Open AccessArticle

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

Ahmed M. A. El-Sayed

^1,†

Antisar A. A. Alhamali

^2,*,† and

Eman M. A. Hamdallah

^1,†

Faculty of Science, Alexandria University, Alexandria 21521, Egypt

Faculty of Science, Al Zaytona University, Misrata 51, Libya

Author to whom correspondence should be addressed.

^†

These authors contributed equally to this work.

Axioms 2023, 12(8), 788; https://doi.org/10.3390/axioms12080788

Submission received: 26 June 2023 / Revised: 4 August 2023 / Accepted: 9 August 2023 / Published: 14 August 2023

(This article belongs to the Special Issue Recent Advances in Fractional Differential Equations and Inequalities)

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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.

Keywords:

Caputo fractional derivative; fractional-order integro-differential equations; conjugate-order integro-differential equations; Schauder fixed-point theorem; maximal and minimal solutions; continuous dependence

MSC:

26A33; 34K45; 47G10

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{d x}{d r} (r) = f_{1} (r, D^{α} x (r) \times I^{β} f_{2} (r, D^{γ} x (r))), α, β, γ \in (0, 1], r \in (0, 1]

(1)

with the nonlocal fractional-order integro-differential condition

x (0) + \int_{0}^{1} h (ς, x (ς), D^{δ} x (ς)) d ς = x_{0}, δ \in (0, 1],

(2)

where

I^{β}

is the fractional Riemann–Liouville integral of order

β

and

D^{α}

is a Caputo fractional derivative of order

α \in (0, 1] .

We investigate the existence of a solution to problem (1) and (2) in

C (I)

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

h \in L^{1} (I), β \in R^{+}

. The Riemann–Liouville fractional integral of

h

of order β is given by

I^{β} h (r) = \int_{0}^{r} \frac{{(r - s)}^{β - 1}}{Γ (β)} h (s) d s,

Definition 2.

Caputo, M. Fractional derivative

D_{a}^{α}

of order

α \in (0, 1]

h (r) \in A C (I)

is given by (see [12])

D^{α} h (r) = \int_{0}^{r} \frac{{(r - s)}^{- α}}{Γ (1 - α)} \frac{d}{d s} h (s) d s = I^{1 - α} \frac{d}{d r} h (r)

2. Existence of Solution

We take into account the following assumptions:

(i): $f_{1} : I \times R \to R$ is measurable in $r \in I \forall y \in R$ and continuous in $y \in R \forall r \in I$ . Furthermore, ∃ a bounded measurable function $a_{1} : I \to R$ and a constant $b_{1} > 0$ where

$\begin{matrix} | f_{1} (r, y) | \leq | a_{1} (r) | + b_{1} | y | \leq a_{1}^{*} + b_{1} | y |, a_{1}^{*} = sup_{r \in I} I^{1 - α} | a_{1} (r) | . \end{matrix}$
(ii): $f_{2} : I \times R \to R$ is measurable in $r \in I \forall y \in R$ and continuous in $y \in R \forall r \in I$ . Furthermore, ∃ a bounded measurable function $a_{2} : I \to R$ and a constant $b_{2} > 0$ where

$\begin{matrix} | f_{2} (r, y) | \leq | a_{2} (r) | + b_{2} | y | \leq a_{2}^{*} + b_{2} | y |, a_{2}^{*} = sup_{r \in I} I^{β} | a_{2} (r) | . \end{matrix}$
(iii): $r_{1}$ is a positive root of the following equation:

$a_{1}^{*} + (\frac{b_{1} a_{2}^{*}}{Γ (2 - α)} - 1) r_{1} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (2 - α) Γ (α + β - γ + 1)} = 0 .$

(3)
(iv): $h : I \times R \times R ⟶ R$ is measurable in $r$ for all $x, y \in R$ and continuous in $x, y$ for $r \in I$ , and there exists a bounded measurable function $a_{3} : I \to R$ and a constant $b_{3} > 0$ where

$| h (r, x, y) | \leq | a_{3} (r) | + b_{3} (| x | + | y |)$

where

$\begin{matrix} sup_{r \in I} \int_{0}^{t} | a_{3} (ς) | d ς \leq N . \end{matrix}$
(v): $∥ x_{0} ∥ > N + \frac{b_{3} r_{1}}{Γ (α - δ + 1)} - \frac{r_{1}}{Γ (α + 1)}$

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 (r) = x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r), δ, γ \leq α .

(4)

where y satisfies the equation

\begin{matrix} y (r) = I^{1 - α} f_{1} (r, y (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) \end{matrix}

(5)

and the two systems (1) and (2), and (4) and (5) are equivalent such that

δ, γ \leq α .

Proof.

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

\begin{matrix} D^{α} x (r) = I^{1 - α} \frac{d x}{d r} = I^{1 - α} f_{1} (r, D^{α} x (r) \times I^{β} f_{2} (r, I^{α - γ} y (r)) . \end{matrix}

We take

D^{α} x (r) = y (r)

; thus

\begin{matrix} y (r) = I^{1 - α} f_{1} (r, y (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) \end{matrix}

and

x (r) = x (0) + I^{α} y (r) .

(6)

Now,

\begin{matrix} I^{α - δ} y (r) & = & I^{α - δ} D^{α} x (r) = I^{α - δ} I^{1 - α} \frac{d x}{d r} \\ = & I^{1 - δ} \frac{d x}{d r} = D^{δ} x (r), \end{matrix}

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

By letting

y (r) = D^{α} x (r)

in (4), we obtain (2), and in (5), we obtain

I^{1 - α} \frac{d}{d r} x (r) = I^{1 - α} f_{1} (r, D^{α} x (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, D^{γ} x (θ) d θ) .

Operating by

I^{α},

then by

\frac{d}{d 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 (I)

Proof.

We characterize a closed ball

Q_{r_{1}}

and an operator

F_{1}

\begin{matrix} Q_{r_{1}} = {y \in C (I) : ∥ y ∥ \leq r_{1}}, r_{1} = a_{1}^{*} + \frac{1}{Γ (2 - α)} (r_{1} b_{1} a_{2}^{*} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (α + β - γ + 1)}) \end{matrix}

and

\begin{matrix} F_{1} y (r) = I^{1 - α} f_{1} (r, y (r) \times \int_{0}^{r} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς . \end{matrix}

For

y \in Q_{r_{1}}

, we can obtain

\begin{matrix} | y (r) | & = & I^{1 - α} | f_{1} (r, y (r) \times I^{β} f_{2} (r, I^{α - γ} y (r))) | \\ \leq & I^{1 - α} [| a_{1} (r) | + b_{1} | y (r) \times I^{β} f_{2} (r, I^{α - γ} y (r)) |] \\ \leq & I^{1 - α} [| a_{1} (r) | + b_{1} ∥ y ∥ (a_{2}^{*} + b_{2} I^{α + β - γ} ∥ y ∥)] \\ \leq & I^{1 - α} [| a_{1} (r) | + b_{1} r_{1} (a_{2}^{*} + \frac{r_{1} b_{2}}{Γ (α + β - γ + 1)})], \end{matrix}

\begin{matrix} I^{1 - α} | f_{1} (r, y (r) \times I^{β} f_{2} (r, I^{α - γ} y (r)) | & \leq & a_{1}^{*} + \frac{1}{Γ (2 - α)} (r_{1} b_{1} a_{2}^{*} + \frac{r_{1}^{2} b_{1} b_{2}}{Γ (α + β - γ + 1)}) \end{matrix}

and we deduce that

\begin{matrix} | F_{1} y (r) | & = & | \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς | \\ \leq & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} (a_{1} + b_{1} | y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ |) d ς \\ \leq & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} (a_{1} + b_{1} ∥ y ∥ \times (a_{2}^{*} + \frac{b_{2} ∥ y ∥}{Γ (α + β - γ + 1)}) d ς \\ \leq & a_{1}^{*} + \frac{1}{Γ (2 - α)} (r_{1} b_{1} a_{2}^{*} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (α + β - γ + 1)}) = r_{1} . \end{matrix}

So, we have

\begin{matrix} ∥ F_{1} y ∥ & \leq & a_{1}^{*} + \frac{1}{Γ (2 - α)} (r_{1} b_{1} a_{2}^{*} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (α + β - γ + 1)}) = r_{1} . \end{matrix}

This proves that

F_{1} : Q_{r_{1}} \to Q_{r_{1}}

and the family

{F_{_{1}} y}

is uniformly bounded on

Q_{r_{1}}

Let

y \in Q_{r_{1}}

and

r_{1}, r_{2} \in I

, where

r_{2} > r_{1}

and

∣ r_{1} - r_{2} ∣ \leq δ

; therefore,

\begin{matrix} | F_{1} y (r_{2}) - F_{1} y (r_{1}) | = \\ | \int_{0}^{r_{2}} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d s \\ - & \int_{0}^{r_{1}} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς | \\ \leq & | \int_{0}^{r_{1}} \frac{{(r_{2} - ς)}^{- α}}{Γ (1 - α)} f (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς \\ + & \int_{r_{1}}^{r_{2}} \frac{{(r_{2} - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς \\ - & \int_{0}^{r_{1}} \frac{{(r_{1} - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς | \\ \leq & | \int_{0}^{r_{1}} \frac{{(r_{2} - ς)}^{- α}}{Γ (1 - α)} - \frac{{(r_{1} - ς)}^{- α}}{Γ (1 - α)} (a_{1}^{*} + r_{1} b_{1} a_{2}^{*} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (α + β - γ + 1)}) d ς \\ + & \int_{r_{1}}^{r_{2}} \frac{{(r_{2} - ς)}^{- α}}{Γ (1 - α)} (a_{1} + \frac{b_{1} a_{2} r_{1}}{Γ (1 + β)} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (1 + β) Γ (1 + α - γ)}) d ς | \\ \leq & \int_{0}^{r_{1}} | \frac{{(r_{2} - ς)}^{α} - {(r_{1} - ς)}^{α}}{Γ (1 - α) {(r_{1} - ς)}^{α} {(r_{2} - ς)}^{α}} | (a_{1}^{*} + r_{1} b_{1} a_{2}^{*} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (α + β - γ + 1)}) d s \\ + & \int_{r_{1}}^{r_{2}} | \frac{{(r_{2} - ς)}^{α}}{Γ (1 - α) {(r_{2} - ς)}^{α}} | (a_{1}^{*} + r_{1} b_{1} a_{2}^{*} + \frac{b_{1} b_{2} r_{1}^{2}}{Γ (α + β - γ + 1)}) d ς . \end{matrix}

This shows that family

{F_{1} y}

is equi-continuous on

Q_{r_{1}}

, and by the Arzela–Ascoli result [13], mapping

F_{1}

is relatively compact.

Let

{y_{n}} \subset Q_{r_{1}}

be such that

y_{n} \to y

; then,

\begin{matrix} F_{1} y_{n} (r) & = & I^{1 - α} f_{1} (r, y_{n} (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{n} (σ) d σ) d θ), \end{matrix}

\begin{matrix} lim_{n \to \infty} F_{1} y_{n} (r) & = & lim_{n \to \infty} I^{1 - α} f_{1} (r, y_{n} (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{n} (σ) d σ) d θ) \\ = & I^{1 - α} f_{1} (r, y (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} lim_{n \to \infty} y_{n} (σ) d σ) d θ) d ς \\ = & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς \\ = & F_{1} y (r) . \end{matrix}

This yields that

F_{1} y_{n} (r) \to F_{1} y (r)

Thus, mapping

F_{1}

is continuous, and by applying the Schauder fixed-point theorem [13],

y \in Q_{r_{1}} \subset C (I)

is a solution of (5).

To prove that

x \in C (I)

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 (I)

and satisfies (4).

Proof.

Let

Q_{r_{2}}

be the closed ball

\begin{matrix} Q_{r_{2}} = {x \in C (I) : ∥ x ∥ \leq r_{2}}, r_{2} = \frac{∥ x_{0} ∥ - N - \frac{b_{3} r_{1}}{Γ (α - δ + 1)} + \frac{r_{1}}{Γ (α + 1)}}{(1 + b_{3})} \end{matrix}

and define operator

F_{2}

\begin{matrix} F_{2} x (r) = x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) . \end{matrix}

Let

x \in Q_{r_{2}}

; then,

\begin{matrix} | F_{2} x (r) | & = & | x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) | \\ \leq & | x_{0} | - \int_{0}^{1} | h (ς, x (ς), I^{α - δ} y (ς)) | d ς + I^{α} | y (r) | \\ \leq & | x_{0} | - \int_{0}^{1} (| a_{3} | + b_{3} (| x (ς) | + \frac{∥ y ∥}{Γ (α - δ + 1)})) d ς + \frac{∥ y ∥}{Γ (α + 1)} \\ \leq & | x_{0} | - \int_{0}^{1} | a_{3} (ς) | d ς - b_{3} \int_{0}^{1} | x (ς) | d ς - \frac{b_{3} r_{1}}{Γ (α - δ + 1)} \int_{0}^{1} d ς + \frac{r_{1}}{Γ (α + 1)} \\ \leq & ∥ x_{0} ∥ - N - b_{3} r_{2} - \frac{b_{3} r_{1}}{Γ (α - δ + 1)} + \frac{r_{1}}{Γ (α + 1)} = r_{2} \end{matrix}

and

\begin{matrix} ∥ F_{2} x ∥ \leq | x_{0} | - N - b_{3} r_{2} - \frac{b_{3} r_{1}}{Γ (α - δ + 1)} + \frac{r_{1}}{Γ (α + 1)} = r_{2} \end{matrix}

Similarly as performed above,

F_{2} : Q_{r_{2}} \to Q_{r_{2}}

, and

{F_{_{2}} x}

is uniformly bounded on

Q_{r_{2}}

Let

x \in Q_{r_{2}}

and

r_{1}, r_{2} \in I

; moreover,

r_{2} > r_{1}

and

∣ r_{1} - r_{2} ∣ \leq δ

. Then,

\begin{matrix} | F_{2} x (r_{2}) - F_{2} x (r_{1}) | & = & | x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r_{2}) \\ - & x_{0} + \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς - I^{α} y (r_{1}) | \\ \leq & \int_{0}^{r_{2}} | f_{1} (ς, y (ς)) \times I^{β} f_{2} (r, I^{α - γ} y (ς)) | d ς \\ - & \int_{0}^{r_{1}} | f_{1} (ς, y (ς)) \times I^{β} f_{2} (r, I^{α - γ} y (ς)) | d ς \\ \leq & \int_{r_{1}}^{r_{2}} | f_{1} (ς, y (ς)) \times I^{β} f_{2} (r, I^{α - γ} y (ς)) | d ς . \end{matrix}

Thus, family

{F_{2} x}

is equicontinuous on

Q_{r_{2}}

, and operator

F_{2}

is relatively compact.

For

{x_{n}} \subset Q_{r_{2}}

, and

x_{n} \to y

; then,

\begin{matrix} F_{2} x_{n} (r) & = & x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) \end{matrix}

and

\begin{matrix} lim_{n \to \infty} F_{2} x_{n} (r) & = & lim_{n \to \infty} (x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r)) \end{matrix}

Using the Lebesgue dominated convergence theorem, we have

\begin{matrix} lim_{n \to \infty} F_{2} x_{n} (r) & = & x_{0} - \int_{0}^{1} h (ς, lim_{n \to \infty} x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) \\ = & x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) = F_{2} x (r) \end{matrix}

This means that

F_{2} x_{n} (r) \to F_{2} x (r)

. Hence, operator

F_{2}

is continuous, and there exists

x \in Q_{r_{2}} \subset C (I)

that satisfies (4).

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

2.1. Conjugate-Order Problem

For

γ = 1 - α, α \in [\frac{1}{2}, 1)

in problem (1) and (2), we have the nonlocal problem of conjugate-orders

(α, 1 - α)

\frac{d x}{d r} (r) = f_{1} (r, D^{α} x (r) \times I^{β} f_{2} (r, D^{1 - α} x (r))), α, β \in (0, 1], r \in (0, 1]

(7)

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

By putting

D^{α} x (r) = y (r),

we can prove that nonlocal problem (7) with condition (2) is equivalent to the functional integral equation

y (r) = I^{1 - α} f_{1} (r, y (r) \times I^{β} f_{2} (r, I^{2 α - 1} y (r))), α \geq \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 (I) .

2.2. Absolutely Continuous Solution

By taking

y (r) = \frac{d x}{d r} (r),

(1) and (2) reduce to the fractional-order quadratic integral equation

y (r) = f_{1} (r, I^{1 - α} y (r) \times I^{β} f_{2} (r, I^{1 - γ} y (r))), r \in (0, 1]

(9)

and the nonlocal integro-differential condition

x (0) + \int_{0}^{1} h (ς, x_{0} + \int_{0}^{ς} y (s) d s, I^{1 - δ} y (ς)) d ς = 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 (r) = x_{0} - \int_{0}^{1} h (ς, x (ς), y (ς)) d ς + \int_{0}^{t} y (s) d s

(11)

With the same technique as in [14], the existence of solution

y \in L^{1} (I)

of (9) can be proved. So, the existence of an absolutely continuous solution

x (r) = x_{0} - \int_{0}^{1} h (ς, x (ς), I^{1 - δ} y (ς)) d ς + \int_{0}^{t} y (s) d s \in A C (I)

(12)

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

2.3. Integer-Order Problem

By letting

α, β, γ \to 1

in problem (1) and (2), we have the integer-order equation

\frac{d x}{d r} (r) = f_{1} (r, \frac{d}{d r} x (r) \times \int_{0}^{r} f_{2} (ς, \frac{d}{d ς} x (ς)) d ς), r \in (0, 1]

(13)

with the nonlocal integro-differential condition

x (0) + \int_{0}^{1} h (ς, x (ς), \frac{d}{d ς} x (ς)) d ς = x_{0} .

(14)

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

y (r) = f_{1} (r, y (r) \times \int_{0}^{r} f_{2} (ς, y (ς)) d ς), r \in (0, 1]

(15)

and

x (r) = x_{0} - \int_{0}^{1} h (ς, x (ς), y (ς)) d ς + \int_{0}^{t} y (s) d s

(16)

and as in [14], the existence of solution

y \in L^{1} (I)

of (15) can be proved. Consequently,

x \in A C (I)

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 (I)

satisfy

\begin{matrix} x (r) & \leq & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, x (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} x (σ) d σ) d θ) d ς, \\ y (r) & \geq & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς \end{matrix}

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{matrix} x (r) & < & y (r), r > 0 . \end{matrix}

(17)

Proof.

Let us assume that result (17) is not true; then,

\exists t_{1}

where

x (r_{1}) = y (r_{1}), r_{1} > 0,

x (r) < y (r) 0 < r < r_{1} .

From the monotonicity of

f_{1}

and

f_{2}

, we obtain

\begin{matrix} x (r_{1}) & \leq & \int_{0}^{r_{1}} \frac{{(r_{1} - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, x (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} x (σ) d σ) d θ) d ς \\ < & \int_{0}^{r_{1}} \frac{{(r_{1} - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς \\ = & y (r_{1}) . \end{matrix}

Hence

x (r_{1}) < y (r_{1}),

which contradicts that

x (r_{1}) = y (t_{1})

; then,

x (r) < y (r), 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{matrix} y_{ϵ} (r) = I^{1 - α} f_{1 ϵ} (r, y_{ϵ} (r) \times \int_{0}^{r} \frac{{(r - θ)}^{β - 1}}{Γ (β)} f_{2 ϵ} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ} (σ) d σ) d θ), \end{matrix}

(18)

where

f_{1 ϵ} (r, y_{ϵ} (r)) = f_{1} (r, y_{ϵ} (r)) + ϵ

f_{2 ϵ} (r, y_{ϵ} (r)) = f_{2} (r, y_{ϵ} (r)) + ϵ .

Obviously,

f_{1 ϵ}

and

f_{2 ϵ}

satisfy the assumptions of Theorem 1; according to Theorem 1, Equation (18) has a continuous solution

y_{ϵ}

and

\begin{matrix} y_{ϵ} (r) & = \\ \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} \\ \times [ϵ + f_{1} (ς, y_{ϵ} (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} \{ϵ + f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ} (σ) d σ) d θ)\} d ς] . \end{matrix}

Next, we take

ϵ_{1}, ϵ_{2} > 0

such that

0 < ϵ_{2} < ϵ_{1} < ϵ

; then,

\begin{matrix} y_{ϵ_{1}} (r) & = & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} [ϵ_{1} + f_{1} (ς, y_{ϵ_{1}} (ς) \\ \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} \{ϵ_{1} + f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ_{1}} (σ) d σ) d θ)} d ς] \\ > & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} [ϵ_{2} + f_{1} (ς, y_{ϵ_{1}} (ς) \\ \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} \{ϵ_{2} + f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ_{1}} (σ) d σ) d θ)} d ς] \end{matrix}

(19)

y_{ϵ_{2}} (r) = \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} [ϵ_{2} + f_{1} (ς, y_{ϵ_{2}} (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} \{ϵ_{2} + f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ_{2}} (σ) d σ) d θ)\} d ς]

(20)

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

y_{ϵ_{2}} (r) < y_{ϵ_{1}} (r), r \in I .

As proved in Theorem 1,

{y_{ϵ} (r)}

is equicontinuous and uniformly bounded on I; then,

{y_{ϵ}}

is relatively compact, and there exists a decreasing sequence

ϵ_{n}

such that

ϵ_{n} \to 0,

n \to \infty

and

lim_{n \to \infty} y_{ϵ_{n}} (r)

exists uniformly on I. Let

\begin{matrix} lim_{n \to \infty} y_{ϵ_{n}} (r) = q (r) . \end{matrix}

Then,

\begin{matrix} \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y_{ϵ_{n}} (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ_{n}} (σ) d σ) d θ) d ς) \to \\ \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, q (ς) \times \int_{0}^{ς} \frac{{(ς) - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} q (σ) d σ) d θ) d ς, \end{matrix}

then,

\begin{matrix} q (r) & = & lim_{n \to \infty} y_{ϵ_{n}} (r) \\ = & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, q (ς) \times \int_{0}^{ς} \frac{{(ς) - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} q (σ) d σ) d θ) d ς \end{matrix}

which yields that

q (r)

is a solution of (5).

Now, if each

y (r)

is a solution of (5), then

\begin{matrix} y_{ϵ} (r) & = & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} [ϵ + f_{1} (ς, y_{ϵ} (ς) \\ \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} \{ϵ + f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ} (σ) d σ) d θ)} d ς] \\ > & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y_{ϵ} (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{ϵ} (σ) d σ) d θ) d ς \end{matrix}

(21)

and

y (r) = \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f_{1} (ς, y (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y (σ) d σ) d θ) d ς

(22)

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

\begin{matrix} y (r) < y_{ϵ} (r), r \in I . \end{matrix}

The uniqueness of the maximal solution yields that

y_{ϵ} (r) \to q (r)

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:

${(i)}^{*}$: $f_{1}, f_{2} : I \times R \to R$ are measurable in $r \in I \forall x \in R$ and satisfy the Lipschitz condition,

$\begin{matrix} | f_{1} (r, x) - f_{1} (r, y) | \leq L_{1} | x - y |, r \in I, x, y \in R . \end{matrix}$

(23)

$\begin{matrix} | f_{2} (r, x) - f_{2} (r, y) | \leq L_{2} | x - y |, r \in I, x, y \in R . \end{matrix}$

(24)

Assumption

(i v)

implies that

| f_{1} (r, y) | \leq | f_{1} (r, 0) | + L_{1} | y |

and

\begin{matrix} | f_{1} (r, y) | \leq a_{1} + L_{1} | y |, s u c h t h a t a_{1} = sup_{r \in I} | f_{1} (r, 0) | . \end{matrix}

Furthermore,

| f_{2} (r, y) | \leq | f_{2} (r, 0) | + L_{2} | y |

and

\begin{matrix} | f_{2} (r, y) | \leq a_{2} + L_{2} | y |, w h e r e a_{2} = sup_{r \in I} | f_{2} (r, 0) | . \end{matrix}

${(i i)}^{*}$: $h : I \times R \times R \to R$ is measurable in $r \in I \forall x, y \in R$ and satisfies the Lipschitz condition,

$\begin{matrix} | h (r, x_{1}, y_{1}) - h (r, x_{2}, y_{2}) | \leq L_{3} (| x_{1} - x_{2} | + | y_{1} - y_{2} |) \end{matrix}$

(25)

with Lipschitz condition $L_{3} > 0$ .

Theorem 4.

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

\begin{matrix} \frac{2 b_{1} b_{2} r_{1}}{Γ (2 - α) Γ (α + β - γ + 1)} + \frac{b_{1} a_{2}^{*}}{Γ (2 - α)} < 1 . \end{matrix}

(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{matrix} | y_{2} (r) - y_{1} (r) | & = & | \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f (ς, y_{2} (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{2} (σ) d σ) d θ) d ς \\ - & \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} f (ς, y_{1} (ς) \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} y_{1} (σ) d σ) d θ) d ς | \\ \leq & b_{1} \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} (| y_{2} (ς) | \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} | y_{2} (σ) | d σ) d θ \\ - & | y_{1} (ς) | \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} | y_{1} (σ) | d σ) d θ) d ς \\ \leq & b_{1} \int_{0}^{t} \frac{{(t - ς)}^{- α}}{Γ (1 - α)} (| y_{2} (ς) | \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} | y_{2} (σ) | d σ) d θ \\ - & | y_{2} (ς) | \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} | y_{1} (σ) | d σ) d θ \\ + & | y_{2} (ς) | \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} | y_{1} (σ) | d σ) d θ \\ - & | y_{1} (ς) | \times \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} f_{2} (θ, \int_{0}^{θ} \frac{{(θ - σ)}^{(α - γ) - 1}}{Γ (α - γ)} | y_{1} (σ) | d σ) d θ) d ς \\ \leq & b_{1} \int_{0}^{r} \frac{{(r - ς)}^{- α}}{Γ (1 - α)} (∥ y_{2} ∥ \times b_{2} \int_{0}^{ς} \frac{{(ς - θ)}^{β - 1}}{Γ (β)} \int_{0}^{θ} \frac{{(θ - σ)}^{α - 1}}{Γ (α)} | y_{2} (σ) - y_{1} (σ) | d σ d θ \\ + & ∥ y_{2} - y_{1} ∥ \times (a_{2}^{*} + \frac{b_{2} r_{1}}{Γ (β + α - γ + 1)})) d ς \\ \leq & \frac{b_{1}}{Γ (2 - α)} (∥ y_{2} - y_{1} ∥ \frac{b_{2} r_{1}}{Γ (β + α - γ + 1)} \\ + & ∥ y_{2} - y_{1} ∥ (a_{2}^{*} + \frac{b_{2} r_{1}}{Γ (β + α - γ + 1)})) . \end{matrix}

Hence,

\begin{matrix} ∥ y_{2} - y_{1} ∥ (1 - (\frac{2 b_{1} b_{2} r_{1}}{Γ (2 - α) Γ (β + α - γ + 1)} + \frac{b_{1} a_{2}^{*}}{Γ (2 - α)})) \leq 0 . \end{matrix}

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 (I)

of Equation (5), there exists a unique solution x of (4).

Proof.

Let

y \in C (I)

satisfy Equation (5) and

x_{1}, x_{2}

satisfy Equation (4); then, we have

\begin{matrix} | x_{2} (r) - x_{1} (r) | & = & | x_{0} - \int_{0}^{1} h (ς, x_{2} (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) \\ - & x_{0} + \int_{0}^{1} h (ς, x_{1} (ς), I^{α - δ} y (ς)) d ς - I^{α} y (r) | \\ \leq & \int_{0}^{1} (| h (ς, x_{2} (ς), I^{α - δ} y (ς)) | - | h (ς, x_{1} (ς), I^{α - δ} y (ς)) |) d ς \\ \leq & L_{3} | x_{2} (r) - x_{1} (r) |, \end{matrix}

then,

\begin{matrix} ∥ x_{2} - x_{1} ∥ (1 - L_{3}) \leq 0 \to x_{2} = x_{1} . \end{matrix}

□

3.3. Continuous Dependency Results

Definition 3.

The unique solution

x \in C (I)

of (4) depends continuously on

x_{0}

, if

\forall ϵ > 0

\exists δ > 0

, where

| x_{0} - x_{0}^{*} | \leq δ \Rightarrow ∥ x - x^{*} ∥ \leq ϵ

where

x^{*}

satisfies

\begin{matrix} x^{*} (r) = x_{0}^{*} - \int_{0}^{1} h (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) . \end{matrix}

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 (r)

and

x^{*} (r)

satisfy (4); then,

\begin{matrix} | x (r) - x^{*} (r) | & = & | x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d s + I^{α} y (r) \\ - & x_{0}^{*} + \int_{0}^{1} h (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς - I^{α} y (r) | \\ \leq & | x_{0} - x_{0}^{*} | + \int_{0}^{1} | h (ς, x (ς), I^{α - δ} y (ς)) - h (ς, x^{*} (ς), I^{α - δ} y (ς)) | d ς \\ \leq & | x_{0} - x_{0}^{*} | + L_{3} | x (ς) - x^{*} (ς) | \\ \leq & δ + L_{3} ∥ x - x^{*} ∥ . \end{matrix}

Hence,

\begin{matrix} ∥ x - x^{*} ∥ (1 - L_{3}) \leq δ \end{matrix}

and

\begin{matrix} ∥ x - x^{*} ∥ \leq \frac{δ}{1 - L_{3}} = ϵ . \end{matrix}

□

Definition 4.

The unique solution

x \in C (I)

of Equation (4) depends continuously on y; if

\forall ϵ > 0

\exists δ > 0

, where

| y - y^{*} | \leq δ \Rightarrow ∥ x - x^{*} ∥ \leq ϵ

where

x^{*}

satisfies the integral equation

\begin{matrix} x^{*} (r) = x_{0}^{*} - \int_{0}^{1} h (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) . \end{matrix}

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 (r)

and

x^{*} (r)

satisfy Equation (4); then,

\begin{matrix} | x (r) - x^{*} (r) | & = & | x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) \\ - & x_{0}^{*} + \int_{0}^{1} h (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς - I^{α} y (r) | \\ = & | \int_{0}^{1} (h (ς, x (ς), I^{α - δ} y (ς)) - h (ς, x^{*} (ς), I^{α - δ} y (ς))) d ς \\ + & I^{α} (y (r) - y^{*} (r)) | \\ \leq & L_{3} | x (ς) - x^{*} (ς) | + L_{3} I^{α - δ} | y (ς) - y^{*} (ς) | + I^{α} | y (r) - y^{*} (r) | \\ \leq & L_{3} ∥ x - x^{*} ∥ + \frac{L_{3} δ}{Γ (α - δ + 1)} + \frac{δ}{Γ (α + 1)} . \end{matrix}

Hence,

\begin{matrix} ∥ x - x^{*} ∥ (1 - L_{3}) \leq \frac{L_{3} δ}{Γ (α - δ + 1)} + \frac{δ}{Γ (α + 1)} \end{matrix}

\begin{matrix} ∥ x - x^{*} ∥ \leq \frac{\frac{L_{3} δ}{Γ (α - δ + 1)} + \frac{δ}{Γ (α + 1)}}{(1 - L_{3})} = ϵ . \end{matrix}

□

Definition 5.

The unique solution

x \in C (I)

of Equation (4) depends continuously on h; if

\forall ϵ > 0

\exists δ > 0

, where

| h (ς, x (ς), I^{α - δ} y (ς)) - h^{*} (ς, x (ς), I^{α - δ} y (ς)) | \leq δ \Rightarrow ∥ x - x^{*} ∥ \leq ϵ

where

x^{*}

satisfies the integral equation

\begin{matrix} x^{*} (r) = x_{0} - \int_{0}^{1} h^{*} (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς + I^{α} y (r) . \end{matrix}

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 (r)

and

x^{*} (r)

satisfy Equation (4); then,

\begin{matrix} | x (r) - x^{*} (r) | & = & | x_{0} - \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d s + I^{α} y (r) \\ - & x_{0} + \int_{0}^{1} h^{*} (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς - I^{α} y (r) | \\ = & | \int_{0}^{1} h (ς, x (ς), I^{α - δ} y (ς)) d ς - \int_{0}^{1} h^{*} (ς, x (ς), I^{α - δ} y (ς)) d ς \\ + & \int_{0}^{1} h^{*} (ς, x (ς), I^{α - δ} y (ς)) d ς - \int_{0}^{1} h^{*} (ς, x^{*} (ς), I^{α - δ} y (ς)) d ς | \\ = & \int_{0}^{1} | h (ς, x (ς), I^{α - δ} y (ς)) - \int_{0}^{1} h^{*} (ς, x (ς), I^{α - δ} y (ς)) | d ς \\ + & \int_{0}^{1} | h^{*} (ς, x (ς), I^{α - δ} y (ς)) - \int_{0}^{1} h^{*} (ς, x^{*} (ς), I^{α - δ} y (ς)) | d ς \\ \leq & δ + L_{3} | x (ς) - x^{*} (ς) | \\ \leq & δ + L_{3} ∥ x - x^{*} ∥ . \end{matrix}

Hence,

\begin{matrix} ∥ x - x^{*} ∥ (1 - L_{3}) \leq δ \end{matrix}

so,

\begin{matrix} ∥ x - x^{*} ∥ \leq \frac{δ}{1 - L_{3}} = ϵ . \end{matrix}

□

Example 1.

We give the quadratic functional integro-differential equation

\begin{matrix} \frac{d x}{d r} (r) = {(\frac{r}{2})}^{2} + \frac{1}{2} (D^{\frac{3}{4}} x (r) \times I^{\frac{1}{2}} (\frac{\sqrt{r}}{3} + \frac{1}{2} D^{\frac{1}{5}} x (ς))) . r \in (0, 1] \end{matrix}

(27)

with the nonlocal integro-differential condition

\begin{matrix} x (0) - 1 = \int_{0}^{1} {(\frac{r}{2})}^{3} + \frac{1}{2} (x (r) + D^{\frac{1}{2}} x (ς)) d ς . \end{matrix}

(28)

Thus,

\begin{matrix} f_{1} (r, D^{α} x (r) \times I^{β} f_{2} (r, D^{γ} x (r))) = {(\frac{r}{2})}^{2} + \frac{1}{2} (D^{\frac{1}{2}} x (r) \times I^{\frac{1}{2}} (\frac{\sqrt{r}}{3} + \frac{1}{2} D^{\frac{1}{5}} x (ς))) r \in I, \\ f_{2} (r, D^{γ} x (r)) = \frac{\sqrt{r}}{3} + \frac{1}{2} D^{\frac{1}{5}} x (ς) r \in I . \end{matrix}

Obviously, Theorem 1 holds, when

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}, δ = \frac{1}{2}, α = \frac{3}{4}, β = \frac{1}{2}, γ = \frac{1}{5} a n d 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

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{matrix} \frac{d x}{d r} (r) = {(\frac{r}{2})}^{2} + \frac{1}{2} (D^{\frac{1}{2}} x (r) \times I^{\frac{1}{2}} (\frac{\sqrt{r}}{3} + \frac{1}{2} D^{\frac{1}{2}} x (ς))) . r \in (0, 1] \end{matrix}

(29)

with the nonlocal integro-differential condition

\begin{matrix} x (0) - 1 = \int_{0}^{1} {(\frac{r}{2})}^{3} + \frac{1}{2} (x (r) + D^{\frac{1}{2}} x (ς)) d ς . \end{matrix}

(30)

we have

\frac{9}{16} r_{1}^{2} - \frac{3}{4} r_{1} + 0.19613558245703772 = 0,

then,

r_{1} = 0.4211280954182888

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 (I)

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 (t)

, on parameter

x_{0}

and on function h are also proved.

Author Contributions

These authors contributed equally to this work. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Acknowledgments

We thank the referee for their remarks and comments that help to improve our manuscript.

Conflicts of Interest

The authors declare no conflict of interest.

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El-Sayed, A.M.A.; Alhamali, A.A.A.; Hamdallah, E.M.A. Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition. Axioms 2023, 12, 788. https://doi.org/10.3390/axioms12080788

AMA Style

El-Sayed AMA, Alhamali AAA, Hamdallah EMA. Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition. Axioms. 2023; 12(8):788. https://doi.org/10.3390/axioms12080788

Chicago/Turabian Style

El-Sayed, Ahmed M. A., Antisar A. A. Alhamali, and Eman M. A. Hamdallah. 2023. "Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition" Axioms 12, no. 8: 788. https://doi.org/10.3390/axioms12080788

APA Style

El-Sayed, A. M. A., Alhamali, A. A. A., & Hamdallah, E. M. A. (2023). Analysis of a Fractional-Order Quadratic Functional Integro-Differential Equation with Nonlocal Fractional-Order Integro-Differential Condition. Axioms, 12(8), 788. https://doi.org/10.3390/axioms12080788

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

Abstract

1. Introduction

2. Existence of Solution

2.1. Conjugate-Order Problem

2.2. Absolutely Continuous Solution

2.3. Integer-Order Problem

3. Some Characteristics of the Solution

3.1. Maximal and Minimal Solutions

3.2. Existence of Unique Solution

3.3. Continuous Dependency Results

4. Conclusions

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Acknowledgments

Conflicts of Interest

References

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