been investigated by some researchers and using different techniques (see [4,5,6,7,8] and the references therein).

Some existence results under mixed Lipschitz and Caratheodory conditions have been discussed for functional integro-differential equations of fractional orders, and the existence of some properties of the solution has been proved in [6].

Subashini et al. [10] discussed the existence of a nonlocal fractional integro-differential problem of the Hilfer type by applying the Monch fixed-point theorem and techniques of measure of noncompactness.

An investigation described the applicability of the a priori estimate method on a nonlocal, nonlinear fractional differential equation for which the weak solution and the unique solution exist [11].

Motivated by the above results, we discuss the solvability of the nonlocal problem of the fractional-order quadratic functional integro-differential equation

$$\frac{dx}{d\mathfrak{r}}\left(\mathfrak{r}\right)=f_1\left(\mathfrak{r},D^{\alpha }x\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},D^{\gamma }x\left(\mathfrak{r}\right))\right),\alpha ,\beta ,\gamma \in (0,1],\mathfrak{r}\in (0,1]$$

(1)

with the nonlocal fractional-order integro-differential condition

$$x\left(0\right)+{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),D^{\delta }x\left(\varsigma \right))d\varsigma =x_0,\delta \in (0,1],$$

(2)

where ${\mathfrak{I}}^{\beta }$ is the fractional Riemann–Liouville integral of order $\beta$ and $D^{\alpha }$ is a Caputo fractional derivative of order $\alpha \in (0,1].$

We investigate the existence of a solution to problem (1) and (2) in $C\left(I\right)$ of all continuous functions on $I=[0,1].$ Moreover, we establish some features and characteristics of these solutions and obtain some particular cases and remarks.

Now, we recall the following definitions.

Definition 1

([12]). Let $\mathfrak{h}\in L^1\left(I\right),\beta \in R^{+}$. The Riemann–Liouville fractional integral of $\mathfrak{h}$ of order β is given by

$${\mathfrak{I}}^{\beta }\mathfrak{h}\left(\mathfrak{r}\right)={\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\mathfrak{s})}^{\beta -1}}{\Gamma \left(\beta \right)}\mathfrak{h}\left(\mathfrak{s}\right)d\mathfrak{s},$$

Definition 2.

Caputo, M. Fractional derivative $D_a^{\alpha }$ of order $\alpha \in (0,1]$ of $\mathfrak{h}\left(\mathfrak{r}\right)\in AC\left(I\right)$ is given by (see [12])

$$D^{\alpha }\mathfrak{h}\left(\mathfrak{r}\right)={\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\mathfrak{s})}^{-\alpha }}{\Gamma (1-\alpha )}\frac{d}{d\mathfrak{s}}\mathfrak{h}\left(\mathfrak{s}\right)d\mathfrak{s}={\mathfrak{I}}^{1-\alpha }\frac{d}{d\mathfrak{r}}\mathfrak{h}\left(\mathfrak{r}\right)$$

2. Existence of Solution

We take into account the following assumptions:

(i)

$f_1:I\times R\to R$ is measurable in $\mathfrak{r}\in I\forall y\in R$ and continuous in $y\in R\forall \mathfrak{r}\in I$. Furthermore, ∃ a bounded measurable function $a_1:I\to R$ and a constant $b_1>0$ where

$$\begin{array}{c}\hfill |f_1(\mathfrak{r},y)|\le |a_1\left(\mathfrak{r}\right)|+b_1|y|\le a_1^{*}+b_1|y|,a_1^{*}=\underset{\mathfrak{r}\in I}{sup}{\mathfrak{I}}^{1-\alpha }|a_1\left(\mathfrak{r}\right)|.\end{array}$$

(ii)

$f_2:I\times R\to R$ is measurable in $\mathfrak{r}\in I\forall y\in R$ and continuous in $y\in R\forall \mathfrak{r}\in I$. Furthermore, ∃ a bounded measurable function $a_2:I\to R$ and a constant $b_2>0$ where

$$\begin{array}{c}\hfill |f_2(\mathfrak{r},y)|\le |a_2\left(\mathfrak{r}\right)|+b_2|y|\le a_2^{*}+b_2|y|,a_2^{*}=\underset{\mathfrak{r}\in I}{sup}{\mathfrak{I}}^{\beta }|a_2\left(\mathfrak{r}\right)|.\end{array}$$

(iii)

$r_1$ is a positive root of the following equation:

$$a_1^{*}+\left(\frac{b_1{a}_2^{*}}{\Gamma (2-\alpha )}-1\right)r_1+\frac{b_1{b}_2{r}_1^2}{\Gamma (2-\alpha )\Gamma (\alpha +\beta -\gamma +1)}=0.$$

(3)

(iv)

$h:I\times R\times R⟶R$ is measurable in $\mathfrak{r}$ for all $x,y\in R$ and continuous in $x,y$ for $\mathfrak{r}\in I$, and there exists a bounded measurable function $a_3:I\to R$ and a constant $b_3>0$ where

$$|h(\mathfrak{r},x,y)|\le |a_3\left(\mathfrak{r}\right)|+b_3(|x|+|y|)$$

where

$$\begin{array}{c}\hfill \underset{\mathfrak{r}\in I}{sup}{\int }_0^t|a_3\left(\varsigma \right)|d\varsigma \le N.\end{array}$$

(v)

$\parallel x_0\parallel >N+\frac{b_3{r}_1}{\Gamma (\alpha -\delta +1)}-\frac{r_1}{\Gamma (\alpha +1)}$

Now, we have to prove the following auxiliary result.

Lemma 1.

The solution to the problem (1) and (2) is given in the form, if it exists,

$$x\left(\mathfrak{r}\right)=x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right),\delta ,\gamma \le \alpha .$$

(4)

where y satisfies the equation

$$\begin{array}{c}\hfill y\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)\end{array}$$

(5)

and the two systems (1) and (2), and (4) and (5) are equivalent such that $\delta ,\gamma \le \alpha .$

Proof. 

x is a solution of (1) and (2). Then, we obtain

$$\begin{array}{ccc}& & D^{\alpha }x\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }\frac{dx}{d\mathfrak{r}}={\mathfrak{I}}^{1-\alpha }{f}_1(\mathfrak{r},D^{\alpha }x\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\mathfrak{r}\right)).\hfill \end{array}$$

We take $D^{\alpha }x\left(\mathfrak{r}\right)=y\left(\mathfrak{r}\right)$; thus

$$\begin{array}{c}\hfill y\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)\end{array}$$

and

$$x\left(\mathfrak{r}\right)=x\left(0\right)+{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right).$$

(6)

Now,

$$\begin{array}{ccc}\hfill {\mathfrak{I}}^{\alpha -\delta }y\left(\mathfrak{r}\right)& =& {\mathfrak{I}}^{\alpha -\delta }{D}^{\alpha }x\left(\mathfrak{r}\right)={\mathfrak{I}}^{\alpha -\delta }{\mathfrak{I}}^{1-\alpha }\frac{dx}{d\mathfrak{r}}\hfill \\ & =& {\mathfrak{I}}^{1-\delta }\frac{dx}{d\mathfrak{r}}=D^{\delta }x\left(\mathfrak{r}\right),\hfill \end{array}$$

then from (2) and (6), we obtain (4).

By letting $y\left(\mathfrak{r}\right)=D^{\alpha }x\left(\mathfrak{r}\right)$ in (4), we obtain (2), and in (5), we obtain

$${\mathfrak{I}}^{1-\alpha }\frac{d}{d\mathfrak{r}}x\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},D^{\alpha }x\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,D^{\gamma }x\left(\theta \right)d\theta \right).$$

Operating by ${\mathfrak{I}}^{\alpha },$ then by $\frac{d}{d\mathfrak{r}}$, respectively, we obtain (1). This proves the equivalence of the two systems (1) and (2), and (4) and (5). □

Theorem 1.

Let us assume that the conditions (i)–(iv) hold; then, problem (1) and (2) has a solution in $C\left(I\right)$.

Proof. 

We characterize a closed ball $Q_{r_1}$ and an operator $F_1$ as

$$\begin{array}{c}\hfill Q_{r_1}=\{y\in C\left(I\right):\parallel y\parallel \le r_1\},r_1=a_1^{*}+\frac{1}{\Gamma (2-\alpha )}(r_1{b}_1{a}_2^{*}+\frac{b_1{b}_2{r}_1^2}{\Gamma (\alpha +\beta -\gamma +1)})\end{array}$$

and

$$\begin{array}{c}\hfill F_{1}y\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma .\end{array}$$

For $y\in Q_{r_1}$, we can obtain

$$\begin{array}{ccc}\hfill |y(\mathfrak{r})|& =& {\mathfrak{I}}^{1-\alpha }|f_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\mathfrak{r}\right))\right)|\hfill \\ & \le & {\mathfrak{I}}^{1-\alpha }\left[|a_1\left(\mathfrak{r}\right)|+b_1|y\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\mathfrak{r}\right))|\right]\hfill \\ & \le & {\mathfrak{I}}^{1-\alpha }\left[|a_1\left(\mathfrak{r}\right)|+b_1\parallel y\parallel (a_2^{*}+b_2{\mathfrak{I}}^{\alpha +\beta -\gamma }\parallel y\parallel )\right]\hfill \\ & \le & {\mathfrak{I}}^{1-\alpha }\left[|a_1\left(\mathfrak{r}\right)|+b_1{r}_1(a_2^{*}+\frac{r_1{b}_2}{\Gamma (\alpha +\beta -\gamma +1)})\right],\hfill \end{array}$$

$$\begin{array}{ccc}\hfill {\mathfrak{I}}^{1-\alpha }|f_1(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\mathfrak{r}\right))|& \le & a_1^{*}+\frac{1}{\Gamma (2-\alpha )}(r_1{b}_1{a}_2^{*}+\frac{r_1^2{b}_1{b}_2}{\Gamma (\alpha +\beta -\gamma +1)})\hfill \end{array}$$

and we deduce that

$$\begin{array}{ccc}\hfill |F_{1}y\left(\mathfrak{r}\right)|& =& \left|{\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \right|\hfill \\ & \le & {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}\left(a_1+b_1|y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta |\right)d\varsigma \hfill \\ & \le & {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}\left(a_1+b_1\parallel y\parallel \times (a_2^{*}+\frac{b_2\parallel y\parallel }{\Gamma (\alpha +\beta -\gamma +1)}\right)d\varsigma \hfill \\ & \le & a_1^{*}+\frac{1}{\Gamma (2-\alpha )}(r_1{b}_1{a}_2^{*}+\frac{b_1{b}_2{r}_1^2}{\Gamma (\alpha +\beta -\gamma +1)})=r_1.\hfill \end{array}$$

So, we have

$$\begin{array}{ccc}\hfill \parallel F_{1}y\parallel & \le & a_1^{*}+\frac{1}{\Gamma (2-\alpha )}(r_1{b}_1{a}_2^{*}+\frac{b_1{b}_2{r}_1^2}{\Gamma (\alpha +\beta -\gamma +1)})=r_1.\hfill \end{array}$$

This proves that $F_1:Q_{r_1}\to Q_{r_1}$ and the family $\left\{F_{{}_1}y\right\}$ is uniformly bounded on $Q_{r_1}$.

Let $y\in Q_{r_1}$ and ${\mathfrak{r}}_1,{\mathfrak{r}}_2\in I$, where ${\mathfrak{r}}_2>{\mathfrak{r}}_1$ and $\mid {\mathfrak{r}}_1-{\mathfrak{r}}_2\mid \le \delta$; therefore,

$$\begin{array}{ccc}& & |F_{1}y\left({\mathfrak{r}}_2\right)-F_{1}y\left({\mathfrak{r}}_1\right)|=\hfill \\ \\ & & |{\int }_0^{{\mathfrak{r}}_2}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)ds\hfill \\ & -& {\int }_0^{{\mathfrak{r}}_1}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}f\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma |\hfill \\ & \le & |{\int }_0^{{\mathfrak{r}}_1}\frac{{({\mathfrak{r}}_2-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}f\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & +& {\int }_{{\mathfrak{r}}_1}^{{\mathfrak{r}}_2}\frac{{({\mathfrak{r}}_2-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & -& {\int }_0^{{\mathfrak{r}}_1}\frac{{({\mathfrak{r}}_1-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma |\hfill \\ & \le & |{\int }_0^{{\mathfrak{r}}_1}\frac{{({\mathfrak{r}}_2-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}-\frac{{({\mathfrak{r}}_1-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}(a_1^{*}+r_1{b}_1{a}_2^{*}+\frac{b_1{b}_2{r}_1^2}{\Gamma (\alpha +\beta -\gamma +1)})d\varsigma \hfill \\ & +& {\int }_{{\mathfrak{r}}_1}^{{\mathfrak{r}}_2}\frac{{({\mathfrak{r}}_2-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}(a_1+\frac{b_1{a}_2{r}_1}{\Gamma (1+\beta )}+\frac{b_1{b}_2{r}_1^2}{\Gamma (1+\beta )\Gamma (1+\alpha -\gamma )})d\varsigma |\hfill \\ & \le & {\int }_0^{{\mathfrak{r}}_1}|\frac{{({\mathfrak{r}}_2-\varsigma )}^{\alpha }-{({\mathfrak{r}}_1-\varsigma )}^{\alpha }}{\Gamma (1-\alpha ){({\mathfrak{r}}_1-\varsigma )}^{\alpha }{({\mathfrak{r}}_2-\varsigma )}^{\alpha }}|(a_1^{*}+r_1{b}_1{a}_2^{*}+\frac{b_1{b}_2{r}_1^2}{\Gamma (\alpha +\beta -\gamma +1)})ds\hfill \\ & +& {\int }_{{\mathfrak{r}}_1}^{{\mathfrak{r}}_2}|\frac{{({\mathfrak{r}}_2-\varsigma )}^{\alpha }}{\Gamma (1-\alpha ){({\mathfrak{r}}_2-\varsigma )}^{\alpha }}|(a_1^{*}+r_1{b}_1{a}_2^{*}+\frac{b_1{b}_2{r}_1^2}{\Gamma (\alpha +\beta -\gamma +1)})d\varsigma .\hfill \end{array}$$

This shows that family $\left\{F_{1}y\right\}$ is equi-continuous on $Q_{r_1}$, and by the Arzela–Ascoli result [13], mapping $F_1$ is relatively compact.

Let $\left\{y_n\right\}\subset Q_{r_1}$ be such that $y_n\to y$; then,

$$\begin{array}{ccc}\hfill F_1{y}_n\left(\mathfrak{r}\right)& =& {\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y_n\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_n\left(\sigma \right)d\sigma )d\theta \right),\hfill \end{array}$$

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{F}_1{y}_n\left(\mathfrak{r}\right)& =& \underset{n\to \infty }{lim}{\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y_n\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_n\left(\sigma \right)d\sigma )d\theta \right)\hfill \\ & =& {\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}\underset{n\to \infty }{lim}{y}_n\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & =& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & =& F_{1}y\left(\mathfrak{r}\right).\hfill \end{array}$$

This yields that $F_1{y}_n\left(\mathfrak{r}\right)\to F_{1}y\left(\mathfrak{r}\right)$.

Thus, mapping $F_1$ is continuous, and by applying the Schauder fixed-point theorem [13], $y\in Q_{r_1}\subset C\left(I\right)$ is a solution of (5).

To prove that $x\in C\left(I\right)$ exists and satisfies Equation (4), we prove the next result. □

Theorem 2.

Let us suppose that (i)–(iv) are satisfied; then, for a fixed function y (given by Theorem 1), $\exists x\in C\left(I\right)$ and satisfies (4).

Proof. 

Let $Q_{r_2}$ be the closed ball

$$\begin{array}{c}\hfill Q_{r_2}=\{x\in C\left(I\right):\parallel x\parallel \le r_2\},r_2=\frac{\parallel x_0\parallel -N-\frac{b_3{r}_1}{\Gamma (\alpha -\delta +1)}+\frac{r_1}{\Gamma (\alpha +1)}}{(1+b_3)}\end{array}$$

and define operator $F_2$

$$\begin{array}{c}\hfill F_{2}x\left(\mathfrak{r}\right)=x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right).\end{array}$$

Let $x\in Q_{r_2}$; then,

$$\begin{array}{ccc}\hfill |F_{2}x\left(\mathfrak{r}\right)|& =& |x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)|\hfill \\ & \le & |x_0|-{\int }_0^1|h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|d\varsigma +{\mathfrak{I}}^{\alpha }|y\left(\mathfrak{r}\right)|\hfill \\ & \le & |x_0|-{\int }_0^1\left(|a_3|+b_3(|x\left(\varsigma \right)|+\frac{\parallel y\parallel }{\Gamma (\alpha -\delta +1)})\right)d\varsigma +\frac{\parallel y\parallel }{\Gamma (\alpha +1)}\hfill \\ & \le & |x_0|-{\int }_0^1|a_3\left(\varsigma \right)|d\varsigma -b_3{\int }_0^1|x\left(\varsigma \right)|d\varsigma -\frac{b_3{r}_1}{\Gamma (\alpha -\delta +1)}{\int }_0^{1}d\varsigma +\frac{r_1}{\Gamma (\alpha +1)}\hfill \\ & \le & \parallel x_0\parallel -N-b_3{r}_2-\frac{b_3{r}_1}{\Gamma (\alpha -\delta +1)}+\frac{r_1}{\Gamma (\alpha +1)}=r_2\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \parallel F_{2}x\parallel \le |x_0|-N-b_3{r}_2-\frac{b_3{r}_1}{\Gamma (\alpha -\delta +1)}+\frac{r_1}{\Gamma (\alpha +1)}=r_2\end{array}$$

Similarly as performed above, $F_2:Q_{r_2}\to Q_{r_2}$, and $\left\{F_{{}_2}x\right\}$ is uniformly bounded on $Q_{r_2}$.

Let $x\in Q_{r_2}$ and ${\mathfrak{r}}_1,{\mathfrak{r}}_2\in I$; moreover, ${\mathfrak{r}}_2>{\mathfrak{r}}_1$ and $\mid {\mathfrak{r}}_1-{\mathfrak{r}}_2\mid \le \delta$. Then,

$$\begin{array}{ccc}\hfill |F_{2}x\left({\mathfrak{r}}_2\right)-F_{2}x\left({\mathfrak{r}}_1\right)|& =& |x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left({\mathfrak{r}}_2\right)\hfill \\ & -& x_0+{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\mathfrak{I}}^{\alpha }y\left({\mathfrak{r}}_1\right)|\hfill \\ & \le & {\int }_0^{{\mathfrak{r}}_2}|f_1(\varsigma ,y\left(\varsigma \right))\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\varsigma \right))|d\varsigma \hfill \\ & -& {\int }_0^{{\mathfrak{r}}_1}|f_1(\varsigma ,y\left(\varsigma \right))\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\varsigma \right))|d\varsigma \hfill \\ & \le & {\int }_{{\mathfrak{r}}_1}^{{\mathfrak{r}}_2}|f_1(\varsigma ,y\left(\varsigma \right))\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{\alpha -\gamma }y\left(\varsigma \right))|d\varsigma .\hfill \end{array}$$

Thus, family $\left\{F_{2}x\right\}$ is equicontinuous on $Q_{r_2}$, and operator $F_2$ is relatively compact.

For $\left\{x_n\right\}\subset Q_{r_2}$, and $x_n\to y$; then,

$$\begin{array}{ccc}\hfill F_2{x}_n\left(\mathfrak{r}\right)& =& x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{F}_2{x}_n\left(\mathfrak{r}\right)& =& \underset{n\to \infty }{lim}\left(x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\right)\hfill \end{array}$$

Using the Lebesgue dominated convergence theorem, we have

$$\begin{array}{ccc}\hfill \underset{n\to \infty }{lim}{F}_2{x}_n\left(\mathfrak{r}\right)& =& x_0-{\int }_0^{1}h(\varsigma ,\underset{n\to \infty }{lim}x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\hfill \\ & =& x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)=F_{2}x\left(\mathfrak{r}\right)\hfill \end{array}$$

This means that $F_2{x}_n\left(\mathfrak{r}\right)\to F_{2}x\left(\mathfrak{r}\right)$. Hence, operator $F_2$ is continuous, and there exists $x\in Q_{r_2}\subset C\left(I\right)$ that satisfies (4).

Therefore, we prove the existence of a continuous solution x of Equation (4). □

2.1. Conjugate-Order Problem

For $\gamma =1-\alpha ,\alpha \in [\frac{1}{2},1)$ in problem (1) and (2), we have the nonlocal problem of conjugate-orders $(\alpha ,1-\alpha )$

$$\frac{dx}{d\mathfrak{r}}\left(\mathfrak{r}\right)=f_1\left(\mathfrak{r},D^{\alpha }x\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},D^{1-\alpha }x\left(\mathfrak{r}\right))\right),\alpha ,\beta \in (0,1],\mathfrak{r}\in (0,1]$$

(7)

with nonlocal fractional-order integro-differential condition (2).

By putting $D^{\alpha }x\left(\mathfrak{r}\right)=y\left(\mathfrak{r}\right),$ we can prove that nonlocal problem (7) with condition (2) is equivalent to the functional integral equation

$$y\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }{f}_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{2\alpha -1}y\left(\mathfrak{r}\right))\right),\alpha \ge \frac{1}{2}$$

(8)

with (4). Then, by applying Theorems 1 and 2, we can deduce the solvability of nonlocal problem (7) and (2) in $C\left(I\right).$

2.2. Absolutely Continuous Solution

By taking $y\left(\mathfrak{r}\right)=\frac{dx}{d\mathfrak{r}}\left(\mathfrak{r}\right),$ (1) and (2) reduce to the fractional-order quadratic integral equation

$$y\left(\mathfrak{r}\right)=f_1\left(\mathfrak{r},{\mathfrak{I}}^{1-\alpha }y\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},{\mathfrak{I}}^{1-\gamma }y\left(\mathfrak{r}\right))\right),\mathfrak{r}\in (0,1]$$

(9)

and the nonlocal integro-differential condition

$$x\left(0\right)+{\int }_0^{1}h(\varsigma ,x_0+{\int }_0^{\varsigma }y\left(s\right)ds,{\mathfrak{I}}^{1-\delta }y\left(\varsigma \right))d\varsigma =x_0.$$

(10)

In the same way as in Lemma 1, we can prove the equivalence between (1) and (2), and system (9) and

$$x\left(\mathfrak{r}\right)=x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),y\left(\varsigma \right))d\varsigma +{\int }_0^{t}y\left(s\right)ds$$

(11)

With the same technique as in [14], the existence of solution $y\in L^1\left(I\right)$ of (9) can be proved. So, the existence of an absolutely continuous solution

$$x\left(\mathfrak{r}\right)=x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{1-\delta }y\left(\varsigma \right))d\varsigma +{\int }_0^{t}y\left(s\right)ds\in AC\left(I\right)$$

(12)

of problem (1) and (2) can be proved.

2.3. Integer-Order Problem

By letting $\alpha ,\beta ,\gamma \to 1$ in problem (1) and (2), we have the integer-order equation

$$\frac{dx}{d\mathfrak{r}}\left(\mathfrak{r}\right)=f_1\left(\mathfrak{r},\frac{d}{d\mathfrak{r}}x\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}{f}_2(\varsigma ,\frac{d}{d\varsigma }x\left(\varsigma \right))d\varsigma \right),\mathfrak{r}\in (0,1]$$

(13)

with the nonlocal integro-differential condition

$$x\left(0\right)+{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),\frac{d}{d\varsigma }x\left(\varsigma \right))d\varsigma =x_0.$$

(14)

In the same way as in Lemma 1, we can prove the equivalence between (13) and (14), and the system

$$y\left(\mathfrak{r}\right)=f_1\left(\mathfrak{r},y\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}{f}_2(\varsigma ,y\left(\varsigma \right))d\varsigma \right),\mathfrak{r}\in (0,1]$$

(15)

and

$$x\left(\mathfrak{r}\right)=x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),y\left(\varsigma \right))d\varsigma +{\int }_0^{t}y\left(s\right)ds$$

(16)

and as in [14], the existence of solution $y\in L^1\left(I\right)$ of (15) can be proved. Consequently,$x\in AC\left(I\right)$ of integer-order problem (13) and (14) can be proved.

3. Some Characteristics of the Solution

3.1. Maximal and Minimal Solutions

Lemma 2.

Let us assume that the conditions of Theorem 1 hold. Let $x,y\in C\left(I\right)$ satisfy

$$\begin{array}{ccc}\hfill x\left(\mathfrak{r}\right)& \le & {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,x\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}x\left(\sigma \right)d\sigma )d\theta \right)d\varsigma ,\hfill \\ \hfill y\left(\mathfrak{r}\right)& \ge & {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \end{array}$$

such that one of them is strict. Let us assume that $f_1$ and $f_2$ are monotonic non-decreasing in the second argument; thus,

$$\begin{array}{ccc}\hfill x\left(\mathfrak{r}\right)& <& y\left(\mathfrak{r}\right),\mathfrak{r}>0.\hfill \end{array}$$

(17)

Proof. 

Let us assume that result (17) is not true; then, $\exists t_1$ where

$x\left({\mathfrak{r}}_1\right)=y\left({\mathfrak{r}}_1\right),{\mathfrak{r}}_1>0,$$x\left(\mathfrak{r}\right)<y\left(\mathfrak{r}\right)0<\mathfrak{r}<{\mathfrak{r}}_1.$

From the monotonicity of $f_1$ and $f_2$, we obtain

$$\begin{array}{ccc}\hfill x\left({\mathfrak{r}}_1\right)& \le & {\int }_0^{{\mathfrak{r}}_1}\frac{{({\mathfrak{r}}_1-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,x\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}x\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & <& {\int }_0^{{\mathfrak{r}}_1}\frac{{({\mathfrak{r}}_1-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & =& y\left({\mathfrak{r}}_1\right).\hfill \end{array}$$

Hence $x\left({\mathfrak{r}}_1\right)<y\left({\mathfrak{r}}_1\right),$ which contradicts that $x\left({\mathfrak{r}}_1\right)=y\left(t_1\right)$; then,

$x\left(\mathfrak{r}\right)<y\left(\mathfrak{r}\right),\mathfrak{r}\in I.$ □

Theorem 3.

Let us assume that the conditions of Theorem 1 hold. For monotonic non-decreasing functions $f_1$ and $f_2$ in the second argument, problem (1) and (2) has maximal and minimal solutions.

Proof. 

To investigate the existence of the maximal solution of Equation (5), we take $ϵ>0$ and consider the following integral equations:

$$\begin{array}{c}\hfill y_{ϵ}\left(\mathfrak{r}\right)={\mathfrak{I}}^{1-\alpha }{f}_{1ϵ}\left(\mathfrak{r},y_{ϵ}\left(\mathfrak{r}\right)\times {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_{2ϵ}(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{ϵ}\left(\sigma \right)d\sigma )d\theta \right),\end{array}$$

(18)

where

$$f_{1ϵ}(\mathfrak{r},y_{ϵ}\left(\mathfrak{r}\right))=f_1(\mathfrak{r},y_{ϵ}\left(\mathfrak{r}\right))+ϵ$$

$$f_{2ϵ}(\mathfrak{r},y_{ϵ}\left(\mathfrak{r}\right))=f_2(\mathfrak{r},y_{ϵ}\left(\mathfrak{r}\right))+ϵ.$$

Obviously, $f_{1ϵ}$ and $f_{2ϵ}$ satisfy the assumptions of Theorem 1; according to Theorem 1, Equation (18) has a continuous solution $y_{ϵ}$ and

$$\begin{array}{ccc}\hfill y_{ϵ}\left(\mathfrak{r}\right)& & =\hfill \\ \hfill & & {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}\hfill \\ \hfill & & \times \left[ϵ+f_1\left(\varsigma ,y_{ϵ}\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}\left\{ϵ+f_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{ϵ}\left(\sigma \right)d\sigma )d\theta \right)\right\}d\varsigma \right].\hfill \end{array}$$

Next, we take ${ϵ}_1,{ϵ}_2>0$ such that $0<{ϵ}_2<{ϵ}_1<ϵ$; then,

$$\begin{array}{ccc}\hfill y_{{ϵ}_1}\left(\mathfrak{r}\right)& =& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}[{ϵ}_1+f_1(\varsigma ,y_{{ϵ}_1}\left(\varsigma \right)\hfill \\ & & \times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}\left\{{ϵ}_1+f_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{{ϵ}_1}\left(\sigma \right)d\sigma )d\theta \right)\left\}d\varsigma \right]\hfill \\ & >& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}[{ϵ}_2+f_1(\varsigma ,y_{{ϵ}_1}\left(\varsigma \right)\hfill \\ & & \times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}\left\{{ϵ}_2+f_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{{ϵ}_1}\left(\sigma \right)d\sigma )d\theta \right)\left\}d\varsigma \right]\hfill \end{array}$$

(19)

$$y_{{ϵ}_2}\left(\mathfrak{r}\right)={\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}\left[{ϵ}_2+f_1\left(\varsigma ,y_{{ϵ}_2}\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}\left\{{ϵ}_2+f_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{{ϵ}_2}\left(\sigma \right)d\sigma )d\theta \right)\right\}d\varsigma \right]$$

(20)

By applying Lemma 2 on estimates (19) and (20), we obtain

$$y_{{ϵ}_2}\left(\mathfrak{r}\right)<y_{{ϵ}_1}\left(\mathfrak{r}\right),\mathfrak{r}\in I.$$

As proved in Theorem 1, $\left\{y_{ϵ}\left(\mathfrak{r}\right)\right\}$ is equicontinuous and uniformly bounded on I; then, $\left\{y_{ϵ}\right\}$ is relatively compact, and there exists a decreasing sequence ${ϵ}_n$ such that ${ϵ}_n\to 0,$ $n\to \infty$ and $\underset{n\to \infty }{lim}{y}_{{ϵ}_n}\left(\mathfrak{r}\right)$ exists uniformly on I. Let

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{y}_{{ϵ}_n}\left(\mathfrak{r}\right)=q\left(\mathfrak{r}\right).\end{array}$$

Then,

$$\begin{array}{c}\hfill {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y_{{ϵ}_n}\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{{ϵ}_n}\left(\sigma \right)d\sigma )d\theta \right)d\varsigma )\to \\ \hfill {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,q\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{\left(\varsigma \right)-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}q\left(\sigma \right)d\sigma )d\theta \right)d\varsigma ,\end{array}$$

then,

$$\begin{array}{ccc}\hfill q\left(\mathfrak{r}\right)& =& \underset{n\to \infty }{lim}{y}_{{ϵ}_n}\left(\mathfrak{r}\right)\hfill \\ & =& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,q\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{\left(\varsigma \right)-\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}q\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \end{array}$$

which yields that $q\left(\mathfrak{r}\right)$ is a solution of (5).

Now, if each $y\left(\mathfrak{r}\right)$ is a solution of (5), then

$$\begin{array}{ccc}\hfill y_{ϵ}\left(\mathfrak{r}\right)& =& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}[ϵ+f_1(\varsigma ,y_{ϵ}\left(\varsigma \right)\hfill \\ & & \times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}\left\{ϵ+f_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{ϵ}\left(\sigma \right)d\sigma )d\theta \right)\left\}d\varsigma \right]\hfill \\ & >& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y_{ϵ}\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_{ϵ}\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \end{array}$$

(21)

and

$$y\left(\mathfrak{r}\right)={\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}{f}_1\left(\varsigma ,y\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}y\left(\sigma \right)d\sigma )d\theta \right)d\varsigma$$

(22)

Using Lemma 2 on estimates (21) and (22), we obtain

$$\begin{array}{c}\hfill y\left(\mathfrak{r}\right)<y_{ϵ}\left(\mathfrak{r}\right),\mathfrak{r}\in I.\end{array}$$

The uniqueness of the maximal solution yields that $y_{ϵ}\left(\mathfrak{r}\right)\to q\left(\mathfrak{r}\right)$ uniformly on I as $ϵ\to 0$. Hence, q is the maximal solution of (5).

Similarly, we study the existence of the minimal solution of problem (1) and (2). □

3.2. Existence of Unique Solution

Let us assume the following conditions:

${\left(i\right)}^{*}$

$f_1,f_2:I\times R\to R$ are measurable in $\mathfrak{r}\in I\forall x\in R$ and satisfy the Lipschitz condition,

$$\begin{array}{c}\hfill |f_1(\mathfrak{r},x)-f_1(\mathfrak{r},y)|\le L_1|x-y|,\mathfrak{r}\in I,x,y\in R.\end{array}$$

(23)

$$\begin{array}{c}\hfill |f_2(\mathfrak{r},x)-f_2(\mathfrak{r},y)|\le L_2|x-y|,\mathfrak{r}\in I,x,y\in R.\end{array}$$

(24)

Assumption $\left(iv\right)$ implies that

$$|f_1(\mathfrak{r},y)|\le |f_1(\mathfrak{r},0)|+L_1|y|$$

and

$$\begin{array}{c}\hfill |f_1(\mathfrak{r},y)|\le a_1+L_1|y|,suchthat{a}_1=\underset{\mathfrak{r}\in I}{sup}|f_1(\mathfrak{r},0)|.\end{array}$$

Furthermore,

$$|f_2(\mathfrak{r},y)|\le |f_2(\mathfrak{r},0)|+L_2|y|$$

and

$$\begin{array}{c}\hfill |f_2(\mathfrak{r},y)|\le a_2+L_2|y|,where{a}_2=\underset{\mathfrak{r}\in I}{sup}|f_2(\mathfrak{r},0)|.\end{array}$$

${\left(ii\right)}^{*}$

$h:I\times R\times R\to R$ is measurable in $\mathfrak{r}\in I\forall x,y\in R$ and satisfies the Lipschitz condition,

$$\begin{array}{c}\hfill |h(\mathfrak{r},x_1,y_1)-h(\mathfrak{r},x_2,y_2)|\le L_3(|x_1-x_2|+|y_1-y_2|)\end{array}$$

(25)

with Lipschitz condition $L_3>0$.

Theorem 4.

Let us assume that (iii), (i*) are verified. If

$$\begin{array}{c}\hfill \frac{2b_1{b}_2{r}_1}{\Gamma (2-\alpha )\Gamma (\alpha +\beta -\gamma +1)}+\frac{b_1{a}_2^{*}}{\Gamma (2-\alpha )}<1.\end{array}$$

(26)

then the solution of (5) is unique.

Proof. 

Let us consider that assumptions (iii), (i*) are satisfied; then, Theorem 1 is verified, and the solution of Equation (5) exists. Let $y_1,y_2$ be two solutions of (5), where

$$\begin{array}{ccc}\hfill |y_2\left(\mathfrak{r}\right)-y_1\left(\mathfrak{r}\right)|& =& |{\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}f\left(\varsigma ,y_2\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_2\left(\sigma \right)d\sigma )d\theta \right)d\varsigma \hfill \\ & -& {\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}f\left(\varsigma ,y_1\left(\varsigma \right)\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}{y}_1\left(\sigma \right)d\sigma )d\theta \right)d\varsigma |\hfill \\ & \le & b_1{\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}(|y_2\left(\varsigma \right)|\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}|y_2\left(\sigma \right)|d\sigma )d\theta \hfill \\ & -& |y_1\left(\varsigma \right)|\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}|y_1\left(\sigma \right)|d\sigma )d\theta )d\varsigma \hfill \\ & \le & b_1{\int }_0^t\frac{{(t-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}(|y_2\left(\varsigma \right)|\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}|y_2\left(\sigma \right)|d\sigma )d\theta \hfill \\ & -& |y_2\left(\varsigma \right)|\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}|y_1\left(\sigma \right)|d\sigma )d\theta \hfill \\ & +& |y_2\left(\varsigma \right)|\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}|y_1\left(\sigma \right)|d\sigma )d\theta \hfill \\ & -& |y_1\left(\varsigma \right)|\times {\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{f}_2(\theta ,{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{(\alpha -\gamma )-1}}{\Gamma (\alpha -\gamma )}|y_1\left(\sigma \right)|d\sigma )d\theta )d\varsigma \hfill \\ & \le & b_1{\int }_0^{\mathfrak{r}}\frac{{(\mathfrak{r}-\varsigma )}^{-\alpha }}{\Gamma (1-\alpha )}(\parallel y_2\parallel \times b_2{\int }_0^{\varsigma }\frac{{(\varsigma -\theta )}^{\beta -1}}{\Gamma \left(\beta \right)}{\int }_0^{\theta }\frac{{(\theta -\sigma )}^{\alpha -1}}{\Gamma \left(\alpha \right)}|y_2\left(\sigma \right)-y_1\left(\sigma \right)|d\sigma d\theta \hfill \\ & +& \parallel y_2-y_1\parallel \times (a_2^{*}+\frac{b_2{r}_1}{\Gamma (\beta +\alpha -\gamma +1)}))d\varsigma \hfill \\ & \le & \frac{b_1}{\Gamma (2-\alpha )}(\parallel y_2-y_1\parallel \frac{b_2{r}_1}{\Gamma (\beta +\alpha -\gamma +1)}\hfill \\ & +& \parallel y_2-y_1\parallel (a_2^{*}+\frac{b_2{r}_1}{\Gamma (\beta +\alpha -\gamma +1)})).\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel y_2-y_1\parallel \left(1-(\frac{2b_1{b}_2{r}_1}{\Gamma (2-\alpha )\Gamma (\beta +\alpha -\gamma +1)}+\frac{b_1{a}_2^{*}}{\Gamma (2-\alpha )})\right)\le 0.\end{array}$$

which implies the uniqueness of the solution of (5). □

Corollary 1.

Let us suppose that Theorem (4) is verified. If $L_3<1,$ then for every solution $y\in C\left(I\right)$ of Equation (5), there exists a unique solution x of (4).

Proof. 

Let $y\in C\left(I\right)$ satisfy Equation (5) and $x_1,x_2$ satisfy Equation (4); then, we have

$$\begin{array}{ccc}\hfill |x_2\left(\mathfrak{r}\right)-x_1\left(\mathfrak{r}\right)|& =& |x_0-{\int }_0^{1}h(\varsigma ,x_2\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\hfill \\ & -& x_0+{\int }_0^{1}h(\varsigma ,x_1\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)|\hfill \\ & \le & {\int }_0^1\left(|h(\varsigma ,x_2\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|-|h(\varsigma ,x_1\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|\right)d\varsigma \hfill \\ \\ & \le & L_3|x_2\left(\mathfrak{r}\right)-x_1\left(\mathfrak{r}\right)|,\hfill \end{array}$$

then,

$$\begin{array}{c}\hfill \parallel x_2-x_1\parallel (1-L_3)\le 0\to x_2=x_1.\end{array}$$

 □

3.3. Continuous Dependency Results

Definition 3.

The unique solution $x\in C\left(I\right)$ of (4) depends continuously on $x_0$, if $\forall ϵ>0$, $\exists \delta >0$, where

$$|x_0-x_0^{*}|\le \delta \Rightarrow \parallel x-x^{*}\parallel \le ϵ$$

where $x^{*}$ satisfies

$$\begin{array}{c}\hfill x^{*}\left(\mathfrak{r}\right)=x_0^{*}-{\int }_0^{1}h(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right).\end{array}$$

Theorem 5.

Let assumption (iv) hold; then, the unique solution of (4) depends continuously on $x_0$.

Proof. 

Let us assume that the two functions $x\left(\mathfrak{r}\right)$ and $x^{*}\left(\mathfrak{r}\right)$ satisfy (4); then,

$$\begin{array}{ccc}\hfill |x\left(\mathfrak{r}\right)-x^{*}\left(\mathfrak{r}\right)|& =& |x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))ds+{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\hfill \\ & -& x_0^{*}+{\int }_0^{1}h(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)|\hfill \\ & \le & |x_0-x_0^{*}|+{\int }_0^1|h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))-h(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|d\varsigma \hfill \\ & \le & |x_0-x_0^{*}|+L_3|x\left(\varsigma \right)-x^{*}\left(\varsigma \right)|\hfill \\ & \le & \delta +L_3\parallel x-x^{*}\parallel .\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel (1-L_3)\le \delta \end{array}$$

and

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel \le \frac{\delta }{1-L_3}=ϵ.\end{array}$$

 □

Definition 4.

The unique solution $x\in C\left(I\right)$ of Equation (4) depends continuously on y; if $\forall ϵ>0$, $\exists \delta >0$, where

$$|y-y^{*}|\le \delta \Rightarrow \parallel x-x^{*}\parallel \le ϵ$$

where $x^{*}$ satisfies the integral equation

$$\begin{array}{c}\hfill x^{*}\left(\mathfrak{r}\right)=x_0^{*}-{\int }_0^{1}h(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right).\end{array}$$

Theorem 6.

Let assumption (iv) hold; then, the unique solution of the integral Equation (4) depends continuously on y.

Proof. 

Let us assume that the two functions $x\left(\mathfrak{r}\right)$ and $x^{*}\left(\mathfrak{r}\right)$ satisfy Equation (4); then,

$$\begin{array}{ccc}\hfill |x\left(\mathfrak{r}\right)-x^{*}\left(\mathfrak{r}\right)|& =& |x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\hfill \\ & -& x_0^{*}+{\int }_0^{1}h(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)|\hfill \\ & =& |{\int }_0^1(h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))-h(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right)))d\varsigma \hfill \\ & +& {\mathfrak{I}}^{\alpha }(y\left(\mathfrak{r}\right)-y^{*}\left(\mathfrak{r}\right))|\hfill \\ & \le & L_3|x\left(\varsigma \right)-x^{*}\left(\varsigma \right)|+L_3{\mathfrak{I}}^{\alpha -\delta }|y\left(\varsigma \right)-y^{*}\left(\varsigma \right)|+{\mathfrak{I}}^{\alpha }|y\left(\mathfrak{r}\right)-y^{*}\left(\mathfrak{r}\right)|\hfill \\ & \le & L_3\parallel x-x^{*}\parallel +\frac{L_3\delta }{\Gamma (\alpha -\delta +1)}+\frac{\delta }{\Gamma (\alpha +1)}.\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel (1-L_3)\le \frac{L_3\delta }{\Gamma (\alpha -\delta +1)}+\frac{\delta }{\Gamma (\alpha +1)}\end{array}$$

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel \le \frac{\frac{L_3\delta }{\Gamma (\alpha -\delta +1)}+\frac{\delta }{\Gamma (\alpha +1)}}{(1-L_3)}=ϵ.\end{array}$$

 □

Definition 5.

The unique solution $x\in C\left(I\right)$ of Equation (4) depends continuously on h; if $\forall ϵ>0$, $\exists \delta >0$, where

$$|h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))-h^{*}(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|\le \delta \Rightarrow \parallel x-x^{*}\parallel \le ϵ$$

where $x^{*}$ satisfies the integral equation

$$\begin{array}{c}\hfill x^{*}\left(\mathfrak{r}\right)=x_0-{\int }_0^1{h}^{*}(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma +{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right).\end{array}$$

Theorem 7.

Let assumption (iv) hold; then, the unique solution of (4) depends continuously on h.

Proof. 

Let us assume that the two functions $x\left(\mathfrak{r}\right)$ and $x^{*}\left(\mathfrak{r}\right)$ satisfy Equation (4); then,

$$\begin{array}{ccc}\hfill |x\left(\mathfrak{r}\right)-x^{*}\left(\mathfrak{r}\right)|& =& |x_0-{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))ds+{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)\hfill \\ & -& x_0+{\int }_0^1{h}^{*}(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\mathfrak{I}}^{\alpha }y\left(\mathfrak{r}\right)|\hfill \\ & =& |{\int }_0^{1}h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\int }_0^1{h}^{*}(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma \hfill \\ & +& {\int }_0^1{h}^{*}(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma -{\int }_0^1{h}^{*}(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))d\varsigma |\hfill \\ & =& {\int }_0^1|h(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))-{\int }_0^1{h}^{*}(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|d\varsigma \hfill \\ & +& {\int }_0^1|h^{*}(\varsigma ,x\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))-{\int }_0^1{h}^{*}(\varsigma ,x^{*}\left(\varsigma \right),{\mathfrak{I}}^{\alpha -\delta }y\left(\varsigma \right))|d\varsigma \hfill \\ \\ & \le & \delta +L_3|x\left(\varsigma \right)-x^{*}\left(\varsigma \right)|\hfill \\ & \le & \delta +L_3\parallel x-x^{*}\parallel .\hfill \end{array}$$

Hence,

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel (1-L_3)\le \delta \end{array}$$

so,

$$\begin{array}{c}\hfill \parallel x-x^{*}\parallel \le \frac{\delta }{1-L_3}=ϵ.\end{array}$$

 □

Example 1.

We give the quadratic functional integro-differential equation

$$\begin{array}{c}\hfill \frac{dx}{d\mathfrak{r}}\left(\mathfrak{r}\right)={(\frac{\mathfrak{r}}{2})}^2+\frac{1}{2}(D^{\frac{3}{4}}x\left(\mathfrak{r}\right)\times {\mathfrak{I}}^{\frac{1}{2}}(\frac{\sqrt{\mathfrak{r}}}{3}+\frac{1}{2}{D}^{\frac{1}{5}}x\left(\varsigma \right))).\mathfrak{r}\in (0,1]\end{array}$$

(27)

with the nonlocal integro-differential condition

$$\begin{array}{c}\hfill x\left(0\right)-1={\int }_0^1{(\frac{\mathfrak{r}}{2})}^3+\frac{1}{2}(x\left(\mathfrak{r}\right)+D^{\frac{1}{2}}x\left(\varsigma \right))d\varsigma .\end{array}$$

(28)

Thus,

$$\begin{array}{ccc}& & f_1(\mathfrak{r},D^{\alpha }x(\mathfrak{r})\times {\mathfrak{I}}^{\beta }{f}_2(\mathfrak{r},D^{\gamma }x(\mathfrak{r})))={(\frac{\mathfrak{r}}{2})}^2+\frac{1}{2}(D^{\frac{1}{2}}x(\mathfrak{r})\times {\mathfrak{I}}^{\frac{1}{2}}(\frac{\sqrt{\mathfrak{r}}}{3}+\frac{1}{2}{D}^{\frac{1}{5}}x(\varsigma )))\mathfrak{r}\in I,\hfill \\ & & f_2(\mathfrak{r},D^{\gamma }x(\mathfrak{r}))=\frac{\sqrt{\mathfrak{r}}}{3}+\frac{1}{2}{D}^{\frac{1}{5}}x(\varsigma )\mathfrak{r}\in I.\hfill \end{array}$$

Obviously, Theorem 1 holds, when $\mathfrak{r}=1$,

$$a_1^{*}=0.196136,a_2^{*}=\frac{1}{3},a_3^{*}=\frac{1}{8},b_1=b_2=b_3=\frac{1}{2},\delta =\frac{1}{2},\alpha =\frac{3}{4},\beta =\frac{1}{2},\gamma =\frac{1}{5}and{x}_0=1.$$

From (3), we have

$$\frac{41}{64}{r}_1^2-\frac{19}{24}{r}_1+0.19613558245703772=0,$$

then, $r_1=0.342894$ or $r_1=0.892878$ is a positive number. Then, problem (27) and (28) has at least one solution.

Example 2.

Given the problem of conjugate order

$$\begin{array}{c}\hfill \frac{dx}{d\mathfrak{r}}(\mathfrak{r})={(\frac{\mathfrak{r}}{2})}^2+\frac{1}{2}(D^{\frac{1}{2}}x(\mathfrak{r})\times {\mathfrak{I}}^{\frac{1}{2}}(\frac{\sqrt{\mathfrak{r}}}{3}+\frac{1}{2}{D}^{\frac{1}{2}}x(\varsigma ))).\mathfrak{r}\in (0,1]\end{array}$$

(29)

with the nonlocal integro-differential condition

$$\begin{array}{c}\hfill x(0)-1={\int }_0^1{(\frac{\mathfrak{r}}{2})}^3+\frac{1}{2}(x(\mathfrak{r})+D^{\frac{1}{2}}x(\varsigma ))d\varsigma .\end{array}$$

(30)

we have

$$\frac{9}{16}{r}_1^2-\frac{3}{4}{r}_1+0.19613558245703772=0,$$

then, $r_1=0.4211280954182888$ or $r_1=0.6899830156928223.$

4. Conclusions

Integral and differential equations contribute significantly to mathematical analysis and have numerous applications to issues in the real world. It has a wide range of uses in mechanics, population dynamics, mathematical biology, engineering, mathematical physics and other fields [1,2,3,8,15,16,17,18].

In this study, we establish the solvability of nonlocal problem (1) and (2). The existence of solution $x\in C\left(I\right)$ of problem (1) and (2) is investigated, and the existence of the maximal and minimal solutions of problem (1) and (2) is proved. Furthermore, some continuous dependency results of solution x on fractional-order derivative $y\left(t\right)$, on parameter $x_0$ and on function h are also proved.