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In this paper, we investigate the nonlinear neutral fractional integral-differential equation involving conformable fractional derivative and integral. First of all, we give the form of the solution by lemma. Furthermore, existence results for the solution and sufficient conditions for uniqueness solution are given by the Leray-Schauder nonlinear alternative and Banach contraction mapping principle. Finally, an example is provided to show the application of results.

The theory of fractional calculus has played a major role in control theory, fluid dynamics, biological systems, economics and other fields [

The definition of the conformable fractional derivative and integral was introduced in 2014 by Khalil et al. [

Over these years, there has been a significant development in fractional functional differential equations. Among them, Li, Liu and Jiang gave sufficient conditions of the existence of positive solutions for a class of nonlinear fractional differential equations with caputo derivative [

Time-delay is part of the theoretical fields investigated by many authors, including unbounded time-delay, bounded time-delay, state-dependent time-delay and others. In 2009, the existence and uniqueness of solutions of the Caputo fractional neutral differential equations with unbounded delays were discussed [

Based on the above research background and relevant discussions, we found that few people used conformable derivative to study fractional differential equations with time-delay. In 2019, Mohamed I. Abbas gave the existence of solutions and uniqueness of solution for fractional neutral integro-differential equations by the Hadamard fractional derivative of order α ∈ ( 0,1 ) and the Riemann-Liouville integral [

{ T α [ w ( t ) − ∑ i = 1 p I β i u i ( t , w t ) ] = l ( t , w t ) , t ∈ [ 0 , ρ ] , w ( t ) = ψ ( t ) , t ∈ [ − υ , 0 ] , (1.1)

where T α denotes the conformable fractional derivative of order α , 1 < α < 2 , I β i denotes the conformable fractional integral with order β i , β i ∈ ( 0,1 ) , i = 1,2,3 , ⋯ , p , p ∈ N + . ρ , υ > 0 are constants. And for any t ∈ [ 0, ρ ] , we denote by w t the element of C ( [ − υ ,0 ] , R ) and is defined by w t ( θ ) = w ( t + θ ) , θ ∈ [ − υ ,0 ] . Here w t ( ⋅ ) represents the history of the state from time t − υ up to the present time t. l , u i : [ 0, ρ ] × C ( [ − υ ,0 ] , R ) → R are continuous functions that satisfy some hypotheses given later, ψ ∈ C ( [ − υ ,0 ] , R ) .

The rest of this paper is organized as follows: In Section 2, we introduce the concepts and basic properties of conformable fractional integral and derivative. In Section 3, we give existence results for the solution and sufficient conditions for uniqueness solution by Leray-Schauder nonlinear alternative and Banach contraction mapping principle. In Section 4, the numerical simulation is showed to illustrate the results.

Notations: C ( [ 0, ρ ] , R ) denotes all continuous functions that mapped from [ 0, ρ ] to R and R denotes all real numbers. R + denotes all positive real numbers.

In this section, we present some necessary definitions and lemmas to establish our main results.

Definition 2.1. ( [

I a α w ( t ) = I a n + 1 [ ( t − a ) α − n − 1 w ( t ) ] = 1 n ! ∫ a t ( t − x ) n ( x − a ) α − n − 1 w ( x ) d x .

If a = 0 , I a α w ( t ) can be written as I α w ( t ) .

Definition 2.2. ( [

T a α w ( t ) = l i m ε → 0 w [ α ] − 1 ( t + ε ( t − a ) [ α ] − α ) − w [ α ] − 1 ( t ) ε ,

where [ α ] denotes the smallest integer greater than or equal to α . If a = 0 , T a α w ( t ) can be written as T α w ( t ) .

Lemma 2.3 ( [

1) T α l ( t ) = 0 , for all constant functions l ( t ) = λ ;

2) l is n + 1 times differential simultaneously, then T α ( l ) ( t ) = t [ α ] − α l [ α ] ( t ) .

Lemma 2.4. ( [

1) For a function w : [ a , + ∞ ) → R , if w n ( t ) is continuous, then for any t > a , we have

T a α I a α w ( t ) = w ( t ) , α ∈ ( n , n + 1 ] ;

2) For a function w : [ a , + ∞ ) → R , if w is n + 1 times differentiable, then for any t > a , we have

I a α T a α w ( t ) = w ( t ) − ∑ i = 0 n w ( i ) ( a ) ( t − a ) i i ! , α ∈ ( n , n + 1 ] .

Lemma 2.5. ( [

1) T α ( a 1 w 1 + a 2 w 2 ) = a 1 T α ( w 1 ) + a 2 T α ( w 2 ) ;

2) T α ( t p ) = p t p − α ;

3) T α ( w 1 w 2 ) = w 1 T α ( w 2 ) + w 2 T α ( w 1 ) ;

4) T α ( w 1 w 2 ) = w 2 T α ( w 1 ) − w 1 T α ( w 2 ) ( w 2 ) 2 .

Lemma 2.6. ( [

1) A has a fixed point in P, or

2) There is a w ∈ ∂ P (the boundary of P in C 1 ) and λ ∈ ( 0,1 ) , such that w = λ A ( w ) .

Lemma 2.7. Let l ( t ) be a continuous function, then the fractional differential equation

{ T α [ w ( t ) − η ( t ) ] = l ( t ) , t ∈ [ 0 , ρ ] , 1 < α < 2 , w ( 0 ) = ψ 0 , w ′ ( 0 ) = ψ ′ 0 , (2.1)

is equivalent to the integral-differential equation

w ( t ) = ψ 0 − η ( 0 ) + [ ψ ′ 0 − η ′ ( 0 ) ] t + η ( t ) + ∫ 0 t ( t − s ) s α − 2 l ( s ) d s , (2.2)

where ψ 0 = w ( 0 ) , ψ ′ 0 = w ′ ( 0 ) .

Proof. Consider Equation (2.1), for any t ∈ [ 0, ρ ] ,

T α [ w ( t ) − η ( t ) ] = l ( t ) .

Transforming α times conformable fractional integral on both sides of the equation and using Lemma 2.4, since 1 < α < 2 , we have

w ( t ) = ψ 0 − η ( 0 ) + [ ψ ′ 0 − η ′ ( 0 ) ] t + η ( t ) + ∫ 0 t ( t − s ) s α − 2 l ( s ) d s ,

where ψ 0 = w ( 0 ) , ψ ′ 0 = w ′ ( 0 ) . The proof is completed. □

In this section, we give several results about the fractional integral-differential Equation (1.1).

If X = { w | w ∈ C ( [ − υ , ρ ] , ℝ ) } is Banach Space, the norm is defined as ‖ w ‖ = sup { | w ( t ) | , t ∈ [ − υ , ρ ] } .

Let η ( t ) = ∑ i = 1 p I β i u i ( t , w t ) . Giving the definition of the operator A : X → X ,

A w ( t ) = { ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + ∑ i = 1 p I β j u i ( t , w t ) + ∫ 0 t ( t − s ) s α − 2 l ( s , w s ) d s , t ∈ [ 0 , ρ ] , ψ ( t ) , t ∈ [ − υ , 0 ] . (3.1)

where ψ 0 = w ( 0 ) , ψ ′ 0 = w ′ ( 0 ) , w 0 = w ( 0 + θ ) = ψ ( θ ) , θ ∈ [ − υ , 0 ] , η ( 0 ) = ∑ i = 1 p I β i u i ( t , w t ) | t = 0 = 0 .

It should be noticed that Equation (1.1) has solutions if and only if the operator A has fixed points. So as to achieve the desired goals, we impose the following assumptions for the Equation (1.1).

(H_{1}) There exist functions γ ( t ) , δ i ( t ) : [ 0, ρ ] → R + and continuous non decreasing functions ζ ( t ) , ϕ i ( t ) : [ 0, ∞ ] → [ 0, ∞ ] , such that, for any ( t , w t ) ∈ [ 0, ρ ] × C ( [ − υ ,0 ] , R ) ,

| l ( t , w t ) | ≤ γ ( t ) ζ ( ‖ w t ‖ ) ,

| u i ( t , w t ) | ≤ δ i ( t ) ϕ i ( ‖ w t ‖ ) .

(H_{2}) Functions l , u i : [ 0, ρ ] × C ( [ − υ ,0 ] , R ) → R are continuous. There exist positive functions λ i , μ with bounds ‖ λ i ‖ , ‖ μ ‖ , i = 1 , 2 , ⋯ , p , p ∈ N + , respectively such that

| u i ( t , x ) − u i ( t , y ) | ≤ λ i ( t ) ‖ x − y ‖ ,

| l ( t , x ) − l ( t , y ) | ≤ μ ( t ) ‖ x − y ‖ .

(H_{3}) There exists constant M > 0 , υ 1 < υ , such that

M | ψ 0 | + [ ψ ′ 0 − η ′ ( 0 ) ] ρ + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ρ β i β i + ‖ γ ‖ ζ ( υ 1 ) ρ α α ( α − 1 ) > 1.

(H_{4})

∑ i = 1 p ‖ λ i ‖ ρ β i β i + ‖ μ ‖ ρ α α ( α − 1 ) < 1.

We give an existence result based on the nonlinear alternative of Leray-Schauder type applied to a completely continuous operator.

Theorem 3.1. Suppose that the assumptions (H_{1})-(H_{3}) are satisfied, then the Equation (1.1) has at least one solution.

Proof. The operator A is defined as (3.1). Define

D = { w ∈ C ( [ − υ , ρ ] , R ) : ‖ w t ‖ ≤ υ 1 } .

Firstly, we prove that operator A is uniformly bounded. For any ( t , w t ) ∈ [ 0, ρ ] × C ( [ − υ ,0 ] , R ) , i = 1 , 2 , ⋯ , p , p ∈ N + , and w ∈ D , by (H_{1}), we have

| A w ( t ) | = | ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + ∑ i = 1 p I β i u i ( s , w s ) + ∫ 0 t ( t − s ) s α − 2 l ( s , w s ) d s | ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p I β i | u i ( s , w s ) | + ∫ 0 t ( t − s ) s α − 2 | l ( s , w s ) | d s ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p I β i δ i ( t ) ϕ i ( ‖ w t ‖ ) + ∫ 0 t ( t − s ) s α − 2 γ ( t ) ζ ( ‖ w t ‖ ) d s ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ∫ 0 t s β i − 1 d s + ‖ γ ‖ ζ ( υ 1 ) ∫ 0 t ( t − s ) s α − 2 d s ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) t β i β i + ‖ γ ‖ ζ ( υ 1 ) t α α ( α − 1 ) ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | ρ + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ρ β i β i + ‖ γ ‖ ζ ( υ 1 ) ρ α α ( α − 1 ) < M .

For any t ∈ [ − υ , 0 ] and w ∈ D , we have

| A w ( t ) | = | ψ ( t ) | ≤ ‖ ψ ‖ .

Denote M 1 = max { M , ‖ ψ ‖ } , then

‖ A w ( t ) ‖ ≤ M 1 , t ∈ [ − υ , ρ ] , w ∈ D .

This implies that the operator A is uniformly bounded in D.

Besides, we need to prove that AD is an equicontinuous set. Let w n be a sequence such that lim n → ∞ w n = w in D. Then, for any t ∈ [ 0, ρ ] , we have

lim n → ∞ A w n ( t ) = ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + lim n → ∞ ∑ i = 1 p ∫ 0 t s β i − 1 u i ( s , w s n ) d s + lim n → ∞ ∫ 0 t ( t − s ) s α − 2 l ( s , w s n ) d s .

By (H_{2}), functions l , u i are uniformly continuous, thus, we have

lim n → ∞ A w n ( t ) = ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + ∑ i = 1 p ∫ 0 t lim n → ∞ s β i − 1 u i ( s , w s n ) d s + ∫ 0 t lim n → ∞ ( t − s ) s α − 2 l ( s , w s n ) d s = ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + ∑ i = 1 p ∫ 0 t s β i − 1 u i ( s , w s ) d s + ∫ 0 t ( t − s ) s α − 2 l ( s , w s ) d s = A w ( t ) .

For any t ∈ [ − υ , 0 ] , it is obvious that

lim n → ∞ A w n ( t ) = ψ ( t ) = A w ( t ) .

Therefore, A w ( t ) is continuous and uniformly continuous for t ∈ [ − υ , ρ ] , which implies that A w ( t ) is equicontinuous for t ∈ [ − υ , ρ ] , A is continuous in C ( [ − υ , ρ ] , R ) .

Furthermore, we consider | A w ( t 2 ) − A w ( t 1 ) | , for any t 1 , t 2 ∈ [ − υ , ρ ] , t 1 < t 2 .

Case 1. If 0 ≤ t 1 < t 2 ≤ ρ , for any ( t , w t ) ∈ [ 0, ρ ] × C ( [ − υ ,0 ] , R ) and w ∈ D , i = 1 , 2 , ⋯ , p , we have

| A w ( t 2 ) − A w ( t 1 ) | = | ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t 2 + ∑ i = 1 p ∫ 0 t 2 s β i − 1 u i ( s , w s ) d s + ∫ 0 t 2 ( t 2 − s ) s α − 2 l ( s , w s ) d s − ψ 0 − [ ψ ′ 0 − η ′ ( 0 ) ] t 1 − ∑ i = 1 p ∫ 0 t 1 s β i − 1 u i ( s , w s ) d s − ∫ 0 t 1 ( t 1 − s ) s α − 2 l ( s , w s ) d s |

≤ | ψ ′ 0 − η ′ ( 0 ) | ( t 2 − t 1 ) + ∑ i = 1 p ∫ t 1 t 2 s β i − 1 | u i ( s , w s ) | d s + ∫ 0 t 1 ( t 2 − t 1 ) s α − 2 | l ( s , w s ) | d s + ∫ t 1 t 2 ( t 2 − s ) s α − 2 | l ( s , w s ) | d s ≤ | ψ ′ 0 − η ′ ( 0 ) | ( t 2 − t 1 ) + ∑ i = 1 p ∫ t 1 t 2 s β i − 1 δ i ( t ) ϕ i ( ‖ w t ‖ ) d s + ∫ 0 t 1 ( t 2 − t 1 ) s α − 2 γ ( t ) ζ ( ‖ w t ‖ ) d s + ∫ t 1 t 2 ( t 2 − s ) s α − 2 γ ( t ) ζ ( ‖ w t ‖ ) d s

≤ | ψ ′ 0 − η ′ ( 0 ) | ( t 2 − t 1 ) + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ∫ t 1 t 2 s β i − 1 d s + ‖ γ ‖ ζ ( υ 1 ) ∫ 0 t 1 ( t 2 − t 1 ) s α − 2 d s + ‖ γ ‖ ζ ( υ 1 ) ∫ t 1 t 2 ( t 2 − s ) s α − 2 d s ≤ | ψ ′ 0 − η ′ ( 0 ) | ( t 2 − t 1 ) + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) t 2 β i − t 1 β i β i + ‖ γ ‖ ζ ( υ 1 ) t 2 α − t 1 α α ( α − 1 ) .

If t 2 − t 1 → 0 , then | A w ( t 2 ) − A w ( t 1 ) | → 0 .

Case 2. If − υ < t 1 < 0 ≤ t 2 < ρ , for any ( t , w t ) ∈ [ 0, ρ ] × C ( [ − υ ,0 ] , R ) and w ∈ D , i = 1 , 2 , ⋯ , p , A w ( t 1 ) = ψ ( t 1 ) holds for any − υ < t 1 < 0 , we have

| A w ( t 2 ) − A w ( t 1 ) | = | ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t 2 + ∑ i = 1 p ∫ 0 t 2 s β i − 1 u i ( s , w s ) d s + ∫ 0 t 2 ( t 2 − s ) s α − 2 l ( s , w s ) d s − ψ ( t 1 ) | ≤ [ ψ ′ 0 − η ′ ( 0 ) ] t 2 + | ψ 0 − ψ ( t 1 ) | + ∑ i = 1 p ∫ 0 t 2 s β i − 1 | u i ( s , w s ) | d s + ∫ 0 t 2 ( t 2 − s ) s α − 2 | l ( s , w s ) | d s

≤ [ ψ ′ 0 − η ′ ( 0 ) ] t 2 + | ψ 0 − ψ ( t 1 ) | + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ∫ 0 t 2 s β i − 1 d s + ‖ γ ‖ ζ ( υ 1 ) ∫ 0 t 2 ( t 2 − s ) s α − 2 d s ≤ [ ψ ′ 0 − η ′ ( 0 ) ] t 2 + | ψ 0 − ψ ( t 1 ) | + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) t 2 β i β i + ‖ δ ‖ ζ ( υ 1 ) t 2 α α ( α − 1 ) .

Since ψ ( t ) is a continuous function, if t 1 → 0 , then we have ψ ( t 1 ) → ψ 0 . If t 2 − t 1 → 0 , then | A w ( t 2 ) − A w ( t 1 ) | → 0 .

Case 3. If − υ ≤ t 1 < t 2 < 0 , for any − υ ≤ t 1 < t 2 < 0 and w ∈ D ,

| A w ( t 2 ) − A w ( t 1 ) | = | ψ ( t 2 ) − ψ ( t 1 ) | .

Since ψ ( t ) is continuous function, if t 2 − t 1 → 0 , then | A w ( t 2 ) − A w ( t 1 ) | → 0 .

From what has been discussed above, AD is equicontinuous. By Arzelá-Ascoli theorem, AN is compact, then A is completely continuous on X.

For any ( t , w t ) ∈ [ 0, ρ ] × C ( [ − υ ,0 ] , R ) and t ∈ [ 0, ρ ] , i = 1 , 2 , ⋯ , p , we have

w ( t ) = ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + ∑ i = 1 p I β i u i ( s , w s ) + ∫ 0 t ( t − s ) s α − 2 l ( s , w s ) d s ,

For any t ∈ [ 0, ρ ] , by assumption (H_{1}), we have

| w ( t ) | ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p I β i | u i ( s , w s ) | + ∫ 0 t ( t − s ) s α − 2 | l ( s , w s ) | d s ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p I β i δ i ( t ) ϕ i ( ‖ w t ‖ ) + ∫ 0 t ( t − s ) s α − 2 γ ( t ) ζ ( ‖ w t ‖ ) d s ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ∫ 0 t s β i − 1 d s + ‖ γ ‖ ζ ( υ 1 ) ∫ 0 t ( t − s ) s α − 2 d s

≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | t + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) t β i β i + ‖ γ ‖ ζ ( υ 1 ) t α α ( α − 1 ) ≤ | ψ 0 | + | ψ ′ 0 − η ′ ( 0 ) | ρ + ∑ i = 1 p ‖ δ i ‖ ϕ i ( υ 1 ) ρ β i β i + ‖ γ ‖ ζ ( υ 1 ) ρ α α ( α − 1 ) .

Consider the assumption (H_{3}), there exists M ≠ ‖ w ( t ) ‖ . Define set P = { w ∈ C ( [ − υ , ρ ] , R ) : ‖ w ( t ) ‖ < M } . We can show that A : P ¯ → D is continuous and completely continuous. Assuming there exists w ∈ ∂ P and λ ∈ ( 0,1 ) , such that w = λ A ( w ) . Then we have | λ | = ‖ w ‖ ‖ A ( w ) ‖ ≥ 1 .

By Lemma 2.6, A has a fixed point in P, which implies that there exists at least one solution to the Equation (1.1). The proof is completed. □

We will give the uniqueness result of solutions of Equation (1.1):

Theorem 3.2. Suppose that the assumptions (H_{2}) and (H_{4}) are satisfied, then the Equation (1.1) has a unique solution.

Proof. The operator A is defined as (3.1). For any t ∈ [ 0, ρ ] and w 1 , w 2 ∈ C ( [ − υ , ρ ] , R ) , by (H_{2}), we have

| A w 1 ( t ) − A w 2 ( t ) | = | ψ 0 + [ ψ ′ 0 − η ′ ( 0 ) ] t + ∑ i = 1 p I β i u i ( s , w 1 s ) + ∫ 0 t ( t − s ) s α − 2 l ( s , w s 1 ) d s − ψ 0 − [ ψ ′ 0 − η ′ ( 0 ) ] t − ∑ i = 1 p I β i u i ( s , w 2 s ) − ∫ 0 t ( t − s ) s α − 2 l ( s , w s 2 ) d s | ≤ ∑ i = 1 p I β i | u i ( s , w s 1 ) − u i ( s , w s 2 ) | + ∫ 0 t ( t − s ) s α − 2 | l ( s , w s 1 ) − l ( s , w s 2 ) | d s

≤ ∑ i = 1 p I β i λ i ( t ) ‖ w 1 − w 2 ‖ + ∫ 0 t ( t − s ) s α − 2 μ ( t ) ‖ w 1 − w 2 ‖ d s ≤ ‖ w 1 − w 2 ‖ ( ∑ i = 1 p ∫ 0 t s β i − 1 ‖ λ i ‖ d s + ∫ 0 t ( t − s ) s α − 2 ‖ μ ‖ d s ) ≤ ‖ w 1 − w 2 ‖ ( ∑ i = 1 p ‖ λ i ‖ t β i β i + ‖ μ ‖ t α α ( α − 1 ) ) ≤ ‖ w 1 − w 2 ‖ ( ∑ i = 1 p ‖ λ i ‖ ρ β i β i + ‖ μ ‖ ρ α α ( α − 1 ) ) .

For any t ∈ [ − υ ,0 ] ,

| A w 1 ( t ) − A w 2 ( t ) | = | ψ ( t ) − ψ ( t ) | = 0.

By (H_{4}), A is a contraction mapping, then Equation (1.1) has a unique solution. The proof is completed. □

This section presents an example where we apply Theorems 3.1 and 3.2 to some particular cases.

Example 4.1. Consider the fractional integral-differential equation

{ T 3 2 [ w ( t ) − ∑ i = 1 3 I i 4 | w t | ( 20 i + 6 t ) ( 1 + | w t | ) ] = 1 16 − t 2 ( | w t | 3 ( 1 + | w t | ) + 1 5 ) , t ∈ [ 0 , 2 ] , w ( t ) = ψ ( t ) , t ∈ [ − 1 2 , 0 ] . (4.1)

where T 3 2 denotes the conformable fractional derivative of order 3 2 , I i 4 denotes the conformable fractional integral of the order i 4 , i = 1 , 2 , 3 . If w ( t ) : [ − 1 2 ,2 ] → R , then for any t ∈ [ 0,2 ] , we define w t ( θ ) = w ( t + θ ) , θ ∈ [ − 1 2 ,0 ] . Functions

u i ( t , w t ) = | w t | ( 20 i + 6 t ) ( 1 + | w t | ) ,

l ( t , w t ) = 1 16 − t 2 ( | w t | 3 ( 1 + | w t | ) + 1 5 ) .

The continuous function ψ ( t ) satisfies the condition that ψ ( 0 ) = ψ ′ ( 0 ) = 0 .

For any ( t , w t ) ∈ [ 0,2 ] × C ( [ − 1 2 ,0 ] , R ) , i = 1 , 2 , 3 , we have

| u i ( t , w t ) | = | | w t | ( 20 i + 6 t ) ( 1 + | w t | ) | ≤ 1 20 i + 6 t ≤ 1 20 .

For any ( t , w t ) ∈ [ 0,2 ] × C ( [ − 1 2 ,0 ] , R ) , we have

| l ( t , w t ) | = | 1 16 − t 2 ( | w t | 3 ( 1 + | w t | ) + 1 5 ) | ≤ 2 45 .

For any ( t , w t ) ∈ [ 0,2 ] × C ( [ − 1 2 ,0 ] , R ) , ψ ( 0 ) = ψ ′ ( 0 ) = η ( 0 ) = η ′ ( 0 ) = 0 , if M > 0.659 , we have

M ∑ i = 1 3 1 20 2 i 4 i 4 + 2 45 2 3 2 3 2 ( 3 2 − 1 ) > 1 , i = 1 , 2 , 3.

Consequently, by Theorem 3.1, the Equation (4.1) has at least one solution.

For continuous functions column u i ( t , w t 1 ) , u i ( t , w t 2 ) : [ 0,2 ] × C ( [ − 1 2 ,0 ] , R ) , i = 1 , 2 , 3 , we have

| u i ( t , w t 1 ) − u i ( t , w t 2 ) | = | | w t 1 | ( 20 i + 6 t ) ( 1 + | w t 1 | ) − | w t 2 | ( 20 i + 6 t ) ( 1 + | w t 2 | ) |

≤ ‖ w 1 − w 2 ‖ ( 20 i + 6 t ) ( 1 + | w t 1 | ) ( 1 + | w t 2 | ) ≤ 1 20 i + 6 t ‖ w 1 − w 2 ‖ .

For continuous functions l ( t , w t 1 ) , l ( t , w t 2 ) : [ 0,2 ] × C ( [ − 1 2 ,0 ] , R ) , we have

| l ( t , w t 1 ) − l ( t , w t 2 ) | = | 1 16 − t 2 ( | w t 1 | 3 ( 1 + | w t 1 | ) + 1 5 ) − 1 16 − t 2 ( | w t 2 | 3 ( 1 + | w t 2 | ) + 1 5 ) | ≤ ‖ w 1 − w 2 ‖ 3 ( 16 − t 2 ) ( 1 + | w t 1 | ) ( 1 + | w t 2 | ) ≤ ‖ w 1 − w 2 ‖ 3 ( 16 − t 2 ) .

For any ( t , w t ) ∈ [ 0,2 ] × C ( [ − 1 2 ,0 ] , R ) , we have

∑ i = 1 3 1 20 i 2 i 4 i 4 + 1 36 2 3 2 3 2 ( 3 2 − 1 ) < 0.596 < 1 , i = 1 , 2 , 3.

Thus, by Theorem 3.2, the Equation (4.1) has a unique solution.

The conformable fractional derivative brings great convenience to the study of fractional functional differential equations due to its unique properties. This paper uses conformable derivative to study the fractional neutral integro-differential equations, and obtains the results of the existence of the solution and the sufficient conditions for the uniqueness of the solution.

The authors declare no conflicts of interest regarding the publication of this paper.

Li, R., Jiang, W., Sheng, J.L. and Wang, S. (2020) On the Nonlinear Neutral Conformable Fractional Integral-Differential Equation. Applied Mathematics, 11, 1041-1051. https://doi.org/10.4236/am.2020.1110069