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Article

# On the Nonlinear Integro-Differential Equations

by
Chenkuan Li
* and
Joshua Beaudin
Department of Mathematics and Computer Science, Brandon University, Brandon, MB R7A 6A9, Canada
*
Author to whom correspondence should be addressed.
Fractal Fract. 2021, 5(3), 82; https://doi.org/10.3390/fractalfract5030082
Submission received: 14 July 2021 / Revised: 25 July 2021 / Accepted: 28 July 2021 / Published: 30 July 2021

## Abstract

:
The goal of this paper is to study the uniqueness of solutions of several nonlinear Liouville–Caputo integro-differential equations with variable coefficients and initial conditions, as well as an associated coupled system in Banach spaces. The results derived are new and based on Banach’s contractive principle, the multivariate Mittag–Leffler function and Babenko’s approach. We also provide a few examples to demonstrate the use of our main theorems by convolutions and the gamma function.

## 1. Introduction

Let $T > 0$. The space $C [ 0 , T ]$ of continuous functions on $[ 0 , T ]$ is given by
$C [ 0 , T ] = u ( x ) : u C = max x ∈ [ 0 , T ] | u ( x ) | < ∞ .$
Clearly, $C [ 0 , T ]$ is a Banach space.
We further define the space $C n [ 0 , T ]$, for $n ∈ N = { 1 , 2 , ⋯ }$, of those functions on $[ 0 , T ]$ with up to nth order continuous derivatives by
$C n [ 0 , T ] = u ( x ) : d n d x n u ( x ) ∈ C [ 0 , T ] and u n < ∞ ,$
where
$u n = max u ( x ) C , d d x u ( x ) C , ⋯ , d n d x n u ( x ) C .$
Obviously,
$C n [ 0 , T ] ⊂ C [ 0 , T ]$
for all $n = 1 , 2 , ⋯$. Furthermore, $C n [ 0 , T ]$ is a Banach space using Theorem 7.17 in [1] stated as follows:
Theorem 1.
If ${ u n }$ is a sequence of differentiable functions on $[ a , b ]$ such that $lim n → ∞ u n ( x 0 )$ exists (and is finite) for some $x 0 ∈ [ a , b ]$, and the sequence ${ u n ′ }$ converges uniformly on $[ a , b ]$, then $u n$ converges uniformly to a function u on $[ a , b ]$, and $u ′ ( x ) = lim n → ∞ u n ′ ( x )$ for $x ∈ [ a , b ]$.
In addition, the product space $C n [ 0 , T ] × C m [ 0 , T ]$ ($m ∈ N$) is given by
$C n [ 0 , T ] × C m [ 0 , T ] = ( u , v ) | u ∈ C n [ 0 , T ] , v ∈ C m [ 0 , T ] and ( u , v ) < ∞ ,$
where
$( u , v ) = u n + v m .$
Clearly, $C n [ 0 , T ] × C m [ 0 , T ]$ is also a Banach space. The fractional integral (or the Riemann–Liouville) $I α$ of fractional order $α ∈ R +$ of function $u ( x )$ [2,3] is defined by
$( I α u ) ( x ) = 1 Γ ( α ) ∫ 0 x ( x − t ) α − 1 u ( t ) d t .$
In particular,
$( I 0 u ) ( x ) = u ( x ) ,$
from [2]. Indeed,
$( I 0 u ) ( x ) = δ ( x ) ∗ u ( x ) = u ( x ) ,$
where $δ ( x )$ is the Dirac delta function (or distribution) given by
$( δ ( x ) , ϕ ( x ) ) = ∫ − ∞ ∞ δ ( x ) ϕ ( x ) d x = ϕ ( 0 ) ,$
for any sufficiently smooth function $ϕ ( x )$ with a compact support.
Let $u ( x )$ and $a ( x )$ be in $C [ 0 , T ]$. Then,
$I α u C = max x ∈ [ 0 , T ] 1 Γ ( α ) ∫ 0 x ( x − t ) α − 1 u ( t ) d t ≤ T α Γ ( α + 1 ) u C , a I α u C = max x ∈ [ 0 , T ] a ( x ) 1 Γ ( α ) ∫ 0 x ( x − t ) α − 1 u ( t ) d t ≤ T α Γ ( α + 1 ) a C u C .$
The Liouville–Caputo derivative of fractional order $α ∈ R +$ of function $u ( x )$ is defined as
$( C D α u ) ( x ) = I n − α d n d x n u ( x ) = 1 Γ ( n − α ) ∫ 0 x ( x − t ) n − α − 1 u ( n ) ( t ) d t ,$
where $n − 1 < α ≤ n ∈ N$.
Assume that
$u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 ,$
then integration by parts infers that
$I n d n d x n u ( x ) = 1 ( n − 1 ) ! ∫ 0 x ( x − t ) n − 1 u ( n ) ( t ) d t = 1 ( n − 1 ) ! ( x − t ) n − 1 u ( n − 1 ) ( t ) 0 x + 1 ( n − 2 ) ! ∫ 0 x ( x − t ) n − 2 u ( n − 1 ) ( t ) d t = ⋯ = u ( x ) .$
Assume that $a i ( x )$ and $b j ( x )$ are in $C [ 0 , T ]$ for all $i = 1 , 2 , ⋯ , n 1 ∈ N$ and $j = 1 , 2 , ⋯ , n 2 ∈ N$. In this paper, we begin to establish a unique and global solution in the space $C n [ 0 , T ]$ using Babenko’s method and the multivariate Mittag–Leffler function for the following integro-differential equation for $f ∈ C [ 0 , T ]$,
$u ( n ) ( x ) + a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + a 1 ( x ) C D β 1 u ( x ) + b 1 ( x ) I α 1 u ( x ) + ⋯ + b n 2 ( x ) I α n 2 u ( x ) = f ( x ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 ,$
where $0 ≤ β 1 < β 2 < ⋯ < β n 1 < n$ and $0 < α 1 < α 2 < ⋯ < α n 2$. Then, we further study the uniqueness of solutions in $C n [ 0 , T ]$ for
$u ( n ) ( x ) + a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + a 1 ( x ) C D β 1 u ( x ) + b 1 ( x ) I α 1 u ( x ) + ⋯ + b n 2 ( x ) I α n 2 u ( x ) = g x , u ( x ) , u ′ ( x ) , ⋯ , u ( n − 1 ) ( x ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 ,$
where $g ( x , y 1 , ⋯ , y n )$ is a mapping from $[ 0 , T ] × R n$ to R and satisfies certain conditions. Finally, the sufficient conditions are provided for the uniqueness of solutions in $C n [ 0 , T ] × C m [ 0 , T ]$ to the coupled system
$u ( n ) ( x ) + a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + a 1 ( x ) C D β 1 u ( x ) + b 1 ( x ) I α 1 u ( x ) + ⋯ + b n 2 ( x ) I α n 2 u ( x ) = g 1 x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) , v ( x ) , ⋯ , v ( m − 1 ) ( x ) , v ( m ) ( x ) + c m 1 ( x ) C D μ m 1 v ( x ) + ⋯ + c 1 ( x ) C D μ 1 v ( x ) + d 1 ( x ) I γ 1 v ( x ) + ⋯ + d m 2 ( x ) I γ m 2 v ( x ) = g 2 x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) , v ( x ) , ⋯ , v ( m − 1 ) ( x ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 , v ( 0 ) = u ′ ( 0 ) = ⋯ = v ( m − 1 ) ( 0 ) = 0 ,$
where both $g 1 ( x , y 1 , ⋯ , y n + m )$ and $g 2 ( x , y 1 , ⋯ , y n + m )$ are mappings from $[ 0 , T ] × R n + m$ to R, and all coefficient functions $c i ( x )$ and $d j ( x )$ are in $C [ 0 , T ]$. As far as we know, Equations (1)–(3) are new and have never been investigated before.
There have been intensive studies on the existence and uniqueness of different types of integral and fractional differential equations based on fixed point theory [4,5,6,7,8,9]. Marasi et al. [6] studied the existence and multiplicity of positive solutions for the following initial value problem using the fixed point index theory:
$( C D α ) u ( x ) = f x , u ( x ) , ( C D β ) u ( x ) , x ∈ ( 0 , 1 ] , u ( k ) ( 0 ) = η k , k = 0 , 1 , ⋯ , n − 1 ,$
where $n − 1 < β < α < n ∈ N$. J. Deng and Z. Deng [10] considered the existence of the following initial value problems based on the Schauder fixed point theorem:
$( C D α ) u ( x ) = f x , ( C D β ) u ( x ) , x ∈ ( 0 , 1 ] , u ( k ) ( 0 ) = η k , k = 0 , 1 , ⋯ , m − 1 ,$
where $m − 1 < α < m ∈ N$, $n − 1 < β < n ∈ N$ ($m − 1 ≥ n$) and $f ∈ C [ 0 , 1 ] × R$ satisfies certain conditions.
The Hadamard derivative of fractional order $α$ of a function $u : [ 1 , ∞ ) → R$ is defined as
$D α u ( x ) = 1 Γ ( n − α ) x d d x n ∫ 1 x log x s n − α − 1 u ( s ) s d s , n − 1 < α < n , n = [ α ] + 1 ,$
where $[ α ]$ denotes the integer part of the real number $α$ and $log ( · ) = log e ( · )$.
Ahmad and Ntouyas [11] studied the existence and uniqueness of solutions to the following initial value problem of Hadamard sequential fractional order neutral functional differential equations using fixed point theory
$D α [ D β u ( x ) − g ( x , u x ) ] = f ( x , u x ) , x ∈ J = [ 1 , b ] , u ( x ) = ϕ ( x ) , x ∈ [ 1 − r , 1 ] , D β u ( 1 ) = η ∈ R ,$
where $0 < α , β < 1$, $f , g : J × C ( [ − r , 0 ] , R ) → R$ are given functions and $ϕ ∈ C ( [ 1 − r , 1 ] , R )$. For any function u defined on $[ 1 − r , b ]$ and $x ∈ [ 1 , b ]$, we denote $u x$ as the element of $C ( [ − r , 0 ] , R )$ defined by
$u x ( θ ) = u ( x + θ ) , θ ∈ [ − r , 0 ] .$
Let $A C [ a , b ]$ denote the set of all absolutely continuous functions on $[ a , b ]$. The Banach space $A C 0 [ a , b ]$ is defined as
$A C 0 [ a , b ] = u : u ( x ) ∈ A C [ a , b ] with u ( a ) = 0 and u 0 = ∫ a b | u ′ ( x ) | d x < ∞ .$
The fractional version of the Hadamard-type integral and derivative is given by
$( J a + , μ α u ) ( x ) = 1 Γ ( α ) ∫ a x t x μ log x t α − 1 u ( t ) d t t α > 0 , 0 < a < x < b ,$
and
$( D a + , μ α u ) ( x ) = x − μ x d d x [ α ] + 1 x μ ( J a + , μ n − α u ) ( x ) .$
Let $0 < α 0 < α 1 < ⋯ < α n < 1$ and $0 ≤ β n + 1 < ⋯ < β m ∈ R$ where $n = 0 , 1 , ⋯$ and $m > n$. Very recently, Li [4] considered the uniqueness of solutions for the following nonlinear Hadamard-type integro-differential equation with constant coefficients for all $μ ∈ R$, in the space $A C 0 [ a , b ]$:
$D a + , μ α n u + a n − 1 D a + , μ α n − 1 u + ⋯ + a 0 D a + , μ α 0 u + b n + 1 J a + , μ β n + 1 u + ⋯ + b m J a + , μ β m u = ∫ a x f ( τ , u ′ ( τ ) ) d τ ,$
according to Banach’s contractive principle and Babenko’s approach [12]. Babenko’s approach is a very useful method in solving differential and integral equations by treating integral operators as variables and derives convergent infinite series as solutions in spaces under consideration. Li also investigated Abel’s integral equations of the first [13] and second kind with variable coefficients in distribution by Babenko’s technique [14,15].
The following multivariate Mittag–Leffler function was introduced by Hadid and Luchko [16,17] for solving linear fractional differential equations with constant coefficients by the operational calculus:
$E ( α 1 , ⋯ , α m ) , β ( z 1 , ⋯ , z m ) = ∑ k = 0 ∞ ∑ k 1 + ⋯ + k m = k k k 1 , ⋯ , k m z 1 k 1 ⋯ z m k m Γ ( α 1 k 1 + ⋯ + α m k m + β ) ,$
where $α i , β > 0$ for $i = 1 , 2 , ⋯ , m$.

## 2. The Main Results

In this section, we begin to construct a unique and global solution to Equation (1) in space $C n [ 0 , T ]$ using Babenko’s method and the multivariate Mittag–Leffler function.
Theorem 2.
Assume that $a i ( x )$ and $b j ( x )$ are continuous functions on $[ 0 , T ]$ for all $i = 1 , 2 , ⋯ , n 1 ∈ N$ and $j = 1 , 2 , ⋯ , n 2 ∈ N$, and $0 ≤ β 1 < β 2 < ⋯ < β n 1 < n$ and $0 < α 1 < α 2 < ⋯ < α n 2$. Then, Equation (1) has a unique, convergent, and global solution in $C n [ 0 , T ]$ for $f ∈ C [ 0 , T ]$.
$u ( x ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) .$
Proof.
Equation (1) turns out to be
$u ( x ) + I n a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + I n a 1 ( x ) C D β 1 u ( x ) + I n b 1 ( x ) I α 1 u ( x ) + ⋯ + I n b n 2 ( x ) I α n 2 u ( x ) = I n f ( x ) ,$
by applying the operator $I n$ to both sides of Equation (1) and using initial conditions
$u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 .$
Hence,
$( 1 + I n a n 1 ( x ) C D β n 1 + ⋯ + I n a 1 ( x ) C D β 1 + I n b 1 ( x ) I α 1 + ⋯ + I n b n 2 ( x ) I α n 2 ) u ( x ) = I n f ( x ) .$
According to Babenko’s approach,
$u ( x ) = 1 + I n a n 1 ( x ) C D β n 1 + ⋯ + I n a 1 ( x ) C D β 1 + I n b 1 ( x ) I α 1 + ⋯ + I n b n 2 ( x ) I α n 2 − 1 · I n f ( x ) = ∑ k = 0 ∞ ( − 1 ) k I n a n 1 ( x ) C D β n 1 + ⋯ + I n a 1 ( x ) C D β 1 + ⋯ + I n b n 2 ( x ) I α n 2 k I n f ( x ) = ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( I n a n 1 ( x ) C D β n 1 ) k 1 ⋯ ( I n a 1 ( x ) C D β 1 ) k n 1 I n b 1 ( x ) I α 1 k n 1 + 1 ⋯ I n b n 2 ( x ) I α n 2 k n 1 + n 2 I n f ( x ) .$
Clearly,
$I n b n 2 ( x ) I α n 2 k n 1 + n 2 I n f ( x ) = I n b n 2 ( x ) I α n 2 ⋯ I n b n 2 ( x ) I α n 2 I n f ( x ) = I n b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) .$
Therefore,
$u ( x ) = ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 I n ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) .$
Since $a i ( x )$ and $b j ( x )$ are continuous functions on $[ 0 , T ]$ for all $i = 1 , 2 , ⋯ , n 1 ∈ N$ and $j = 1 , 2 , ⋯ , n 2 ∈ N$, there exist $A i > 0$ and $B j > 0$ such that
$max x ∈ [ 0 , T ] | a i ( x ) | ≤ A i , max x ∈ [ 0 , T ] | b j ( x ) | ≤ B j .$
Thus,
$u ( x ) C ≤ ∑ k = 0 ∞ ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 A n 1 k 1 ⋯ A 1 k n 1 B 1 k n 1 + 1 ⋯ B n 2 k n 1 + n 2 · I n + k 1 ( n − β n 1 ) + ⋯ + k n 1 ( n − β 1 ) + k n 1 + 1 ( n + α 1 ) + ⋯ + k n 1 + n 2 ( n + α n 2 ) C f C ≤ ∑ k = 0 ∞ ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 A n 1 k 1 ⋯ A 1 k n 1 B 1 k n 1 + 1 ⋯ B n 2 k n 1 + n 2 · T n + k 1 ( n − β n 1 ) + ⋯ + k n 1 ( n − β 1 ) + ⋯ + k n 1 + n 2 ( n + α n 2 ) Γ ( n + 1 + k 1 ( n − β n 1 ) + ⋯ + k n 1 ( n − β 1 ) + ⋯ + k n 1 + n 2 ( n + α n 2 ) ) f C = T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 f C < ∞ .$
This implies that the series on the right-hand side of Equation (4) is uniformly convergent on $[ 0 , T ]$ with respect to x and
$u ( x ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) ,$
which is well defined over $[ 0 , T ]$. Clearly,
$u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 ,$
due to the integral operator $I n$.
It remains to show that $u ( x ) ∈ C n [ 0 , T ]$ and is a unique solution. Obviously,
$u ′ ( x ) = I n − 1 ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) ,$
and
$u ′ ( x ) C ≤ T n − 1 E ( n − β n 1 , ⋯ , n + α n 2 ) , n T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 f C < ∞ .$
Similarly,
$u ( n ) ( x ) C ≤ E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 f C < ∞ .$
So $u ( x ) ∈ C n [ 0 , T ]$. To see $u ( x )$ is a solution to Equation (1), we have
$u ( n ) ( x ) = ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) = f ( x ) + ∑ k = 1 ∞ ( − 1 ) k ∑ k 1 + ⋯ + k n 1 + n 2 = k k k 1 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) = f ( x ) + X ,$
where
$X = ∑ k = 1 ∞ ( − 1 ) k ∑ k 1 + ⋯ + k n 1 + n 2 = k k k 1 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) .$
On the other hand,
$Y = a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + a 1 ( x ) C D β 1 u ( x ) + b 1 ( x ) I α 1 u ( x ) + ⋯ + b n 2 ( x ) I α n 2 u ( x ) = ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 + 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 + 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 + 1 f ( x ) .$
This implies that
$X + Y = 0 ,$
due to the sign change and the fact that all series above are absolutely convergent.
To illustrate more in detail, we can easily deduce that
$− ∑ k 1 + ⋯ + k n 1 + n 2 = 1 1 k 1 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = 0 0 k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 + 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = 0 0 k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 + 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = 0 0 k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 f ( x ) + ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = 0 0 k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 + 1 f ( x ) = 0 .$
The pattern follows similarly for other values of k. The uniqueness can be derived from the fact that the integro-differential equation
$u ( n ) ( x ) + a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + a 1 ( x ) C D β 1 u ( x ) + b 1 ( x ) I α 1 u ( x ) + ⋯ + b n 2 ( x ) I α n 2 u ( x ) = 0 ,$
has only the zero solution by Babenko’s method. This completes the proof of Theorem 2. □
Remark 1.
(i) If all coefficient functions $a i ( x ) = b j ( x ) = 0$ over $[ 0 , T ]$, then Equation (1) becomes
$u ( n ) ( x ) = f ( x ) , u ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 ,$
with the solution
$u ( x ) = I n f ( x ) = 1 ( n − 1 ) ! ∫ 0 x ( x − t ) n − 1 f ( t ) d t .$
(ii) If all coefficient functions $a i ( x ) = b j ( x ) = 1$ over $[ 0 , T ]$, then Equation (1) has the solution for $f ∈ C [ 0 , T ]$:
$u ( x ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 · I k 1 ( n − β n 1 ) + ⋯ + k n 1 + n 2 ( n + α n 2 ) f ( x )$
in $C n [ 0 , T ]$.
(iii) If all coefficient functions $b j ( x ) = 0$ on $[ 0 , T ]$, then Equation (1) turns out to be the fractional differential equation with the following initial conditions:
$u ( n ) ( x ) + a n 1 ( x ) C D β n 1 u ( x ) + ⋯ + a 1 ( x ) C D β 1 u ( x ) = f ( x ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 .$
The solution is given as
$u ( x ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 = k k k 1 , k 2 , ⋯ , k n 1 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 f ( x ) .$
(iv) If all coefficient functions $a i ( x ) = 0$ on $[ 0 , T ]$, then Equation (1) turns out to be the integro-differential equation with the following initial conditions:
$u ( n ) ( x ) + b 1 ( x ) I α 1 u ( x ) + ⋯ + b n 2 ( x ) I α n 2 u ( x ) = f ( x ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 1 ) ( 0 ) = 0 .$
The solution is given as
$u ( x ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 2 = k k k 1 , k 2 , ⋯ , k n 2 b 1 ( x ) I α 1 + n k 1 ⋯ b n 2 ( x ) I α n 2 + n k n 2 f ( x ) .$
Let $α ∈ R$. Define
and
$Φ α ( x ) = x + α − 1 Γ ( α ) .$
Then, it follows that $Φ 0 ( x ) = δ ( x )$ (the Dirac delta function) and
$Φ α ∗ Φ β = Φ α + β ,$
for $α , β ∈ R$ [18].
Example 1.
The following integro-differential equation
$u ′ ( x ) + x 0.5 C D 0.5 u ( x ) + I 0.5 u ( x ) = x , u ( 0 ) = 0$
has a unique and global solution in $C 1 [ 0 , T ]$
$u ( x ) = ∑ k = 0 ∞ ( − 1 ) k ∑ j = 0 k k j C j , k Φ 2 + j + 1.5 ( k − j ) ,$
where the coefficient $C j , k$ is given below.
Proof.
Based on Theorem 2,
$u ( x ) = I ∑ k = 0 ∞ ( − 1 ) k ∑ j = 0 k k j x 0.5 I 0.5 j I 1.5 ( k − j ) x .$
Obviously from Equation (5),
$I 1.5 ( k − j ) x = Φ 1.5 ( k − j ) ∗ x Γ ( 2 ) = Φ 1.5 ( k − j ) ∗ Φ 2 = Φ 2 + 1.5 ( k − j ) , x 0.5 I 0.5 0 I 1.5 ( k − j ) x = I 1.5 ( k − j ) x = Φ 2 + 1.5 ( k − j ) , x 0.5 I 0.5 1 I 1.5 ( k − j ) x = x 0.5 Φ 2.5 + 1.5 ( k − j ) = x 2 + 1.5 ( k − j ) Γ ( 2.5 + 1.5 ( k − j ) ) = Γ ( 3 + 1.5 ( k − j ) ) Γ ( 2.5 + 1.5 ( k − j ) ) Φ 3 + 1.5 ( k − j ) , x 0.5 I 0.5 2 I 1.5 ( k − j ) x = Γ ( 3 + 1.5 ( k − j ) ) Γ ( 2.5 + 1.5 ( k − j ) ) x 0.5 I 0.5 Φ 3 + 1.5 ( k − j ) = Γ ( 3 + 1.5 ( k − j ) ) Γ ( 2.5 + 1.5 ( k − j ) ) x 0.5 Φ 3.5 + 1.5 ( k − j ) = Γ ( 3 + 1.5 ( k − j ) ) Γ ( 4 + 1.5 ( k − j ) ) Γ ( 2.5 + 1.5 ( k − j ) ) Γ ( 3.5 + 1.5 ( k − j ) ) Φ 4 + 1.5 ( k − j ) , ⋯ , x 0.5 I 0.5 j I 1.5 ( k − j ) x = Γ ( 2 + 1 + 1.5 ( k − j ) ) ⋯ Γ ( 2 + j + 1.5 ( k − j ) ) Γ ( 1.5 + 1 + 1.5 ( k − j ) ) ⋯ Γ ( 1.5 + j + 1.5 ( k − j ) ) Φ 2 + j + 1.5 ( k − j ) = C j , k Φ 2 + j + 1.5 ( k − j ) , I x 0.5 I 0.5 j I 1.5 ( k − j ) x = C j , k Φ 3 + j + 1.5 ( k − j ) ,$
where
This completes the proof of Example 1. □
Using Banach’s contractive principle, we are now ready to show the uniqueness of solutions to Equation (2) in space $C n [ 0 , T ]$.
Theorem 3.
Assume that $a i ( x )$ and $b j ( x )$ are continuous functions on $[ 0 , T ]$ for all $i = 1 , 2 , ⋯ , n 1 ∈ N$ and $j = 1 , 2 , ⋯ , n 2 ∈ N$, and $0 ≤ β 1 < β 2 < ⋯ < β n 1 < n$ and $0 < α 1 < α 2 < ⋯ < α n 2$. Furthermore, suppose $g ( x , y 1 , ⋯ , y n ) ∈ C ( [ 0 , T ] × R n )$ and that there exist constants $C 1 , ⋯ , C n$ such that
$| g ( x , y 1 , ⋯ , y n ) − g ( x , z 1 , ⋯ , z n ) | ≤ C 1 | y 1 − z 1 | + ⋯ + C n | y n − z n | ,$
and
$q = n max { C 1 , ⋯ , C n } · max { T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , ⋯ , E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 } < 1 ,$
where
$max x ∈ [ 0 , T ] | a i ( x ) | ≤ A i , max x ∈ [ 0 , T ] | b j ( x ) | ≤ B j .$
Then Equation (2) has a unique solution in $C n [ 0 , T ]$.
Proof.
Clearly, $g ( x , u , ⋯ , u ( n − 1 ) ) ∈ C [ 0 , T ]$ for $u ∈ C n [ 0 , T ]$. Define a mapping on $C n [ 0 , T ]$ as
$T ( u ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 g ( x , u , ⋯ , u ( n − 1 ) ) .$
It follows from the proof of Theorem 2 that
$T ( u ) C ≤ T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 · g ( x , u , ⋯ , u ( n − 1 ) ) C < ∞ , ⋯ , d n d x n T ( u ) C ≤ E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 · g ( x , u , ⋯ , u ( n − 1 ) ) C < ∞ .$
This infers that T is a mapping from $C n [ 0 , T ]$ to itself. It remains to prove that T is contractive. Indeed,
$g ( x , u , ⋯ , u ( n − 1 ) ) − g ( x , v , ⋯ , v ( n − 1 ) ) C = max x ∈ [ 0 , T ] | g ( x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) ) − g ( x , v ( x ) , ⋯ , v ( n − 1 ) ( x ) ) | ≤ C 1 u − v C + ⋯ + C n u ( n − 1 ) ( x ) − v ( n − 1 ) ( x ) C ≤ n max { C 1 , C 2 , ⋯ , C n } u − v n ,$
and
$T u − T v C ≤ n max { C 1 , C 2 , ⋯ , C n } T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 u − v n , ⋯ , T ( n ) u − T ( n ) v C ≤ n max { C 1 , C 2 , ⋯ , C n } E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 u − v n .$
Hence,
$T u − T v n = max T u − T v C , ⋯ , T ( n ) u − T ( n ) v C ≤ n max { C 1 , ⋯ , C n } max { T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , ⋯ , E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 } u − v n = q u − v n ,$
where $q < 1$. This completes the proof of Theorem 3. □
Remark 2.
If all coefficient functions $a i ( x ) = b j ( x ) = 0$ over $[ 0 , T ]$, then Equation (2) becomes the integral equation
$u ( x ) = 1 ( n − 1 ) ! ∫ 0 x ( x − t ) n − 1 g ( x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) ) d x .$
Example 2.
The following integro-differential equation with initial conditions:
$u ( 2 ) ( x ) + sin x 2 C D 0.8 u ( x ) − x I 0.5 u ( x ) + I u ( x ) x 2 + 1 = 1 42 ( ( u ′ ) 2 ( x ) + 1 ) + 1 43 cos u ( x ) , u ( 0 ) = u ′ ( 0 ) = 0 ,$
has a unique solution in $C 2 [ 0 , 1 ]$.
Proof.
Clearly,
$g ( x , y 1 , y 2 ) = 1 42 ( y 2 2 + 1 ) + 1 43 cos y 1 ,$
and
$| g ( x , y 1 , y 2 ) − g ( x , z 1 , z 2 ) | ≤ 1 43 | y 1 − z 1 | + 1 42 | y 2 − z 2 | .$
Thus,
$max { C 1 , C 2 } = max 1 43 , 1 42 = 1 42 .$
Obviously,
$max x ∈ [ 0 , 1 ] | sin x 2 | ≤ 1 , max x ∈ [ 0 , 1 ] | − x | ≤ 1 , max x ∈ [ 0 , 1 ] 1 x 2 + 1 ≤ 1 .$
Therefore,
$q = 2 · 1 42 max E ( 1.2 , 2.5 , 3 ) , 3 ( 1 , 1 , 1 ) , E ( 1.2 , 2.5 , 3 ) , 2 ( 1 , 1 , 1 ) , E ( 1.2 , 2.5 , 3 ) , 1 ( 1 , 1 , 1 ) = 1 21 E ( 1.2 , 2.5 , 3 ) , 1 ( 1 , 1 , 1 ) ,$
since
$E ( 1.2 , 2.5 , 3 ) , 3 ( 1 , 1 , 1 ) ≤ E ( 1.2 , 2.5 , 3 ) , 2 ( 1 , 1 , 1 ) ≤ E ( 1.2 , 2.5 , 3 ) , 1 ( 1 , 1 , 1 ) .$
Furthermore,
$E ( 1.2 , 2.5 , 3 ) , 1 ( 1 , 1 , 1 ) = ∑ k = 0 ∞ ∑ k 1 + k 2 + k 3 = k k k 1 , k 2 , k 3 1 Γ ( 1.2 k 1 + 2.5 k 2 + 3 k 3 + 1 ) , ∑ k 1 + k 2 + k 3 = k k k 1 , k 2 , k 3 = 3 k , 1 Γ ( 1.2 k 1 + 2.5 k 2 + 3 k 3 + 1 ) ≤ 1 Γ ( k + 1 ) , k = 0 , 1 , ⋯ ,$
by noting that $Γ ( x + 1 )$ is an increasing function if $x ≥ 1$ [19].
It follows that
$E ( 1.2 , 2.5 , 3 ) , 1 ( 1 , 1 , 1 ) ≤ ∑ k = 0 ∞ 3 k k ! = 1 + ∑ k = 1 ∞ 3 k k ! = 1 + 3 1 + 3 · 3 1 · 2 + 3 · 3 · 3 1 · 2 · 3 + 3 · 3 · 3 · 3 1 · 2 · 3 · 4 + 3 · 3 · 3 · 3 · 3 1 · 2 · 3 · 4 · 5 + 3 · 3 · 3 · 3 · 3 · 3 1 · 2 · 3 · 4 · 5 · 6 + 3 · 3 · 3 · 3 · 3 · 3 · 3 1 · 2 · 3 · 4 · 5 · 6 · 7 + 3 · 3 · 3 · 3 · 3 · 3 · 3 · 3 1 · 2 · 3 · 4 · 5 · 6 · 7 · 8 + ⋯ < 1 + 3 + 4.5 + 4.5 + 3.375 + 2.025 + 1 80 + 1 + 1 2 + 1 2 2 + 1 2 3 + ⋯ = 20.4125 .$
This implies that
$q < 20.4125 21 < 1 .$
This completes the proof of Example 2. □
Finally, we provide sufficient conditions for the uniqueness of system (3) in the product space $C n [ 0 , T ] × C m [ 0 , T ]$.
Theorem 4.
Assume that all coefficient functions $a i ( x )$, $b j ( x )$, $c i 1 ( x )$ and $d j 1 ( x )$ in system (3) are continuous functions on $[ 0 , T ]$ for all $i = 1 , 2 , ⋯ , n 1 ∈ N$, $j = 1 , 2 , ⋯ , n 2 ∈ N$, $i 1 = 1 , 2 , ⋯ , m 1 ∈ N$ and $j 1 = 1 , 2 , ⋯ , m 2 ∈ N$. Furthermore, $0 ≤ β 1 < β 2 < ⋯ < β n 1 < n$, $0 < α 1 < α 2 < ⋯ < α n 2$, $0 ≤ μ 1 < μ 2 < ⋯ < μ m 1 < m$ and $0 < γ 1 < γ 2 < ⋯ < γ m 2$. In addition, suppose $g 1 ( x , y 1 , ⋯ , y n + m ) , g 2 ( x , y 1 , ⋯ , y n + m ) ∈ C ( [ 0 , T ] × R n + m )$ and that there exist constants $K 1 , ⋯ , K n + m$ such that
$| g 1 ( x , y 1 , ⋯ , y n + m ) − g 1 ( x , z 1 , ⋯ , z n + m ) | ≤ K 1 | y 1 − z 1 | + ⋯ + K n + m | y n + m − z n + m | , | g 2 ( x , y 1 , ⋯ , y n + m ) − g 2 ( x , z 1 , ⋯ , z n + m ) | ≤ K 1 | y 1 − z 1 | + ⋯ + K n + m | y n + m − z n + m | ,$
and
$Q = ( n + m ) max { K 1 , ⋯ , K n + m } · max { T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , ⋯ , E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , T m E ( m − μ m 1 , ⋯ , m + γ m 2 ) , m + 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 , ⋯ , E ( m − μ m 1 , ⋯ , m + γ m 2 ) , 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 } < 1 ,$
where
$max x ∈ [ 0 , T ] | a i ( x ) | ≤ A i , max x ∈ [ 0 , T ] | b j ( x ) | ≤ B j , max x ∈ [ 0 , T ] | c i 1 ( x ) | ≤ C i 1 , max x ∈ [ 0 , T ] | d j 1 ( x ) | ≤ D j 1 .$
Then, system (3) has a unique solution in $C n [ 0 , T ] × C m [ 0 , T ]$.
Proof.
Clearly,
$g 1 x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) , v ( x ) , ⋯ , v ( m − 1 ) ( x ) ∈ C [ 0 , T ] , g 2 x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) , v ( x ) , ⋯ , v ( m − 1 ) ( x ) ∈ C [ 0 , T ] ,$
for $( u , v ) ∈ C n [ 0 , T ] × C m [ 0 , T ]$. Define a mapping T on $C n [ 0 , T ] × C m [ 0 , T ]$ as
$T ( u , v ) = ( T 1 ( u , v ) , T 2 ( u , v ) ) ,$
where
$T 1 ( u , v ) = I n ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k n 1 + n 2 = k k k 1 , k 2 , ⋯ , k n 1 + n 2 ( a n 1 ( x ) I n − β n 1 ) k 1 ⋯ ( a 1 ( x ) I n − β 1 ) k n 1 b 1 ( x ) I α 1 + n k n 1 + 1 ⋯ b n 2 ( x ) I α n 2 + n k n 1 + n 2 g 1 ( x , u , ⋯ , u ( n − 1 ) , v , ⋯ , v ( m − 1 ) ) .$
Clearly,
$T 1 ( u , v ) C ≤ T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 · g 1 ( x , u , ⋯ , u ( n − 1 ) , v , ⋯ , v ( m − 1 ) ) C < ∞ , ⋯ , d n d x n T 1 ( u , v ) C ≤ E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 · g 1 ( x , u , ⋯ , u ( n − 1 ) , v , ⋯ , v ( m − 1 ) ) C < ∞ .$
Hence,
$T 1 ( u , v ) ∈ C n [ 0 , T ] .$
Similarly,
$T 2 ( u , v ) = I m ∑ k = 0 ∞ ( − 1 ) k ∑ k 1 + k 2 + ⋯ + k m 1 + m 2 = k k k 1 , k 2 , ⋯ , k m 1 + m 2 ( c m 1 ( x ) I m − μ m 1 ) k 1 ⋯ ( c 1 ( x ) I m − μ 1 ) k m 1 d 1 ( x ) I γ 1 + m k m 1 + 1 ⋯ d m 2 ( x ) I α m 2 + m k m 1 + m 2 g 2 ( x , u , ⋯ , u ( n − 1 ) , v , ⋯ , v ( m − 1 ) ) .$
Clearly,
$T 2 ( u , v ) C ≤ T m E ( m − μ m 1 , ⋯ , m + γ m 2 ) , m + 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 · g 2 ( x , u , ⋯ , u ( n − 1 ) , v , ⋯ , v ( m − 1 ) ) C < ∞ , ⋯ , d m d x m T 2 ( u , v ) C ≤ E ( m − μ m 1 , ⋯ , m + γ m 2 ) , 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 · g 2 ( x , u , ⋯ , u ( n − 1 ) , v , ⋯ , v ( m − 1 ) ) C < ∞ .$
Hence,
$T 2 ( u , v ) ∈ C m [ 0 , T ] ,$
and T is a mapping from $C n [ 0 , T ] × C m [ 0 , T ]$ to itself. We need to show that T is contractive. In fact,
$g 1 ( x , u 1 , ⋯ , u 1 ( n − 1 ) , v 1 , ⋯ , v 1 ( m − 1 ) ) − g 1 ( x , u 2 , ⋯ , u 2 ( n − 1 ) , v 2 , ⋯ , v 2 ( m − 1 ) ) C = max x ∈ [ 0 , T ] | g 1 ( x , u 1 ( x ) , ⋯ , u 1 ( n − 1 ) ( x ) , v 1 ( x ) , ⋯ , v 1 ( m − 1 ) ( x ) ) − g 1 ( x , u 2 ( x ) , ⋯ , u 2 ( n − 1 ) ( x ) , v 2 ( x ) , ⋯ , v 2 ( m − 1 ) ( x ) ) | ≤ K 1 u 1 − u 2 C + ⋯ + K n + m v 1 ( m − 1 ) ( x ) − v 2 ( m − 1 ) ( x ) C ≤ ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } ( u 1 , v 1 ) − ( u 2 , v 2 ) ,$
by noting that
$( u 1 , v 1 ) − ( u 2 , v 2 ) = ( u 1 − u 2 , v 1 − v 2 ) = u 1 − u 2 n + v 1 − v 2 m , u 1 − u 2 C ≤ u 1 − u 2 n ≤ ( u 1 , v 1 ) − ( u 2 , v 2 ) , ⋯ , v 1 ( m − 1 ) ( x ) − v 2 ( m − 1 ) ( x ) C ≤ v 1 − v 2 m ≤ ( u 1 , v 1 ) − ( u 2 , v 2 ) .$
Furthermore,
$T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) C ≤ ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } · T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 · ( u 1 , v 1 ) − ( u 2 , v 2 ) , ⋯ , T 1 ( n ) ( u 1 , v 1 ) − T 1 ( n ) ( u 2 , v 2 ) C ≤ ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } · E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 · ( u 1 , v 1 ) − ( u 2 , v 2 ) .$
This deduces that
$T 1 ( u 1 , v 1 ) − T 1 ( u 2 , v 2 ) n ≤ ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } · max { T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , ⋯ , E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 } · ( u 1 , v 1 ) − ( u 2 , v 2 ) .$
Similarly,
$T 2 ( u 1 , v 1 ) − T 2 ( u 2 , v 2 ) m ≤ ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } · max { T m E ( m − μ m 1 , ⋯ , m + γ m 2 ) , m + 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 , ⋯ , E ( m − μ m 1 , ⋯ , m + γ m 2 ) , 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 } · ( u 1 , v 1 ) − ( u 2 , v 2 ) .$
In summary,
$T ( u 1 , v 1 ) − T ( u 2 , v 2 ) = ( T 1 u 1 , T 2 v 1 ) − ( T 1 u 2 , T 2 v 2 = T 1 u 1 − T 1 u 2 n + T 2 v 1 − T 2 v 2 m ≤ ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } · max { T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , ⋯ , E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , T m E ( m − μ m 1 , ⋯ , m + γ m 2 ) , m + 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 , ⋯ , E ( m − μ m 1 , ⋯ , m + γ m 2 ) , 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 } · ( u 1 , v 1 ) − ( u 2 , v 2 ) = Q ( u 1 , v 1 ) − ( u 2 , v 2 ) ,$
where
$Q = ( n + m ) max { K 1 , K 2 , ⋯ , K n + m } · max { T n E ( n − β n 1 , ⋯ , n + α n 2 ) , n + 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , ⋯ , E ( n − β n 1 , ⋯ , n + α n 2 ) , 1 T n − β n 1 A n 1 , ⋯ , T n + α n 2 B n 2 , T m E ( m − μ m 1 , ⋯ , m + γ m 2 ) , m + 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 , ⋯ , E ( m − μ m 1 , ⋯ , m + γ m 2 ) , 1 T m − μ m 1 C m 1 , ⋯ , T m + γ m 2 D m 2 } < 1 .$
This completes the proof of Theorem 4. □
Remark 3.
If all coefficient functions $a i ( x ) = b j ( x ) = c i 1 ( x ) = d j 1 ( x ) = 0$ in system (3), then it is equivalent to
$u ( x ) = 1 ( n − 1 ) ! ∫ 0 x ( x − t ) n − 1 g 1 ( x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) , v ( x ) , ⋯ , v ( m − 1 ) ( x ) ) d x , v ( x ) = 1 ( m − 1 ) ! ∫ 0 x ( x − t ) m − 1 g 2 ( x , u ( x ) , ⋯ , u ( n − 1 ) ( x ) , v ( x ) , ⋯ , v ( m − 1 ) ( x ) ) d x .$
Example 3.
The following integro-differential system with initial conditions
$u ( 2 ) ( x ) + x C D 0.5 u ( x ) − 1 x + 1 I 0.5 u ( x ) = 1 51 sin v ( x ) , u ( 0 ) = u ′ ( 0 ) = 0 , v ( 3 ) ( x ) = 1 42 cos u ( x ) , v ( 0 ) = v ′ ( 0 ) = v ″ ( 0 ) = 0$
has a unique nonzero solution in $C 2 [ 0 , 1 ] × C 3 [ 0 , 1 ]$.
Proof.
Clearly,
$g 1 ( x , y 1 , ⋯ , y 5 ) = 1 51 sin y 3 , g 2 ( x , y 1 , ⋯ , y 5 ) = 1 42 cos y 1$
and
$| g 1 ( x , y 1 , ⋯ , y 5 ) − g 1 ( x , z 1 , ⋯ , z 5 ) | ≤ 1 51 | y 3 − z 3 | , | g 2 ( x , y 1 , ⋯ , y 5 ) − g 1 ( x , z 1 , ⋯ , z 5 ) | ≤ 1 42 | y 1 − z 1 | .$
Thus,
$( n + m ) max { K 1 , K 2 , ⋯ , K n + m } = ( 2 + 3 ) max 1 42 , 1 51 = 5 42 .$
Obviously,
$max x ∈ [ 0 , 1 ] | x | ≤ 1 , max x ∈ [ 0 , 1 ] − 1 x + 1 ≤ 1 .$
Using Theorem 4, we derive that
$Q = 5 42 max { E ( 1.5 , 2.5 ) , 3 ( 1 , 1 ) , E ( 1.5 , 2.5 ) , 2 ( 1 , 1 ) , E ( 1.5 , 2.5 ) , 1 ( 1 , 1 ) , 1 Γ ( 3 + 1 ) , 1 Γ ( 2 + 1 ) , 1 Γ ( 1 + 1 ) , 1 Γ ( 1 ) } = 5 42 E ( 1.5 , 2.5 ) , 1 ( 1 , 1 ) ,$
by noting that
$c i 1 ( x ) = d j 1 ( x ) = 0$
over $[ 0 , 1 ]$. Furthermore,
$E ( 1.5 , 2.5 ) , 1 ( 1 , 1 ) = ∑ k = 0 ∞ ∑ k 1 + k 2 = k k k 1 , k 2 1 Γ ( 1.5 k 1 + 2.5 k 2 + 1 ) ≤ ∑ k = 0 ∞ 2 k Γ ( k + 1 ) = ∑ k = 0 ∞ 2 k k ! = 1 + 2 1 ! + 2 · 2 1 · 2 + 2 · 2 · 2 1 · 2 · 3 + 2 · 2 · 2 · 2 1 · 2 · 3 · 4 + 2 · 2 · 2 · 2 · 2 1 · 2 · 3 · 4 · 5 + ⋯ ≤ 1 + 2 + 2 + 1 3 + 2 3 0 + 2 3 1 + 2 3 2 + ⋯ = 25 3 .$
Therefore,
$Q ≤ 5 42 · 25 3 = 125 126 < 1 ,$
and $( u , v ) = 0$ is clearly not a solution. This completes the proof of Example 3. □

## 3. Conclusions

Using Banach’s contractive principle, the multivariate Mittag–Leffler function and Babenko’s approach, we studied the uniqueness of solutions of several nonlinear Liouville–Caputo integro-differential equations with variable coefficients and initial conditions, as well as the associated coupled system in Banach spaces. The results obtained are new and original. We also presented three examples to demonstrate the use of our main theorems.

## Author Contributions

Conceptualization, C.L. and J.B.; methodology, C.L.; software, C.L. and J.B.; validation, C.L. and J.B.; formal analysis, C.L.; investigation, C.L. and J.B.; resources, C.L.; data curation, C.L. and J.B.; writing—original draft preparation, C.L.; writing—review and editing, C.L.; visualization, C.L.; supervision, C.L.; project administration, C.L. and J.B.; funding acquisition, C.L. All authors have read and agreed to the published version of the manuscript.

## Funding

This work is supported by the Natural Sciences and Engineering Research Council of Canada (Grant No. 2019-03907).

Not applicable.

Not applicable.

Not applicable.

## Acknowledgments

The authors are grateful to the two reviewers for their careful reading of the paper with productive comments and suggestions.

## Conflicts of Interest

The authors declare no conflict of interest.

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Li, C.; Beaudin, J. On the Nonlinear Integro-Differential Equations. Fractal Fract. 2021, 5, 82. https://doi.org/10.3390/fractalfract5030082

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Li C, Beaudin J. On the Nonlinear Integro-Differential Equations. Fractal and Fractional. 2021; 5(3):82. https://doi.org/10.3390/fractalfract5030082

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Li, Chenkuan, and Joshua Beaudin. 2021. "On the Nonlinear Integro-Differential Equations" Fractal and Fractional 5, no. 3: 82. https://doi.org/10.3390/fractalfract5030082