Existence of Solutions for Boundary Value Problems of Conformable Fractional Differential Equations

(1)

ISSN Online: 2327-4379 ISSN Print: 2327-4352

DOI: 10.4236/jamp.2019.71019 Jan. 29, 2019 233 Journal of Applied Mathematics and Physics

Existence of Solutions for Boundary Value

Problems of Conformable Fractional

Differential Equations

Xingyue Jian

College of Science, University of Shanghai for Science and Technology, Shanghai, China

Abstract

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 contra-diction. Finally, a concrete example is given to illustrate the wide applicability of our main results.

Keywords

Boundary Value Problems, Conformable Fractional Differential Equations, The Method of Coupled Upper and Lower Solution, Coupled Solution, Monotone Iteration

1. Introduction

In recent years, there are few studies on boundary value problems of conforma-ble fractional differential equations under new definitions [1] [2] [3]. And con-formable fractional derivatives not only have good operational properties (Four Operational Rules of Derivatives, Chain Rule and Leibniz Rule), this definition can also construct fractional Newton equation and Euler-Lagrange equation from fractional variational method, this is of great significance to the study of uniform or uniformly accelerated motion of particles and to the solution of Newton’s fractional-order mechanical problems [4] [5] (fractional-order har-monic oscillator, fractional-order damped oscillator and forced oscillator). And the method of upper and lower solution for monotone iteration can not only How to cite this paper: Jian, X.Y. (2019)

Existence of Solutions for Boundary Value Problems of Conformable Fractional Diffe-rential Equations. Journal of Applied Ma-thematics and Physics, 7, 233-242.

https://doi.org/10.4236/jamp.2019.71019

Received: January 7, 2019 Accepted: January 26, 2019 Published: January 29, 2019

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DOI: 10.4236/jamp.2019.71019 234 Journal of Applied Mathematics and Physics

gives the existence theorem, but also determines the value range of the solution. Therefore, this method has gradually become an important method for studying nonlinear differential equations [6] [7] [8] [9]. In addition, with the application of anti-periodic boundary value problems in various mathematical models and physical processes has been widely applied, the integral boundaries are also widely used in heat conduction, chemical engineering, groundwater flow, ther-moelasticity, plasma physics and other fields. As a result, more and more studies have been made on this kind of problems [10] [11] [12] (anti-periodic boundary value problems, anti-periodic boundary value problems with integral bounda-ries). However, the indefinite sign of solutions of nonlinear differential equa-tions determines that some problems (anti-periodic boundary value problems and their generalizations) cannot be studied directly by the method of upper and lower solutions for monotone iteration. But the development of nonlinear analy-sis theory provides a powerful tool for the study of these problems. In the gene-ralized monotone iteration process, the method of coupled upper and lower so-lution becomes an important method to study this kind of problem by the flexi-ble construction of the comparison theorem [13] [14] [15] [16]. Motivated by the above work, in this paper, the existence of solutions for a class of boundary value problems of conformable fractional differential equations under a new de-finition is proved by using the method of coupled upper and lower solution, and the range of solutions is obtained. Throughout this paper, we consider the exis-tence of solutions of boundary value problems for the following uniform frac-tional differential equations

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

1

( )

0

, , 0,1 ,

0 1 d

x t f t x t t

x rx x s s

δ

λ

 = ∈

 

= − +



∫

(1)

where x( )δ

( )

t

is the conformable fractional derivatives of order δ for

( )

0,1

t∈ _{which is defined in}_[1]_{, and}δ ∈

(

0,1

]

_, r>0, r>0, = −∞ +∞

(

,

)

,

[ ]

: 0,1

f × →_ __{is continuous.}

2. Preliminaries

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

Definition 2.1. (See [1]) Given a function x: 0,

[

+∞ →

)

__{. Then the}

con-formable fractional derivative of x of order δ is defined by

( )

_{( )}

(

1

)

( )

0

: limx t t x t ,

x t

δ δ

ε

ε 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 lim₀ ( )

( )

t x t

δ

→ exists, then we define

( )

_{( )}

( )

_{( )}

0

0 : lim .

t

xδ xδ t

→ =

(3)

are said to be coupled lower and upper solutions of (1), respectively, if

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

0 0

1

0 0 ₀ 0

0 0

1

0 0 ₀ 0

, , 0,1 ,

0 1 d .

, , 0,1 ,

0 1 d .

y t f t y t t

y rz y s s

z t f t z t t

z ry z s s

δ

λ

 ≤ ∈



 + ≤

 

≥ ∈



 ₊ _≥



∫

Definition 2.3. Let y z C, ∈

(

[ ]

0,1 ,_

)

, then the function pair

( )

y z, is said to be coupled solutions of (1), if

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

1 0

, , 0,1 ,

0 1 d .

, , 0,1 ,

0 1 d .

y t f t y t t

y rz y s s

z t f t z t t

z ry z s s

δ

λ

 ₌ _∈



 + =

 

= ∈



 ₊ ₌



∫

Let γ ρ ∈, C

(

[ ]

0,1 ,

)

, then

(

ρ γ,

)

_{is said to be minimum and maximum}

coupled solutions of (1), if

(

ρ γ,

)

_{are coupled solutions of (1), and}

( )

t y t z t

( ) ( )

,

( )

t

ρ ≤ ≤γ _{for any coupled solution}

( )

y z, .

Lemma 2.1. (See [1]) Let δ ∈

(

0,1

]

_,_{and assume} x x₁, ₂ to be δ-differentiable,

then

1)

(

ax bx1 2

)

( )

t ax1( )

( )

t bx2( )

( )

t

δ δ δ

+ = + ;

2)

(

x x1 2

)

( )

t x t x1

( )

2( )

( )

t x t x2

( )

1( )

( )

t

δ ₌ δ ₊ δ _;

3) ( )

( )

2

( )

1( )

( )

_{( )}

1

( )

2( )

( )

1

2

2 2

x t x t x t x t

x _t

x x t

δ δ ₋ δ

 

=

 

  .

for t∈

( )

0,1 _, a b, ∈.

Lemma 2.2. (See[1]) If x is differentiable, t>0, then ( )

( )

1 d

( )

d

x

x t t t

t

δ ₌ −δ _.

Lemma. 2.3 (See [3]) If x( )δ

( )

t

exists, then for t≠0_,_{we have}

( )

_{( )}

1

_{( )}

xδ t ₌t x t−δ _′ _.

Lemma 2.4. Assume that g C∈

(

[ ]

0,1 ,_

)

_,_and δ ∈

(

0,1

]

_, m∈__, M>0_,

Define function p: 0,1

[ ]

→__{as follows:}

( )

1

( )

(

)

0

e Mt t eM s t d .

p t m δ δ s g sδ δ δ δ s

− ₋

−

= +

_∫

(2) Then p t

( )

is the solution of the initial value problem as follows

( )

_{( )}

₍

_]

( )

, 0,1 , 0

p t Mp t g t t

p m

δ

 ₊ ₌ _∈

 

= 

Proof Assume that p t

( )

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

( )

_{( )}

( )

(

)

( )

1 1 1

0

1 0

e e e

e e d

.

Mt Mt Ms

t

M_t M _s _t

t

p t t t mM t M s g s g t

M m s g s s g t

Mp t g t

δ δ δ

δ δ δ δ δ δ δ δ

δ

δ δ

− −

− − −

− ₋

−

 

 

= − − +

 

 

 

= − _ + _+

 

= − +

∫

(4)

from Lemma 2.2, and p t

( )

subject to the condition

( )

0 .

p =m



Lemma 2.5. (Comparison Theorem) Let p C∈

(

[ ]

0,1 ,

)

,M >0, and the following inequalities hold true

( )

_{( )}

_{[ ]}

( )

0, 0,1 , 0 0.

p t Mp t t

p

δ

 + ≤ ∈

 

≤  then p t

( )

≤0_{, for} t∈

[ ]

0,1 _.

Proof Let p( )δ

( )

t ₊Mp 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.}

3. Conclusions

Theorem 3.1. Assume that y t z t0

( ) ( )

, 0 are coupled lower and upper solutions of (1.1) with y t0

( )

≤z t0

( )

for t∈

[ ]

0,1 , let

[ ]

(

)

( )

{

0,1 , | 0 0

}

D= x C∈  y t ≤ ≤x z t . And if y t0

( )

≤x2≤x1≤z t0

( )

, then the following inequalities hold true

( )

, 1

(

, 2

)

(

1 2

)

.

f t x − f t x ≥ −M x x−

(3) for t∈

[ ]

0,1 _and M >0_{. If we take} y t z t₀

( ) ( )

, ₀ as initial elements, the itera-tive sequences defined by

( )

(

( )

)

( )

(

)

[ ]

( )

(

( )

)

( )

(

)

[ ]

1

1 ₁

1 1

0 0

1 ₁

1 1

0 0

d 1 e e d , 0,1

n

M_t M _{s t}

t

n n n y

M_t M _{s t}

t

n n n z

y t y s s rz s f s s t

z t z s s ry s f s s t

δ δ δ

δ

δ δ

δ

δ δ

λ

−

− ₋

−

− −

− ₋

−

− −



= − + ∈

 

 ₌ ₋ ₊ _∈



∫

(4)

are

{

y tn

( )

}

and

{

z tn

( )

}

, then

1)

( )

*

( )

n

y t →y t _and

( )

*

( )

n

z 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

( )

_{( )}

₍

_{( )}

₎

₍

_{( ) ( )}

₎

_{( )}

( )

_{( )}

₍

_{( )}

₎

₍

_{( ) ( )}

₎

_{( )}

( )

1 0

, , 0,1 ,

0 1 d .

, , 0,1 ,

0 1 d ,

y t f t u t M y t u t t

y rv u s s

z t f t v t M z t v t t

z ru v s s

δ

λ

 = − − ∈



 ₊ ₌

 

= − − ∈

 

 + =



∫

(5)

( )

(

( )

)

( )

(

)

[ ]

( )

(

( )

)

( )

(

)

[ ]

1 ₁ 0 0 1 ₁ 0 0

d 1 e e d , 0,1 ,

d 1 e e d , 0,1 .

M_t M _s _t

t u

M_t M _s _t

t v

y t u s s rv s f s s t

z t v s s ru s f s s t

δ δ δ

δ δ δ δ δ δ λ λ − ₋ − − ₋ −  = − + ∈    ₌ ₋ ₊ _∈ 

∫

for u v D, ∈ and u v≤ from Lemma 2.2 and Lemma 2.3. Where

( )

(

,

( )

)

( )

v

f t = f t v t +Mv t _, f tu

( )

= f t u t

(

,

( )

)

+Mu t

( )

. Define operator

[ ]

(

)

(

[ ]

)

: 0,1 , 0,1 ,

T D D× →C  ×C 

( )( )

,

(

1

( ) ( )

, , 2 ,

)

,

T u v t = T u v T u v

where operators T T1, 2 are given by

( )

(

( )

)

( )

(

)

[ ]

( )

(

( )

)

( )

(

)

[ ]

1 ₁

1 ₀ ₀

1 ₁

2 ₀ ₀

, d 1 e e d , 0,1 ,

, d 1 e e d , 0,1 .

M_t M _{s t}

t u

M_t M _{s t}

t v

T u v u s s rv s f s s t

T u v v s s ru s f s s t

δ δ δ

δ δ δ δ δ δ λ λ − ₋ − − ₋ −  = − + ∈    ₌ ₋ ₊ _∈ 

∫

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

Let y T y z1= 1

(

0, 0

)

,z T y z1= 2

(

0, 0

)

.

Here we prove that y0≤ y z1, 1≤z y0, 1≤z1, and y z1, 1 are coupled lower and

upper solutions of (1). Whereas

( )

_{( )}

₍

_{( )}

₎

₍

_{( )}

₎

_{( )}

( )

_{( )}

₍

_{( )}

₎

₍

_{( )}

₎

_{( )}

( )

1 0 1 0

1

1 0 ₀ 0

1 0 1 0

1

1 0 ₀ 0

, , 0,1 ,

0 1 d .

, , 0,1 ,

0 1 d .

y t f t y t M y t y t t

y rz y s s

z t f t z t M z t z t t

z ry z s s

δ δ λ λ  ₌ ₋ ₋ _∈   ₊ ₌   = − − ∈    + = 

∫

(5)

And y z0, 0 are coupled lower and upper solutions of (1), then we have

( )

_{( )}

( )

_{( )}

(

)

(

( )

)

( )

_{( )}

( )

_{( )}

(

)

(

( )

)

( )

0 1 0 1

0 1

1 0 1 0

1 0

0,

0 0 0,

0,

0 0 0.

y t y t M y t y t

y y

z t z t M z t z t

z z δ δ δ δ  ₋ ₊ ₋ _≤   ₋ _≤   − + − ≤   − ≤ 

for t∈

[ ]

0,1 _{. And by Lemma 2.5, we have}

( )

( ) ( )

( )

[ ]

0 1 , 1 0 , 0,1 .

y t ≤ y t z t ≤z t t∈

So we can easily get that

( )

_{( )}

₍

_{( )}

₎

₍

_{( )}

₎

₍

_{( )}

₎

( )

_{( )}

₍

_{( )}

₎

₍

_{( )}

₎

₍

_{( )}

₎

( )

1 0 1 0 1

1

1 1 ₀ 1

1 0 1 0 1

1

1 1 ₀ 1

, , ,

0 1 d ,

, , ,

0 1 d .

y t f t y t M y t y t f t y t

y rz y s s

z t f t z t M z t z t f t z t

z ry z s s

δ δ λ λ  = − − ≤   ₊ _≤   = − − ≥    + ≥ 

∫

(6)

(1).

We also get that

( )

_{( )}

( )

_{( )}

₍

_{( )}

₎

( )

(

( )

)

(

( )

)

1 1 1 1

1

1 1 ₀ 0 0 0 0

,

0 0 d 1 1 0.

y t z t M y t z t

y z y s z s s r y z

δ δ

λ

 − ≤ − −

 

− = − + − ≤



∫

from formula (5) and y0≤z0. Similarly, we have y1≤z1. by Lemma 2.5.

Let yn=T y1

(

n−1,zn−1

)

,zn=T y2

(

n−1,zn−1

)

, then from formula (4), we have that

, n n

y z are coupled lower and upper solutions of (1) for any n≥2_{, which is}

similar to the proof above. And

1 1.

n n n n

y− ≤ y ≤z ≤z−

In summary, we have

( )

0 1 n n 1 0 ;

y t ≤ y t ≤≤y t ≤≤z t ≤≤z t ≤z t

for t∈

[ ]

0,1 _{. Therefore, function sequences}

{

y tn

( )

}

,

{

z tn

( )

}

are uniformly

bounded, i.e.,

0, 0.

n n

y ≤M z ≤M

for n=0,1,2, and M0>0. Because f is continuous, we have

( )n1

( )

1.

y

f ₋ t ≤M

for n=1,2,3, , t∈

[ ]

0,1 and M1>0. In addition, because that functions

(

)

eM s tδ δ−δ _and _eδM tδ −

are continuous, we have

( )

(

)

( )

(

)

( )

(

)

( )

(

)

( )

(

)

(

)

( )

2 1

1 1

2 1

1 2

1 ₁ 1

2 1

1

1 1

0

1 1

0 0

1

1 1

0

1 1

0

d 1 e e

e d e d

d 1 e e

e e d e

n n

M_t M_t

n n

M _s _t M _s _t

t t

y y

M_t M_t

n n

M _s _t M _s _t M

t t

y _t y

y t y t

y s s rz

s f s s s f s s

y s s rz

s f s s s f s

δ δ

δ δ δ δ

δ δ

δ δ δ δ

δ δ

δ δ δ δ

δ δ

δ δ δ δ δ

λ

− −

−

 

= − _ − _

 

+ −

≤ − −

 

+ _ − _ +

 

∫

(

2

)

d s t

s

δ₋δ

( )

(

)

( )

(

)

( )

(

)

2 1

2 1 2

1 2

1 ₁ 1

1

1 1

0

( )

1 1

0

d 1 e e

e e d e d

0

n n

M_t M_t

n n

M _s _t M _s _t M _s _t

t t

y _t y

y s s rz

s f s s s f s s

δ δ

δ δ δ δ δ δ

δ δ

δ δ δ δ δ

λ

− −

− − −

− −

≤ − −

+ − +

→

∫

if 0≤ < ≤t t1 2 1 and t2→t1. Hence,

{

y tn

( )

}

is equicontinuous, we can also

get that

{

z tn

( )

}

is equicontinuous similarly.

In summary, by Ascoli-Arzela theorem [17], we can prove that

{ } { }

yn , zn

(7)

functions _{y z}*_, *_{, such that}

* *

lim n 0, lim n 0. n→∞ y −y = n→∞ z −z =

and _{y z}*_, *_∈_D_{. Next we take limits on both sides of (4), then from Lebesgue}

Dominated Convergence Theorem, we have

(

* *

)

(

* *

) (

* *

)

(

* *

)

1 2

, , , , , .

T y z = T y z T y z = y z

if n→ ∞. i.e.,

(

_{y z}*_, *

)

_{are coupled solutions of (1).}

2) Here we prove that

(

_{y z}*_, *

)

_{are coupled minimal and maximal solutions}

of (1) respectively in D.

Assume that

(

x x1, 2

)

are a set of coupled solutions of (1), then the above

problem is equivalent to prove that

* *

1, 2 .

y ≤x x ≤z

Whereas x x1, 2∈D, therefore y0≤x x1, 2≤z0. Assume that yk ≤x x1, 2≤zk

for k >1_{, here we prove that} yk+₁≤x x₁, ₂≤zk+₁.

Consider that

( )

_{( )}

₍

_{( )}

₎

₍

_{( )}

₎

_{( )}

( )

_{( )}

₍

_{( )}

₎

₍

_{( )}

₎

_{( )}

( )

1 1

1

1 ₀

1 1

1

1 ₀

, , 0,1 ,

0 1 d .

, , 0,1 ,

0 1 d .

k k k k

k k k

k k k k

k k k

y t f t y t M y t y t t

y rz y s s

z t f t z t M z t z t t

z ry z s s

δ

λ

+ +

+

+ +

+

 ₌ ₋ ₋ _∈



 ₊ ₌

 

= − − ∈

 

 + =



∫

And from Definition 2.3, we have that

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

1 1

1

1 2 ₀ 1

2 2

1

2 1 ₀ 2

, , 0,1 ,

0 1 d ,

, , 0,1 ,

0 1 d .

x t f t x t t

x rx x s s

x t f t x t t

x rx x s s

δ

λ

 ₌ _∈



 ₊ ₌

 

= ∈

 

 + =



∫

Then from (3), we get that

( )

_{( )}

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

_{( )}

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

1 1 1 1

1 1

1 2 1 2

1 2

0, 0,1

0 0 0,

0, 0,1

0 0 0.

k k

k

k k

k

x t y t M x t y t t

x y

z t x t M z t x t t

z x

δ δ

+ +

+

+ +

+

 − + − ≥ ∈



 − ≥

 

− + − ≥ ∈

 

− ≥



In that way, we have

1 1, 2 1,

k k

y+ ≤x x ≤z+

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

1, 2 .

n n

y ≤x x ≤z

(8)

* *

1, 2 ,

y ≤x x ≤z

if n→ ∞. i.e.,

( )

( ) ( )

( )

* *

1 , 2 ,

y t ≤x t x t ≤z t

for t∈

[ ]

0,1 _{. Therefore,}

(

_{y z}*_, *

)

_{are coupled minimal and maximal solutions}

of (1) respectively in D from Definition 2.3.

3) Here we prove that if x is the solution of (1) in D, then _y*_{≤ ≤}_{x z}*_{. In}

conclusion (2) above, let x t1

( )

=x t2

( )

=x t

( )

, because that x is the solution of

(1) in D, therefore,

(

x x1, 2

)

are a set of coupled solutions of (1). Obviously, x

subject to

* *_.

y ≤ ≤x z

In summary, Theorem 3.1 is proved.

Theorem 3.2. Assume that f t x

( )

, is increasing in x on

[ ]

(

)

( )

{

0,1 , | 0 0

}

D= x C∈  y t ≤ ≤x z t , and 1> >r 0,λ>0, then there exists a unique solution of (1) in D=

{

x C∈

(

[ ]

0,1 ,

)

|y t0

( )

≤ ≤x z t0

( )

}

.

Proof By Theorem 3.1, we get that

( )

*

( )

n

y t →y t _and

( )

*

( )

n

z t →z t _{. And}

we have

( )

*

( )

*

( )

0 0

y t ≤y t ≤z t ≤z t for t∈

[ ]

0,1 . Then we have that

( )

* * ₀

y t −z t ≤ _for t∈

[ ]

0,1 _{. Here we prove that} _{y t}*

( )

₌_{z t}*

( )

_.

If f t x

( )

, is increasing in x on D=

{

x C∈

(

[ ]

0,1 ,

)

|y t0

( )

≤ ≤x z t0

( )

}

,

as-sume that _{y t}*

( )

₋_{z t}*

( )

_<₀_{, then we have}

( )

_{( )}

( )

_{( )}

(

_{( )}

)

(

_{( )}

)

( )

(

( )

)

(

( )

)

* *

1

* * * * * *

0

, , 0,

0 0 (1) 1 d .

y t z t f t y t f t z t

y z r y z y s z s s

δ δ

λ

∗ ∗

 − = − ≤

 

− = − + −



∫

considering the convergence of iterative sequences. Let _{h t}

( )

₌_{y t}*

( )

₋_{z t}*

( )

_,

then we have h( )δ

( )

t ₌y∗( )δ

( )

t ₋z∗( )δ

( )

t _≤0

by Lemma 2.2, i.e., the function

( )

*

( )

*

( )

h t =y t −z t is monotonically decreasing. Hence, h

( )

0 ≥h

( )

1 _,

there-fore, we draw a contradiction from the conclusion that h

( )

0 <h

( )

1 _{, which can}

be obtained from the condition 1> >r 0,λ>0 and the boundary value condi-tions above. Therefore, we have _{h t}

( )

₌ _{y t}*

( )

₋_{z t}*

( )

₌₀_,_i.e._, _{y t}*

( )

₌_{z t}*

( )

_is

the solution of (1).

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

( )

_{( )}

₍

_{( )}

₎

_{( )}

( )

_{( )}

( )

_{( )}

1

_{( )}

0

, , 0,1 ,

0 1 d , 1,2, , .

k k

x t f t x t t

x rx x s s k n

δ

λ

 ₌ _∈

 

= − + =



∫

(6)

where x( )δ

( )

t _{is the conformable fractional derivatives of order} _δ_for

( )

0,1

t∈ _{which is defined in}_[1]_{, and} δ ∈

(

n n, +1

]

_, n≥1,r>0,λ>0,

(

,

)

= −∞ +∞

, f : 0,1

[ ]

× →_ __{is continuous. Similarly, the existence of the}

solution can be proved by the method of coupled upper and lower solution, and the range of the solution can be obtained. Due to δ ∈

(

n n, +1

]

_{, so the original}

(9)

4. Examples

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

( )

(

( )

)

( )

1

2 2

1 0

1 3 , 0,1 ,

1 1

0 1 d .

3 2

x t t x t x t t

x x x s s

      

= − − ∈

 

 _{= −} ₊



∫

(7)

It is obvious that y t0

( )

= −1,z t0

( )

=1 are coupled lower and upper solutions

of (7), and from the condition _{f t x}

( )

_, ₌_t

(

₁₋_x2

)

₋₃_x_{, we can get that there}

ex-ists a constant M t x x≥

(

1+ 2

)

+ >3 0 for − ≤1 x2≤x1≤1, such that the

for-mula (4) of Theorem 3.1 holds. Hence, problem (4) has at least one solution

[

1,1

]

x∈ − _for t∈

[ ]

0,1 _{by Theorem 3.2.} 

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

( )

1 3

1 0

arctan , 0,1 ,

0 1 d .

x t x t t

x ax b x s s

      

= ∈

 

 ₊ ₌



∫

(8)

where 1> >a 0,b>0_{, it is easy to get that}

( )

[ ]

( )

[ ]

1 3 0

3_π 3 π 3 _, _{0,1 ,}

2 2 1 4

3_π 3 π 3 _, _{0,1 .}

2 2 1 4

y t t a b t

a b

z t t a b t

a b

 _ _

= − + + ∈

 _{+ −}  

  



 

 ₌ ₋ ₊ _∈

 

 _{+ − } _



which yield to

( )

1 3

0 0

1

0 0 ₀ 0

1 3

0 0

1

0 0 ₀ 0

π _arctan _, _{0,1 ,}

2

0 1 d ,

π _arctan _, _{0,1 ,}

2

0 1 d .

y t y t t

y az b y s s

z t z t t

z ay b z s s

     

      

− = ≤ ∈

 

 ₊ _≤

  

= ≥ ∈

 

 ₊ _≥



∫

Therefore, y t z t0

( ) ( )

, 0 are coupled lower and upper solutions of (8), it is

obvious that the formula (4) of Theorem 3.1 holds. Hence, problem (8) has at least one solution x∈ y t z t0

( ) ( )

, 0  for t∈

[ ]

0,1 by Theorem 3.2.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this pa-per.

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