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Theory and Modern Applications

Solvability for a coupled system of fractional differential equations with integral boundary conditions at resonance

Abstract

By constructing suitable operators, we investigate the existence of solutions for a coupled system of fractional differential equations with integral boundary conditions at resonance. Our analysis relies on the coincidence degree theory due to Mawhin. An example is given to illustrate our main result.

MSC:34A08, 70K30, 34B10.

1 Introduction

Fractional differential equations arise in a variety of different areas such as rheology, fluid flows, electrical networks, viscoelasticity, chemical physics, electron-analytical chemistry, biology, control theory etc. (see [1, 2]). Recently, more and more authors have paid their close attention to them (see [324]). The existence of solutions for differential equations at resonance has been studied by many authors (see [1923, 2529] and references cited therein). In papers [1922], the authors investigated the fractional differential equations with multi-point boundary conditions at resonance. In paper [23], the authors discussed a coupled system of fractional differential equations with two-point boundary condition at resonance. In paper [24], the authors showed the existence of solutions for higher-order fractional differential inclusions with multi-strip fractional integral boundary conditions. In paper [26], the authors studied solvability of integer-order differential equations with integral boundary conditions at resonance, which was the generalization of two, three, multi-point and nonlocal boundary value problems.

Motivated by the excellent results mentioned above, in this paper, we discuss the existence of solutions for a coupled system of fractional differential equations with integral boundary conditions at resonance

{ D 0 + α x ( t ) = f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) , 0 < t < 1 , D 0 + β y ( t ) = f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) , 0 < t < 1 , x ( 0 ) = 0 , D 0 + α 1 x ( 0 ) = 0 1 h 1 ( t ) D 0 + α 1 x ( t ) d t , D 0 + α 1 x ( 1 ) = 0 1 h 2 ( t ) D 0 + α 1 x ( t ) d t , y ( 0 ) = 0 , D 0 + β 1 y ( 0 ) = 0 1 g 1 ( t ) D 0 + β 1 y ( t ) d t , D 0 + β 1 y ( 1 ) = 0 1 g 2 ( t ) D 0 + β 1 y ( t ) d t ,
(1.1)

where 2<α,β3, D 0 + α is the standard Riemann-Liouville fractional derivative, D 0 + γ u(ξ):= D 0 + γ u(t) | t = ξ . To the best of our knowledge, this is the first paper to study the boundary value problems of a coupled system of fractional differential equations with integral boundary conditions at resonance with dimKerL=4.

In this paper, we will always suppose that the following conditions hold.

( H 1 ) 2<α,β3, h i , g i L[0,1], 0 1 h i (t)dt=1, 0 1 g i (t)dt=1, i=1,2.

( H 2 )

Δ 1 = | 0 1 t h 1 ( t ) d t 1 0 1 t h 2 ( t ) d t 1 2 0 1 t 2 h 1 ( t ) d t 1 2 ( 1 0 1 t 2 h 2 ( t ) d t ) | : = | Δ 11 Δ 12 Δ 21 Δ 22 | 0 , Δ 2 = | 0 1 t g 1 ( t ) d t 1 0 1 t g 2 ( t ) d t 1 2 0 1 t 2 g 1 ( t ) d t 1 2 ( 1 0 1 t 2 g 2 ( t ) d t ) | : = | δ 11 δ 12 δ 21 δ 22 | 0 .

( H 3 ) f i :[0,1]× R 3 R satisfies the Carathéodory conditions and there exist functions a 0 i (t), b 0 i (t), c 0 i (t), d 0 i (t), r i (t)L[0,1] and constants η 1 , η 2 (0,1) with c 0 <1, c 0 <1, 1 Γ ( α ) ( 1 c 0 ) (2+ 1 η 1 α 2 ) a 0 <1, 1 Γ ( β ) ( 1 c 0 ) (2+ 1 η 2 β 2 ) b 0 <1, A 1 A 2 a 0 b 0 <1 such that

| f 1 ( t , x , y , z ) | a 01 ( t ) | x | + b 01 ( t ) | y | + c 01 ( t ) | z | + d 01 ( t ) | x | θ 1 + r 1 ( t ) , | f 2 ( t , x , y , z ) | a 02 ( t ) | x | + b 02 ( t ) | y | + c 02 ( t ) | z | + d 02 ( t ) | y | θ 2 + r 2 ( t ) ,

where a 0 = 0 1 a 01 (t)dt, b 0 = 0 1 b 01 (t)dt, c 0 = 0 1 c 01 (t)dt, d 0 = 0 1 d 01 (t)dt, r 0 = 0 1 r 1 (t)dt, a 0 = 0 1 a 02 (t)dt, b 0 = 0 1 b 02 (t)dt, c 0 = 0 1 c 02 (t)dt, d 0 = 0 1 d 02 (t)dt, r 0 = 0 1 r 2 (t)dt, 0 θ 1 , θ 2 <1, A 1 = 2 η 1 α 2 + 1 Γ ( α ) ( 1 c 0 ) η 1 α 2 a 0 ( 2 η 1 α 2 + 1 ) , A 2 = 2 η 2 β 2 + 1 Γ ( β ) ( 1 c 0 ) η 2 β 2 b 0 ( 2 η 2 β 2 + 1 ) .

2 Preliminaries

For convenience, we introduce some notations and a theorem. For more details, see [30].

Let X and Y be real Banach spaces and L:dom(L)XY be a Fredholm operator with index zero, let P:XX, Q:YY be projectors such that

ImP=KerL,KerQ=ImL,X=KerLKerP,Y=ImLImQ.

It follows that

L | dom L Ker P :domLKerPImL

is invertible. We denote the inverse by K P .

Assume that Ω is an open bounded subset of X, domL Ω ¯ . The map N:XY will be called L-compact on Ω ¯ if QN( Ω ¯ ) is bounded and K P (IQ)N: Ω ¯ X is compact.

Theorem 2.1 [30]

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

  1. (1)

    LxλNx for every (x,λ)[(domLKerL)Ω]×(0,1);

  2. (2)

    NxImL for every xKerLΩ;

  3. (3)

    deg(QN | Ker L ,ΩKerL,0)0, where Q:YY is a projection such that ImL=KerQ.

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

The following definitions and lemmas can be found in [1, 2].

Definition 2.1 The 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,
(2.1)

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

Definition 2.2 The 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 n d t n 0 t ( t s ) n α 1 y(s)ds,
(2.2)

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

Lemma 2.1 Assume fL[0,1], qp0, q>1, then

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

Lemma 2.2 Assume α>0, λ>1, then

D 0 + α t λ = Γ ( λ + 1 ) Γ ( n + λ α + 1 ) d n d t n ( t n + λ α ) ,

where n is the smallest integer greater than or equal to α.

Lemma 2.3 D 0 + α u(t)=0 if and only if

u(t)= c 1 t α 1 + c 2 t α 2 ++ c n t α n ,

where n is the smallest integer greater than or equal to α, c i R, i=1,2,,n.

Take X= C α 1 [0,1]× C β 1 [0,1] with the norm

( x , y ) =max { x , y , D 0 + α 1 x , D 0 + β 1 y } ,

where C α 1 [0,1]={xx, D 0 + α 1 xC[0,1]}, x = max t [ 0 , 1 ] |x(t)|. Set Y=L[0.1]×L[0.1] with the norm

( f , g ) =max { 0 1 | f ( x ) | d x , 0 1 | g ( x ) | d x } .

Define operators L:domLXY, N:XY as follows:

L(x,y)= ( D 0 + α x , D 0 + β y ) ,(x,y)domL,N(x,y)= ( N 1 ( x , y ) , N 2 ( x , y ) ) ,(x,y)X,

where

dom L = { ( x , y ) | ( x , y ) X , ( D 0 + α x , D 0 + β y ) Y , x ( 0 ) = y ( 0 ) = 0 , D 0 + α 1 x ( 0 ) = 0 1 h 1 ( t ) D 0 + α 1 x ( t ) d t , D 0 + α 1 x ( 1 ) = 0 1 h 2 ( t ) D 0 + α 1 x ( t ) d t , D 0 + β 1 y ( 0 ) = 0 1 g 1 ( t ) D 0 + β 1 y ( t ) d t , D 0 + β 1 y ( 1 ) = 0 1 g 2 ( t ) D 0 + β 1 y ( t ) d t } ,

N 1 (x,y)= f 1 (t,x(t),y(t), D 0 + α 1 x(t)), N 2 (x,y)= f 2 (t,x(t),y(t), D 0 + β 1 y(t)). Then problem (1.1) is L(x,y)=N(x,y).

By Lemma 2.3 in [20], we get that X is a Banach space.

Definition 2.3 (x,y)domL is a solution of problem (1.1) if it satisfies (1.1), i.e., L(x,y)=N(x,y).

3 Main result

Define operators T i :L[0,1]R, i=1,2,3,4, and Q j :L[0,1]L[0,1], j=1,2 as follows:

T 1 u = 0 1 u ( t ) t 1 h 1 ( s ) d s d t , T 2 u = 0 1 u ( t ) 0 t h 2 ( s ) d s d t , T 3 u = 0 1 u ( t ) t 1 g 1 ( s ) d s d t , T 4 u = 0 1 u ( t ) 0 t g 2 ( s ) d s d t , Q 1 u = 1 Δ 1 ( Δ 22 T 1 u Δ 21 T 2 u ) + 1 Δ 1 ( Δ 11 T 2 u Δ 12 T 1 u ) t , Q 2 u = 1 Δ 2 ( δ 22 T 3 u δ 21 T 4 u ) + 1 Δ 2 ( δ 11 T 4 u δ 12 T 3 u ) t .

It is clear that Δ 11 = T 1 1, Δ 12 = T 2 1, Δ 21 = T 1 t, Δ 22 = T 2 t.

Lemma 3.1 If ( H 1 ) and ( H 2 ) hold, then L:domLXY is a Fredholm operator of index zero, the linear continuous projectors P:XX and Q:YY can be defined as

P ( x , y ) = ( D 0 + α 1 x ( 0 ) Γ ( α ) t α 1 + D 0 + α 2 x ( 0 ) Γ ( α 1 ) t α 2 , D 0 + β 1 y ( 0 ) Γ ( β ) t β 1 + D 0 + β 2 y ( 0 ) Γ ( β 1 ) t β 2 ) , Q ( u , v ) = ( Q 1 u , Q 2 v ) ,

respectively, and the linear operator K P :ImLdomLKerP can be written by

K P (u,v)= ( I 0 + α u , I 0 + β v ) .

Proof We can easily get that

KerL= { ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) c 1 , c 2 , d 1 , d 2 R } .

Obviously, ImP=KerL, P 2 (u,v)=P(u,v).

By a simple calculation, we obtain that

ImL= { ( u , v ) Y T 1 u = T 2 u = T 3 v = T 4 v = 0 }

and Q 2 (u,v)=Q(u,v). By ( H 2 ), we have ImL=KerQ. It is clear that

X=KerPKerL,Y=ImLImQ.

This means that L is a Fredholm operator of index zero.

For (u,v)ImL, we can easily get that K P (u,v)=( I 0 + α u, I 0 + β v)domLKerP. Obviously, L K P (u,v)=(u,v), (u,v)ImL. For (x,y)domLKerP, by Lemma 2.3 and K P L(x,y)domL, we get that

K P L(x,y)= ( x ( t ) + c 1 t α 1 + c 2 t α 2 , y ( t ) + d 1 t β 1 + d 2 t β 2 ) .

It follows from (x,y)KerP that D 0 + α 1 x(0)= D 0 + α 2 x(0)= D 0 + β 1 y(0)= D 0 + β 2 y(0)=0. This, together with K P L(x,y)KerP, means that c 1 = c 2 = d 1 = d 2 =0. So, K P L(x,y)=(x,y). Therefore, K P = ( L | dom L Ker P ) 1 . The proof is completed. □

Lemma 3.2 Suppose that ( H 1 ), ( H 2 ) and ( H 3 ) hold. If ΩX is an open bounded subset and domL Ω ¯ , then N is L-compact on Ω ¯ .

Proof Since Ω is bounded, there exists a constant r>0 such that (x,y)<r, (x,y) Ω ¯ . It follows from condition ( H 3 ) that there exist functions Φ i L[0,1] such that | f i (t,x,y,z)| Φ i (t) for all |x|,|y|,|z|[0,r], a.e. t[0,1], i=1,2. Thus,

| T 1 N 1 ( x , y ) | = | 0 1 N 1 ( x , y ) t 1 h 1 ( s ) d s d t | | T 1 N 1 ( x , y ) | 0 1 Φ 1 ( t ) d t 0 1 | h 1 ( s ) | d s < + , ( x , y ) Ω ¯ , | T 2 N 1 ( x , y ) | = | 0 1 N 1 ( x , y ) 0 t h 2 ( s ) d s d t | | T 2 N 1 ( x , y ) | 0 1 Φ 1 ( t ) d t 0 1 | h 2 ( s ) | d s < + , ( x , y ) Ω ¯ , | T 3 N 2 ( x , y ) | = | 0 1 N 2 ( x , y ) t 1 g 1 ( s ) d s d t | | T 3 N 2 ( x , y ) | 0 1 Φ 2 ( t ) d t 0 1 | g 1 ( s ) | d s < + , ( x , y ) Ω ¯ , | T 4 N 2 ( x , y ) | = | 0 1 N 2 ( x , y ) 0 t g 2 ( s ) d s d t | | T 4 N 2 ( x , y ) | 0 1 Φ 2 ( t ) d t 0 1 | g 2 ( s ) | d s < + , ( x , y ) Ω ¯ .

These mean that there exist constants a i >0, b i >0, i=1,2, such that

| Q 1 N 1 ( x , y ) | a 1 + b 1 t, | Q 2 N 2 ( x , y ) | a 2 + b 2 t,(x,y) Ω ¯ ,t[0,1],

i.e., QN( Ω ¯ )Y is bounded. Now we will prove that K P (IQ)N: Ω ¯ X is compact.

Obviously, K P (IQ)N( Ω ¯ ) is bounded. For 0 t 1 < t 2 1, (x,y) Ω ¯ , we have

K P ( I Q ) N ( x , y ) ( t 2 ) K P ( I Q ) N ( x , y ) ( t 1 ) = ( I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 2 ) , I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 2 ) ) ( I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 1 ) , I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 1 ) ) = ( I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 2 ) I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 1 ) , I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 2 ) I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 1 ) ) ,

where I 0 :L[0,1]L[0,1] is an identical mapping.

It follows from

| I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 2 ) I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 1 ) | = 1 Γ ( α ) | 0 t 2 ( t 2 s ) α 1 ( I 0 Q 1 ) N 1 ( x ( s ) , y ( s ) ) d s 0 t 1 ( t 1 s ) α 1 ( I 0 Q 1 ) N 1 ( x ( s ) , y ( s ) ) d s | 1 Γ ( α ) [ 0 t 1 ( ( t 2 s ) α 1 ( t 1 s ) α 1 ) ( Φ 1 ( s ) + a 1 + b 1 s ) d s + t 1 t 2 ( Φ 1 ( s ) + a 1 + b 1 s ) d s ] , | D 0 + α 1 I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 2 ) D 0 + α 1 I 0 + α ( I 0 Q 1 ) N 1 ( x , y ) ( t 1 ) | = | 0 t 2 ( I 0 Q 1 ) N 1 ( x ( s ) , y ( s ) ) d s 0 t 1 ( I 0 Q 1 ) N 1 ( x ( s ) , y ( s ) ) d s | t 1 t 2 ( Φ 1 ( s ) + a 1 + b 1 s ) d s , | I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 2 ) I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 1 ) | = 1 Γ ( β ) | 0 t 2 ( t 2 s ) β 1 ( I 0 Q 2 ) N 2 ( x ( s ) , y ( s ) ) d s 0 t 1 ( t 1 s ) β 1 ( I 0 Q 2 ) N 2 ( x ( s ) , y ( s ) ) d s | 1 Γ ( β ) [ 0 t 1 ( ( t 2 s ) β 1 ( t 1 s ) β 1 ) ( Φ 2 ( s ) + a 2 + b 2 s ) d s + t 1 t 2 ( Φ 2 ( s ) + a 2 + b 2 s ) d s ] , | D 0 + β 1 I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 2 ) D 0 + β 1 I 0 + β ( I 0 Q 2 ) N 2 ( x , y ) ( t 1 ) | = | 0 t 2 ( I 0 Q 2 ) N 2 ( x ( s ) , y ( s ) ) d s 0 t 1 ( I 0 Q 2 ) N 2 ( x ( s ) , y ( s ) ) d s | t 1 t 2 ( Φ 2 ( s ) + a 2 + b 2 s ) d s ,

the uniform continuity of ( t s ) α 1 and ( t s ) β 1 on [0,1]×[0,1], the absolute continuity of integral of Φ i + a i + b i t on [0,1], i=1,2, and the Ascoli-Arzela theorem that K P (IQ)N: Ω ¯ X is compact. The proof is completed. □

In order to obtain our main results, we present the following conditions.

( H 4 ) There exist constants M i >0, L i >0, i=1,2, such that if either

min t [ η 1 , 1 ] | x ( t ) | > M 1 or min t [ η 1 , 1 ] | D 0 + α 1 x ( t ) | > L 1 ,

then either

0 1 f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) t 1 h 1 (s)dsdt0

or

0 1 f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) 0 t h 2 (s)dsdt0,

and if either

min t [ η 2 , 1 ] | y ( t ) | > M 2 or min t [ η 2 , 1 ] | D 0 + β 1 y ( t ) | > L 2 ,

then either

0 1 f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) t 1 g 1 (s)dsdt0

or

0 1 f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) 0 t g 2 (s)dsdt0,

where η i , i=1,2, are the same as in ( H 3 ).

( H 5 ) For ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 )KerL, there exist constants k 1 , k 2 , l 1 , l 2 such that either (1) or (2) holds, where

( 1 ) c 1 T 1 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) > 0 , if  | c 1 | > k 1 , c 2 T 2 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) > 0 , if  | c 1 | k 1 , | c 2 | > k 2 , d 1 T 3 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) > 0 , if  | d 1 | > l 1 , d 2 T 4 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) > 0 , if  | d 1 | l 1 , | d 2 | > l 2 . ( 2 ) c 1 T 1 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) < 0 , if  | c 1 | > k 1 , c 2 T 2 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) < 0 , if  | c 1 | k 1 , | c 2 | > k 2 , d 1 T 3 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) < 0 , if  | d 1 | > l 1 , d 2 T 4 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) < 0 , if  | d 1 | l 1 , | d 2 | > l 2 .

Lemma 3.3 Suppose that ( H 1 )-( H 4 ) hold, then the set

Ω 1 = { ( x , y ) dom L Ker L L ( x , y ) = λ N ( x , y ) , λ ( 0 , 1 ) }

is bounded in X.

Proof Take (x,y) Ω 1 . By L(x,y)=λN(x,y), we get

{ x ( t ) = λ Γ ( α ) 0 t ( t s ) α 1 f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) d s + a 1 t α 1 + a 2 t α 2 , y ( t ) = λ Γ ( β ) 0 t ( t s ) β 1 f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) d s + b 1 t β 1 + b 2 t β 2 .
(3.1)

By Lemmas 2.1, 2.2 and (3.1), we have

{ D 0 + α 1 x ( t ) = λ 0 t f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) d s + a 1 Γ ( α ) , D 0 + β 1 y ( t ) = λ 0 t f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) d s + b 1 Γ ( β ) .
(3.2)

It follows from N(x,y)ImL that

0 1 f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) t 1 h 1 ( s ) d s d t = 0 , 0 1 f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) 0 t h 2 ( s ) d s d t = 0 , 0 1 f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) t 1 g 1 ( s ) d s d t = 0 , 0 1 f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) 0 t g 2 ( s ) d s d t = 0 .

These, together with ( H 4 ), mean that there exist constants t 0 , t 1 [ η 1 ,1] and t 0 , t 1 [ η 2 ,1] such that

| x ( t 0 ) | M 1 , | D 0 + α 1 x ( t 1 ) | L 1 , | y ( t 0 ) | M 2 , | D 0 + β 1 y ( t 1 ) | L 2 .
(3.3)

By (3.2), we get

D 0 + α 1 x ( t ) = λ t 1 t f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) d s + D 0 + α 1 x ( t 1 ) , D 0 + β 1 y ( t ) = λ t 1 t f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) d s + D 0 + β 1 y ( t 1 ) .

By (3.3) and ( H 3 ), we obtain that

{ D 0 + α 1 x 1 1 c 0 ( a 0 x + b 0 y + d 0 x θ 1 + r 0 + L 1 ) , D 0 + β 1 y 1 1 c 0 ( a 0 x + b 0 y + d 0 y θ 2 + r 0 + L 2 ) .
(3.4)

Instead of t by t 0 , t 0 in (3.1) and t 1 , t 1 in (3.2), respectively, we get

{ x ( t ) = λ Γ ( α ) [ 0 t ( t s ) α 1 f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) d s x ( t ) = + t α 2 ( t 0 t ) 0 t 1 f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) d s x ( t ) = t α 2 t 0 α 2 0 t 0 ( t 0 s ) α 1 f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) d s ] x ( t ) = + t α 2 Γ ( α ) D 0 + α 1 x ( t 1 ) ( t t 0 ) + t α 2 t 0 α 2 x ( t 0 ) , y ( t ) = λ Γ ( β ) [ 0 t ( t s ) β 1 f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) d s y ( t ) = + t β 2 ( t 0 t ) 0 t 1 f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) d s y ( t ) = t β 2 t 0 β 2 0 t 0 ( t 0 s ) β 1 f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) d s ] y ( t ) = + t β 2 Γ ( β ) D 0 + β 1 y ( t 1 ) ( t t 0 ) + t β 2 t 0 β 2 y ( t 0 ) .
(3.5)

It follows from (3.4), (3.5) and ( H 3 ) that

| x ( t ) | 1 Γ ( α ) ( 2 + 1 η 1 α 2 ) 0 1 | f 1 ( s , x ( s ) , y ( s ) , D 0 + α 1 x ( s ) ) | d s + ( L 1 Γ ( α ) + M 1 η 1 α 2 ) 1 Γ ( α ) ( 2 + 1 η 1 α 2 ) ( a 0 x + b 0 y + c 0 D 0 + α 1 x + d 0 x θ 1 + r 0 ) + ( L 1 Γ ( α ) + M 1 η 1 α 2 ) 1 Γ ( α ) ( 1 c 0 ) ( 2 + 1 η 1 α 2 ) ( a 0 x + b 0 y + d 0 x θ 1 + r 0 + c 0 L 1 ) + ( L 1 Γ ( α ) + M 1 η 1 α 2 )

and

| y ( t ) | 1 Γ ( β ) ( 2 + 1 η 2 β 2 ) 0 1 | f 2 ( s , x ( s ) , y ( s ) , D 0 + β 1 y ( s ) ) | d s + ( L 2 Γ ( β ) + M 2 η 2 β 2 ) 1 Γ ( β ) ( 2 + 1 η 2 β 2 ) ( a 0 x + b 0 y + c 0 D 0 + β 1 y + d 0 y θ 2 + r 0 ) + ( L 2 Γ ( β ) + M 2 η 2 β 2 ) 1 Γ ( β ) ( 1 c 0 ) ( 2 + 1 η 2 β 2 ) ( a 0 x + b 0 y + d 0 y θ 2 + r 0 + c 0 L 2 ) + ( L 2 Γ ( β ) + M 2 η 2 β 2 ) .

Thus,

x A 1 [ b 0 y + d 0 x θ 1 ] + M 0 ,
(3.6)
y A 2 [ a 0 x + d 0 y θ 2 ] + M 0 ,
(3.7)

where M 0 = A 1 [( r 0 + c 0 L 1 )+( L 1 Γ ( α ) + M 1 η 1 α 2 )/ 1 Γ ( α ) ( 1 c 0 ) (2+ 1 η 1 α 2 )], M 0 = A 2 [( r 0 + c 0 L 2 )+( L 2 Γ ( β ) + M 2 η 2 β 2 )/ 1 Γ ( β ) ( 1 c 0 ) (2+ 1 η 2 β 2 )].

By ( H 3 ), (3.4), (3.6) and (3.7), we can get that Ω 1 is bounded in X. The proof is completed. □

Lemma 3.4 Suppose that ( H 1 ), ( H 2 ), ( H 3 ) and ( H 5 ) hold, then the set

Ω 2 = { ( x , y ) ( x , y ) Ker L , N ( x , y ) Im L }

is bounded in X.

Proof For (x,y) Ω 2 , we have (x,y)=( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ), c 1 , c 2 , d 1 , d 2 R and T 1 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) = T 2 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) = T 3 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) = T 4 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) = 0. By ( H 5 ), we get that | c 1 | k 1 , | c 2 | k 2 , | d 1 | l 1 , | d 2 | l 2 . These imply that Ω 2 is bounded in X. □

Lemma 3.5 Suppose that ( H 1 ), ( H 2 ), ( H 3 ) and ( H 5 ) hold. The set

Ω 3 = { ( x , y ) Ker L λ J ( x , y ) + ( 1 λ ) θ Q N ( x , y ) = ( 0 , 0 ) , λ [ 0 , 1 ] }

is bounded in X, where J:KerLImQ is a linear isomorphism given by

J ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) = ( 1 Δ 1 ( Δ 22 c 1 Δ 21 c 2 ) + 1 Δ 1 ( Δ 11 c 2 Δ 12 c 1 ) t , 1 Δ 2 ( δ 22 d 1 δ 21 d 2 ) + 1 Δ 2 ( δ 11 d 2 δ 12 d 1 ) t ) , θ = { 1 , if ( H 5 ) ( 1 ) holds , 1 , if ( H 5 ) ( 2 ) holds .

Proof For ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) Ω 3 , there exists λ[0,1] such that

λJ ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) =(1λ)θQN ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) .

This means that

λ c 1 = ( 1 λ ) θ T 1 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) , λ c 2 = ( 1 λ ) θ T 2 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) , λ d 1 = ( 1 λ ) θ T 3 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) , λ d 2 = ( 1 λ ) θ T 4 N 2 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) .

If λ=0, by ( H 5 ), we get | c 1 | k 1 , | c 2 | k 2 , | d 1 | l 1 , | d 2 | l 2 . If λ=1, then c 1 = c 2 = d 1 = d 2 =0. For λ(0,1), if | c 1 |> k 1 , we can get

λ c 1 2 =(1λ)θ c 1 T 1 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) <0,

a contradiction. If | c 1 | k 1 and | c 2 |> k 2 , we can get

λ c 2 2 =(1λ)θ c 2 T 2 N 1 ( c 1 t α 1 + c 2 t α 2 , d 1 t β 1 + d 2 t β 2 ) <0.

This is a contradiction, too. Thus, | c i | k i , i=1,2. By the same methods, we can obtain that | d i | l i , i=1,2. This means that Ω 3 is bounded in X. □

Theorem 3.1 Suppose that ( H 1 )-( H 5 ) hold. Then problem (1.1) has at least one solution in X.

Proof Let Ω i = 1 3 Ω i ¯ {(0,0)} be a bounded open subset of X. It follows from Lemma 3.2 that N is L-compact on Ω ¯ . By Lemmas 3.3 and 3.4, we get

  1. (1)

    L(x,y)λN(x,y) for every (x,y,λ)[(domLKerL)Ω]×(0,1),

  2. (2)

    N(x,y)ImL for every (x,y)KerLΩ.

    We need only to prove

  3. (3)

    deg(QN | Ker L ,ΩKerL,(0,0))0.

Take

H(x,y,λ)=λJ(x,y)+θ(1λ)QN(x,y).

According to Lemma 3.5, we know H(x,y,λ)(0,0) for (x,y)ΩKerL. By the homotopy of degree, we get that

deg ( Q N | Ker L , Ω Ker L , ( 0 , 0 ) ) = deg ( θ H ( , 0 ) , Ω Ker L , ( 0 , 0 ) ) = deg ( θ H ( , 1 ) , Ω Ker L , ( 0 , 0 ) ) = deg ( θ J , Ω Ker L , ( 0 , 0 ) ) 0 .

By Theorem 2.1, we can get that L(x,y)=N(x,y) has at least one solution in domL Ω ¯ , i.e., (1.1) has at least one solution in X. The proof is completed. □

4 Example

Let us consider the following system of fractional differential equations at resonance:

{ D 0 + 5 2 x ( t ) = f 1 ( t , x ( t ) , y ( t ) , D 0 + 3 2 x ( t ) ) , 0 < t < 1 , D 0 + 5 2 y ( t ) = f 2 ( t , x ( t ) , y ( t ) , D 0 + 3 2 y ( t ) ) , 0 < t < 1 , x ( 0 ) = 0 , D 0 + 3 2 x ( 0 ) = 0 1 h 1 ( t ) D 0 + 3 2 x ( t ) d t , D 0 + 3 2 x ( 1 ) = 0 1 h 2 ( t ) D 0 + 3 2 x ( t ) d t , y ( 0 ) = 0 , D 0 + 3 2 y ( 0 ) = 0 1 g 1 ( t ) D 0 + 3 2 y ( t ) d t , D 0 + 3 2 y ( 1 ) = 0 1 g 2 ( t ) D 0 + 3 2 y ( t ) d t ,
(4.1)

where

f 1 ( t , x , y , z ) = { 1 4 t sin x + 1 8 t 3 sin y , t [ 0 , 1 4 ) , 1 4 t sin x + 1 8 t 3 sin y + t z , t [ 1 4 , 1 2 ) , 1 4 t x + 1 8 t 3 sin y + t sin z , t [ 1 2 , 1 ] , f 2 ( t , x , y , z ) = { 1 8 t 3 sin x + 1 10 sin y , t [ 0 , 1 9 ) , 1 8 t 3 sin x + 1 10 sin y + t z , t [ 1 9 , 1 4 ) , 1 8 t 3 sin x + 1 10 y + t sin z , t [ 1 4 , 1 ] , h 1 ( t ) = { 2 , t [ 0 , 1 2 ) , 0 , t [ 1 2 , 1 ] , h 2 ( t ) = { 0 , t [ 0 , 1 2 ) , 2 , t [ 1 2 , 1 ] , g 1 ( t ) = { 4 , t [ 0 , 1 4 ) , 0 , t [ 1 4 , 1 ] , g 2 ( t ) = { 0 , t [ 0 , 1 4 ) , 4 3 , t [ 1 4 , 1 ] .

Corresponding to problem (1.1), we have α=β= 5 2 ,

Δ 1 = | 1 4 1 4 1 24 5 24 | = 1 24 0, Δ 2 = | 1 8 3 8 1 96 9 32 | = 1 32 0.

Obviously, 0 1 h i (t)dt=1, 0 1 g i (t)dt=1, i=1,2. Thus, conditions ( H 1 ) and ( H 2 ) are satisfied. It is easy to get that a 0 = 1 8 , b 0 = 1 32 , c 0 = 15 32 , a 0 = 1 32 , b 0 = 1 10 , c 0 = 40 81 . Take M 1 =8, L 1 =1, η 1 = 1 4 , M 2 =20, L 2 =4, η 2 = 1 9 . By a simple calculation, we can get that ( H 3 ) is satisfied and the following inequations hold

0 1 f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) t 1 h 1 ( s ) d s d t 0 , if  min t [ η 1 , 1 ] | D 0 + α 1 x ( t ) | > L 1 , 0 1 f 1 ( t , x ( t ) , y ( t ) , D 0 + α 1 x ( t ) ) 0 t h 2 ( s ) d s d t 0 , if  min t [ η 1 , 1 ] | x ( t ) | > M 1 , 0 1 f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) t 1 g 1 ( s ) d s d t 0 , if  min t [ η 2 , 1 ] | D 0 + β 1 y ( t ) | > L 2 ,

and

0 1 f 2 ( t , x ( t ) , y ( t ) , D 0 + β 1 y ( t ) ) 0 t g 2 (s)dsdt0,if  min t [ η 2 , 1 ] | y ( t ) | > M 2 .

So, ( H 4 ) holds. Set k 1 =1, k 2 =20, l 1 =4, l 2 =140. By a simple calculation, we can obtain that condition ( H 5 ) is satisfied.

By Theorem 3.1, problem (4.1) has at least one solution.

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Acknowledgements

This work is supported by the National Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108). The author is grateful to anonymous referees for their constructive comments and suggestions which led to improvement of the original manuscript.

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Jiang, W. Solvability for a coupled system of fractional differential equations with integral boundary conditions at resonance. Adv Differ Equ 2013, 324 (2013). https://doi.org/10.1186/1687-1847-2013-324

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