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Solvability of differential equations of order 2<α3 involving the p-Laplacian operator with boundary conditions

Abstract

In this paper, we study the existence of solutions for non-linear fractional differential equations of order 2<α3 involving the p-Laplacian operator with various boundary value conditions including an anti-periodic case. By using the Banach contraction mapping principle, we prove that, under certain conditions, the suggested non-linear fractional boundary value problem involving the p-Laplacian operator has a unique solution for both cases of 0<p<1 and p2. Finally, we illustrate our results with some examples.

1 Introduction

Recently, boundary value problems for fractional differential equations have gained popularity among researchers since they have many applications in biophysics, blood flow phenomena, aerodynamics, polymer rheology, viscoelasticity, thermodynamics, electro-dynamics of complex medium, capacitor theory, electrical circuits, electro-analytical chemistry, control theory (see [18] and [9]). Various results on boundary value problems for fractional differential equations have appeared in the literature (see [1013] and [14]). Also, solvability of fractional differential equations with anti-periodic boundary value problems have been considered by different authors ([15, 16] and [17]). On the other hand, in studying turbulent flow in a porous medium, Leibenson introduced the concept of p-Laplacian operator [18] and it was used for fractional boundary value problems in [19] and [20]. In this paper, we consider boundary value problems for fractional differential equations of order 2<α3 involving the p-Laplacian operator with various boundary conditions including an anti-periodic case.

Now, we present basic definitions and results that will be needed in the rest of the paper. More detailed information about the theory of fractional calculus and fractional differential equations can be found in [1, 2, 4] and [21].

It is well known that the beta function B(t,s) has the following integral representation:

B(t,s)= 0 1 τ t 1 ( 1 τ ) s 1 dτ,t,s>0.

Moreover, B(t,s) can be expressed in terms of Γ(t), the gamma function, as follows:

B(t,s)= Γ ( t ) Γ ( s ) Γ ( t + s ) .
(1)

Definition 1 Let α be a positive real number. Then the fractional integral of x:(0,)R is defined by

I 0 + α x(t)= 1 Γ ( α ) 0 t x ( s ) ( t s ) 1 α ds.

Definition 2 Let α be a positive real number. Then the Caputo fractional derivative of x:(0,)R is defined by

C 0 + α x(t)= 1 Γ ( n α ) 0 t x ( n ) ( s ) ( t s ) α + 1 n ds,

where

n= { [ α ] + 1 if  α N 0 , α if  α N 0
(2)

and [α] denotes the greatest integer less than or equal to α.

Recall that C n ([a,b],R) is the space of all real-valued functions x(t) which have continuous derivatives up to order n1 on [a,b].

In the following lemmas, we give some auxiliary results which will be used in the sequel.

Lemma 1 [1]

Let α>0 and y(t) C n ([0,1],R). Then

( I 0 + α C 0 + α x ) (t)=x(t) k = 0 n 1 x ( k ) ( 0 ) k ! t k ,

where n is given in (2).

On the other hand, the operator φ p (s)=|s | p 2 s, where p>1, is called the p-Laplacian operator. It is easy to see that φ p 1 = φ q , where 1 p + 1 q =1. The following properties of the p-Laplacian operator will play an important role in the rest of the paper.

Lemma 2 Let φ p be a p-Laplacian operator.

  1. (i)

    If 1<p<2, xy>0, and |x|,|y|m>0, then

    | φ p (x) φ p (y)|(p1) m p 2 |xy|.
    (3)
  2. (ii)

    If p2 and |x|,|y|M, then

    | φ p (x) φ p (y)|(p1) M p 2 |xy|.
    (4)

In this paper, we focus on the solvability of the following non-linear fractional differential equations of order α(2,3] involving the p-Laplacian operator with boundary conditions:

{ ( φ p ( C 0 + α x ( t ) ) ) = f ( t , x ( t ) ) , x ( 0 ) = a 0 x ( 1 ) , x ( 0 ) = a 1 x ( 1 ) , x ( 0 ) = a 2 x ( 1 ) ,
(5)

where a i 1, i=0,1,2, t[0,1], fC([0,1]×R,R) and x(t) C 2 ([0,1],R).

Remark 1 For a i =1, i=0,1,2, the boundary value problem given in (5) becomes anti-periodic.

Lemma 3 Assume that α(2,3], a 0 , a 1 , a 2 1, t[0,1] and hC([0,1]). Then the solution x(t) of the boundary value problem

{ ( φ p ( C 0 + α x ( t ) ) ) = h ( t ) , x ( 0 ) = a 0 x ( 1 ) , x ( 0 ) = a 1 x ( 1 ) , x ( 0 ) = a 2 x ( 1 )
(6)

can be represented by the following integral equation:

x ( t ) = 1 Γ ( α ) 0 t ( t τ ) α 1 φ q ( 0 τ h ( s ) d s ) d τ × A 0 Γ ( α ) 0 1 ( 1 τ ) α 1 φ q ( 0 τ h ( s ) d s ) d τ + [ A 0 A 1 + t A 1 Γ ( α 1 ) ] 0 1 ( 1 τ ) α 2 φ q ( 0 τ h ( s ) d s ) d τ + [ 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) 2 Γ ( α 2 ) ] 0 1 ( 1 τ ) α 3 φ q ( 0 τ h ( s ) d s ) d τ ,
(7)

where A 0 = a 0 1 a 0 , A 1 = a 1 1 a 1 and A 2 = a 2 1 a 2 .

Proof Using (6) and the fact that φ p ( C 0 + α x(0))=0, we have

φ p ( C 0 + α x ( t ) ) = 0 t h(s)ds,

or equivalently,

C 0 + α x(t)= φ q ( 0 t h ( s ) d s ) ,
(8)

where 1 p + 1 q =1. Applying the fractional integral operator I 0 + α to both sides of (8), we get

x(t)x(0) x (0)t x ( 0 ) 2 t 2 = 1 Γ ( α ) 0 t ( t τ ) α 1 φ q ( 0 τ h ( s ) d s ) dτ,

or equivalently,

x(t)= 1 Γ ( α ) 0 t ( t τ ) α 1 φ q ( 0 τ h ( s ) d s ) dτ+x(0)+ x (0)t+ x ( 0 ) 2 t 2 ,
(9)
x (t)= 1 Γ ( α 1 ) 0 t ( t τ ) α 2 φ q ( 0 τ h ( s ) d s ) dτ+ x (0)+ x (0)t,
(10)

and

x (t)= 1 Γ ( α 2 ) 0 t ( t τ ) α 3 φ q ( 0 τ h ( s ) d s ) dτ+ x (0).
(11)

Taking t=1 on both sides of (9), (10) and (11), we have

x(1)= 1 Γ ( α ) 0 1 ( 1 τ ) α 1 φ q ( 0 τ h ( s ) d s ) dτ+x(0)+ x (0)+ x ( 0 ) 2 ,
(12)
x (1)= 1 Γ ( α 1 ) 0 1 ( 1 τ ) α 2 φ q ( 0 τ h ( s ) d s ) dτ+ x (0)+ x (0),
(13)
x (1)= 1 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 φ q ( 0 τ h ( s ) d s ) dτ+ x (0).
(14)

Using equations (12), (13), (14) and the boundary value conditions

x ( 0 ) = a 0 x ( 1 ) , x ( 0 ) = a 1 x ( 1 ) , x ( 0 ) = a 2 x ( 1 ) ,

we can get that

x (0)= A 2 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 φ q ( 0 τ h ( s ) d s ) dτ,
(15)
x ( 0 ) = A 1 Γ ( α 1 ) 0 1 ( 1 τ ) α 2 φ q ( 0 τ h ( s ) d s ) d τ + A 2 A 1 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 φ q ( 0 τ h ( s ) d s ) d τ
(16)

and

x ( 0 ) = A 0 Γ ( α ) 0 1 ( 1 τ ) α 1 φ q ( 0 τ h ( s ) d s ) d τ + A 0 A 1 Γ ( α 1 ) 0 1 ( 1 τ ) α 2 φ q ( 0 τ h ( s ) d s ) d τ + A 0 A 1 A 2 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 φ q ( 0 τ h ( s ) d s ) d τ × A 0 A 2 2 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 φ q ( 0 τ h ( s ) d s ) d τ .
(17)

Substituting (15), (16) and (17) into (9) gives (7) and this completes the proof. □

2 Solvability of the fractional boundary value problem

This section is devoted to the solvability of the fractional boundary value problem given in (5). First, we obtain conditions for existence and uniqueness of the solution x(t) of the fractional boundary value problem given in (5). Then, each result obtained here is illustrated by examples.

Recall that C[0,1], the space of continuous functions on [0,1] is a Banach space with the norm x= max t [ 0 , 1 ] |x(t)|. Now consider T i :C[0,1]C[0,1], i=0,1, with

T 0 x(t):= φ q ( 0 t f ( s , x ( s ) ) d s )

and

T 1 x ( t ) = 1 Γ ( α ) 0 t ( t τ ) α 1 x ( τ ) d τ + A 0 Γ ( α ) 0 1 ( 1 τ ) α 1 x ( τ ) d τ + ( A 0 A 1 + t A 1 Γ ( α 1 ) ) 0 1 ( 1 τ ) α 2 x ( τ ) d τ + ( 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) 2 Γ ( α 2 ) ) 0 1 ( 1 τ ) α 3 x ( τ ) d τ .

Then the operator T:C[0,1]C[0,1], defined by T= T 1 T 0 , is continuous and compact.

Theorem 1 Suppose 1<q<2, a 0 , a 1 , a 2 1, and the following conditions hold: λ>0, 0<δ< 2 2 q and d with

0 < d < [ ( 2 λ 2 q Γ ( δ ( q 2 ) + α + 2 ) ) / ( ( q 1 ) Γ ( δ ( q 2 ) + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( δ ( q 2 ) + α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] ) ]
(18)

such that

λδ t δ 1 f(t,x)for any (t,x)(0,1]×R
(19)

and

|f(t,x)f(t,y)|d|xy|for t[0,1] and x,yR.
(20)

Then boundary value problem (5) has a unique solution.

Proof Using inequality (19), we get

λ t δ 0 t f(s,x)dsfor any (t,x)[0,1]×R.

By (3) and (20), we have

| T 0 x ( t ) T 0 y ( t ) | = | φ q ( 0 t f ( s , x ( s ) ) d s ) φ q ( 0 t f ( s , y ( s ) ) d s ) | ( q 1 ) ( λ t δ ) q 2 | 0 t f ( s , x ( s ) ) d s 0 t f ( s , y ( s ) ) d s | ( q 1 ) λ q 2 t δ ( q 2 ) 0 t | f ( s , x ( s ) ) f ( s , y ( s ) ) | d s d ( q 1 ) λ q 2 t δ ( q 2 ) 0 t | x ( s ) y ( s ) | d s d ( q 1 ) λ q 2 t δ ( q 2 ) + 1 x y .
(21)

Moreover,

| T x ( t ) T y ( t ) | = | T 1 ( T 0 ( x ( t ) ) ) T 1 ( T 0 ( y ( t ) ) ) | = | 1 Γ ( α ) 0 t ( t τ ) α 1 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ + A 0 Γ ( α ) 0 1 ( 1 τ ) α 1 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ + ( A 1 ( A 0 + t ) Γ ( α 1 ) ) 0 1 ( 1 τ ) α 2 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ + ( 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) 2 Γ ( α 2 ) ) × 0 1 ( 1 τ ) α 3 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ | .
(22)

Finally, substituting (21) in (22), we get

| T x ( t ) T y ( t ) | d ( q 1 ) λ q 2 x y [ 1 Γ ( α ) 0 t ( t τ ) α 1 τ δ ( q 2 ) + 1 d τ + | A 0 Γ ( α ) | 0 1 ( 1 τ ) α 1 τ δ ( q 2 ) + 1 d τ + | A 1 ( A 0 + t ) Γ ( α 1 ) | 0 1 ( 1 τ ) α 2 τ δ ( q 2 ) + 1 d τ + | 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) 2 Γ ( α 2 ) | 0 1 ( 1 τ ) α 3 τ δ ( q 2 ) + 1 d τ ] .
(23)

Using the equality

0 t ( t τ ) α 1 τ δ ( q 2 ) + 1 dτ= t δ ( q 2 ) + α + 1 0 1 ( 1 τ ) α 1 τ δ ( q 2 ) + 1 dτ

in (23) one can write that

| T x ( t ) T y ( t ) | d ( q 1 ) λ q 2 x y [ t δ ( q 2 ) + α + 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ δ ( q 2 ) + 1 d τ + | A 0 Γ ( α ) | 0 1 ( 1 τ ) α 1 τ δ ( q 2 ) + 1 d τ + | ( A 0 A 1 + t A 1 ) Γ ( α 1 ) | 0 1 ( 1 τ ) α 2 τ δ ( q 2 ) + 1 d τ + | 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) 2 Γ ( α 2 ) | 0 1 ( 1 τ ) α 3 τ δ ( q 2 ) + 1 d τ ] = d ( q 1 ) λ q 2 x y B ( δ ( q 2 ) + 2 , α ) × [ t δ ( q 2 ) + α + 1 Γ ( α ) + | A 0 | Γ ( α ) + | ( A 0 A 1 + t A 1 ) | Γ ( α ) ( δ ( q 2 ) + α + 1 ) + | 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) | 2 Γ ( α ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] .

Using (1), we get that

| T x ( t ) T y ( t ) | d ( q 1 ) λ q 2 x y Γ ( δ ( q 2 ) + 2 ) Γ ( α ) Γ ( δ ( q 2 ) + α + 2 ) × [ 1 Γ ( α ) + | A 0 | Γ ( α ) + ( | A 0 | | A 1 | + | A 1 | ) Γ ( α ) ( δ ( q 2 ) + α + 1 ) + ( 2 | A 1 | | A 2 | ( | A 0 | + 1 ) + | A 2 | ( 1 + | A 0 | ) ) 2 Γ ( α ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] d ( q 1 ) λ q 2 x y Γ ( δ ( q 2 ) + 2 ) 2 Γ ( δ ( q 2 ) + 2 + α ) × [ 2 ( 1 + | A 0 | ) + 2 | A 1 | ( 1 + | A 0 | ) ( δ ( q 2 ) + α + 1 ) + | A 2 | ( 1 + | A 0 | ) ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] d ( q 1 ) λ q 2 x y Γ ( δ ( q 2 ) + 2 ) 2 Γ ( δ ( q 2 ) + 2 + α ) × ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( δ ( q 2 ) + α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] = K x y ,

where

K = d ( q 1 ) λ q 2 Γ ( δ ( q 2 ) + 2 ) 2 Γ ( δ ( q 2 ) + α + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( δ ( q 2 ) + α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] .
(24)

Combining (24) with (18) implies that 0<K<1, therefore T is a contraction. As a consequence of the Banach contraction mapping theorem [22] the boundary value problem given in (5) has a unique solution. □

Theorem 2 Suppose 1<q<2, a 0 , a 1 , a 2 1 and the following conditions hold for a fixed, λ>0, 0<δ< 2 2 q and d with

0 < d < [ ( 2 λ 2 q Γ ( δ ( q 2 ) + α + 2 ) ) / ( ( q 1 ) Γ ( δ ( q 2 ) + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( δ ( q 2 ) + α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] ) ]

such that

f(t,x)λδ t δ 1 for any (t,x)(0,1]×R

and

|f(t,x)f(t,y)|d|xy|for t[0,1] and x,yR.

Then boundary value problem (5) has a unique solution.

Proof The inequality f(t,x)λδ t δ 1 implies that λδ t δ 1 f(t,x). Therefore replace f(t,x) by f(t,x) in the proof of Theorem 1. □

Theorem 3 Suppose q>2, a 0 , a 1 , a 2 1, there exists a non-negative function g(x)L[0,1] with M:= 0 1 g(τ)dτ0 such that

|f(t,x)|g(t)for any (t,x)[0,1]×R
(25)

and there exists a constant d with

0<d< 2 Γ ( α + 2 ) ( q 1 ) M q 2 ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) α ( α + 1 ) ]
(26)

and

|f(t,x)f(t,y)|d|xy|for t[0,1] and x,yR.

Then boundary value problem (5) has a unique solution.

Proof Using (25), we get that

0 t |f ( τ , x ( τ ) ) |dτ 0 1 g(τ)dτ=M
(27)

for all t[0,1]. By the definition of the operator T 0 , one can write that

| T 0 x(t) T 0 y(t)|=| φ q ( 0 t f ( s , x ( s ) ) d s ) φ q ( 0 t f ( s , y ( s ) ) d s ) |.
(28)

As a consequence of (4), (27) and (28), we have

| T 0 x ( t ) T 0 y ( t ) | ( q 1 ) M q 2 | 0 t f ( s , x ( s ) ) d s 0 t f ( s , y ( s ) ) d s | ( q 1 ) M q 2 0 t | f ( s , x ( s ) ) f ( s , y ( s ) ) | d s d ( q 1 ) M q 2 0 t | x ( s ) y ( s ) | d s d ( q 1 ) M q 2 t x y .

Moreover,

| T x ( t ) T y ( t ) | = | T 1 ( T 0 ( x ( t ) ) ) T 1 ( T 0 ( y ( t ) ) ) | = | 1 Γ ( α ) 0 t ( t τ ) α 1 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ + A 0 Γ ( α ) 0 1 ( 1 τ ) α 1 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ + A 1 ( A 0 + t ) Γ ( α 1 ) 0 1 ( 1 τ ) α 2 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ + | 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) 2 Γ ( α 2 ) | × 0 1 ( 1 τ ) α 3 ( ( T 0 x ) ( τ ) ( T 0 y ) ( τ ) ) d τ | , | T x ( t ) T y ( t ) | d ( q 1 ) M q 2 x y [ 1 Γ ( α ) 0 t ( t τ ) α 1 τ d τ | T x ( t ) T y ( t ) | + | A 0 | Γ ( α ) 0 1 ( 1 τ ) α 1 τ d τ + | A 1 ( A 0 + 1 ) Γ ( α 1 ) | 0 1 ( 1 τ ) α 2 τ d τ | T x ( t ) T y ( t ) | + | 2 A 1 A 2 ( A 0 + t ) + A 2 ( t 2 + A 0 ) | 2 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 τ d τ ] | T x ( t ) T y ( t ) | d ( q 1 ) M q 2 x y [ t α + 1 Γ ( α ) 0 1 ( 1 τ ) α 1 τ d τ | T x ( t ) T y ( t ) | + | A 0 | Γ ( α ) 0 1 ( 1 τ ) α 1 τ d τ + | A 1 | ( | A 0 | + 1 ) Γ ( α 1 ) 0 1 ( 1 τ ) α 2 τ d τ | T x ( t ) T y ( t ) | + | A 2 | ( | A 0 | + 1 ) ( 1 + 2 | A 1 | ) 2 Γ ( α 2 ) 0 1 ( 1 τ ) α 3 τ d τ ] | T x ( t ) T y ( t ) | d ( q 1 ) M q 2 B ( α , 2 ) x y [ 1 Γ ( α ) + | A 0 | Γ ( α ) + | A 1 | ( | A 0 | + 1 ) ( α + 1 ) Γ ( α ) | T x ( t ) T y ( t ) | + | A 2 | ( | A 0 | + 1 ) ( 1 + 2 | A 1 | ) α ( α + 1 ) 2 Γ ( α ) ] | T x ( t ) T y ( t ) | d ( q 1 ) M q 2 x y Γ ( α + 2 ) [ ( 1 + | A 0 | ) ( 1 + | A 1 | ( α + 1 ) ) | T x ( t ) T y ( t ) | + | A 2 | ( | A 0 | + 1 ) ( 1 + 2 | A 1 | ) α ( α + 1 ) 2 ] | T x ( t ) T y ( t ) | d ( q 1 ) M q 2 Γ ( α + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( α + 1 ) ) 2 | T x ( t ) T y ( t ) | + | A 2 | ( 1 + 2 | A 1 | ) α ( α + 1 ) 2 ] x y | T x ( t ) T y ( t ) | K x y ,

where

K= d ( q 1 ) M q 2 2 Γ ( α + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) α ( α + 1 ) ] .

By (26), we get K<1, which implies that T is a contraction, therefore the boundary value problem given in (5) has a unique solution. □

In the present part, we illustrate our results by examples.

Example 1 Consider the following anti-periodic boundary value problem:

{ ( φ 7 3 ( C 0 5 2 x ( t ) ) ) = 4 t 2 ( 2 + sin ( π x 16 + ω ) ) , t ( 0 , 1 ) , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 1 ) ,
(29)

where

p= 7 3 ,α= 5 2 ,and a 0 = a 1 = a 2 =1.

Then q= 7 4 , | A 0 |=| A 1 |=| A 2 |= 1 2 and take δ=4, λ=1 and d= π 4 . Obviously,

[ ( 2 λ 2 q Γ ( δ ( q 2 ) + α + 2 ) ) / ( ( q 1 ) Γ ( δ ( q 2 ) + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( δ ( q 2 ) + α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] ) ] = 64 Γ ( 7 2 ) 297 = 40 π 99 > π 4 = d .

On the other hand,

δλ t δ 1 =4 t 3 4 t 2 ( 2 + sin ( π x 16 + ω ) ) =f(t,x).

Finally,

| f ( t , x ) f ( t , y ) | = | 4 t 2 ( 2 + sin ( π x 16 + ω ) ) 4 t 2 ( 2 + sin ( π y 16 + ω ) ) | = 4 t 2 | sin ( π x 16 + ω ) sin ( π y 16 + ω ) | 4 | ( π x 16 + ω ) ( π y 16 + ω ) | = π 4 | x y | .

Therefore, as a consequence of Theorem 1, the boundary value problem given in (29) has a unique solution.

Example 2 Consider the following anti-periodic boundary value problem:

{ ( φ 7 3 ( C 0 5 2 x ( t ) ) ) = 2 t ( 2 + sin ( x 2 π + ω ) ) , t ( 0 , 1 ) , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 1 ) , x ( 0 ) = x ( 1 ) ,
(30)

where

p= 7 3 ,α= 5 2 and a 0 = a 1 = a 2 =1.

Then, obviously, q= 7 4 , | A 0 |=| A 1 |=| A 2 |= 1 2 . Taking δ=2, λ=1 and d= 1 π , we have

[ ( 2 λ 2 q Γ ( δ ( q 2 ) + α + 2 ) ) / ( ( q 1 ) Γ ( δ ( q 2 ) + 2 ) ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( δ ( q 2 ) + α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) ( δ ( q 2 ) + α ) ( δ ( q 2 ) + α + 1 ) ] ) ] = 16 Γ ( 4 ) 9 Γ ( 3 2 ) [ 2 ( 1 + 3 2 ) + 6 ] = 64 33 π > 1 π = d .

On the other hand,

f(t,x)=2t ( 2 + sin ( x 2 π + ω ) ) 2t=δλ t δ 1 .

Finally,

| f ( t , x ) f ( t , y ) | = | 2 t ( 2 + sin ( x 2 π + ω ) ) + 2 t ( 2 + sin ( y 2 π + ω ) ) | 2 t | sin ( x 2 π + ω ) sin ( y 2 π + ω ) | 2 | ( x 2 π + ω ) ( y 2 π + ω ) | = 1 π | x y | .

Therefore, as a consequence of Theorem 2, the boundary value problem given in (30) has a unique solution.

Example 3 Now consider the following boundary value problem:

{ ( φ 9 5 ( C 0 7 3 x ( t ) ) ) = sin 2 ( π x 40 + ω ) , t ( 0 , 1 ) , x ( 0 ) = 1 5 x ( 1 ) , x ( 0 ) = 1 2 x ( 1 ) , x ( 0 ) = 1 2 x ( 1 ) ,
(31)

where

p= 9 5 ,α= 7 3 , a 0 = 1 5 and a 1 = a 2 = 1 2 .

Then q= 9 4 , | A 0 |= 1 6 and | A 1 |=| A 1 |=1. Also, taking d= π 20 and g(t)=1, we have

M=1

and

[ 2 Γ ( α + 2 ) ( q 1 ) M q 2 ( 1 + | A 0 | ) [ 2 ( 1 + | A 1 | ( α + 1 ) ) + | A 2 | ( 1 + 2 | A 1 | ) α ( α + 1 ) ] ] = 48 Γ ( 13 3 ) 35 [ 2 ( 1 + 10 3 ) + ( 3 7 3 10 3 ) ] = 4 Γ ( 4 3 ) 9 > π 20 .

On the other hand,

| f ( t , x ) f ( t , y ) | | sin 2 ( π x 40 + ω ) sin 2 ( π y 40 + ω ) | π 20 | x y | for  t [ 0 , 1 ]  and  x , y R ;

therefore, by Theorem 3, the boundary value problem given in (31) has a unique solution.

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Acknowledgements

We would like to thank the anonymous referees for their valuable suggestions.

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Correspondence to Mehmet Ali Özarslan.

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Aktuğlu, H., Özarslan, M.A. Solvability of differential equations of order 2<α3 involving the p-Laplacian operator with boundary conditions. Adv Differ Equ 2013, 358 (2013). https://doi.org/10.1186/1687-1847-2013-358

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Keywords

  • p-Laplacian operators
  • fractional derivative
  • fractional integral
  • Caputo fractional derivative
  • boundary value problem
  • Caputo fractional boundary value problem
  • anti-periodic boundary value problem