Open Access

Solvability for a coupled system of fractional differential equations with impulses at resonance

Boundary Value Problems20132013:80

DOI: 10.1186/1687-2770-2013-80

Received: 9 November 2012

Accepted: 18 March 2013

Published: 8 April 2013

Abstract

In this paper, some Banach spaces are introduced. Based on these spaces and the coincidence degree theory, a 2m-point boundary value problem for a coupled system of impulsive fractional differential equations at resonance is considered, and the new criterion on existence is obtained. Finally, an example is also given to illustrate the availability of our main results.

MSC:34A08, 34B10, 34B37.

Keywords

coupled system impulsive fractional differential equations at resonance coincidence degree

1 Introduction

Recently, Wang et al. [1] presented a counterexample to show an error formula of solutions to the traditional boundary value problem for impulsive differential equations with fractional derivative in [25]. Meanwhile, they introduced the correct formula of solutions for an impulsive Cauchy problem with the Caputo fractional derivative. Shortly afterwards, many works on the better formula of solutions to the Cauchy problem for impulsive fractional differential equations have been reported by Li et al. [6], Wang et al. [7], Fečkan [8], etc.

Fractional differential equations have been paid much attention to in recent years due to their wide applications such as nonlinear oscillations of earthquakes, Nutting’s law, charge transport in amorphous semiconductors, fluid dynamic traffic model, non-Markovian diffusion process with memory etc. [911]. For more details, see the monographs of Hilfer [12], Miller and Ross [13], Podlubny [14], Lakshmikantham et al. [15], Samko et al. [16], and the papers of [2, 1719] and the references therein.

In recent years, many researchers paid much attention to the coupled system of fractional differential equations due to its applications in different fields [2025]. Zhang et al. [25] investigated a three-point boundary value problem at resonance for a coupled system of nonlinear fractional differential equations given by
{ D 0 + α u ( t ) = f ( t , v ( t ) , D 0 + β 1 v ( t ) ) , 0 < t < 1 ; D 0 + β v ( t ) = g ( t , u ( t ) , D 0 + α 1 u ( t ) ) , 0 < t < 1 ; u ( 0 ) = v ( 0 ) = 0 , u ( 1 ) = σ 1 u ( η 1 ) , v ( 1 ) = σ 2 v ( η 2 ) ,
where 1 < α , β 2 , 0 < η 1 , η 2 < 1 , σ 1 , σ 2 > 0 , σ 1 η 1 α 1 = σ 2 η 2 β 1 = 1 , D 0 + α is the standard Riemann-Liouville fractional derivative and f , g : [ 0 , 1 ] × R 2 R are continuous. And Wang et al. [23] considered a 2m-point boundary value problem (BVP) at resonance for a coupled system as follows:
{ D 0 + α u ( t ) = f ( t , v ( t ) , D 0 + β 1 v ( t ) , D 0 + β 2 v ( t ) ) , 0 < t < 1 ; D 0 + β v ( t ) = g ( t , u ( t ) , D 0 + α 1 u ( t ) , D 0 + α 2 u ( t ) ) , 0 < t < 1 ; I 0 + 3 α u ( 0 ) = 0 , D 0 + α 2 u ( 1 ) = i = 1 m a i D 0 + α 2 u ( ξ i ) , u ( 1 ) = i = 1 m b i u ( η i ) ; I 0 + 3 β v ( 0 ) = 0 , D 0 + β 2 v ( 1 ) = i = 1 m c i D 0 + β 2 v ( γ i ) , v ( 1 ) = i = 1 m d i v ( δ i ) ,

where 1 < α , β 2 . With the help of the coincidence degree theory, many existence results have been given in the above literatures. It is worth mentioning that the orders of derivative in the nonlinear function on the right-hand of equal signs are all fixed in the above works, but the opposite case is more difficult and complicated, then this work attempts to deal exactly with this case. What is more, this case of arbitrary order derivative included in the nonlinear functions is very important in many aspects [20, 22].

There are significant developments in the theory of impulses especially in the area of impulsive differential equations with fixed moments, which provided a natural description of observed evolution processes, regarding as important tools for better understanding several real word phenomena in applied sciences [1, 7, 2629]. In addition, motivated by the better formula of solutions cited by the work of Zhou et al. [1, 7, 8], the aim of this work is to discuss a boundary value problem for a coupled system of impulsive fractional differential equation. Exactly, this paper deals with the 2m-point boundary value problem of the following coupled system of impulsive fractional differential equations at resonance:
{ D 0 + α u ( t ) = f ( t , v ( t ) , D 0 + p v ( t ) ) , D 0 + β v ( t ) = g ( t , u ( t ) , D 0 + q u ( t ) ) , 0 < t < 1 ; Δ u ( t i ) = A i ( v ( t i ) , D 0 + p v ( t i ) ) , Δ D 0 + q u ( t i ) = B i ( v ( t i ) , D 0 + p v ( t i ) ) , i = 1 , 2 , , k ; Δ v ( t i ) = C i ( u ( t i ) , D 0 + q u ( t i ) ) , Δ D 0 + p v ( t i ) = D i ( u ( t i ) , D 0 + q u ( t i ) ) , i = 1 , 2 , , k ; D 0 + α 1 u ( 0 ) = i = 1 m a i D 0 + α 1 u ( ξ i ) , u ( 1 ) = i = 1 m b i η i 2 α u ( η i ) ; D 0 + β 1 v ( 0 ) = i = 1 m c i D 0 + β 1 v ( ζ i ) , v ( 1 ) = i = 1 m d i θ i 2 β v ( θ i ) ,
(1.1)

where 1 < α , β < 2 , α q 1 , β p 1 and 0 < ξ 1 < ξ 2 < < ξ m < 1 , 0 < η 1 < η 2 < < η m < 1 , 0 < ζ 1 < ζ 2 < < ζ m < 1 , 0 < θ 1 < θ 2 < < θ m < 1 . f , g : [ 0 , 1 ] × R 2 R satisfy Carathéodory conditions, A i , B i , C i , D i : R × R R . Δ w ( t i ) = w ( t i + ) w ( t i ) , Δ D 0 + r w ( t i ) = D 0 + r w ( t i + ) D 0 + r w ( t i ) , here w { u , v } , r { p , q } , w ( t i + ) and w ( t i ) denote the right and left limits of w ( t ) at t = t i , respectively, and the fractional derivative is understood in the Riemann-Liouville sense. k, m, a i , b i , c i , d i ( i = 1 , 2 , , m ) are fixed constant satisfying i = 1 m a i = i = 1 m b i = i = 1 m c i = i = 1 m d i = 1 and i = 1 m b i η i = i = 1 m d i θ i = 1 .

The coupled system (1.1) happens to be at resonance in the sense that the associated linear homogeneous coupled system
{ D 0 + α u ( t ) = 0 , D 0 + β v ( t ) = 0 , 0 < t < 1 ; D 0 + α 1 u ( 0 ) = i = 1 m a i D 0 + α 1 u ( ξ i ) , u ( 1 ) = i = 1 m b i η i 2 α u ( η i ) ; D 0 + β 1 v ( 0 ) = i = 1 m c i D 0 + β 1 v ( ζ i ) , v ( 1 ) = i = 1 m d i θ i 2 β v ( θ i )

has ( u ( t ) , v ( t ) ) = ( h 1 t α 1 + h 2 t α 2 , h 3 t β 1 + h 4 t β 2 ) , c i R , i = 1 , 2 , 3 , 4 as a nontrivial solution. To solve this interesting and important problem and to overcome the difficulties caused by the impulses, we will construct some Banach spaces, then we shall obtain the new solvability results for the coupled system (1.1) with the help of a coincidence degree continuation theorem. The main contributions of this work are Lemma 2.1 and Lemma 3.1 in Section 3 since the calculations are disposed well.

The plan of this work is organized as follows. Section 2 contains some necessary notations, definitions and lemmas that will be used in the sequel. In Section 3, we establish a theorem on the existence of solutions for the coupled system (1.1) based on the coincidence degree theory due to Mawhin [30, 31].

2 Background materials and preliminaries

For the convenience of the readers, we recall some notations and an abstract existence theorem [30, 31].

Let Y, Z be real Banach spaces, L : dom ( L ) Y Z be a Fredholm map of index zero and P : Y Y , Q : Z Z be continuous projectors such that Im ( P ) = Ker ( L ) , Ker ( Q ) = Im ( L ) and Y = Ker ( L ) Ker ( P ) , Z = Im ( L ) Im ( Q ) . It follows that L | dom ( L ) Ker ( P ) : dom ( L ) Ker ( P ) Im ( L ) is invertible. We denote the inverse of the map by K P . If Ω is an open bounded subset of Y such that dom ( L ) Ω , the map N : Y Z will be called L-compact on Ω ¯ if Q N ( Ω ¯ ) is bounded and K P ( I Q ) N : Ω ¯ Y is compact.

The main tool we used is Theorem 2.4 of [30].

Theorem 2.1 Let L be a Fredholm operator of index zero, and let N be L-compact on Ω ¯ . Assume that the following conditions are satisfied:
  1. (i)

    L x λ N x for every ( x , λ ) [ ( dom ( L ) Ker ( L ) ) Ω ] × ( 0 , 1 ) ;

     
  2. (ii)

    N x Im ( L ) for every x Ker ( L ) Ω ;

     
  3. (iii)

    deg ( Q N | Ker ( L ) , Ω Ker ( L ) , 0 ) 0 , where Q : Z Z is a projection as above with Im ( L ) = Ker ( Q ) .

     

Then the equation L x = N x has at least one solution in dom ( L ) Ω ¯ .

Now, we present some basic knowledge and definitions about fractional calculus theory, which can be found in the recent works [13, 16, 32].

Definition 2.1 The fractional integral of order α > 0 of a function y : ( 0 , ) R is defined by
I 0 + α y ( t ) = 0 t ( t s ) α 1 Γ ( α ) y ( s ) d s ,

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

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

Remark 2.1 It can be directly verified that the Riemann-Liouville fractional integration and fractional differentiation operators of the power functions t μ yield power functions of the same form. For α 0 , μ 1 , we have
I 0 + α t μ = Γ ( μ + 1 ) Γ ( μ + α + 1 ) t μ + α , D 0 + α t μ = Γ ( μ + 1 ) Γ ( μ α + 1 ) t μ α ( μ α ) .
(2.1)

Proposition 2.1 [17]

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

for some c i R , i = 1 , 2 , , N , where N is the smallest integer grater than or equal to α.

Proposition 2.2 [32]

If α > 0 , β > 0 , then the equation
( I 0 + α I 0 + β y ) ( t ) = ( I 0 + α + β y ) ( t )

is satisfied for a continuous function y.

If α > 0 , m N and D = d / d t , the fractional derivatives ( D 0 + α y ) ( t ) and ( D 0 + α + m y ) ( t ) exist, then
( D m D 0 + α y ) ( t ) = ( D 0 + α + m y ) ( t ) .
If α > 0 , then the equation
( D 0 + α I 0 + α y ) ( t ) = y ( t )

is satisfied for a continuous function y.

If α > β > 0 , then the relation
( D 0 + β I 0 + α y ) ( t ) = ( I 0 + α β y ) ( t )

holds for a continuous function y.

Let C [ 0 , 1 ] = { u | u  is continuous in  [ 0 , 1 ] } with the norm u = max t [ 0 , 1 ] | u ( t ) | and
P C [ 0 , 1 ] = { x : x C ( t i , t i + 1 ] ,  there exist  x ( t i )  and x ( t i + )  with  x ( t i ) = x ( t i ) , i = 1 , 2 , , k 1 }
with the norm x P C = sup t [ 0 , 1 ] | x ( t ) | . Denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equk_HTML.gif
where u α ( t ) = t 2 α u ( t ) , v β ( t ) = t 2 β v ( t ) with the norm
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equl_HTML.gif

Thus, Y = Y 1 × Y 2 is a Banach space with the norm defined by ( u , v ) Y = max { u Y 1 , v Y 2 } .

Set Z 1 = Z 2 = P C [ 0 , 1 ] × R 2 k equipped with the norm
x Z 1 = max { y P C , | c | } , x = ( y , c ) Z 1 ,

thus Z = Z 1 × Z 2 is a Banach space with the norm defined by ( x , y ) Z = max { x Z 1 , y Z 2 } .

Define the operator L : Y Z , L ( u , v ) = ( L 1 u , L 2 v ) , dom ( L ) = dom ( L 1 ) × dom ( L 2 ) , where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equn_HTML.gif
with
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equo_HTML.gif
Let N : Y Z be defined as N ( u , v ) = ( N 1 v , N 2 u ) , where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equp_HTML.gif
Then the coupled system of boundary value problem (1.1) can be written as
L ( u , v ) = N ( u , v ) .
For the sake of simplicity, we define the operators T 1 , T 2 : Z 1 Z 1 for X = ( x , δ 1 , , δ k , ω 1 , , ω k ) as follows:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ4_HTML.gif
(2.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ5_HTML.gif
(2.4)
By the same way, we define the operators T 3 , T 4 : Z 2 Z 2 for Y = ( y , ρ 1 , , ρ k , τ 1 , , τ k ) as follows:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ6_HTML.gif
(2.5)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ7_HTML.gif
(2.6)

In what follows, we present the following lemmas which will be used to prove our main results.

Lemma 2.1 If the following condition is satisfied:

(H1) σ 1 = | σ 11 σ 12 σ 13 σ 14 | 0 , σ 2 = | σ 21 σ 22 σ 23 σ 24 | 0 , where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equr_HTML.gif
then L : dom ( L ) Y Z is a Fredholm operator of index zero. Moreover, Ker ( L ) = Ker ( L 1 ) × Ker ( L 2 ) , where
Ker ( L 1 ) = { h 1 t α 1 + h 2 t α 2 , h 1 , h 2 R } , Ker ( L 2 ) = { h 3 t β 1 + h 4 t β 2 , h 3 , h 4 R }
(2.7)
and Im ( L ) = Im ( L 1 ) × Im ( L 2 ) , here
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ9_HTML.gif
(2.8)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ10_HTML.gif
(2.9)

Proof It is clear that (2.7) holds. For ( u , v ) Ker ( L ) , we have L ( u , v ) = ( L 1 u , L 2 v ) = ( 0 , 0 ) , i.e., L 1 u = 0 , L 2 v = 0 , then u Ker ( L 1 ) , v Ker ( L 2 ) , so Ker ( L ) = Ker ( L 1 ) × Ker ( L 2 ) . Similarly, it is not difficult to see that Im ( L ) = Im ( L 1 ) × Im ( L 2 ) . Next, we will show that (2.8) and (2.9) hold.

If Z 1 = ( z 1 , δ 1 , , δ k , ω 1 , , ω k ) Im ( L 1 ) , Z 2 = ( z 2 , ρ 1 , , ρ k , τ 1 , , τ k ) Im ( L 2 ) , then there exist u dom ( L 1 ) and v dom ( L 2 ) such that
{ D 0 + α u ( t ) = z 1 ( t ) , Δ u ( t i ) = δ i , Δ D 0 + q u ( t i ) = ω i , { D 0 + β v ( t ) = z 2 ( t ) , Δ v ( t i ) = ρ i , Δ D 0 + p v ( t i ) = τ i
(2.10)
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ12_HTML.gif
(2.11)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ13_HTML.gif
(2.12)
Proposition 2.1 together with (2.10)-(2.12) gives that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ14_HTML.gif
(2.13)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ15_HTML.gif
(2.14)
Substituting the boundary condition D 0 + α 1 u ( 0 ) = i = 1 m a i D 0 + α 1 u ( ξ i ) into (2.13), one has
i = 1 m a i ( 0 ξ i z 1 ( s ) d s + Γ ( α q ) t i < ξ i ω i t i q + 1 α ) = 0 ,
(2.15)
and substituting the boundary condition u ( 1 ) = i = 1 m b i η i 2 α u ( η i ) into (2.13), one has
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ17_HTML.gif
(2.16)
By the same way, if we substitute the condition (2.12) into (2.14), then we can obtain that
i = 1 m c i ( 0 ζ i z 2 ( s ) d s + Γ ( β p ) t i < ζ i τ i t i p + 1 β ) = 0 ,
(2.17)
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ19_HTML.gif
(2.18)
Conversely, if (2.15)-(2.18) hold, set
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equs_HTML.gif

It is easy to check that the above u, v satisfy equation (2.10)-(2.12). Thus, (2.8) and (2.9) hold.

Define the operator Q : Z Z , Q ( x , y ) = ( Q 1 x , Q 2 y ) with Q 1 X = Q 11 X + Q 12 X t , Q 2 Y = Q 21 Y + Q 22 Y t , here
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equt_HTML.gif
In what follows, we will show that Q 1 and Q 2 are linear projectors. By some direct computations, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equu_HTML.gif
As a result,
Q 1 ( Q 1 X ) = Q 1 ( Q 11 X + Q 12 X t ) = Q 11 ( Q 11 X + Q 12 X t ) + Q 12 ( Q 11 X + Q 12 X t ) t = Q 11 2 X + Q 11 ( Q 12 X t ) + [ Q 12 ( Q 11 X ) + Q 12 ( Q 12 X t ) ] t = Q 11 X + Q 12 X t = Q 1 X .

Similarly, we can see that Q 2 ( Q 2 Y ) = Q 2 Y . Then for ( X , Y ) Z , we have Q 2 ( X , Y ) = Q ( Q 1 X , Q 2 Y ) = ( Q 1 2 X , Q 2 2 Y ) = ( Q 1 X , Q 2 Y ) = Q ( X , Y ) . It means that the operator Q : Z Z is a projector.

Now, we show that Ker ( Q ) = Im ( L ) . Obviously, Im ( L ) Ker ( Q ) . On the other hand, for ( X , Y ) Ker ( Q ) , then Q ( X , Y ) = ( 0 , 0 ) implies that
{ σ 11 T 1 X σ 12 T 2 X = ( 0 , 0 , , 0 ) , σ 13 T 1 X σ 14 T 2 X = ( 0 , 0 , , 0 ) , { σ 21 T 1 Y σ 22 T 2 Y = ( 0 , 0 , , 0 ) , σ 23 T 1 Y σ 24 T 2 Y = ( 0 , 0 , , 0 ) .

The condition (H1) guarantees that T 1 X = T 2 X = ( 0 , 0 , , 0 ) , T 3 Y = T 4 Y = ( 0 , 0 , , 0 ) , then ( X , Y ) Im ( L ) . Hence, Ker ( Q ) = Im ( L ) .

For W Z , let W = ( W Q W ) + Q W . Then W Q W Ker ( Q ) = Im ( L ) , Q W Im ( Q ) , it means that Z = Im ( L ) + Im ( Q ) . Moreover, Ker ( Q ) = Im ( L ) gives that Im ( L ) Im ( Q ) = ( 0 , 0 ) . Thus, Z = Im ( L ) Im ( Q ) . Then dim Ker ( L ) = dim Im ( Q ) = codim Im ( L ) = 4 , L is a Fredholm map of index zero. □

Define the operator P : Y Y with P ( u , v ) = ( P 1 u , P 2 v ) , here P 1 : Y 1 Y 1 , P 2 : Y 2 Y 2 are defined as follows:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equx_HTML.gif
Moreover, we define K P : Im ( L ) dom ( L ) Ker ( P ) as K P ( X , Y ) = ( K P 1 X , K P 2 Y ) , where K P i : Im ( L i ) dom ( L i ) Ker ( P i ) , i = 1 , 2 is defined as follows:
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equy_HTML.gif

Lemma 2.2 Assume that Ω Y is an open bounded subset with dom ( L ) Ω ¯ , then N is L-compact on Ω ¯ .

Proof Obviously, Im ( P ) = Ker ( L ) . By a direct computation, we have that
P 1 2 u = 1 Γ ( α ) D 0 + α 1 P 1 u ( 0 ) t α 1 + lim t 0 t 2 α P 1 u ( t ) t α 2 = 1 Γ ( α ) D 0 + α 1 u ( 0 ) t α 1 + lim t 0 t 2 α u ( t ) t α 2 = P 1 u .

Similarly, P 2 2 v = P 2 v . This gives that P 2 ( u , v ) = P ( P 1 u , P 2 v ) = ( P 1 2 u , P 2 2 v ) = ( P 1 u , P 2 v ) = P ( u , v ) , that is to say, the operator P is a linear projector. It is easy to check from w = ( w P w ) + P w that Y = Ker ( P ) + Ker ( L ) . Moreover, we can see that Ker ( P ) Ker ( L ) = ( 0 , 0 ) . Thus, Y = Ker ( P ) Ker ( L ) .

In what follows, we will show that K P defined above is the inverse of L | dom ( L ) Ker ( P ) .

If ( X , Y ) Im ( L ) , then L 1 K P 1 X = X , L 2 K P 2 Y = Y , which gives that
L K P ( X , Y ) = ( L 1 K P 1 X , L 2 K P 2 Y ) = ( X , Y ) .
On the other hand, for ( u , v ) dom ( L ) Ker ( P ) , we have
( K P 1 L 1 ) u ( t ) = K P 1 ( D 0 + α u ( t ) , δ 1 , , δ k , ω 1 , , ω k ) = u ( t ) + ( h 1 + Γ ( α q ) Γ ( α ) t i < t ω i t i q + 1 α ) t α 1 + ( h 2 + t i < t δ i t i 2 α Γ ( α q ) Γ ( α ) t i < t ω i t i q + 2 α ) t α 2 .
Since u K P 1 and K P 1 L 1 u Ker ( P 1 ) , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ20_HTML.gif
(2.19)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ21_HTML.gif
(2.20)
By some calculations, (2.19) and (2.20) imply that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equac_HTML.gif

It means that K P 1 L 1 u = u . Analogously, K P 2 L 2 v = v . Thus, K P L ( u , v ) = ( K P 1 L 1 u , K P 2 L 2 v ) = ( u , v ) . So, K P is the inverse of L | dom ( L ) Ker ( P ) .

Finally, we show that N is L-compact on Ω ¯ . Denote Q 1 N 1 v = ( v , 0 , , 0 ) , Q 2 N 2 u = ( u , 0 , , 0 ) , where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equad_HTML.gif
Then we can see that
K P ( I Q ) N ( u , v ) = K P ( I Q ) ( N 1 v , N 2 u ) = ( K P 1 ( I Q 1 ) N 1 v , K P 2 ( I Q 2 ) N 2 u ) ,
where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equaf_HTML.gif

So, we can see that Q 1 N 1 is bounded and K P 1 ( I Q 1 ) N 1 is uniformly bounded.

For 0 t 1 < t 2 1 , we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ22_HTML.gif
(2.21)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ23_HTML.gif
(2.22)

The equicontinuity of t α , t α + 1 together with (2.21) and (2.22) gives that | K P 1 ( I Q 1 ) N 1 v ( t 2 ) K P 1 ( I Q 1 ) N 1 v ( t 1 ) | 0 as t 2 t 1 , which yields that K P 1 ( I Q 1 ) N 1 is equicontinuous. By the Ascoli-Arzela theorem, we can see that K P 1 ( I Q 1 ) N 1 is compact. By the same way, Q 2 N 2 is bounded and K P 2 ( I Q 2 ) N 2 is compact. Since Q N ( u , v ) = Q ( N 1 v , N 2 u ) = ( Q 1 N 1 v , Q 2 N 2 u ) and K P ( I Q ) N ( u , v ) = ( K P 1 ( I Q 1 ) N 1 v , K P 2 ( I Q 2 ) N 2 u ) , then QN is bounded and K P ( I Q ) N is compact. This means that N is L-compact on Ω ¯ . □

3 Main results

In this section, we present the existence results of the coupled system (1.1). To do this, we need the following hypotheses.

(H2) There exist functions φ i , ψ i , γ i C [ 0 , 1 ] , i = 1 , 2 , such that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equag_HTML.gif
where ψ i , γ i ( i = 1 , 2 ) satisfy
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equah_HTML.gif
here
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equai_HTML.gif
(H3) For ( u , v ) dom ( L ) , there exist constants e i ( 0 , 1 ) ( i = 0 , 1 , 2 ), M i > 0 ( i = 1 , 2 ) such that
  1. (1)

    if either | u ( t ) | > M 1 or | v ( t ) | > M 1 for t [ e 0 , e 1 ] , then either T 2 N 1 v ( t ) 0 or T 4 N 2 u ( t ) 0 ;

     
  2. (2)

    if either | D 0 + q u ( t ) | > M 2 or | D 0 + p v ( t ) | > M 2 , t [ e 2 , 1 ] , then either T 1 N 1 v ( t ) 0 or T 3 N 2 u ( t ) 0 .

     
(H4) For ( u , v ) Ker ( L ) , there exist constants g i > 0 ( i = 1 , 2 ) such that if either | h 1 | g 1 or | h 2 | g 1 , either | h 3 | g 2 or | h 4 | g 2 , then either (1) or (2) holds, where
  1. (1)
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equaj_HTML.gif
     
here s 1 , s 2 are positive constants;
  1. (2)
    https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equak_HTML.gif
     

here s 3 , s 4 are negative constants.

Lemma 3.1 Suppose that (H2)-(H3) hold. Then the set
Ω 1 = { ( u , v ) dom ( L ) Ker ( L ) | L ( u , v ) = λ N ( u , v ) , λ ( 0 , 1 ) }

is bounded in Y.

Proof For ( u , v ) Ω 1 , by L ( u , v ) = ( L 1 u , L 2 v ) = λ N ( u , v ) = ( λ N 1 v , λ N 2 u ) and ( u , v ) dom ( L ) , we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ24_HTML.gif
(3.1)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ25_HTML.gif
(3.2)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ26_HTML.gif
(3.3)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ27_HTML.gif
(3.4)
Since N 1 v Im ( L 1 ) , N 2 u Im ( L 2 ) , then T 1 ( N 1 v ) = T 2 ( N 1 v ) = 0 , T 3 ( N 2 u ) = T 4 ( N 2 u ) = 0 . Then we can see, from the condition (H3), that there exist constants e 0 , e , e ( 0 , 1 ) such that | u ( t ) | M 1 , | v ( t ) | M 1 for t [ e 0 , e ] and | D 0 + q u ( t ) | M 2 , | D 0 + p v ( t ) | M 2 for t [ e , 1 ] . So, we can see from (3.1) and (3.2) that
| h 1 | Γ ( α q ) Γ ( α ) t α q 1 M 2 + 1 Γ ( α ) sup t [ 0 , 1 ] | f ( t , v ( t ) , D 0 + p v ( t ) ) | + Γ ( α q ) Γ ( α ) t i < t | B i | t i q + 1 α Γ ( α q ) Γ ( α ) e α q 1 M 2 + 1 Γ ( α ) sup t [ 0 , 1 ] | f ( t , v ( t ) , D 0 + p v ( t ) ) | + Γ ( α q ) Γ ( α ) i = 1 k | B i | t i q + 1 α
(3.5)
and
| h 2 | M 1 + 1 Γ ( α ) sup t [ 0 , 1 ] | f ( t , v ( t ) , D 0 + p v ( t ) ) | + | h 1 | + Γ ( α q ) Γ ( α ) ( t i < t | B i | t i q + 1 α ) t + t i < t | A i | t i 2 α + Γ ( α q ) Γ ( α ) t i < t | B i | t i q + 2 α M 1 + 1 Γ ( α ) sup t [ 0 , 1 ] | f ( t , v ( t ) , D 0 + p v ( t ) ) | + | h 1 | + i = 1 k | A i | t i 2 α + Γ ( α q ) Γ ( α ) i = 1 k | B i | t i q + 1 α ( 1 + t i ) .
(3.6)
Then for t [ 0 , 1 ] and u dom ( L 2 ) , we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ30_HTML.gif
(3.7)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ31_HTML.gif
(3.8)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ32_HTML.gif
(3.9)
Similarly, for u dom ( L 2 ) , we have that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ33_HTML.gif
(3.10)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ34_HTML.gif
(3.11)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ35_HTML.gif
(3.12)
Substitute (3.11) and (3.12) into (3.8), then we have
u α P C 4 Γ ( α ) φ 1 + 4 Γ ( α ) ψ 1 × ( 4 Γ ( β ) [ φ 2 + ψ 2 u α P C + γ 2 D 0 + q u P C ] + R 2 ) + 4 Γ ( α ) γ 1 ( 2 Γ ( β p ) [ φ 2 + ψ 2 u α P C + γ 2 D 0 + q u P C ] + R 3 ) = ( 16 ψ 1 Γ ( α ) Γ ( β ) + 8 γ 1 Γ ( α ) Γ ( β p ) ) [ φ 2 + ψ 2 u α P C + γ 2 D 0 + q u P C ] + 4 Γ ( α ) ( φ 1 + ψ 1 R 2 + γ 1 R 3 ) .
(3.13)
It means that
u α P C A D 0 + q u P C + B ,
similarly,
v β P C A D 0 + p v P C + B .
Substituting the above two into (3.9) and (3.12), we can see that
D 0 + q u P C 2 Γ ( α q ) ( ψ 1 A + γ 1 ) D 0 + p v P C + 2 Γ ( α q ) ( φ 1 + ψ 1 B ) R 3
(3.14)
and
D 0 + p v P C 2 Γ ( β p ) ( ψ 2 A + γ 2 ) D 0 + q u P C + 2 Γ ( β p ) ( φ 2 + ψ 2 B ) R 3 .
(3.15)

From the condition (H2), (3.14) and (3.15) give that D 0 + q u P C and D 0 + p v P C are bounded, then u α P C and v β P C are also bounded. Thus, by the definition of the norm on Y, u Y 1 and v Y 2 are bounded. That is, Ω 1 is bounded in Y. □

Lemma 3.2 Suppose that the condition (H3) holds. Then the set
Ω 2 = { ( u , v ) | ( u , v ) Ker ( L ) , N ( u , v ) Im ( L ) }

is bounded in Y.

Proof For ( u , v ) Ker ( L ) , we have that ( u , v ) = ( h 1 t α 1 + h 2 t α 2 , h 3 t β 1 + h 4 t β 2 ) , where h i , i { 1 , 2 , 3 , 4 } . Since N ( u , v ) Im ( L ) , so we have
T 1 N 1 ( h 3 t β 1 + h 4 t β 2 ) = T 2 N 1 ( h 3 t β 1 + h 4 t β 2 ) = 0
and
T 3 N 2 ( h 1 t α 1 + h 2 t α 2 ) = T 4 N 2 ( h 1 t α 1 + h 2 t α 2 ) = 0 .
From (H3), there exist positive constants M , M , e 0 , e , e such that for t [ e , 1 ] ,
| D 0 + q u ( t ) | = Γ ( α ) Γ ( α q ) | h 1 | t α q 1 M ,
which means that | h 1 | M Γ ( α q ) Γ ( α ) e α q 1 . And for t [ e 0 , e ] ,
| u ( t ) | = | h 1 t α 1 + h 2 t α 2 | M ,
which means that | h 2 | = | u ( t ) t 2 α + h 1 t | | u ( t ) | + | h 1 | M + | h 1 | . So, we can see that for t [ 0 , 1 ] ,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equat_HTML.gif

The above two arguments imply that | u | Y 1 is bounded. In the same way, | v | Y 2 is bounded. Thus, Ω 2 is bounded in Y. □

Lemma 3.3 The set
Ω 3 = { ( u , v ) Ker ( L ) | λ J ( u , v ) + ( 1 λ ) θ Q N ( u , v ) = ( 0 , 0 , , 0 ) , λ [ 0 , 1 ] }
is bounded in Y, where J : Ker ( L ) Im ( Q ) is the linear isomorphism given by
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equav_HTML.gif
and
θ = { 1 , if  (H ) (1)  hold 4 , 1 , if  (H ) (2)  hold 4 .
Proof For ( u , v ) Ker ( L ) , set u = h 1 t α 1 + h 2 t α 2 , v = h 3 t β 1 + h 4 t β 2 , then λ J ( u , v ) + ( 1 λ ) θ Q N ( u , v ) = ( 0 , 0 , , 0 ) implies that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ39_HTML.gif
(3.16)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ40_HTML.gif
(3.17)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ41_HTML.gif
(3.18)
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equ42_HTML.gif
(3.19)
From (3.16) and (3.17), we have
λ ( h 1 2 + h 2 2 , 0 , , 0 ) + ( 1 λ ) θ [ h 1 T 1 N 1 ( v ) + h 2 T 2 N 1 ( v ) ] = ( 0 , 0 , , 0 ) ,
the condition (H4) gives that
λ ( h 1 2 + h 2 2 ) = ( 1 λ ) θ s < 0 ,
where
s = { s 1 , if (H ) (1) hold 4 , s 3 , if (H ) (2) hold 4 ,

which is a contradiction. As a result, there exist positive constants g 1 , g 2 such that | h 1 | g 1 , | h 2 | g 2 . Similarly, from (3.18)-(3.19) and the second part of (1) or (2) of (H4), there exist two positive constants g 3 , g 4 such that | h 3 | g 3 , | h 4 | g 4 . It follows that u Y 1 , v Y 1 are bounded, that is, Ω 3 is bounded in Y. □

Theorem 3.1 Suppose that (H1)-(H4) hold. Then the 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 2.2 that N is L-compact on Ω ¯ . By means of above Lemmas 3.1-3.3, one obtains that
  1. (i)

    L ( u , v ) λ N ( u , v ) for every ( ( u , v ) , λ ) [ ( dom ( L ) Ker ( L ) ) Ω ] × ( 0 , 1 ) ;

     
  2. (ii)

    N ( u , v ) Im ( L ) for every ( u , v ) Ker ( L ) Ω .

     
Then we need only to prove
  1. (iii)

    deg ( Q N | Ker ( L ) , Ω Ker ( L ) , ( 0 , 0 , , 0 ) ) 0 .

     
Take
H ( u , v , λ ) = ± λ J + ( 1 λ ) N ( u , v ) .
According to Lemma 3.3, we know H ( ( u , v ) , λ ) ( 0 , 0 , , 0 ) for all ( u , v ) Ω Ker ( L ) . Thus, the homotopy invariance property of degree theory gives that
deg ( Q N | Ker ( L ) , Ω Ker ( L ) , ( 0 , 0 , , 0 ) ) = deg ( H ( , 0 ) , Ω Ker ( L ) , ( 0 , 0 , , 0 ) ) = deg ( H ( , 1 ) , Ω Ker ( L ) , ( 0 , 0 , , 0 ) ) = deg ( ± J , Ω Ker ( L ) , ( 0 , 0 , , 0 ) ) 0 .

Then, by Theorem 2.1, L ( u , v ) = N ( u , v ) has at least one solution in dom ( L ) Ω ¯ , i.e., the problem (1.1) has at least one solution in Y, which completes the proof. □

4 An example

Example 4.1

Consider the following boundary value problem for coupled systems of impulsive fractional differential equations:
{ D 0 + 3 2 u ( t ) = f ( t , v ( t ) , D 0 + 1 6 v ( t ) ) , D 0 + 4 3 v ( t ) = g ( t , u ( t ) , D 0 + 1 4 u ( t ) ) , 0 < t < 1 ; Δ u ( 3 4 ) = A 1 ( v ( 3 4 ) , D 0 + 1 6 v ( 3 4 ) ) , Δ D 0 + 1 4 u ( 3 4 ) = B 1 ( v ( 3 4 ) , D 0 + 1 6 v ( 3 4 ) ) ; Δ v ( 2 3 ) = C 1 ( u ( 2 3 ) , D 0 + 1 4 u ( 2 3 ) ) , Δ D 0 + 1 6 v ( 2 3 ) = D 1 ( u ( 2 3 ) , D 0 + 1 4 u ( 2 3 ) ) ; D 0 + 1 2 u ( 0 ) = 3 D 0 + 1 2 u ( 1 6 ) 2 D 0 + 1 2 u ( 1 4 ) , u ( 1 ) = 5 u ( 1 5 ) + 2 3 u ( 1 3 ) ; D 0 + 1 3 v ( 0 ) = 4 D 0 + 1 3 v ( 1 5 ) 3 D 0 + 1 3 v ( 1 4 ) , v ( 1 ) = 3 ( 1 3 ) 2 3 v ( 1 3 ) + 4 ( 1 2 ) 2 3 v ( 1 2 ) ,
(4.1)
where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbc_HTML.gif
and
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbd_HTML.gif
Due to the coupled problem (1.1), we have that α = 3 2 , β = 4 3 , p = 1 6 , q = 1 4 , a 1 = 3 , a 2 = 2 , b 1 = 5 , b 2 = 6 , c 1 = 4 , c 2 = 3 , d 1 = 3 , d 2 = 4 . ξ 1 = 1 6 , ξ 2 = 1 4 ; η 1 = 1 5 , η 2 = 1 3 ; ζ 1 = 1 5 , ζ 2 = 1 4 ; θ 1 = 1 3 , θ 2 = 1 2 . Obviously, a 1 + a 2 = b 1 + b 2 = c 1 + c 2 = d 1 + d 2 = 1 and b 1 η 1 + b 2 η 2 = d 1 θ 1 + d 2 θ 2 = 1 . By direct calculation, we obtain that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Eqube_HTML.gif
It is easy to see that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbf_HTML.gif
where
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbg_HTML.gif
So, ψ 1 = 1 24 , γ 1 = 1 4 π , ψ 2 = 1 20 , γ 2 = 3 100 . And
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbh_HTML.gif

where A = 0.0527672 , A = 0.0110034 . Thus, the condition (H2) holds.

Taking M 1 = 1 , for any v dom ( L 2 ) , assume that | D 0 + 1 6 v ( t ) | > 1 holds for any t [ 1 12 , 1 6 ] . Thus either D 0 + 1 6 v ( t ) > 1 or D 0 + 1 6 v ( t ) < 1 for any t [ 1 12 , 1 6 ] . If D 0 + 1 6 v ( t ) > 1 , t [ 1 12 , 1 6 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbi_HTML.gif
If D 0 + v ( t ) < 1 , t [ 1 12 , 1 6 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbj_HTML.gif
Similarly, assume that | D 0 + 1 4 v ( t ) | > 1 holds for any t [ 1 10 , 1 5 ] . Thus either D 0 + 1 4 v ( t ) > 1 or D 0 + 1 4 v ( t ) < 1 for any t [ 1 10 , 1 5 ] . If D 0 + 1 6 v ( t ) > 1 , t [ 1 10 , 1 5 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbk_HTML.gif
If D 0 + 1 6 v ( t ) < 1 , t [ 1 10 , 1 5 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbl_HTML.gif

So, from the above arguments, the first part of the condition (H3) is true for M 1 = 1 , t [ 1 12 , 1 6 ] .

Taking M 21 = 16 , assume that | v | > 16 holds for any t [ 1 3 , 1 ] . Then either v > 16 or v < 16 for t [ 1 3 , 1 ] . If v > 16 for t [ 1 3 , 1 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbm_HTML.gif
If v < 16 for t [ 1 3 , 1 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbn_HTML.gif
By the same way, taking M 22 = 10 / 3 3 , assume that | u ( t ) | > M 22 holds for any t [ 1 2 , 1 ] . Then either v > 10 / 3 3 or v < 10 / 3 3 for t [ 1 2 , 1 ] . If v > 10 / 3 3 for t [ 1 2 , 1 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbo_HTML.gif
If v > 10 / 3 3 for t [ 1 2 , 1 ] , then
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbp_HTML.gif

So, from the above arguments, the second part of the condition (H3) holds for M 2 = max { M 21 , M 22 } = 16 , t [ 1 3 , 1 ] .

On the other hand, for ( u , v ) = ( h 1 t α 1 + h 2 t α 2 , h 3 t β 1 + h 4 t β 2 ) Ker ( L ) , taking g = 8 , assume that h i < 8 , i = 1 , 2 , 3 , 4 , then h 3 t β 1 + h 4 t β 2 < 16 for t [ 1 3 , 1 ] , h 1 t α 1 + h 2 t α 2 < 10 3 3 for t [ 1 2 , 1 ] . And D 0 + 1 4 u < 1 for t [ 1 10 , 1 5 ] , D 0 + 1 6 u < 1 for t [ 1 12 , 1 6 ] . Then we can see, from the above arguments, that T 1 N 1 v = ( r 1 , 0 , , 0 ) , T 2 N 1 v = ( r 2 , 0 , , 0 ) , T 1 N 2 u = ( r 3 , 0 , , 0 ) , T 2 N 2 u = ( r 4 , 0 , , 0 ) , where r i < 0 , i = 1 , 2 , 3 , 4 . Thus,
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-80/MediaObjects/13661_2012_Article_326_Equbq_HTML.gif

where s 1 > 0 , s 2 > 0 . So, the condition (H4) holds. Hence, from Theorem 3.1, the coupled problem (4.1) has at least one solution in { u 3 2 , D 0 + 1 2 u P C [ 0 , 1 ] } × { v 4 3 , D 0 + 1 3 v P C [ 0 , 1 ] } .

Declarations

Acknowledgements

The authors would like to thank the editor and referee for their valuable comments and remarks which lead to a great improvement of the article. This research is supported by the National Natural Science Foundation of China (11071108), the Provincial Natural Science Foundation of Jiangxi, China (2010GZS0147, 20114BAB201007).

Authors’ Affiliations

(1)
Department of Mathematics, Nanchang University

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