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

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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.

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<α2, 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, n2 is an integer. g:[0,1]R is a nondecreasing function with 0 1 dg(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 dimKerM=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 dimKerM=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<α3, 0<β1, 3<α+β4, 0< ξ 1 < ξ 2 << ξ m <1, 0< η 1 < η 2 << η m <1, a i R, b i R, 1<m, mN, 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 :XdomMY

is said to be quasi-linear if

  1. (i)

    ImM:=M(XdomM) is a closed subset of Y,

  2. (ii)

    KerM:={uXdomM:Mu=0} is linearly homeomorphic to R n , n<.

Let X 1 =KerM 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 ΩX an open and bounded set with origin θΩ. 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 ¯ :Mu= 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],

(IQ) N λ ( Ω ¯ )ImM(IQ)Y,
(2.1)
Q N λ x=θ,λ(0,1)QNx=θ,
(2.2)
R(,λ) | Σ λ =(IP) | Σ λ
(2.3)

and R(,0) is the zero operator,

M [ P + R ( , λ ) ] =(IQ) N λ .
(2.4)

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

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

  1. (i)

    Lu N λ u, (u,λ)(domMΩ)×(0,1),

  2. (ii)

    deg(JQN,KerMΩ,0)0,

where J:ImQKerM is a homeomorphism with J(θ)=θ and N= N 1 , then the equation Mu=Nu has at least one solution in domM 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)ds,

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 ds,

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=Qx, xX,

  2. (ii)

    Q(μx)=μQx, xX, μR.

Lemma 2.2 Assume that uC(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], 0pq, then

D 0 + q I 0 + p u(t)= I 0 + p q u(t).

Lemma 2.4 Assume that α0, then:

  1. (i)

    If λ>1, λαi, 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={uu, D 0 + α 2 u, D 0 + α 1 u, D 0 + α uC[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:domMY by

Mu= 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 domM such as u(t)=c t α 1 domM, cR.

Define the operator N λ :XY, λ[0,1],

N λ u(t)=λf ( 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 Mu=Nu, 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 :YY, 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,kY 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 udomM, 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 udomM, 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:domMXY is quasi-linear.

KerM= { u X u ( t ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R } ,
(3.2)
ImM={yY T j y=0,j=1,2}.
(3.3)

Proof Suppose that u(t)domM, 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

KerM= { u X u ( t ) = c 1 t α 1 + c 2 t α 2 , c 1 , c 2 R } .

It is clear that dimKerM=2. So, KerM is linearly homeomorphic to R 2 .

If yImM, then there exists a function udomM 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 yY and satisfies (3.3), and let u(t)= I 0 + α φ q ( I 0 + β y(t)), then udomM and Mu(t)= D 0 + β φ p ( D 0 + α u(t))=y, so yImM and ImM:=M(domM) is a closed subset of Y. Thus, M is a quasi-linear operator. □

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

Proof Define the continuous projector P:X X 1 by

Pu(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

Qy(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 =KerM=ImP and Y 1 = ImQ. Thus, we have dim Y 1 = dim X 1 =2. For any yY, 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(μy)=μQy, for all yY, μR. Thus, Q is a semi-projector. For any yImM, we can get that Qy=0 and yKerQ, conversely, if yKerQ, we can obtain that Qy=0, that is to say, yImM. Thus, KerQ=ImM. Let ΩX be an open and bounded set with θΩ, for each u Ω ¯ , we can get Q[(IQ) N λ (u)]=0. Thus, (IQ) N λ (u)ImM=KerQ. Take any yImM in the type y=(yQy)+Qy, since Qy=0, we can get y(IQ)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 ) ds.

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 (IQ) 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<α3, 0<β1, 3<α+β4, 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 ((IQ) 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 ((IQ) 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))ImM. 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 =(IP)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 udomM, 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 NuImM=KerQ and QNu=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 Nu(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 ={uuKerM,NuImM}

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 QNu=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 :KerMImQ 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 yY 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<α3, 0<β1, 3<α+β4, q>1, we have (α+β1)(q1)>0 and (α+β2)(q1)>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 cR, 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 yY such that ( T 1 y(t), T 2 y(t))=( m 1 , m 2 ). So J 1 :KerLImQ 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 ) +(1λ) ( 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 +(1λ) 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 +(1λ) 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:domMXY 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 Mu=Nu has at least one solution in domM Ω ¯ . Namely, BVP (1.3) have at least one solution in the space X. □

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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).

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Correspondence to Wenbin Liu.

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Keywords

  • fractional differential equation
  • boundary value problem
  • p-Laplacian operator
  • coincidence degree theory
  • resonance