Open Access

Solvability of fractional boundary value problem with p-Laplacian operator at resonance

Advances in Difference Equations20132013:295

https://doi.org/10.1186/1687-1847-2013-295

Received: 1 January 2013

Accepted: 2 September 2013

Published: 7 November 2013

Abstract

In this paper, a class of multi-point boundary value problems for nonlinear fractional differential equations at resonance with p-Laplacian operator is considered. By using the extension of Mawhin’s continuation theorem due to Ge, the existence of solutions is obtained, which enriches previous results.

MSC:34A08, 34B15.

Keywords

fractional differential equationboundary value problemp-Laplacian operatorcoincidence degree theoryresonance

1 Introduction

In the recent years, fractional differential equations played an important role in many fields such as physics, electrical circuits, biology, control theory, etc. (see [17]). Thus, many scholars have paid more attention to fractional differential equations and gained some achievements (see [822]). For example, Wang [21] considered a class of fractional multi-point boundary value problems at resonance by Mawhin’s continuation theorem (see [23]):
{ D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) ) , t ( 0 , 1 ) ,  a.e.  t ( 0 , 1 ) , u ( 0 ) = 0 , D 0 + α 1 u ( 1 ) = i = 1 m a i D 0 + α 1 u ( ξ i ) , D 0 + α 2 u ( 1 ) = i = 1 m b i D 0 + α 2 u ( η i ) ,
(1.1)

where 1 < α 2 , 0 < ξ 1 < ξ 2 < < ξ m < 1 , 0 < η 1 < η 2 < < η n < 1 , i = 1 m a i = 1 , i = 1 n b i = 1 , i = 1 n b i η i = 1 , D 0 + α is the standard fractional derivative, f : [ 0 , 1 ] × R 2 R satisfies the Carathéodory condition.

But Mawhin’s continuation theorem is not suitable for quasi-linear operators. In [24], Ge and Ren had extended Mawhin’s continuation theorem, which was used to deal with more general abstract operator equations. In [25], Pang et al. considered a higher order nonlinear differential equation with a p-Laplacian operator at resonance:
{ ( φ p ( u ( n 1 ) ( t ) ) ) = f ( t , u ( t ) , , u ( n 1 ) ( t ) ) + e ( t ) , t ( 0 , 1 ) , u ( i ) ( 0 ) = 0 , i = 1 , 2 , , n 1 , u ( 1 ) = 0 1 u ( s ) d g ( s ) ,
(1.2)

where φ p ( s ) = | s | p 2 s , p > 1 , f : [ 0 , 1 ] × R n R n and e : [ 0 , 1 ] R are continuous, n 2 is an integer. g : [ 0 , 1 ] R is a nondecreasing function with 0 1 d g ( s ) = 1 , the integral in the second part of (1.2) is meant in the Riemann-Stieltjes sense.

However, there are few articles which consider the fractional multi-point boundary value problem at resonance with p-Laplacian operator and dim Ker M = 2 , because p-Laplacian operator is a nonlinear operator, and it is hard to construct suitable continuous projectors. In this paper, we will improve and generalize some known results.

Motivated by the work above, our article is to investigate the multi-point boundary value problem at resonance for a class of Riemanne-Liouville fractional differential equations with p-Laplacian operator and dim Ker M = 2 by constructing suitable continuous projectors and using the extension of Mawhin’s continuation theorem:
{ D 0 + β φ p ( D 0 + α u ( t ) ) = f ( t , u ( t ) , D 0 + α 2 u ( t ) , D 0 + α 1 u ( t ) , D 0 + α u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = D 0 + α u ( 0 ) = 0 , u ( 1 ) = i = 1 m a i u ( ξ i ) , D 0 + α 1 u ( 1 ) = i = 1 m b i D 0 + α 1 u ( η i ) ,
(1.3)

where 2 < α 3 , 0 < β 1 , 3 < α + β 4 , 0 < ξ 1 < ξ 2 < < ξ m < 1 , 0 < η 1 < η 2 < < η m < 1 , a i R , b i R , 1 < m , m N , i = 1 m a i ξ i α 1 = 1 , i = 1 m a i ξ i α 2 = 1 , i = 1 m b i = 1 , φ p ( s ) = | s | p 2 s , φ p ( 0 ) = 0 , 1 < p , 1 / p + 1 / q = 1 , φ p is invertible and its inverse operator is φ q , D 0 + α is Riemann-Liouville standard fractional derivative, f : [ 0 , 1 ] × R 4 R is continuous.

In order to investigate the problem, we need to suppose that the following conditions hold:
Λ = Λ 1 Λ 4 Λ 2 Λ 3 0 ,
where
Λ 1 = Γ ( α ) q Γ ( α q + β q q α β + 2 ) Γ ( α + β ) q 1 Γ ( α q + β q q β + 2 ) ( 1 i = 1 m a i ξ i α q + β q q β + 1 ) , Λ 2 = Γ ( α 1 ) q 1 Γ ( α ) Γ ( α q + β q 2 q α β + 3 ) Γ ( α + β 1 ) q 1 Γ ( α q + β q 2 q β + 3 ) ( 1 i = 1 m a i ξ i α q + β q 2 q β + 2 ) , Λ 3 = Γ ( α ) q 1 Γ ( α + β ) q 1 ( α q + β q q α β + 2 ) ( 1 i = 1 m b i η i α q + β q q α β + 2 ) , Λ 4 = Γ ( α 1 ) q 1 Γ ( α + β 1 ) q 1 ( α q + β q 2 q α β + 3 ) ( 1 i = 1 m b i η i α q + β q 2 q α β + 3 ) .

The rest of this article is organized as follows: In Section 2, we give some notations, definitions and lemmas. In Section 3, based on the extension of Mawhin’s continuation theorem due to Ge, we establish a theorem on existence of solutions for BVP (1.3).

2 Preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory that can be found in the recent literature (see [1, 3, 14, 21, 24, 25]).

Let X and Y be two Banach spaces with norms X and u Y , respectively. A continuous operator
M | dom M X : X dom M Y
is said to be quasi-linear if
  1. (i)

    Im M : = M ( X dom M ) is a closed subset of Y,

     
  2. (ii)

    Ker M : = { u X dom M : M u = 0 } is linearly homeomorphic to R n , n < .

     

Let X 1 = Ker M and X 2 be the complement space of X 1 in X, then X = X 1 X 2 . On the other hand, suppose that Y 1 is a subspace of Y, and Y 2 is the complement space of Y 1 in Y, so that Y = Y 1 Y 2 . Let P : X X 1 be a projector and Q : Y Y 1 a semi-projector, and Ω X an open and bounded set with origin θ Ω . θ is the origin of a linear space.

Suppose that N λ : Ω ¯ Y , λ [ 0 , 1 ] is a continuous operator. Denote N 1 by N. Let Σ λ = { u Ω ¯ : M u = N λ u } . N λ is said to be M-compact in Ω ¯ if there is an Y 1 Y with dim Y 1 = dim X 1 and an operator R : Ω ¯ × [ 0 , 1 ] X continuous and compact such that for λ [ 0 , 1 ] ,
( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y ,
(2.1)
Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ ,
(2.2)
R ( , λ ) | Σ λ = ( I P ) | Σ λ
(2.3)
and R ( , 0 ) is the zero operator,
M [ P + R ( , λ ) ] = ( I Q ) N λ .
(2.4)

Lemma 2.1 (Ge-Mawhin’s continuation theorem [24])

Let ( X , X ) and ( Y , Y ) be two Banach spaces, and Ω X an open and bounded nonempty set. Suppose that M : X dom M Y is a quasi-linear operator N λ : Ω ¯ Y , λ [ 0 , 1 ] is M-compact in Ω ¯ . In addition, if
  1. (i)

    L u N λ u , ( u , λ ) ( dom M Ω ) × ( 0 , 1 ) ,

     
  2. (ii)

    deg ( J Q N , Ker M Ω , 0 ) 0 ,

     

where J : Im Q Ker M is a homeomorphism with J ( θ ) = θ and N = N 1 , then the equation M u = N u has at least one solution in dom M Ω ¯ .

Definition 2.1 The Riemann-Liouville fractional integral of order α > 0 of a function u is given by
I 0 + α u ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 u ( s ) d s ,

provided the right-hand side integral is pointwise almost everywhere defined on ( 0 , + ) .

Definition 2.2 The Riemann-Liouville fractional derivative of order α > 0 of a function u is given by
D 0 + α u ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t u ( s ) ( t s ) α n + 1 d s ,

provided the right-hand side integral is pointwise everywhere defined on ( 0 , + ) , where n = [ α ] + 1 .

Definition 2.3 Let X be a Banach space, and X 1 X is a subspace. A mapping Q : X X 1 is a semi-projector if Q satisfies
  1. (i)

    Q 2 x = Q x , x X ,

     
  2. (ii)

    Q ( μ x ) = μ Q x , x X , μ R .

     
Lemma 2.2 Assume that u C ( 0 , 1 ) L 1 ( 0 , 1 ) with a fractional derivative of order α > 0 that belongs to C ( 0 , 1 ) L 1 ( 0 , 1 ) . Then
I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α 1 + c 2 t α 2 + + c N t α N

for some c i R , i = 1 , 2 , , N , where N = [ α ] + 1 .

Lemma 2.3 Assume that u ( t ) C [ 0 , 1 ] , 0 p q , then
D 0 + q I 0 + p u ( t ) = I 0 + p q u ( t ) .
Lemma 2.4 Assume that α 0 , then:
  1. (i)
    If λ > 1 , λ α i , i = 1 , 2 , , [ α ] + 1 , we have that
    D 0 + α t λ = Γ ( λ + 1 ) Γ ( λ α + 1 ) t λ α .
     
  2. (ii)

    D 0 + α t α i = 0 , i = 1 , 2 , , [ α ] + 1 .

     

In this paper, we take X = { u u , D 0 + α 2 u , D 0 + α 1 u , D 0 + α u C [ 0 , 1 ] } with the norm u X = max { u , D 0 + α 2 u D 0 + α 1 u , D 0 + α u } , where u = max t [ 0 , 1 ] | u ( t ) | , and Y = C [ 0 , 1 ] with the norm y Y = y . By means of the linear functional analysis theory, it is easy to prove that X and Y are Banach spaces, so we omit it.

Define the operator M : dom M Y by
M u = D 0 + β φ p ( D 0 + α u ( t ) ) ,
(2.5)
dom M = { u X | D 0 + β φ p ( D 0 + α u ) Y , u ( 0 ) = D 0 + α u ( 0 ) = 0 , dom M = u ( 1 ) = i = 1 m a i u ( ξ i ) , D 0 + α 1 u ( 1 ) = i = 1 m b i D 0 + α 1 u ( η i ) } .
(2.6)

Based on the definition of domM, it is easy to find that dom M such as u ( t ) = c t α 1 dom M , c R .

Define the operator N λ : X Y , λ [ 0 , 1 ] ,
N λ u ( t ) = λ f ( t , u ( t ) , D 0 + α 2 u ( t ) , D 0 + α 1 u ( t ) , D 0 + α u ( t ) ) , t [ 0 , 1 ] .

Then BVP (1.3) is equivalent to the operator equation M u = N u , where N = N 1 .

3 Main result

In this section, a theorem on existence of solutions for BVP (1.3) will be given.

Define operators T j : Y Y , j = 1 , 2 as follows:
T 1 y = 0 1 ( 1 s ) α 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 y ( τ ) d τ ) d s T 1 y = i = 1 m a i 0 ξ i ( ξ i s ) α 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 y ( τ ) d τ ) d s , T 2 y = 0 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 y ( τ ) d τ ) d s T 2 y = i = 1 m b i 0 η i φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 y ( τ ) d τ ) d s .

Let us make some assumptions, which will be used in the sequel.

(H1) There exist nonnegative functions r , d , e , h , k Y such that for all t [ 0 , 1 ] , ( u , v , w , z ) R 4 ,
| f ( t , u , v , w , z ) | r ( t ) + d ( t ) | u | p 1 + e ( t ) | v | p 1 + h ( t ) | w | p 1 + k ( t ) | z | p 1 .
(H2) There exists a constant A > 0 such that for u dom M , if | D 0 + α 1 u ( t ) | > A for all t [ 0 , 1 ] , then
sgn { D 0 + α 1 u ( t ) } 1 Λ ( Λ 4 T 1 N u ( t ) Λ 3 T 2 N u ( t ) ) > 0
or
sgn { D 0 + α 1 u ( t ) } 1 Λ ( Λ 4 T 1 N u ( t ) Λ 3 T 2 N u ( t ) ) < 0 .
(H3) There exists a constant B > 0 such that for u dom M , if | D 0 + α 2 u ( t ) | > B for all t [ 0 , 1 ] , then
sgn { D 0 + α 2 u ( t ) } 1 Λ ( Λ 2 T 1 N u ( t ) + Λ 1 T 2 N u ( t ) ) > 0
or
sgn { D 0 + α 2 u ( t ) } 1 Λ ( Λ 2 T 1 N u ( t ) + Λ 1 T 2 N u ( t ) ) < 0 .
Theorem 3.1 Let f : [ 0 , 1 ] × R 4 R be continuous and condition (H1)-(H3) hold, then BVP (1.3) has at least one solution, provided that
1 Γ ( β + 1 ) ( A 1 d + e + h + k ) < 1 ,
(3.1)

where A 1 = 1 Γ ( α + 1 ) + 2 Γ ( α ) + 7 2 Γ ( α 1 ) .

In order to prove Theorem 3.1, we need to prove some lemmas below.

Lemma 3.1 The operator M : dom M X Y is quasi-linear.
Ker M = { u X u ( t ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R } ,
(3.2)
Im M = { y Y T j y = 0 , j = 1 , 2 } .
(3.3)
Proof Suppose that u ( t ) dom M , by D 0 + β φ p ( D 0 + α u ( t ) ) = 0 , we have
D 0 + α u ( t ) = φ q ( c 0 t β 1 ) .
Based on D 0 + α u ( 0 ) = 0 , one has
u ( t ) = c 1 t α 1 + c 2 t α 2 + c 3 t α 3 ,
which together with u ( 0 ) = 0 yields that
Ker M = { u X u ( t ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R } .

It is clear that dim Ker M = 2 . So, KerM is linearly homeomorphic to R 2 .

If y Im M , then there exists a function u dom M such that y ( t ) = D 0 + β φ p ( D 0 + α u ( t ) ) . Based on Lemmas 2.2 and 2.3, we have
u ( t ) = I 0 + α φ q ( I 0 + β y ( s ) ) + c 1 t α 1 + c 2 t α 2 , D 0 + α 1 u ( t ) = D 0 + α 1 I 0 + α φ q ( I 0 + β y ( s ) ) + c 1 Γ ( α ) ,

which together with i = 1 m a i ξ i α 1 = 1 , i = 1 m a i ξ i α 2 = 1 and i = 1 m b i = 1 yields that T j y ( t ) = 0 , j = 1 , 2 .

On the other hand, suppose that y Y and satisfies (3.3), and let u ( t ) = I 0 + α φ q ( I 0 + β y ( t ) ) , then u dom M and M u ( t ) = D 0 + β φ p ( D 0 + α u ( t ) ) = y , so y Im M and Im M : = M ( dom M ) is a closed subset of Y. Thus, M is a quasi-linear operator. □

Lemma 3.2 Let Ω X be an open and bounded set, then N λ is M-compact in Ω ¯ .

Proof Define the continuous projector P : X X 1 by
P u ( t ) = 1 Γ ( α ) D 0 + α 1 u ( 0 ) t α 1 + 1 Γ ( α 1 ) D 0 + α 2 u ( 0 ) t α 2 , t [ 0 , 1 ] .
Define the continuous projector Q : Y Y 1 , by
Q y ( t ) = ( Q 1 y ( t ) ) t α 1 + ( Q 2 y ( t ) ) t α 2 , t [ 0 , 1 ] ,
where
Q 1 y ( t ) = φ p ( 1 Λ ( Λ 4 T 1 y ( t ) Λ 3 T 2 y ( t ) ) ) , Q 2 y ( t ) = φ p ( 1 Λ ( Λ 2 T 1 y ( t ) + Λ 1 T 2 y ( t ) ) ) .
Obviously, X 1 = Ker M = Im P and Y 1 =  Im Q . Thus, we have dim Y 1 = dim X 1 = 2 . For any y Y , we have
Q 1 ( Q 1 y ( t ) t α 1 ) = φ p ( 1 Λ ( Λ 4 T 1 ( Q 1 y ( t ) t α 1 ) Λ 3 T 2 ( Q 1 y ( t ) t α 1 ) ) ) = Q 1 y ( t ) φ p ( 1 Λ ( Λ 4 Λ 1 Λ 3 Λ 2 ) ) = Q 1 y ( t ) .
Similarly, we can get
Q 1 ( Q 2 y ( t ) t α 2 ) = 0 , Q 2 ( Q 1 y ( t ) t α 1 ) = 0 , Q 2 ( Q 2 y ( t ) t α 2 ) = Q 2 y ( t ) .

Hence, the map Q is idempotent. Similarly, we can get Q ( μ y ) = μ Q y , for all y Y , μ R . Thus, Q is a semi-projector. For any y Im M , we can get that Q y = 0 and y Ker Q , conversely, if y Ker Q , we can obtain that Q y = 0 , that is to say, y Im M . Thus, Ker Q = Im M . Let Ω X be an open and bounded set with θ Ω , for each u Ω ¯ , we can get Q [ ( I Q ) N λ ( u ) ] = 0 . Thus, ( I Q ) N λ ( u ) Im M = Ker Q . Take any y Im M in the type y = ( y Q y ) + Q y , since Q y = 0 , we can get y ( I Q ) Y . So, (2.1) holds. It is easy to verify (2.2).

Define R : Ω ¯ × [ 0 , 1 ] X 2 by
R ( u , λ ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 ( ( I Q ) N λ u ( τ ) ) d τ ) d s .

By the continuity of f, it is easy to get that R ( u , λ ) is continuous on Ω ¯ × [ 0 , 1 ] . Moreover, for all u Ω ¯ , there exists a constant T > 0 such that | I 0 + β ( I Q ) N λ u ( τ ) | T , so, we can easily obtain that R ( Ω ¯ , λ ) , D 0 + α 2 R ( Ω ¯ , λ ) , D 0 + α 1 R ( Ω ¯ , λ ) and D 0 + α R ( Ω ¯ , λ ) are uniformly bounded. By Arzela-Ascoli theorem, we just need to prove that R : Ω ¯ × [ 0 , 1 ] X 2 is equicontinuous.

For u Ω ¯ , 0 < t 1 < t 2 1 , 2 < α 3 , 0 < β 1 , 3 < α + β 4 , we have
| R ( u , λ ) ( t 2 ) R ( u , λ ) ( t 1 ) | = 1 Γ ( α ) | 0 t 2 ( t 2 s ) α 1 φ q ( I 0 + β ( ( I Q ) N λ u ( τ ) ) ) d s 0 t 1 ( t 1 s ) α 1 φ q ( I 0 + β ( ( I Q ) N λ u ( τ ) ) ) d s | φ q ( L ) Γ ( α ) ( 0 t 1 ( ( t 2 s ) α 1 ( t 1 s ) α 1 ) d s + t 1 t 2 ( t 2 s ) α 1 d s ) = φ q ( T ) Γ ( α + 1 ) ( t 2 α t 1 α ) , | D 0 + α 2 R ( u , λ ) ( t 2 ) D 0 + α 2 R ( u , λ ) ( t 1 ) | = | 0 t 2 ( t s ) φ q ( I 0 + β ( ( I Q ) N λ u ( τ ) ) ) d s 0 t 1 ( t s ) φ q ( I 0 + β ( ( I Q ) N λ u ( τ ) ) ) d s | φ q ( T ) ( 0 t 1 ( t 2 s ) ( t 1 s ) d s + t 1 t 2 ( t 2 s ) d s ) = φ q ( T ) 2 ( t 2 2 t 1 2 )
and
| D 0 + α 1 R ( u , λ ) ( t 2 ) D 0 + α 1 R ( u , λ ) ( t 1 ) | = | 0 t 2 φ q ( I 0 + β ( ( I Q ) N λ u ( τ ) ) ) d s 0 t 1 φ q ( I 0 + β ( ( I Q ) N λ u ( τ ) ) ) d s | φ q ( T ) ( t 2 t 1 ) .

Since t α is uniformly continuous on [ 0 , 1 ] , so, R ( Ω ¯ , λ ) , D 0 + α 2 R ( Ω ¯ , λ ) and D 0 + α 1 R ( Ω ¯ , λ ) are equicontinuous. Similarly, we can get that I 0 + β ( ( I Q ) N λ u ( τ ) ) C [ 0 , 1 ] is equicontinuous. Considering that φ q ( s ) is uniformly continuous on [ T , T ] , we have that D 0 + α R ( Ω ¯ , λ ) = I 0 + β ( ( I Q ) N λ ( Ω ¯ ) ) is also equicontinuous. So, we can obtain that R : Ω ¯ × [ 0 , 1 ] X 2 is compact.

For each u Σ λ , we have D 0 + β φ p ( D 0 + α u ( t ) ) = N λ ( u ( t ) ) Im M . Thus,
R ( u , λ ) ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 ( ( I Q ) N λ u ( τ ) ) d τ ) d s = 1 Γ ( α ) 0 t ( t s ) α 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 D 0 + β φ p ( D 0 + α u ( τ ) ) d τ ) d s ,
which together with D 0 + α u ( 0 ) = u ( 0 ) = 0 yields that
R ( u , λ ) ( t ) = u ( t ) 1 Γ ( α ) D 0 + α 1 u ( 0 ) t α 1 1 Γ ( α 1 ) D 0 + α 2 u ( 0 ) t α 2 = ( I P ) u ( t ) .
It is easy to verify that R ( u , 0 ) ( t ) is the zero operator. So, (2.3) holds. Besides, for any u Ω ¯ ,
M [ P u + R ( u , λ ) ] ( t ) = M [ 1 Γ ( α ) 0 t ( t s ) α 1 φ q ( 1 Γ ( β ) 0 s ( s τ ) β 1 ( ( I Q ) N λ u ( τ ) ) d τ ) d s + 1 Γ ( α ) D 0 + α 1 u ( 0 ) t α 1 + 1 Γ ( α 1 ) D 0 + α 2 u ( 0 ) t α 2 ] = ( I Q ) N λ u ( t ) ,

which implies (2.4). So, N λ is M-compact in Ω ¯ . □

Lemma 3.3 Suppose that (H1), (H2) hold, then the set
Ω 1 = { u dom M Ker M M u = N λ u , λ ( 0 , 1 ) }

is bounded.

Proof By Lemma 2.2, for each u dom M , we have
u ( t ) = I 0 + α D 0 + α u ( t ) + c 1 t α 1 + c 2 t α 2 + c 3 t α 3 .
Combined with u ( 0 ) = 0 , we get c 3 = 0 . Thus,
u ( t ) = I 0 + α D 0 + α u ( t ) + c 1 t α 1 + c 2 t α 2 , D 0 + α 1 u ( t ) = I 0 + 1 D 0 + α u ( t ) + c 1 Γ ( α ) , D 0 + α 2 u ( t ) = I 0 + 2 D 0 + α u ( t ) + c 1 Γ ( α ) t + c 2 Γ ( α 1 ) .
By simple calculation, we get
c 1 = 1 Γ ( α ) ( D 0 + α 1 u ( t ) 0 t D 0 + α u ( s ) d s ) , c 2 = 1 Γ ( α 1 ) ( D 0 + α 2 u ( t ) 0 t ( t s ) D 0 + α 1 u ( s ) d s ( D 0 + α 1 u ( t ) 0 t D 0 + α u ( s ) d s ) t ) .
Take any u Ω 1 , then N u Im M = Ker Q and Q N u = 0 . It follows from (H2) and (H3) that there exist ε 1 , ε 2 [ 0 , 1 ] such that | D 0 + α 1 u ( ε 1 ) | A , | D 0 + α 2 u ( ε 2 ) | B . Thus,
D 0 + α 1 u ( t ) = D 0 + α 1 u ( ε 1 ) + ε 1 t D 0 + α u ( t ) d t , D 0 + α 2 u ( t ) = D 0 + α 2 u ( ε 2 ) + ε 2 t D 0 + α 1 u ( t ) d t , D 0 + α 1 u A + D 0 + α u , D 0 + α 2 u B + D 0 + α 1 u A + B + D 0 + α u .
So, we get
| c 1 | 1 Γ ( α ) ( D 0 + α 1 u + D 0 + α u ) 1 Γ ( α ) ( A + 2 D 0 + α u ) , | c 2 | 1 Γ ( α 1 ) ( D 0 + α 2 u + 3 2 D 0 + α 1 u + D 0 + α u ) | c 2 | 1 Γ ( α 1 ) ( 5 A 2 + B + 7 2 D 0 + α u ) , u A 1 D 0 + α u + B 1 ,

where A 1 = 1 Γ ( α + 1 ) + 2 Γ ( α ) + 7 2 Γ ( α 1 ) , B 1 = A Γ ( α ) + 5 A 2 Γ ( α 1 ) + B Γ ( α 1 ) .

Based on D 0 + α u ( 0 ) = 0 , we have
φ p ( D 0 + α u ( t ) ) = λ I 0 + β N u ( t ) .
From (H1) and λ ( 0 , 1 ) , we have
| φ p ( D 0 + α u ( t ) ) | 1 Γ ( β ) 0 t ( t s ) β 1 | f ( s , u ( s ) , D 0 + α 2 u ( t ) , D 0 + α 1 u ( s ) , D 0 + α u ( s ) ) | d s 1 Γ ( β ) 0 t ( t s ) β 1 ( r ( s ) + d ( s ) | u ( s ) | p 1 + e ( s ) | D 0 + α 2 u ( s ) | p 1 + h ( s ) | D 0 + α 1 u ( s ) | p 1 + k ( s ) | D 0 + α u ( s ) | p 1 ) d s 1 Γ ( β + 1 ) ( r + d u p 1 + e D 0 + α 2 u p 1 + h D 0 + α 1 u p 1 + k D 0 + α u p 1 ) ,
which together with | φ p ( D 0 + α u ( t ) ) | = | D 0 + α u ( t ) | p 1 , we can get
D 0 + α u p 1 1 Γ ( β + 1 ) ( r + d u p 1 + e D 0 + α 2 u p 1 + h D 0 + α 1 u p 1 + k D 0 + α u p 1 ) 1 Γ ( β + 1 ) ( r + d ( A 1 D 0 + α u + B 1 ) + e ( 2 A + D 0 + α u ) p 1 + h ( A + D 0 + α u ) p 1 + k D 0 + α u p 1 ) .
In view of (3.1), we can obtain that there exists a constant M 1 > 0 such that
D 0 + α u M 1 , D 0 + α 1 u A + M 1 : = M 2 , D 0 + α 2 u 2 A + M 1 : = M 3 , u A 1 M 1 + B 1 : = M 4 .
Thus, we have
u X = max { u , D 0 + α 2 u , D 0 + α 1 u , D 0 + α u } max { M 1 , M 2 , M 3 , M 4 } : = M .

So, Ω 1 is bounded. □

Lemma 3.4 Suppose that (H2) holds, then the set
Ω 2 = { u u Ker M , N u Im M }

is bounded.

Proof For each u Ω 2 , we have that u ( t ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R and Q N u = 0 . It follows from (H2) and (H3) that there exists an ε 1 , ε 2 [ 0 , 1 ] such that | D 0 + α 1 u ( ε 1 ) | A , | D 0 + α 2 u ( ε 2 ) | A , which implies that | c 1 | A Γ ( α ) and | c 2 | A + B Γ ( α 1 ) . So, Ω 2 is bounded. □

Define the isomorphism J 1 : Ker M Im Q by J 1 ( c 1 t α 1 + c 2 t α 2 ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R , t [ 0 , 1 ] . In fact, for each c 1 , c 2 R , suppose that ( Q 1 y ( t ) , Q 2 y ( t ) ) = ( c 1 , c 2 ) , we have
{ Λ 4 T 1 y ( t ) Λ 3 T 2 y ( t ) = Λ φ q ( c 1 ) : = c 1 ˜ , Λ 2 T 1 y ( t ) + Λ 1 T 2 y ( t ) = Λ φ q ( c 2 ) : = c 2 ˜ ,
(3.4)
where c 1 ˜ , c 2 ˜ R , by the condition Λ 4 Λ 1 Λ 2 Λ 3 0 , there exists a unique solution for (3.4), which is ( T 1 y ( t ) , T 2 y ( t ) ) = ( m 1 , m 2 ) , m 1 , m 2 R . Now, we will prove that there exists y Y such that ( T 1 y ( t ) , T 2 y ( t ) ) = ( m 1 , m 2 ) . Based on y ( t ) C [ 0 , 1 ] , we choose y ( t ) = D 0 + β y ¯ ( t ) , where y ¯ ( t ) = φ p ( l 1 t ( α + β 1 ) ( q 1 ) + l 2 t ( α + β 2 ) ( q 1 ) ) , l 1 = Δ 1 Γ ( α ) q 1 Γ ( α + β ) q 1 , l 2 = Δ 2 Γ ( α 1 ) q 1 Γ ( α + β 1 ) q 1 , Δ 1 = m 1 Λ 4 m 2 Λ 2 Λ , Δ 2 = m 2 Λ 1 m 1 Λ 3 Λ , t [ 0 , 1 ] , which together with 2 < α 3 , 0 < β 1 , 3 < α + β 4 , q > 1 , we have ( α + β 1 ) ( q 1 ) > 0 and ( α + β 2 ) ( q 1 ) > 0 . So, we have y ¯ ( 0 ) = 0 . Thus, we can obtain that
I 0 + β y ( t ) = I 0 + β D 0 + β y ¯ ( t ) = y ¯ ( t ) + c t β 1 ,
where c R , which together with y ¯ ( 0 ) = 0 , we have c = 0 . So, we have
{ T 1 y ( t ) = 0 1 ( 1 s ) α 1 φ q ( y ¯ ( s ) ) d s i = 1 m a i 0 ξ i ( ξ i s ) α 1 φ q ( y ¯ ( s ) ) d s T 1 y ( t ) = Δ 1 Λ 1 + Δ 2 Λ 2 = m 1 , T 2 y ( t ) = 0 1 φ q ( y ¯ ( s ) ) d s i = 1 m b i 0 η i φ q ( y ¯ ( s ) ) d s = Δ 1 Λ 3 + Δ 2 Λ 4 = m 2 .

Thus, there exists y Y such that ( T 1 y ( t ) , T 2 y ( t ) ) = ( m 1 , m 2 ) . So J 1 : Ker L Im Q is well defined.

Lemma 3.5 Suppose that the first part of (H3) holds, then the set
Ω 3 = { u Ker M λ J 1 u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] }

is bounded.

Proof For each u Ω 3 , we can get that u ( t ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R . By the definition of the set Ω 3 , we can obtain that
λ ( c 1 t α 1 + c 2 t α 2 ) + ( 1 λ ) ( Q 1 N ( c 1 t α 1 + c 2 t α 2 ) t α 1 + Q 2 N ( c 1 t α 1 + c 2 t α 2 ) t α 2 ) = 0 .
Thus,
λ c 1 + ( 1 λ ) φ p ( 1 Λ ( Λ 4 T 1 N ( c 1 t α 1 + c 2 t α 2 ) Λ 3 T 2 N ( c 1 t α 1 + c 2 t α 2 ) ) ) = 0 ,
(3.5)
λ c 2 + ( 1 λ ) φ p ( 1 Λ ( Λ 2 T 1 N ( c 1 t α 1 + c 2 t α 2 ) + Λ 1 T 2 N ( c 1 t α 1 + c 2 t α 2 ) ) ) = 0 .
(3.6)

If λ = 0 , then Ω 3 is bounded because of the first part of (H2) and (H3). If λ = 1 , we get c 1 = c 2 = 0 , obviously, Ω 3 is bounded. If λ ( 0 , 1 ) , by the first part of (H2) and (3.5), we can obtain that | c 1 | A Γ ( α ) , by the first part of (H3) and (3.6), we have | c 2 | A + B Γ ( α 1 ) . So, Ω 3 is bounded. □

Remark 3.1 If the second part of (H3) holds, then the set
Ω 3 = { u Ker M λ J 1 u + ( 1 λ ) Q N u = 0 , λ [ 0 , 1 ] }

is bounded.

Proof of Theorem 3.1 Assume that Ω is a bounded open set of X with i = 1 3 Ω ¯ i Ω 3 Ω . By Lemmas 3.1 and 3.2, we can obtain that M : dom M X Y is quasi-linear, and N λ is M-compact on Ω ¯ . By the definition of Ω , we have
L u N λ u , ( u , λ ) ( dom M Ω ) × ( 0 , 1 ) , H ( u , λ ) = ± λ J 1 ( u ) + ( 1 λ ) Q N ( u ) 0 , ( Ω Ker M ) × [ 0 , 1 ] .
Thus, by the homotopic property of degree, we can get
deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( ± I , Ω Ker M , 0 ) 0 .

So Lemma 2.1 is satisfied, and M u = N u has at least one solution in dom M Ω ¯ . Namely, BVP (1.3) have at least one solution in the space X. □

Declarations

Acknowledgements

The authors are grateful to those who gave useful suggestions about the original manuscript. This research is supported by the National Natural Science Foundation of China (No. 11271364) and the Fundamental Research Funds for the Central Universities (2010LKSX09).

Authors’ Affiliations

(1)
College of Sciences, China University of Mining and Technology

References

  1. Kilbas AA, Srivastava HM, Trujillo JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam; 2006.MATHGoogle Scholar
  2. Sabatier J, Agrawal OP, Machado JAT: Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer, Dordrecht; 2007.View ArticleMATHGoogle Scholar
  3. Samko SG, Kilbas AA, Marichev OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, New York; 1993.MATHGoogle Scholar
  4. Magin R: Fractional calculus models of complex dynamics in biological tissues. Comput. Math. Appl. 2010, 59: 1586-1593. 10.1016/j.camwa.2009.08.039MathSciNetView ArticleMATHGoogle Scholar
  5. Metzler R, Klafter J: Boundary value problems for fractional diffusion equations. Physica A 2000, 278: 107-125. 10.1016/S0378-4371(99)00503-8MathSciNetView ArticleMATHGoogle Scholar
  6. Mainardi F: Fractional diffusive waves in viscoelastic solids. In Nonlinear Waves in Solids. Edited by: Wegner JL, Norwood FR. ASME, Fairfield; 1995:93-97.Google Scholar
  7. Bai J, Feng X: Fractional-order anisotropic diffusion for image denoising. IEEE Trans. Image Process. 2007, 16: 2492-2502.MathSciNetView ArticleGoogle Scholar
  8. Agarwal RP, O’Regan D, Stanek S: Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations. J. Math. Anal. Appl. 2010, 371: 57-68. 10.1016/j.jmaa.2010.04.034MathSciNetView ArticleMATHGoogle Scholar
  9. Kosmatov N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ. 2010., 2010: Article ID 135Google Scholar
  10. Jafari H, Gejji VD: Positive solutions of nonlinear fractional boundary value problems using Adomian decomposition method. Appl. Math. Comput. 2006, 180: 700-706. 10.1016/j.amc.2006.01.007MathSciNetView ArticleMATHGoogle Scholar
  11. Bai Z: Solvability for a class of fractional m -point boundary value problem at resonance. Comput. Math. Appl. 2011, 62: 1292-1302. 10.1016/j.camwa.2011.03.003MathSciNetView ArticleMATHGoogle Scholar
  12. Bai Z: On solutions of some fractional m -point boundary value problems at resonance. Electron. J. Qual. Theory Differ. Equ. 2010., 2010: Article ID 37Google Scholar
  13. Bai Z, Zhang Y: Solvability of fractional three-point boundary value problems with nonlinear growth. Appl. Math. Comput. 2011, 218: 1719-1725. 10.1016/j.amc.2011.06.051MathSciNetView ArticleMATHGoogle Scholar
  14. Zhang Y, Bai Z: Existence of positive solutions for nonlinear fractional three-point boundary value problem at resonance. Appl. Math. Comput. 2011, 36: 417-440.MathSciNetMATHGoogle Scholar
  15. Hu Z, Liu W: Solvability for fractional order boundary value problem at resonance. Bound. Value Probl. 2011., 2011: Article ID 20Google Scholar
  16. Rui W: Existence of solutions of nonlinear fractional differential equations at resonance. Electron. J. Qual. Theory Differ. Equ. 2012., 2012: Article ID 66Google Scholar
  17. Wang G, Liu W, Yang J, Zhu S, Zheng T: The existence of solutions for a fractional 2 m -point boundary value problems. J. Appl. Math. 2012. 10.1155/2012/841349Google Scholar
  18. Wang W: Solvability for a coupled system of fractional differential equations at resonance. Nonlinear Anal. 2012, 13: 2285-2292. 10.1016/j.nonrwa.2012.01.023View ArticleMathSciNetMATHGoogle Scholar
  19. Zhou H, Kou C, Xie F: Existence of solutions for fractional differential equations with multi-point boundary conditions at resonance on a half-line. Electron. J. Qual. Theory Differ. Equ. 2011., 2011: Article ID 27Google Scholar
  20. Chen Y, Tang X: Solvability of sequential fractional order multi-point boundary value problems at resonance. Appl. Math. Comput. 2012, 218: 7638-7648. 10.1016/j.amc.2012.01.033MathSciNetView ArticleMATHGoogle Scholar
  21. Wang J: The existence of solutions to boundary value problems of fractional differential equations at resonance. Nonlinear Anal. 2011, 74: 1987-1994. 10.1016/j.na.2010.11.005MathSciNetView ArticleMATHGoogle Scholar
  22. Chen T, Liu W, Hu Z: A boundary value problem for fractional differential equation with p -Laplacian operator at resonance. Nonlinear Anal. 2012, 75: 3210-3217. 10.1016/j.na.2011.12.020MathSciNetView ArticleMATHGoogle Scholar
  23. Mawhin J Lecture Notes in Math. 1537. Topological Degree and Boundary Value Problems for Nonlinear Differential Equations in Topological Methods for Ordinary Differential Equations 1993, 74-142.Google Scholar
  24. Ge W, Ren J: An extension of Mawhin’s continuation theorem and its application to boundary value problems with a p -Laplacian. Nonlinear Anal. 2004, 58: 477-488. 10.1016/j.na.2004.01.007MathSciNetView ArticleMATHGoogle Scholar
  25. Pang H, Ge W, Tian M: Solvability of nonlocal boundary value problems for ordinary differential equation of higher order with a p -Laplacian. Comput. Math. Appl. 2008, 56: 127-142. 10.1016/j.camwa.2007.11.039MathSciNetView ArticleMATHGoogle Scholar

Copyright

© Shen et al.; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an open access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.