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The existence of solutions for nonlinear fractional multipoint boundary value problems at resonance

Abstract

A class of nonlinear fractional multipoint boundary value problems at resonance is considered in this article. The existence results are obtained by the method of the coincidence degree theory of Mawhin. An example is given to illustrate the results.

MSC:34A08.

1 Introduction

The subject of fractional calculus has gained considerable popularity during the past decades, due mainly to its frequent appearance in a variety of different areas such as physics, aerodynamics, polymer rheology, etc. (see [13]). Many methods have been introduced for solving fractional differential equations (FDEs for short in the remaining), such as the Laplace transform method, the iteration method, the Fourier transform method, etc. (see [4]).

Recently, there have been many works related to the existence of solutions for multipoint boundary value problems (BVPs for short in the remaining) at nonresonance of FDEs (see [511]). Motivated by the above articles and recent studies on FDEs (see [1219]), we consider the existence of solutions for a nonlinear fractional multipoint BVPs at resonance in this article.

In [16], Zhang and Bai considered the following fractional three-point boundary value problems at resonance:

{ D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α ( n 1 ) u ( t ) , , D 0 + α 1 u ( t ) ) + e ( t ) , 0 < t < 1 , I 0 + n α u ( 0 ) = D 0 + α ( n 1 ) u ( 0 ) = = D 0 + α 2 u ( 0 ) = 0 , u ( 1 ) = σ u ( η ) ,

where n>2 is a natural number; n1<αn is a real number; D 0 + α and I 0 + α are the standard Riemann-Liouville derivative and integral respectively; f:[0,1]× R n R is continuous; e(t)L[0,1]; σ(0,+)η(0,1) are given constants such that σ η α 1 =1. In their article, they made the operator Lu= D 0 + α u and got dimKerL=1. In [17], Bai discussed fractional m-point boundary value problems at resonance with the case of dimKerL=1.

In 2010, Bai and Jiang studied the fractional differential equation of boundary value problems at resonance with the case of dimKerL=2 respectively (see [18, 19]), and we can see that they obtained the results by the assumption that a specific algebraic expression is not equal to zero; for example,

R= 1 α η α Γ ( α ) Γ ( α 1 ) Γ ( 2 α 1 ) [ 1 i = 1 m α i η i 2 α 2 ] 1 α 1 η α 1 Γ 2 ( α ) Γ ( 2 α ) [ 1 i = 1 m α i η i 2 α 1 ] 0

is referred to as a condition in [18]. We will show that the assumption like above R0 is not necessary.

In this article, we will use the coincidence degree theory to study the existence of solutions for a nonlinear FDEs at resonance which is given by

D 0 + α u(t)=f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) ,0<t<1,
(1.1)

with boundary conditions

I 0 + 3 α u ( t ) | t = 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 ) ,
(1.2)

where 2<α3; 0< ξ 1 << ξ m <1; 0< η 1 << η m <1; a i , b i R; f:[0,1]× R 3 R with satisfying Carathéodory conditions; D 0 + α and I 0 + α are the standard Riemann-Liouville fractional derivative and fractional integral respectively.

BVPs (1.1)-(1.2) being at resonance means that the associated linear homogeneous equation D 0 + α u(t)=0 with boundary conditions (1.2) has u(t)=a t α 1 +b t α 2 as a nontrivial solution, where 0<t<1, a,bR.

We will always suppose that the following conditions hold:

i = 1 m a i =1, i = 1 m b i η i =1, i = 1 m b i =1.
(C)

The rest of this article is organized as follows: In Section 2, we give some definitions, lemmas and notations. In Section 3, we establish theorems of existence result for BVPs (1.1)-(1.2). In Section 4, we give an example to illustrate our result.

2 Preliminaries

We present here some necessary basic knowledge and definitions of the fractional calculus theory, which can be found in [13].

Definition 2.1 The Riemann-Liouville fractional integral of order α>0 of a function y:(0,)R is given by

I 0 + α y(t)= 1 Γ ( α ) 0 t ( t s ) α 1 y(s)ds,

where Γ() is the Gamma function, provided the right side is pointwise defined on (0,).

Definition 2.2 The Riemann-Liouville fractional derivative of order α>0 of a function y:(0,)R is given by

D 0 + α y(t)= 1 Γ ( n α ) ( d d t ) n 0 t y ( s ) ( t s ) α n + 1 ds,

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

Definition 2.3 ([18])

We say that the map f:[0,1]× R n R satisfies Carathéodory conditions with respect to L[0,1] if the following conditions are satisfied:

  1. (i)

    for each z R n , the mapping tf(t,z) is Lebesgue measurable;

  2. (ii)

    for almost every t[0,1], the mapping tf(t,z) is continuous on R n ;

  3. (iii)

    for each E>0, there exists a ρ E L[0,1] such that, for a.e. t[0,1] and every |u|E, we have f(t,u) ρ E (t).

Lemma 2.4 ([2])

Assumey(t)C[0,1]L[0,1], 0βα, then D 0 + β I 0 + α y(t)= I 0 + α β y(t). And, for allα0, β>1, we have that

I 0 + α t β = Γ ( β + 1 ) Γ ( α + β + 1 ) t α + β , D 0 + α t β = Γ ( β + 1 ) Γ ( β α + 1 ) t β α .

Lemma 2.5 ([2])

Letα>0, n=[α]+1and assume thaty, D 0 + α yL(0,1), then the following equality holds almost everywhere on[0,1],

( I 0 + α D 0 + α y ) (t)=y(t) i = 1 n ( ( I 0 + n α y ) ( t ) ) n i | t = 0 Γ ( α i + 1 ) t α i .

Now, we briefly recall some notations and an abstract existence result, which can be found in [20]. Let Y Z be real Banach spaces, L:domLYZ be a Fredholm map of index zero, and P:YYQ:ZZ be continuous projectors such that

ImP=KerL,KerQ=ImL,Y=KerLKerP,Z=ImQImL.

It follows that L | dom L Ker P :domLKerPImL is invertible. We denote the inverse by K p . If Ω is an open bounded subset of Y such that domLΩ, the map N:YZ will be called L-compact on Ω if QN( Ω ¯ ) is bounded and K p (IQ)N: Ω ¯ Y is compact.

Lemma 2.6 ([20])

Let L be a Fredholm operator of index zero and N be L-compact on Ω ¯ . The equationLx=Nxhas at least one solution indomL Ω ¯ if the following conditions are satisfied:

  1. (i)

    LxλNx for each (x,λ)[domLKerLΩ]×[0,1];

  2. (ii)

    NxImL for each xKerLΩ;

  3. (iii)

    deg(JQN | Ker L ,KerLΩ,0)0,

whereQ:ZZis a projection such thatKerQ=ImLandJ:ImQKerLis a any isomorphism.

In this article, we use the Banach space C[0,1] with the norm u = max t [ 0 , 1 ] |u(t)|.

Lemma 2.7 ([16])

Givenμ>0andN=[μ]+1, for anyxC[0,1], c i R (i=1,2,,N1), we can define a Banach space

C μ [0,1]= { u ( t ) | u ( t ) = I 0 + μ x ( t ) + c 1 t μ 1 + c 2 t μ 2 + + c N 1 t μ ( N 1 ) , t ( 0 , 1 ) } ,

with the norm defined by u C μ = D 0 + μ u ++ D 0 + μ ( N 1 ) u + u .

Lemma 2.8 ([16])

E C μ [0,1]is a sequentially compact set if and only if E is uniformly bounded and equicontinuous. Here, a uniform bound means that there exists a constantM>0with eachuE, such that

u C μ = D 0 + μ u ++ D 0 + μ ( N 1 ) u + u <M,

and equicontinuation means that there exists aδ>0with| t 1 t 2 |<δfor any t 1 , t 2 [0,1], uEandε>0, such that

| u ( t 1 ) u ( t 2 ) | <ε, | D 0 + α i u ( t 1 ) D 0 + α i u ( t 2 ) | <ε(i=1,2,,N1).

In this article, let Z=L[0,1] with the norm y 1 = 0 1 |y(s)|ds and Y= C α 1 [0,1] with the norm u Y = D 0 + α 1 u + D 0 + α 2 u + u . Define the operator L:domLYZ by

Lu= D 0 + α u,
(2.1)

where domL={u C α 1 [0,1]| D 0 + α uZ,usatisfies (1.2)}. Define the operator N:YZ by

Nu(t)=f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) ,t[0,1].
(2.2)

Thus, BVP (1.1) can be written as Lu=Nu for each udomL.

3 Main results

First, let us introduce the following notations for convenience, with setting p{1,2,,m1} and q Z + with qp+1,

Λ 1 : = 1 p ( p + 1 ) ( 1 i = 1 m b i η i p + 1 ) , Λ 2 : = 1 q ( q 1 ) ( 1 i = 1 m b i η i q ) , Λ 3 : = 1 p ( 1 i = 1 m a i ξ i p ) , Λ 4 : = 1 q 1 ( 1 i = 1 m a i ξ i q 1 ) , Λ : = Λ 1 Λ 4 Λ 2 Λ 3 , μ : = 3 + 1 Γ ( α ) + 1 Γ ( α 1 ) , ω : = 2 + 1 Γ ( α ) , ρ : = 5 + 2 Γ ( α ) + 1 Γ ( α 1 ) .

Then, let us make some assumptions which will be used throughout the article.

(H1) There exist functions h(t),r(t),s(t),w(t),e(t)L[0,1] and a constant θ[0,1) such that for all (x,y,z) R 3 , t[0,1],

| f ( t , x , y , z ) | h(t)|x|+r(t)|y|+s(t)|z|+w(t) | z | θ +e(t);

(H2) For any udomL, t[0,1], there exists a constant A>0 such that if D 0 + α 1 u(t)>A, then either

Λ 2 T 1 Nu Λ 4 T 2 Nu<0,or Λ 2 T 1 Nu Λ 4 T 2 Nu>0;

(H3) For any udomL, t[0,1], there exists a constant B>0 such that if D 0 + α 2 u(t)>B, then either

Λ 1 T 1 Nu Λ 3 T 2 Nu>0,or Λ 1 T 1 Nu Λ 3 T 2 Nu<0.

Theorem 3.1 If conditions (C), (H 1)-(H 3) hold, then BVPs (1.1)-(1.2) have at least one solution provided thatρ( h 1 + r 1 + s 1 )<1.

In order to obtain our main result, we first present and prove Lemmas 3.2-3.8. Now, let us define operators T j :ZZ (j=1,2) as follows:

T 1 x ( t ) = 0 1 x ( s ) d s i = 1 m a i 0 ξ i x ( s ) d s , t [ 0 , 1 ] , T 2 x ( t ) = 0 1 ( 1 s ) x ( s ) d s i = 1 m b i 0 η i ( η i s ) x ( s ) d s , t [ 0 , 1 ] .

Lemma 3.2 If condition (C) holds and L is defined by (2.1), then

KerL= { a t α 1 + b t α 2 | a , b R } ,ImL={xZ| T j x=0,j=1,2}.

Proof By (2.1) and Lemma 2.5, D 0 + α u(t)=0 has a solution

u(t)= D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 + D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 + I 0 + 3 α u ( t ) | t = 0 Γ ( α 2 ) t α 3 .

Combining with the condition (1.2), we get KerL={a t α 1 +b t α 2 |a,bR} R 2 .

Suppose xImL, then there exists udomL such that x=Lu, i.e., uY, x= D 0 + α u. By Lemma 2.5, we have

I 0 + α x(t)=u(t) D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 I 0 + 3 α u ( t ) | t = 0 Γ ( α 2 ) t α 3 .

Then in view of condition (C), (1.2) and Lemma 2.4, x satisfies

{ 0 1 x ( s ) d s i = 1 m a i 0 ξ i x ( s ) d s = 0 , 0 1 ( 1 s ) x ( s ) d s i = 1 m b i 0 η i ( η i s ) x ( s ) d s = 0 .
(3.1)

On the other hand, suppose xZ and it satisfies (3.1), let u(t)= I 0 + α x(t), then udomL, D 0 + α u(t)=x(t), i.e., xImL. Therefore, we obtain that

ImL={xZ| T j x=0,j=1,2}.

 □

Lemma 3.3 If condition (C) holds, then there exist two constantsq Z + andp{1,2,,m1}withqp+1such thatΛ0.

Proof From i = 1 m a i =1, we obtain that for any nonnegative integer l, there exists k l 1{lm+1,,(l+1)m} such that i = 1 m a i ξ i k l 1 1. If else, we obtain that i = 1 m a i ξ i k l 1 =1, k l 1=0, lm+1,,(l+1)m.

If l=0, we have

( 1 1 1 ξ 1 ξ 2 ξ m ξ 1 m ξ 2 m ξ m m ) ( a 1 a 2 a m ) = ( 1 1 1 ) .

It is equal to

( 1 ξ 1 1 ξ 2 1 ξ m ξ 1 ( 1 ξ 1 ) ξ 2 ( 1 ξ 2 ) ξ m ( 1 ξ m ) ξ 1 m 1 ( 1 ξ 1 ) ξ 2 m 1 ( 1 ξ 2 ) ξ m m 1 ( 1 ξ m ) ) ( a 1 a 2 a m ) = ( 0 0 0 ) .

Since the determinant of coefficients is not equal to zero, we have that a i =0 (i=1,2,,m), which is a contradiction to condition (C).

If l Z + , we get

( 1 1 1 ξ 1 l m + 1 ξ 2 l m + 1 ξ m l m + 1 ξ 1 l m + m ξ 2 l m + m ξ m l m + m ) ( a 1 a 2 a m ) = ( 1 1 1 ) .

Similarly, we can deduce that the determinant of coefficients is not equal to zero, so we have that a i =0 (i=1,2,,m), which is a contradiction to condition (C). Thus, there exists k l 1{lm+1,,(l+1)m} such that i = 1 m a i ξ i k l 1 1.

Similarly, from i = 1 m b i = i = 1 m b i η i =1, we have that there exists a constant p{1,2,,m1} such that

i = 1 m b i η i p + 1 1.
(3.2)

Let

S= { ( k l 1 ) Z + | ( p + 1 ) ( 1 i = 1 m b i η i k l ) ( 1 i = 1 m a i ξ i p ) k l ( 1 i = 1 m a i ξ i k l 1 ) = 1 i = 1 m b i η i p + 1 } ,

we shall prove that S is a finite set. If else, there exists a strict increasing sequence { k l n } n = 1 such that

( p + 1 ) ( 1 i = 1 m b i η i k l n ) ( 1 i = 1 m a i ξ i p ) k l n ( 1 i = 1 m a i ξ i k l n 1 ) =1 i = 1 m b i η i p + 1 .

Since i = 1 m b i η i p + 1 1, we have i = 1 m a i ξ i p 1. Thus,

1 i = 1 m b i η i p + 1 = lim k l n + ( p + 1 ) ( 1 i = 1 m b i η i k l n ) ( 1 i = 1 m a i ξ i p ) k l n ( 1 i = 1 m a i ξ i k l n 1 ) =0,

which is a contradiction to (3.2). Therefore, there exists two constants p{1,2,,m1} and q Z + with qp+1 such that Λ0. □

Lemma 3.4 If the condition (C) holds and L is defined by (2.1), then L is a Fredholm operator of index zero. Define the linear operator K p :ImLdomLKerPwith K P x= I 0 + α x, then it is the inverse of L. Furthermore, we have

K p x Y ω x 1 .

Proof For each p{1,2,,m1} and q Z + with qp+1, define operator Q:ZZ by

Qx(t)= ( Q 1 x ( t ) ) t p 1 + ( Q 2 x ( t ) ) t q 2 ,t[0,1],
(3.3)

where

Q 1 x(t)= 1 Λ [ Λ 2 T 1 x ( t ) + Λ 4 T 2 x ( t ) ] , Q 2 x(t)= 1 Λ [ Λ 1 T 1 x ( t ) Λ 3 T 2 x ( t ) ] .
(3.4)

It is clear that dimIm Q 1 =2. It follows from (3.4), the definition of T 1 and T 2 that

Q 1 ( ( Q 1 x ) t p 1 ) = 1 Λ [ Λ 2 T 1 ( ( Q 1 x ) t p 1 ) + Λ 4 T 2 ( ( Q 1 x ) t p 1 ) ] = 1 Λ [ Λ 2 Λ 3 ( Q 1 x ) + Λ 1 Λ 4 ( Q 1 x ) ] = Q 1 x ,
(3.5)

similarly, we can derive that

Q 1 ( ( Q 2 x ) t q 2 ) =0, Q 2 ( ( Q 1 x ) t p 1 ) =0, Q 2 ( ( Q 2 x ) t q 2 ) = Q 2 x.
(3.6)

Hence, for each xZ and t[0,1], it follows from the (3.3)-(3.6) that

Q 2 x = Q 1 [ ( Q 1 x ) t p 1 + ( Q 2 x ) t q 2 ] t p 1 + Q 2 [ ( Q 1 x ) t p 1 + ( Q 2 x ) t q 2 ] t q 2 = ( Q 1 x ) t p 1 + ( Q 2 x ) t q 2 = Q x .

Furthermore, Q is a continuous linear projector.

For each xImL, we have Qx=0, i.e., xKerQ. On the other hand, for each xKerQ, we have that

{ Λ 2 T 1 x + Λ 4 T 2 x = 0 , Λ 1 T 1 x Λ 3 T 2 x = 0 .

However, the determinant of coefficients is as follows

| Λ 2 Λ 4 Λ 1 Λ 3 | =Λ0

then we have T j x=0 (j=1,2), i.e., xImL. Thus, KerQ=ImL.

Take any xZ in the type x=(xQx)+Qx, obviously, xQxKerQ=ImL and QxImQ, so Z=ImL+ImQ. For any xImLImQ with x=a t p 1 +b t q 2 , by Lemma 3.2, we have

0 1 ( a s p 1 + b s q 2 ) d s i = 1 m a i 0 ξ i ( a s p 1 + b s q 2 ) d s = 0 , 0 1 ( 1 s ) ( a s p 1 + b s q 2 ) d s i = 1 m b i 0 η i ( η i s ) ( a s p 1 + b s q 2 ) d s = 0 .

That is,

{ a Λ 3 + b Λ 4 = 0 , a Λ 1 + b Λ 2 = 0 ,

but the determinant of coefficients is as follows

| Λ 3 Λ 4 Λ 1 Λ 2 | =Λ0,

we can deduce that a=b=0. Hence, ImLImQ=0. Furthermore, we get Z=ImLImQ. Therefore, dimKerL=dimImQ=codimImL=2, which means that L is a Fredholm operator of index zero.

Let operator P:YY and

Pu(t)= D 0 + α 1 u ( t ) | t = 0 Γ ( α ) t α 1 + D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) t α 2 ,t[0,1].
(3.7)

It is easy to calculate that Pu(t)= P 2 u(t); furthermore, P is a continuous linear projector. Obviously

KerP= { u Y | D 0 + α 1 u ( 0 ) = D 0 + α 2 u ( 0 ) = 0 } .

It is clear that Y=KerLKerP.

For any xImL, in view of the definition of operators Kp and L, we have L K P x=L I 0 + α x= D 0 + α I 0 + α x=x. On the other hand, if udomLKerP, we have D 0 + α 1 u(0)= D 0 + α 2 u(0)=0, udomL. Therefore, by Lemma 2.5 and definitions of operators K p and L, we know that ( K p L)u=u, which implies that K p = [ L | dom L Ker P ] 1 . By the definition of K p , we have

K p x(t)= I 0 + α x(t)= 1 Γ ( α ) 0 t ( t s ) α 1 x(s)ds,xImL.

It follows from Lemma 2.4 that

D 0 + α 1 ( K p x)(t)= 0 t (ts)x(s)ds, D 0 + α 2 ( K p x)(t)= 0 t x(s)ds.

Then, we have

K p x 1 Γ ( α ) x 1 , D 0 + α 1 ( K p x ) x 1 , D 0 + α 2 ( K p x ) x 1 .

By the definition of the norm in space Y, we get K p x Y ω x 1 . □

Lemma 3.5 AssumeΩYis an open bounded subset such thatdomL Ω ¯ , and N is defined by (2.2), then N is L-compact on Ω ¯ .

Proof In order to prove N is L-compact, we only need to prove that QN( Ω ¯ ) is bounded and K p (IQ)N(u): Ω ¯ Y is compact. Since the function f satisfies Carathéodory conditions and u Ω ¯ , for each E>0, there exists a ρ E (t)L[0,1] such that, for a.e. t[0,1] and every |u|E, we have f ρ E . By the definition of operators Q and K p on the interval [0,1], it is easy to get that QN( Ω ¯ ) and K p (IQ)N( Ω ¯ ) are bounded. Thus, there exists a constant r>0 with each t[0,1], such that |QNu(t)|r.

For all 0 t 1 < t 2 1, 2<α3, u Ω ¯ , we have

| K p ( I Q ) N u ( t 2 ) K p ( I Q ) N u ( t 1 ) | = 1 Γ ( α ) | 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] ( I Q ) N u ( s ) d s + t 1 t 2 ( t 2 s ) α 1 ( I Q ) N u ( s ) d s | 1 Γ ( α ) { 0 t 1 [ ( t 2 s ) α 1 ( t 1 s ) α 1 ] ( r + | ρ E ( s ) | ) d s + t 1 t 2 ( t 2 s ) α 1 ( r + | ρ E ( s ) | ) d s } 1 Γ ( α ) { 0 t 2 | ρ E ( s ) | ( t 2 s ) α 1 d s 0 t 1 | ρ E ( s ) | ( t 1 s ) α 1 d s } + r Γ ( α + 1 ) ( t 2 α t 1 α ) ,

and

| D α 1 K p ( I Q ) N u ( t 2 ) D α 1 K p ( I Q ) N u ( t 1 ) | = | 0 t 2 ( I Q ) N u ( s ) d s 0 t 1 ( I Q ) N u ( s ) d s | r ( t 2 t 1 ) + t 1 t 2 | ρ E ( s ) | d s .

Since t α is uniformly continuous on [0,1] and ρ E (t)L[0,1], so K p (IQ)N( Ω ¯ ) and D α 1 K p (IQ)N( Ω ¯ ) are equicontinuous. By Lemma 2.8, we get that K p (IQ)N:YY is completely continuous. □

Lemma 3.6 Suppose (H 1)-(H 3) hold, then the set Ω 1 ={udomLKerL:Lu=λNu,λ[0,1]}is bounded.

Proof Taking any u Ω 1 , then we have Lu=λNu, which yields λ0 and NuImL=KerQ, i.e., QNu=0 for all t[0,1]. It follows from (H2) and (H3) that there exists t 0 [0,1] such that | D 0 + α 1 u( t 0 )|+| D 0 + α 2 u( t 0 )|A+B. Then we can get that

D 0 + α 1 u ( t ) = D 0 + α 1 u ( t 0 ) + t 0 t D 0 + α u ( s ) d s , D 0 + α 2 u ( t ) = D 0 + α 2 u ( t 0 ) + t 0 t D 0 + α 1 u ( s ) d s .

Furthermore, we have that, with setting M=A+B,

(3.8)
(3.9)

By (3.7)-(3.9) and Lemma 2.4, we have that

P u Y μ ( M + N u 1 ) .

As before, for any u Ω 1 , we have (IP)(u)domLKerP and LP(u)=0. From Lemma 3.4 and for each λ(0,1], we can get

( I P ) ( u ) Y = K p L ( I P ) ( u ) Y = K p ( L u ) Y ω N u 1 .

Furthermore, we have

u Y ( I P ) ( u ) Y + P ( u ) Y ρ N u 1 +μM.

By (H1) and the definition of N, we have

u Y ρ [ h 1 u + r 1 D 0 + α 1 u s 1 D 0 + α 2 u + w 1 D 0 + α 2 u θ + D ] ,

where D= e 1 +μM/ρ. Since max{ u , D 0 + α 1 u , D 0 + α 2 u } u Y and ρ( h 1 + r 1 + s 1 )<1 hold true, we can get that

u ρ 1 ρ h 1 [ r 1 D 0 + α 1 u + s 1 D 0 + α 2 u + w 1 D 0 + α 2 u θ + D ] ,

which yield that

D 0 + α 1 u ρ 1 ρ h 1 ρ r 1 [ s 1 D 0 + α 2 u + w 1 D 0 + α 2 u θ + D ] .

Furthermore, from the previous inequalities, we know that

D 0 + α 2 u ρ 1 ρ h 1 ρ r 1 ρ s 1 ( w 1 D 0 + α 2 u θ + D ) .

Since θ[0,1), there exist constants m 1 , m 2 , m 3 >0 such that

D 0 + α 2 u m 1 , D 0 + α 1 u m 2 , u m 3 .

Therefore, Ω 1 is bounded. □

Lemma 3.7 Suppose (H 2) and (H 3) hold, then the set Ω 2 ={uKerL:NuImL}is bounded.

Proof For any u Ω 2 and a,bR, then u(t)=a t α 1 +b t α 2 and QNu=0. By (H2), we get that | D 0 + α 1 u(t)|=|aΓ(α)|A, then we have |a|A/Γ(α). By (H3), we have that | D 0 + α 2 u(t)|=|atΓ(α)+bΓ(α1)|B, thus |b|(B+A)/Γ(α1). Therefore, Ω 2 is bounded. □

Lemma 3.8 If the first parts of (H 2) and (H 3) hold, then the set Ω 3 ={uKerL:λ J 1 u+(1λ)QNu=0,λ[0,1]}is bounded.

Proof Taking any u Ω 3 and a,bR, we have u(t)=a t α 1 +b t α 2 . For all t[0,1], we define the isomorphism J 1 :KerLImQ by

J 1 ( a t α 1 + b t α 2 ) = a 2 Λ t p 1 + b 2 Λ t q 2 .

By the definition of the set Ω 3 , we can get that

{ λ a 2 + ( 1 λ ) [ Λ 2 T 1 N ( a t α 1 + b t α 2 ) + Λ 4 T 2 N ( a t α 1 + b t α 2 ) ] = 0 , λ b 2 + ( 1 λ ) [ Λ 1 T 1 N ( a t α 1 + b t α 2 ) Λ 3 T 2 N ( a t α 1 + b t α 2 ) ] = 0 .
(3.10)

If λ=0, we have

T 1 N ( a t α 1 + b t α 2 ) =0, T 2 N ( a t α 1 + b t α 2 ) =0.

By the first parts of (H2) and (H3), similar to the proof of Lemma 3.7, then

|a| A Γ ( α ) ,|b| B + A Γ ( α 1 ) .

Therefore, Ω 3 is bounded.

If λ=1, we have a=b=0.

If 0<λ<1, we get that | D α 1 u(t)|A and | D α 2 u(t)|B, similar to the proof of Lemma 3.7, Ω 3 is bounded. If else, we have that Λ 2 T 1 Nu Λ 4 T 2 Nu<0 and Λ 1 T 1 Nu Λ 3 T 2 Nu>0. It contradicts (3.10), thus Ω 3 is bounded. □

Remark 3.9 If the other parts of (H2) and (H3) hold, then the set Ω 3 ={uKerL:λ J 1 u+(1λ)QNu=0,λ[0,1]} is bounded.

Now with Lemmas 3.2-3.8 in hands, we can begin to prove our main result - Theorem 3.1.

Proof of Theorem 3.1 Assume that Ω is a bounded open set of Y with i = 1 3 Ω ¯ Ω. By Lemma 3.5, N is L-compact on Ω ¯ . Then by Lemmas 3.6 and 3.7, we have

  1. (i)

    LuλNu for every (u,λ)[domLKerLΩ]×[0,1];

  2. (ii)

    NuImL for every uKerLΩ.

Finally, we will prove that (iii) of Lemma 2.6 is satisfied. We let I as the identity operator in the Banach space Y and H(u,λ)=±λ J 1 (u)+(1λ)QN(u), according to Lemma 3.8 (or Remark 3.9) we know that for all uΩKerL, H(u,λ)0. By the homotopic property of degree, we have

deg ( J Q N | Ker L , Ker L Ω , 0 ) = deg ( H ( , 0 ) , Ker L Ω , 0 ) = deg ( H ( , 1 ) , Ker L Ω , 0 ) = deg ( ± I , Ker L Ω , 0 ) 0 ,

so (iii) of Lemma 2.6 is satisfied.

Consequently, by Lemma 2.6, the equation Lu=Nu has at least one solution in domL Ω ¯ . Namely, BVPs (1.1)-(1.2) have at least one solution in the space Y. □

According to Theorem 3.1, we have the following corollary.

Corollary 3.10 Suppose that (H 1) is replaced by the following condition,

(H4) there exist functionsh(t),r(t),s(t),w(t),e(t)L[0,1]and a constantθ[0,1)such that for all(x,y,z) R 3 , t[0,1],

| f ( t , x , y , z ) | h(t)|x|+r(t)|y|+s(t)|z|+w(t) | y | θ +e(t),

or

| f ( t , x , y , z ) | h(t)|x|+r(t)|y|+s(t)|z|+w(t) | x | θ +e(t),

and the others in Theorem 3.1 are not changed, then BVPs (1.1)-(1.2) have at least one solution.

4 An example

Example Consider the following boundary value problem for all t(0,1):

{ D 13 5 u ( t ) = 7 u ( t ) 200 + 9 D 0 + 8 5 u ( t ) 500 + D 0 + 3 5 u ( t ) 60 + 3 sin 1 7 u ( t ) 80 + cos 2 t + 1 , I 0 + 2 5 u ( 0 ) = 0 , D 0 + 8 5 u ( 1 ) = 2 5 D 0 + 8 5 u ( 1 3 ) + 3 5 D 0 + 8 5 u ( 1 2 ) , D 0 + 3 5 u ( 1 ) = 6 D 0 + 3 5 u ( 1 8 ) + 7 D 0 + 3 5 u ( 1 4 ) .
(4.1)

Let α=13/5, and a 1 =2/5, ξ 1 =1/3, a 2 =3/5, ξ 2 =1/2, b 1 =6, η 1 =1/8, b 2 =7, η 2 =1/4. We can get that the condition (C) holds, i.e., i = 1 2 a i =1, i = 1 2 b i η i =1, i = 1 2 b i =1. Moreover,

f(t,x,y,z)= 7 200 x+ 9 500 y+ 1 60 z+ 3 80 sin 1 / 7 x+ cos 2 t+1.

Thus, we have

| f ( t , x , y , z ) | 7 200 |x|+ 9 500 |y|+ 1 60 |z|+ 3 80 | x | 1 / 7 +2.

Taking h=7/200, r=9/500, s=1/60, w=3/80, e=2, Λ 1 =(1 i = 1 2 b i η i 2 )/2, Λ 2 =(1 i = 1 2 b i η i 2 )/2, Λ 3 =1 i = 1 2 a i ξ i , Λ 4 =1 i = 1 2 a i ξ i , μ=3+ Γ 1 (13/5)+ Γ 1 (8/5), ω=2+ Γ 1 (13/5), ρ=5+2 Γ 1 (13/5)+ Γ 1 (8/5), we can calculate that (H1)-(H3) hold. Furthermore, we can get

ρ ( h 1 + r 1 + s 1 ) =0.52754.

By Corollary 3.10, the BVP (4.1) has at least one solution in C 8 / 5 [0,1].

Author’s contributions

NX designed all the steps of proof in this research and also wrote the article. WBL suggested many good ideas in this article. LSX helped to draft the first manuscript and gave an example to illustrate our result. All authors read and approved the final manuscript.

References

  1. Sabatier J, Agrawal O, Machado J: Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer, Berlin; 2007.

    Book  Google Scholar 

  2. Samko S, Kilbas A, Marichev O: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, New York; 1993.

    Google Scholar 

  3. Kilbas A, Srivastava H, Trujillo J: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam; 2006.

    Google Scholar 

  4. Momani S, Odibat Z: Numerical comparison of methods for solving linear differential equations of fractional order. Chaos Solitons Fractals 2007, 31(5):1248-1255. 10.1016/j.chaos.2005.10.068

    Article  MathSciNet  Google Scholar 

  5. Bai Z, Lü H: Positive solutions for boundary value problem of nonlinear fractional differential equation. J. Math. Anal. Appl. 2005, 311(2):495-505. 10.1016/j.jmaa.2005.02.052

    Article  MathSciNet  Google Scholar 

  6. Xu X, Jiang D, Yuan C: Multiple positive solutions for the boundary value problem of a nonlinear fractional differential equation. Nonlinear Anal. 2009, 71(10):4676-4688. 10.1016/j.na.2009.03.030

    Article  MathSciNet  Google Scholar 

  7. Kaufmann E, Mboumi E: Positive solutions of a boundary value problem for a nonlinear fractional differential equation. Electron. J. Qual. Theory Differ. Equ. 2008, 3: 1-11.

    Article  MathSciNet  Google Scholar 

  8. Zhang S: Positive solutions for boundary-value problems of nonlinear fractional differential equations. Electron. J. Differ. Equ. 2006, 2006(36):1-12.

    Article  Google Scholar 

  9. Benchohra M, Hamani S, Ntouyas S: Boundary value problems for differential equations with fractional order and nonlocal conditions. Nonlinear Anal. 2009, 71(7-8):2391-2396. 10.1016/j.na.2009.01.073

    Article  MathSciNet  Google Scholar 

  10. Liang S, Zhang J: Positive solutions for boundary value problems of nonlinear fractional differential equation. Nonlinear Anal. 2009, 71(11):5545-5550. 10.1016/j.na.2009.04.045

    Article  MathSciNet  Google Scholar 

  11. El-Sayed A: Nonlinear functional differential equations of arbitrary orders. Nonlinear Anal. 1998, 33(2):181-186. 10.1016/S0362-546X(97)00525-7

    Article  MathSciNet  Google Scholar 

  12. Agarwal R, Lakshmikantham V, Nieto J: On the concept of solution for fractional differential equations with uncertainty. Nonlinear Anal. 2010, 72(6):2859-2862. 10.1016/j.na.2009.11.029

    Article  MathSciNet  Google Scholar 

  13. Chang Y, Nieto J: Some new existence results for fractional differential inclusions with boundary conditions. Math. Comput. Model. 2009, 49(3-4):605-609. 10.1016/j.mcm.2008.03.014

    Article  MathSciNet  Google Scholar 

  14. Zhang Y, Bai Z, Feng T: Existence results for a coupled system of nonlinear fractional three-point boundary value problems at resonance. Comput. Math. Appl. 2011, 61(4):1032-1047. 10.1016/j.camwa.2010.12.053

    Article  MathSciNet  Google Scholar 

  15. Kosmatov N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ. 2010, 135: 1-10.

    MathSciNet  Google Scholar 

  16. Zhang Y, Bai Z: Existence of solutions for nonlinear fractional three-point boundary value problems at resonance. J. Appl. Math. Comput. 2011, 36: 417-440. 10.1007/s12190-010-0411-x

    Article  MathSciNet  Google Scholar 

  17. Bai Z: On solutions of some fractional m -point boundary value problems at resonance. Electron. J. Qual. Theory Differ. Equ. 2010, 37: 1-15.

    Google Scholar 

  18. Bai Z, Zhang Y: The existence of solutions for a fractional multi-point boundary value problem. Comput. Math. Appl. 2010, 60(8):2364-2372. 10.1016/j.camwa.2010.08.030

    Article  MathSciNet  Google Scholar 

  19. Jiang W: The existence of solutions to boundary value problems of fractional differential equations at resonance. Nonlinear Anal. 2011, 74(5):1987-1994. 10.1016/j.na.2010.11.005

    Article  MathSciNet  Google Scholar 

  20. Mawhin J: Topological degree and boundary value problems for nonlinear differential equations. In Topological Methods for Ordinary Differential Equations. Edited by: Furi M, Zecca P. Springer, Berlin; 1993:74-142.

    Chapter  Google Scholar 

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Acknowledgement

The authors would like to acknowledge the anonymous referee for many helpful comments and valuable suggestions on this article. This work is sponsored by Fundamental Research Funds for the Central Universities (2012LWB44).

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Xu, N., Liu, W. & Xiao, L. The existence of solutions for nonlinear fractional multipoint boundary value problems at resonance. Bound Value Probl 2012, 65 (2012). https://doi.org/10.1186/1687-2770-2012-65

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