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Existence result for a nonlinear fractional differential equation with integral boundary conditions at resonance

Abstract

In this paper, we establish the existence result for nonlinear fractional differential equations with integral boundary conditions at resonance by means of coincidence degree theory. As applications, three examples are presented to illustrate the main results.

1 Introduction

In this paper, we prove the existence of solutions for fractional differential equations at resonance with integral boundary conditions given by

{ D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) , , D 0 + α ( n 1 ) u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u n 2 ( 0 ) = 0 , u ( 1 ) = ( α + m ) 0 1 t m u ( t ) d t ,
(1.1)

where 0<t<1, n1<α<n, mN, D 0 + α is the Riemann-Liouville fractional derivative, and f:[0,1]×RR is a continuous function.

During the last few years, fractional differential equations have been studied extensively due to their significant applications in various sciences such as physics, mechanics, chemistry, phenomena arising in engineering. See [13] and the references therein. Many researchers paid attention to the existence of boundary value problems for nonlinear fractional differential equations, see [48]. The authors obtained the results by using the classical tools for such problems including the Leray-Schauder nonlinear alternative theorem, etc. Recently, there have been some papers devoted to the theory of fractional differential equations with integral boundary conditions, see [911].

In [9], Cabada and Wang studied the integral boundary value problem for nonlinear fractional differential equations:

{ D 0 + α C u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = λ 0 1 u ( s ) d s ,

where 2<α<3, 0<λ<2, D 0 + α C is the Caputo fractional derivative and f:[0,1]×[0,+)[0,+) is a continuous function. The author established sufficient conditions for the existence of nonlinear fractional differential equations with an integral boundary problem.

In [10], Yuan discussed an (n1,1)-type semipositone integral boundary value problem for the coupled system of fractional differential equations

{ D 0 + α u ( t ) + λ f ( t , v ( t ) ) = 0 , 0 < t < 1 , D 0 + α v ( t ) + λ g ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( j ) ( 0 ) = v ( j ) ( 0 ) = 0 , 0 j n 2 , u ( 1 ) = μ 0 1 u ( s ) d s , v ( 1 ) = μ 0 1 v ( s ) d s ,

where λ, μ are parameters and λ>0, 0<μ<α, α(n1,n], n3, D 0 + α is the Riemann-Liouville fractional derivative, and f,g:(0,1)×[0,+)R are sign-changing continuous functions. Some existence results were given by using the Leray-Schauder nonlinear alternative theorem and the Krasnoselskii fixed point theorem.

In [11], Feng et al. considered a higher-order nonlinear nonlinear fractional differential equations with integral boundary conditions given by

{ D 0 + α x ( t ) + g ( t ) f ( t , x ( t ) ) = 0 , 0 < t < 1 , x ( 0 ) = x ( 0 ) = = x ( n 2 ) ( 0 ) = 0 , 0 < t < 1 , x ( 1 ) = 0 1 h ( t ) x ( t ) d t ,

where 0<t<1, n1<αn, n3, D 0 + α is the Riemann-Liouville fractional derivative, and gC((0,1),[0,+)) which may be singular at t=0 or (and) t=1, h L 1 [0,1] is nonnegative, and fC([0,1]×[0,+),[0,+)) is a continuous function. By means of the Guo-Krasnoselskii fixed point theorem, some results on the existence of positive solutions are obtained.

At the same time, we notice that these problems happen to be at resonance if λ=2 in [9], μ=α in [10] or 0 1 h(t) t α 1 dt=1 in [11]. Though there has been an increasing interest in the study of the existence of solutions for fractional differential equations at resonance (see [1217]), the corresponding theorem and applications to the fractional differential equations with integral conditions at resonance have not been investigated now. Motivated by the works above, we discuss the integral boundary value problems of fractional differential equations (1.1), which happen to be at resonance in the sense that the associated linear homogeneous boundary value problem

{ D 0 + α u ( t ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 0 ) = = u n 2 ( 0 ) = 0 , u ( 1 ) = ( α + m ) 0 1 t m u ( t ) d t ,

has u(t)=c t α 1 , cR as a nontrivial solution.

The rest of this paper is organized as follows. In Section 2, we give some necessary notations, definitions and lemmas. In Section 3, we study the existence of solutions of (1.1) by the coincidence degree theory due to Mawhin [18]. Finally, three examples are given to illustrate our results in Section 4.

2 Preliminaries

In this section, we present the necessary definitions and lemmas from fractional calculus theory. These definitions and properties can be found in the literature. For more details, see [13].

Definition 2.1 [1]

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

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

provided that the right-hand side is pointwise defined on (0,).

Definition 2.2 [1]

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

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

where n1<αn, provided that the right-hand side is pointwise defined on (0,).

Lemma 2.1 [1]

Let n1<αn, uC(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 ,

where C i R, i=1,2,,n.

Lemma 2.2 [1]

If α>0, mN and D=d/dx. If the fractional derivatives D 0 + α u(t) and D 0 + α + m u(t) exist, then

D m D 0 + α u(t)= D 0 + α + m u(t).

Lemma 2.3 [1]

The relation

I a + α I a + β f(x)= I a + α + β f(x)

is valid in the case β>0, α+β>0, f(x) L 1 (a,b).

Lemma 2.4 The relation

0 r I 0 + α f(x)dx= I 0 + α + 1 f(r)

is valid in the case α>0, f(x) L 1 (0,r), r R + .

Proof

0 r I 0 + α f ( x ) d x = 1 Γ ( α ) 0 r 0 t ( t s ) α 1 f ( s ) d s d t = 1 Γ ( α ) 0 r s r ( t s ) α 1 f ( s ) d t d s = 1 Γ ( α ) 0 r f ( s ) s r ( t s ) α 1 d t d s = 1 Γ ( α ) 0 r f ( s ) ( r s ) α α d s = I 0 + α + 1 f ( r ) .

 □

Now let us recall some notation about the coincidence degree continuation theorem. Let Y, Z be real Banach spaces, L:domLYZ be a Fredholm map of index zero and P:YY, Q:ZZ be continuous projectors such that kerL=ImP, ImL=kerQ and Y=kerLkerP, Z=ImLImQ. It follows that L | dom L ker P :domLkerPImL is invertible. We denote the inverse of this map by K P . If Ω is an open bounded subset of Y, the map N will be called L-compact on Ω ¯ if QN( Ω ¯ ) is bounded and K P , Q N= K P (IQ)N: Ω ¯ Y is compact.

Theorem 2.1 Let L be a Fredholm operator of index zero and N be L-compact on Ω ¯ . Suppose that the following conditions are satisfied:

  1. (1)

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

  2. (2)

    NxImL for each xkerLΩ;

  3. (3)

    deg(JQN | ker L ,ΩkerL,0)0, where Q:ZZ is a continuous projection as above with ImL=kerQ and J:ImQkerL is any isomorphism.

Then the equation Lx=Nx has at least one solution in domL Ω ¯ .

3 Main results

In this paper, we always assume the following conditions.

(H0) 1 Γ ( α + 1 ) (α+m) i = 0 m ( 1 ) i m ! ( m i ) ! 1 Γ ( α + 2 + i ) .

In this section, we prove the existence results for (1.1). We use the Banach space E=C[0,1] with the norm u = max 0 t 1 |u(t)|. For α>0, n=[α]+1, we define a linear space

Y= { u | u , D 0 + α i u E , i = 1 , 2 , , n 1 } .
(3.1)

By means of the line functional analysis theory, we can prove that Y is a Banach space with the norm

u Y = D 0 + α 1 u ++ D 0 + α ( n 1 ) u + u .

Define L to be the linear operator from Y to E with

dom(L)= { u Y | u ( 0 ) = u ( 0 ) = = u n 2 ( 0 ) = 0 ; u ( 1 ) = ( α + m ) 0 1 t m u ( t ) d t }

and

Lu= D 0 + α u,udom(L).
(3.2)

Define N:YE by

Nu(t)=f ( t , D 0 + α 1 u ( t ) , , D 0 + α ( n 1 ) u ( t ) ) .
(3.3)

Then problem (1.1) can be written by Lu=Nu.

Lemma 3.1 The mapping L:dom(L)Z is a Fredholm operator of index zero.

Proof It is clear that

ker(L)= { c t α 1 , c R } R 1 .

Let xImL, so there exists a function udomL which satisfies Lu=x. By Lemma 2.1, we have

u(t)= I 0 + α x(t)+ c 1 t α 1 + c 2 t α 2 ++ c n t α n .

By u(0)= u (0)== u ( n 2 ) (0)=0, we can obtain c 2 == c n =0. Hence

u(t)= I 0 + α x(t)+ c 1 t α 1 .

By u(1)=(α+m) 0 1 t m u(t)dt, we have

I 0 + α x ( 1 ) + c 1 = u ( 1 ) = ( α + m ) 0 1 t m [ I 0 + α x ( t ) + c 1 t α 1 ] d t = ( α + m ) 0 1 t m I 0 + α x ( t ) d t + c 1 0 1 ( α + m ) t α + m 1 d t = ( α + m ) 0 1 t m I 0 + α x ( t ) d t + c 1 .

Taking into account Lemma 2.2 and using integration by parts, we obtain

I 0 + α x ( 1 ) = ( α + m ) 0 1 t m I 0 + α x ( t ) d t = ( α + m ) 0 1 t m d I 0 + α + 1 x ( t ) = ( α + m ) [ t m I 0 + α + 1 x ( t ) | 0 1 m 0 1 t m 1 I 0 + α + 1 x ( t ) d t ] = ( α + m ) [ I 0 + α + 1 x ( 1 ) m 0 1 t m 1 d I 0 + α + 2 x ( t ) ] = ( α + m ) i = 0 m ( 1 ) i m ! ( m i ) ! I 0 + α + 1 + i x ( 1 ) .

On the other hand, suppose that xE satisfies I 0 + α x(1)(α+m) i = 0 m ( 1 ) i m ! ( m i ) ! I 0 + α + 1 + i x(1)=0. Let u(t)= I 0 + α x(t), we can easily prove u(t)dom(L).

Thus, we conclude that

Im(L)= { x E | I 0 + α x ( 1 ) ( α + m ) i = 0 m ( 1 ) i m ! ( m i ) ! I 0 + α + 1 + i x ( 1 ) = 0 } .

Consider the linear operators Q:ZR defined by

Qx(t)=σ [ I 0 + α x ( 1 ) ( α + m ) i = 0 m ( 1 ) i m ! ( m i ) ! I 0 + α + 1 + i x ( 1 ) ] ,

where σ= [ 1 α Γ ( a ) ( α + m ) i = 0 m ( 1 ) i m ! ( m i ) ! 1 Γ ( α + 2 + i ) ] 1 and Qx(t)R.

Take x(t)E, then

Q ( Q x ( t ) ) = σ [ I 0 + α Q x ( t ) | t = 1 ( α + m ) i = 0 m ( 1 ) i m ! ( m i ) ! I 0 + α + 1 + i Q x ( t ) | t = 1 ] = σ 1 σ Q x ( t ) = Q x ( t ) .

We can see Q 2 =Q.

For x(t)Z, x(t)=x(t)Qx(t)+Qx(t), one has Q[x(t)Qx(t)]=0.

Therefore

x(t)Qx(t)Ker(Q)=Im(L),Qx(t)Im(Q).

That is to say,

Z=Im(L)+Im(Q).

If uIm(L)Im(Q), so we have u=cR.

Because of u=cIm(L), then

I 0 + α c | t = 1 (α+m) i = 0 m ( 1 ) i m ! ( m i ) ! I 0 + α + 1 + i c | t = 1 =0.

As a result cσ=0, we get c=0.

Then

Z=Im(L)Im(Q).

Note that IndL=dimkerLcodimImL=0. Then L is a Fredholm mapping of index zero. □

We can define the operators P:YY, where

Pu= 1 Γ ( α ) D 0 + α 1 u(0) t α 1 .

For uY,

P(Pu)=P ( 1 Γ ( α ) D 0 + α 1 u ( 0 ) t α 1 ) = 1 Γ ( α ) D 0 + α 1 u(0) t α 1 =Pu.

So we have P 2 =P.

Note that

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

Since u=uPu+Pu, it is easy to say uPuKer(P), PuKer(L). And Y=Ker(P)+Ker(L). If uKer(L)Ker(P), so u= c 1 t α 1 , we can derive D α 1 c 1 t α 1 | t = 0 =0, so c 1 =0. Then

Y=Ker(L)Ker(P).

For uY,

P u Y = D 0 + α 1 u + + D 0 + α ( n 1 ) u + u = 1 Γ ( α ) | D 0 + α 1 u ( 0 ) | ( t α 1 + D 0 + α 1 t α 1 + + D 0 + α ( n 1 ) t α 1 ) = ( i = 1 n 1 1 Γ ( i ) + 1 Γ ( α ) ) | D 0 + α 1 u ( 0 ) | = a | D 0 + α 1 u ( 0 ) | ,
(3.4)

where a= 1 Γ ( α ) + i = 1 n 1 1 Γ ( i ) .

We define K P :ImLdomLkerP by K p x= I 0 + α x. For xIm(L), we have L K p x= D 0 + α I 0 + α x=x.

And for udom(L)Ker(P), we have D 0 + α 1 u(0)=0. So the coefficients c 1 ,, c n in the expressions

u= I 0 + α D 0 + α u(t)+ c 1 t α 1 + c 2 t α 2 ++ c n t α n

are all equal to zero. Thus we obtain

K p Lu= I 0 + α D 0 + α u=u,

which shows that K p = ( L dom ( L ) Ker ( P ) ) 1 .

Again, for each xIm(L),

K p x Y = I 0 + α x Y = I 0 + α x + D 0 + α 1 I 0 + α x + + D 0 + α ( n 1 ) I 0 + α x ( i = 1 n 1 1 Γ ( i + 1 ) + 1 Γ ( α + 1 ) ) x ( i = 1 n 1 1 Γ ( i ) + 1 Γ ( α ) ) x = a x .
(3.5)

Lemma 3.2 [17]

F C μ [0,1] is a sequentially compact set if and only if F is uniformly bounded and equicontinuous. Here uniformly bounded means that there exists M>0 such that for every uF,

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

and equicontinuous means that ϵ>0, δ>0 such that

|u( t 1 )u( t 2 )|<ϵ ( t 1 , t 2 [ 0 , 1 ] , | t 1 t 2 | < δ , u F )

and

| D 0 + α i u( t 1 ) D 0 + α i u( t 2 )|<ϵ ( t 1 , t 2 [ 0 , 1 ] , | t 1 t 2 | < δ , u F , i = 0 , 1 , , N 1 ) .

By Lemma 3.2 and standard arguments, the following lemma holds.

Lemma 3.3 K p (IQ)N:YY is completely continuous.

For convenience, we use the following notations: ρ 0 =aA+2a φ ; ρ i =2a ψ i , i=1,2,,n. To obtain our main results, we need the following conditions.

  • (H1) There exist functions φ, ψ i ,E, i=1,2,,n, such that for all t[0,1],

    |f(t, x 1 , x 2 ,, x n )|φ+ ψ 1 | x 1 |+ ψ 2 | x 2 |++ ψ n | x n |.
  • (H2) There exists a constant A>0 such that for every yR, if | D 0 + α 1 u(t)|>A, for all t[0,1], then

    QNu(t)0.
  • (H3) There exists a constant B>0 such that, for each cR satisfying |c|>B, we have at least one of the following:

    cQN ( c t α 1 ) >0,

    or

    cQN ( c t α 1 ) <0.
  • (H4) i = 1 n ρ i <1.

Lemma 3.4 Ω 1 ={udom(L)Ker(L)|Lu=λNu,λ[0,1]} is bounded.

Proof For u Ω 1 , Lu=λNu, thus λ0. And Lu=λNuIm(L)=Ker(Q), that is,

λQNu(t)=0.

By (H2), there exists a constant t 0 [0,1] such that | D 0 + α 1 u( t 0 )|A.

Again, for u Ω 1 , then (IP)udom(L)Ker(L) and LPu=0, thus from (3.5), we have

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

Now, by Lemma 2.2,

| D 0 + α 1 u ( 0 ) | | D 0 + α 1 u ( t 0 ) | + | 0 t 0 D 0 + α u ( s ) d s | | D 0 + α 1 u ( t 0 ) | + | t 0 | max 0 t t 0 | D 0 + α u ( t ) | | D 0 + α 1 u ( t 0 ) | + D 0 + α u ( t ) A + L u A + N u ,

that is,

| D 0 + α 1 u(0)|A+ N u .
(3.6)

From (3.4), we have

u Y = P u + ( I P ) u Y P u Y + ( I P ) u Y a | D 0 + α 1 u ( 0 ) | + a N u a ( A + N u ) + a N u a A + 2 a N u .

By (H1), we have

u Y a A + 2 a N u a A + 2 a f ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) , , D 0 + α ( n 1 ) u ( t ) ) a A + 2 a ( φ + ψ 1 u + ψ 2 D 0 + α 1 u + + ψ n D 0 + α ( n 1 ) u ) = ρ 0 + ρ 1 u + ρ 2 D 0 + α 1 u + ρ 3 D 0 + α 2 u + + ρ n D 0 + α ( n 1 ) u .

By the definition u Y , we can derive

u 1 1 ρ 1 ( ρ 0 + ρ 2 D 0 + α 1 u + ρ 3 D 0 + α 2 u + + ρ n D 0 + α ( n 1 ) u ) .

Hence, we have

D 0 + α 1 u 1 1 ρ 1 ρ 2 ( ρ 0 + ρ 3 D 0 + α 2 u + + ρ n D 0 + α ( n 1 ) u ) , D 0 + α ( n 1 ) u ρ 0 1 ρ 1 ρ 2 ρ n : = M .

It is easy to see D 0 + α 1 v ,, D 0 + α ( n 1 ) v and u are bounded. We can conclude that u Y is bounded. □

Lemma 3.5 Ω 2 ={uKer(L):NuIm(L)} is bounded.

Proof Let uKer(L), so we have u=c t α 1 , cR, t[0,1], and NuIm(L)=Ker(Q), thus

λQNu(t)=0.

By (H2), there exist two constant t 0 [0,1] such that

| D 0 + α 1 u( t 0 )|=| D 0 + α 1 c t α 1 | t 0 |=|cΓ(α)|A.

It follows that |c| A Γ ( α ) . Here, Ω 2 is bounded. □

Lemma 3.6 Ω 3 ={uKer(L):λu+(1λ)QNu=(0,0),λ[0,1]} is bounded.

Proof Let uKer(L), so we have u=c t α 1 , cR, t[0,1].

Thus

λc t α 1 +(1λ)QN ( t α 1 ) =0.

If λ=0, then |c| A Γ ( α ) . If λ=1, we have c=0.

It follows that

λ c 2 t α 1 +(1λ)cQN ( t α 1 ) =0.

Thus from the first part of (H3), then |c|B. Here, Ω 3 is bounded. □

Remark 3.1 If the other parts of (H3) hold, then the set Ω 3 ={uKer(L):λu+(1λ)QNu=0,λ[0,1]} is bounded.

Theorem 3.1 Suppose (H1)-(H3) hold, then problem (1.1) has at least one solution in Y.

Proof Let Ω be a bounded open set of Y such that i = 1 3 Ω ¯ i Ω. It follows from Lemma 3.3 that N is L-compact on Ω. By Lemma 3.4, Lemma 3.5 and Lemma 3.6, we get:

  1. (1)

    LuλNu for every (u,v)[(domLKerL)Ω]×(0,1),

  2. (2)

    NuImL for every uKerLΩ,

  3. (3)

    Let H((u,v),λ)=±λI(u,v)+(1λ)JQN(u,v), where I is the identical operator. Via the homotopy property of degree, we obtain that

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

    Applying Theorem 2.1, we conclude that Lu=Nu has at least one solution in domL Ω ¯ .

 □

4 Some examples

To illustrate how our main results can be used in practice, we present three examples.

Example 4.1 Let us consider the following boundary value problem:

{ D 0 + 2.5 u ( t ) = t 2 + e u 2 ( t ) + 1 6 | D 0 + 1.5 u ( t ) | + cos ( D 0 + 0.5 u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 5 2 0 1 u ( t ) d t .
(4.1)

By the calculation, we can have A=B=7 and (H0)-(H4) are all satisfied. Hence, we obtain that (4.1) has at least one solution.

Example 4.2 Consider the following integral boundary value problem:

{ D 0 + 3.6 u ( t ) = t 2 + arctan u ( t ) + 1 10 D 0 + 2.6 u ( t ) + cos ( D 0 + 0.6 u ( t ) ) + sin ( D 0 + 1.6 u ( t ) ) , 0 < t < 1 , u ( 0 ) = u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 23 5 0 1 t u ( t ) d t .
(4.2)

It is easy to see that (H0)-(H4) hold. By a simple calculation, we can have A=48, B=14. Thus it follows that problem (4.2) has at least one solution.

Example 4.3 We consider the boundary value problem

{ D 0 + 1.2 u ( t ) = cos ( t + 1 ) 2 + 1 8 sin u ( t ) + 1 9 D 0 + 0.2 u ( t ) , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = 16 5 0 1 t 2 u ( t ) d t .
(4.3)

Corresponding to problem (1.1), we have that α=1.2, m=2. By a simple calculation, we can get A=11, B=12, then (H0)-(H4) hold. By Theorem 3.1, we obtain that (4.3) has at least one solution.

5 Conclusion

In this paper, we have obtained the existence of solutions for fractional differential equations with integral boundary conditions at resonance. By using the coincidence degree theory, we have found the existence results. Though the technique applied to establish the existence results for the problem is a standard one, yet its results are new in the context of integral boundary conditions. As applications, three examples are presented to illustrate the main results. In the future, we will consider the uniqueness of solutions for the fractional differential equations at resonance, and also can make further research on fractional differential equations with a singular integral boundary value problem at resonance.

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Acknowledgements

The authors are grateful to the reviewers for their valuable suggestions and detailed comments which have improved the paper very much. This research is supported by the Natural Science Foundation of China (11371364) and 2013 Science and Technology Research Project of Beijing Municipal Education Commission (KM201310016001).

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Correspondence to Lei Hu.

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Zhang, S., Hu, L. & Shi, A. Existence result for a nonlinear fractional differential equation with integral boundary conditions at resonance. Adv Differ Equ 2013, 353 (2013). https://doi.org/10.1186/1687-1847-2013-353

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Keywords

  • Fractional Derivative
  • Existence Result
  • Fixed Point Theorem
  • Fractional Differential Equation
  • Integral Boundary