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In this paper, we study a class of boundary value problems for conformable fractional differential equations under a new definition. Firstly, by using the monotone iterative technique and the method of coupled upper and lower solution, the sufficient condition for the existence of the boundary value problem is obtained, and the range of the solution is determined. Then the existence and uniqueness of the solution are proved by the proof by contradiction. Finally, a concrete example is given to illustrate the wide applicability of our main results.

In recent years, there are few studies on boundary value problems of conformable fractional differential equations under new definitions [

( x ( δ ) ( t ) = f ( t , x ( t ) ) , t ∈ ( 0,1 ) , x ( 0 ) = − r x ( 1 ) + λ ∫ 0 1 x ( s ) d s (1)

where x ( δ ) ( t ) is the conformable fractional derivatives of order δ for t ∈ ( 0,1 ) which is defined in [

In this section, we present some definitions and lemmas which will be used in the proof of our main results.

Definition 2.1. (See [

x ( δ ) ( t ) : = lim ε → 0 x ( t + ε t 1 − δ ) − x ( t ) ε ,

for all t > 0 , δ ∈ ( 0,1 ) . If the conformable fractional derivative of x of order δ exists, then we simply say that x is δ-differentiable. If x is δ-differentiable in some t ∈ ( 0, a ) , a > 0 , and lim t → 0 x ( δ ) ( t ) exists, then we define

x ( δ ) ( 0 ) : = lim t → 0 x ( δ ) ( t ) .

Definition 2.2. Let y 0 ( t ) , z 0 ( t ) ∈ C ( [ 0,1 ] , ℝ ) , then y 0 = y 0 ( t ) , z 0 = z 0 ( t ) are said to be coupled lower and upper solutions of (1), respectively, if

( y 0 ( δ ) ( t ) ≤ f ( t , y 0 ( t ) ) , t ∈ ( 0,1 ) , y 0 ( 0 ) + r z 0 ( 1 ) ≤ λ ∫ 0 1 y 0 ( s ) d s . z 0 ( δ ) ( t ) ≥ f ( t , z 0 ( t ) ) , t ∈ ( 0,1 ) , z 0 ( 0 ) + r y 0 ( 1 ) ≥ λ ∫ 0 1 z 0 ( s ) d s .

Definition 2.3. Let y , z ∈ C ( [ 0,1 ] , ℝ ) , then the function pair ( y , z ) is said to be coupled solutions of (1), if

( y ( δ ) ( t ) = f ( t , y ( t ) ) , t ∈ ( 0,1 ) , y ( 0 ) + r z ( 1 ) = λ ∫ 0 1 y ( s ) d s . z ( δ ) ( t ) = f ( t , z ( t ) ) , t ∈ ( 0,1 ) , z ( 0 ) + r y ( 1 ) = λ ∫ 0 1 z ( s ) d s .

Let γ , ρ ∈ C ( [ 0,1 ] , ℝ ) , then ( ρ , γ ) is said to be minimum and maximum coupled solutions of (1), if ( ρ , γ ) are coupled solutions of (1), and ρ ( t ) ≤ y ( t ) , z ( t ) ≤ γ ( t ) for any coupled solution ( y , z ) .

Lemma 2.1. (See [

1) ( a x 1 + b x 2 ) ( δ ) ( t ) = a x 1 ( δ ) ( t ) + b x 2 ( δ ) ( t ) ;

2) ( x 1 x 2 ) ( δ ) ( t ) = x 1 ( t ) x 2 ( δ ) ( t ) + x 2 ( t ) x 1 ( δ ) ( t ) ;

3) ( x 1 x 2 ) ( δ ) ( t ) = x 2 ( t ) x 1 ( δ ) ( t ) − x 1 ( t ) x 2 ( δ ) ( t ) x 2 2 ( t ) .

for t ∈ ( 0,1 ) , a , b ∈ ℝ .

Lemma 2.2. (See [

Lemma. 2.3 (See [

Lemma 2.4. Assume that g ∈ C ( [ 0,1 ] , ℝ ) , and δ ∈ ( 0,1 ] , m ∈ ℝ , M > 0 , Define function p : [ 0,1 ] → ℝ as follows:

p ( t ) = m e − M δ t δ + ∫ 0 t s δ − 1 g ( s ) e M δ ( s δ − t δ ) d s . (2)

Then p ( t ) is the solution of the initial value problem as follows

( p ( δ ) ( t ) + M p ( t ) = g ( t ) , t ∈ ( 0 , 1 ] , p ( 0 ) = m

Proof Assume that p ( t ) is given by (2), then p is differentiable for t > 0 , therefore we have

p ( δ ) ( t ) = t 1 − δ t δ − 1 ( − m M e − M t δ δ t δ − M e − M t δ δ ∫ 0 t e M s δ δ s δ − 1 g ( s ) + g ( t ) ) = − M ( m e − M δ t δ + ∫ 0 t s δ − 1 g ( s ) e M δ ( s δ − t δ ) d s ) + g ( t ) = − M p ( t ) + g ( t ) .

from Lemma 2.2, and p ( t ) subject to the condition

p ( 0 ) = m .

□

Lemma 2.5. (Comparison Theorem) Let p ∈ C ( [ 0 , 1 ] , ℝ ) , M > 0 , and the following inequalities hold true

( p ( δ ) ( t ) + M p ( t ) ≤ 0 , t ∈ [ 0 , 1 ] , p ( 0 ) ≤ 0.

then p ( t ) ≤ 0 , for t ∈ [ 0,1 ] .

Proof Let p ( δ ) ( t ) + M p ( t ) = g ( t ) , p ( 0 ) = m , then we have g ( t ) ≥ 0, m ≥ 0 for t ∈ [ 0,1 ] , and we can draw a conclusion from (2.1) and Lemma 2.3. □

Theorem 3.1. Assume that y 0 ( t ) , z 0 ( t ) are coupled lower and upper solutions of (1.1) with y 0 ( t ) ≤ z 0 ( t ) for t ∈ [ 0,1 ] , let D = { x ∈ C ( [ 0,1 ] , ℝ ) | y 0 ( t ) ≤ x ≤ z 0 ( t ) } . And if y 0 ( t ) ≤ x 2 ≤ x 1 ≤ z 0 ( t ) , then the following inequalities hold true

f ( t , x 1 ) − f ( t , x 2 ) ≥ − M ( x 1 − x 2 ) . (3)

for t ∈ [ 0,1 ] and M > 0 . If we take y 0 ( t ) , z 0 ( t ) as initial elements, the iterative sequences defined by

( y n ( t ) = ( λ ∫ 0 1 y n − 1 ( s ) d s − r z n − 1 ( 1 ) ) e − M δ t δ + ∫ 0 t s δ − 1 f y n − 1 ( s ) e M δ ( s δ − t δ ) d s , t ∈ [ 0,1 ] z n ( t ) = ( λ ∫ 0 1 z n − 1 ( s ) d s − r y n − 1 ( 1 ) ) e − M δ t δ + ∫ 0 t s δ − 1 f z n − 1 ( s ) e M δ ( s δ − t δ ) d s , t ∈ [ 0,1 ] (4)

are { y n ( t ) } and { z n ( t ) } , then

1) y n ( t ) → y * ( t ) and z n ( t ) → z * ( t ) uniformly and y * , z * ∈ D ;

2) ( y * , z * ) are coupled minimal and maximal solutions of (1.1) respectively in D;

3) If x ( t ) is the solution of (1.1) in D, then we have y * ≤ x ≤ z * ; i.e., we have

y * ( t ) ≤ x ( t ) ≤ z * ( t ) ,

for t ∈ [ 0,1 ] .

Proof 1). There is a unique solution to the boundary value problem as follows

( y ( δ ) ( t ) = f ( t , u ( t ) ) − M ( y ( t ) − u ( t ) ) , t ∈ ( 0,1 ) , y ( 0 ) + r v ( 1 ) = λ ∫ 0 1 u ( s ) d s . z ( δ ) ( t ) = f ( t , v ( t ) ) − M ( z ( t ) − v ( t ) ) , t ∈ ( 0,1 ) , z ( 0 ) + r u ( 1 ) = λ ∫ 0 1 v ( s ) d s ,

which is given by

( y ( t ) = ( λ ∫ 0 1 u ( s ) d s − r v ( 1 ) ) e − M δ t δ + ∫ 0 t s δ − 1 f u ( s ) e M δ ( s δ − t δ ) d s , t ∈ [ 0,1 ] , z ( t ) = ( λ ∫ 0 1 v ( s ) d s − r u ( 1 ) ) e − M δ t δ + ∫ 0 t s δ − 1 f v ( s ) e M δ ( s δ − t δ ) d s , t ∈ [ 0,1 ] .

for u , v ∈ D and u ≤ v from Lemma 2.2 and Lemma 2.3. Where f v ( t ) = f ( t , v ( t ) ) + M v ( t ) , f u ( t ) = f ( t , u ( t ) ) + M u ( t ) . Define operator T : D × D → C ( [ 0,1 ] , ℝ ) × C ( [ 0,1 ] , ℝ )

T ( u , v ) ( t ) = ( T 1 ( u , v ) , T 2 ( u , v ) ) ,

where operators T 1 , T 2 are given by

( T 1 ( u , v ) = ( λ ∫ 0 1 u ( s ) d s − r v ( 1 ) ) e − M δ t δ + ∫ 0 t s δ − 1 f u ( s ) e M δ ( s δ − t δ ) d s , t ∈ [ 0,1 ] , T 2 ( u , v ) = ( λ ∫ 0 1 v ( s ) d s − r u ( 1 ) ) e − M δ t δ + ∫ 0 t s δ − 1 f v ( s ) e M δ ( s δ − t δ ) d s , t ∈ [ 0,1 ] .

respectively. Then the fixed point of operator T in D × D means the coupled solutions of (1).

Let y 1 = T 1 ( y 0 , z 0 ) , z 1 = T 2 ( y 0 , z 0 ) .

Here we prove that y 0 ≤ y 1 , z 1 ≤ z 0 , y 1 ≤ z 1 , and y 1 , z 1 are coupled lower and upper solutions of (1).

Whereas

( y 1 ( δ ) ( t ) = f ( t , y 0 ( t ) ) − M ( y 1 ( t ) − y 0 ( t ) ) , t ∈ ( 0,1 ) , y 1 ( 0 ) + r z 0 ( 1 ) = λ ∫ 0 1 y 0 ( s ) d s . z 1 ( δ ) ( t ) = f ( t , z 0 ( t ) ) − M ( z 1 ( t ) − z 0 ( t ) ) , t ∈ ( 0,1 ) , z 1 ( 0 ) + r y 0 ( 1 ) = λ ∫ 0 1 z 0 ( s ) d s . (5)

And y 0 , z 0 are coupled lower and upper solutions of (1), then we have

( ( y 0 ( δ ) ( t ) − y 1 ( δ ) ( t ) ) + M ( y 0 ( t ) − y 1 ( t ) ) ≤ 0, y 0 ( 0 ) − y 1 ( 0 ) ≤ 0, ( z 1 ( δ ) ( t ) − z 0 ( δ ) ( t ) ) + M ( z 1 ( t ) − z 0 ( t ) ) ≤ 0, z 1 ( 0 ) − z 0 ( 0 ) ≤ 0.

for t ∈ [ 0,1 ] . And by Lemma 2.5, we have

y 0 ( t ) ≤ y 1 ( t ) , z 1 ( t ) ≤ z 0 ( t ) , t ∈ [ 0,1 ] .

So we can easily get that

( y 1 ( δ ) ( t ) = f ( t , y 0 ( t ) ) − M ( y 1 ( t ) − y 0 ( t ) ) ≤ f ( t , y 1 ( t ) ) , y 1 ( 0 ) + r z 1 ( 1 ) ≤ λ ∫ 0 1 y 1 ( s ) d s , z 1 ( δ ) ( t ) = f ( t , z 0 ( t ) ) − M ( z 1 ( t ) − z 0 ( t ) ) ≥ f ( t , z 1 ( t ) ) , z 1 ( 0 ) + r y 1 ( 1 ) ≥ λ ∫ 0 1 z 1 ( s ) d s .

from formula (3) and (5). i.e., y 1 , z 1 are coupled lower and upper solutions of (1).

We also get that

( y 1 ( δ ) ( t ) − z 1 ( δ ) ( t ) ≤ − M ( y 1 ( t ) − z 1 ( t ) ) , y 1 ( 0 ) − z 1 ( 0 ) = λ ∫ 0 1 ( y 0 ( s ) − z 0 ( s ) ) d s + r ( y 0 ( 1 ) − z 0 ( 1 ) ) ≤ 0.

from formula (5) and y 0 ≤ z 0 . Similarly, we have y 1 ≤ z 1 . by Lemma 2.5.

Let y n = T 1 ( y n − 1 , z n − 1 ) , z n = T 2 ( y n − 1 , z n − 1 ) , then from formula (4), we have that y n , z n are coupled lower and upper solutions of (1) for any n ≥ 2 , which is similar to the proof above. And

y n − 1 ≤ y n ≤ z n ≤ z n − 1 .

In summary, we have

y 0 ( t ) ≤ y 1 ( t ) ≤ ⋯ ≤ y n ( t ) ≤ ⋯ ≤ z n ( t ) ≤ ⋯ ≤ z 1 ( t ) ≤ z 0 ( t ) ;

for t ∈ [ 0,1 ] . Therefore, function sequences { y n ( t ) } , { z n ( t ) } are uniformly bounded, i.e.,

‖ y n ‖ ≤ M 0 , ‖ z n ‖ ≤ M 0 .

for n = 0 , 1 , 2 , ⋯ and M 0 > 0 . Because f is continuous, we have

| f y ( n − 1 ) ( t ) | ≤ M 1 .

for n = 1 , 2 , 3 , ⋯ , t ∈ [ 0 , 1 ] and M 1 > 0 . In addition, because that functions e M δ ( s δ − t δ ) and e − M δ t δ are continuous, we have

| y n ( t 2 ) − y n ( t 1 ) | = | ( λ ∫ 0 1 y n − 1 ( s ) d s − r z n − 1 ( 1 ) ) ( e − M δ t 2 δ − e − M δ t 1 δ ) + ∫ 0 t 2 s δ − 1 f y n − 1 ( s ) e M δ ( s δ − t 2 δ ) d s − ∫ 0 t 1 s δ − 1 f y n − 1 ( s ) e M δ ( s δ − t 1 δ ) d s | ≤ | ( λ ∫ 0 1 y n − 1 ( s ) d s − r z n − 1 ( 1 ) ) | | e − M δ t 2 δ − e − M δ t 1 δ | + | ∫ 0 t 1 s δ − 1 f y n − 1 ( s ) ( e M δ ( s δ − t 2 δ ) − e M δ ( s δ − t 1 δ ) ) d s + ∫ t 1 t 2 s δ − 1 f y n − 1 ( s ) e M δ ( s δ − t 2 δ ) d s |

≤ | ( λ ∫ 0 1 y n − 1 ( s ) d s − r z n − 1 ( 1 ) ) | | e − M δ t 2 δ − e − M δ t 1 δ | + | ∫ 0 t 1 s δ − 1 f y n − 1 ( s ) | e M δ ( s δ − t 2 δ ) − e M δ ( s δ − t 1 δ ) | d s | + | ∫ t 1 t 2 s δ − 1 f y n − 1 ( s ) e M δ ( s δ − t 2 δ ) d s | → 0

if 0 ≤ t 1 < t 2 ≤ 1 and t 2 → t 1 . Hence, { y n ( t ) } is equicontinuous, we can also get that { z n ( t ) } is equicontinuous similarly.

In summary, by Ascoli-Arzela theorem [

and

if

2) Here we prove that

Assume that

Whereas

Consider that

And from Definition 2.3, we have that

Then from (3), we get that

In that way, we have

according to Lemma 2.5. By Mathematical Induction, we can get

for

if

for

3) Here we prove that if x is the solution of (1) in D, then

In summary, Theorem 3.1 is proved.

Theorem 3.2. Assume that

Proof By Theorem 3.1, we get that

If

considering the convergence of iterative sequences. Let

On the basis of (1), we can also consider the existence of solutions of boundary value problems for the following uniform fractional differential equations:

where

To illustrate our main results, we present the following example.

Example 4.1. Consider the boundary value problem of conformable fractional differential equations under the following new definitions

It is obvious that

Example 4.2. Consider the boundary value problem of conformable fractional differential equations under the following new definitions

where

which yield to

Therefore,

The author declares no conflicts of interest regarding the publication of this paper.

Jian, X.Y. (2019) Existence of Solutions for Boundary Value Problems of Conformable Fractional Differential Equations. Journal of Applied Mathematics and Physics, 7, 233-242. https://doi.org/10.4236/jamp.2019.71019