Skip to content

Advertisement

  • Research
  • Open Access

Existence results for a coupled system of nonlinear fractional 2m-point boundary value problems at resonance

Advances in Difference Equations20112011:44

https://doi.org/10.1186/1687-1847-2011-44

  • Received: 14 June 2011
  • Accepted: 12 October 2011
  • Published:

Abstract

A 2m-point boundary value problem for a coupled system of nonlinear fractional differential equations is considered in this article. An existence result is obtained with the use of the coincidence degree theory.

MSC: 34B17; 34L09.

Keywords

  • coincidence degree
  • fractional nonlinear differential equation
  • 2m-point boundary conditions

1. Introduction

In this article, we will consider a 2m-point boundary value problem (BVP) 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 ) , 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 ,
(1.1)
I 0 + 3 - α u ( t ) | t = 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 ) ,
(1.2)
I 0 + 3 - β v ( t ) | t = 0 = 0 , D 0 + β - 2 v ( 1 ) = j = 1 m c j D 0 + β - 2 v ( γ j ) , v ( 1 ) = j = 1 m d j v ( δ j ) ,
(1.3)

where 2 < α, β ≤ 3, 0 < ξ 1 < < ξ m < 1, 0 < η 1 < < η m < 1, 0 < γ 1 < < γ m < 1, 0 < δ 1 < < δ m < 1, a i , b i , c j , d j R, f, g : [0, 1] × R 3R, f, g satisfies Carathéodory conditions, D 0 + α and I 0 + α are the standard Riemann-Liouville fractional derivative and fractional integral, respectively.

Setting:
Λ 1 = 1 α ( α - 1 ) 1 - i = 1 m a i ξ i α + 1 , Λ 2 = 1 α ( α - 1 ) 1 - i = 1 m a i ξ i α , Λ 3 = ( Γ ( α ) ) 2 Γ ( 2 α ) 1 - i = 1 m b i η i 2 α - 1 , Λ 4 = Γ ( α ) Γ ( α - 1 ) Γ ( 2 α - 1 ) 1 - i = 1 m b i η i 2 α - 2 , Δ 1 = 1 β ( β - 1 ) 1 - j = 1 n - 2 c j γ j β + 1 , Δ 2 = 1 β ( β - 1 ) 1 - j = 1 n - 2 c j γ j β , Δ 3 = ( Γ ( β ) ) 2 Γ ( 2 β ) 1 - j = 1 m d i δ j 2 β - 1 , Δ 4 = Γ ( β ) Γ ( β - 1 ) Γ ( 2 β - 1 ) 1 - j = 1 m d j δ j 2 β - 2 .

In this article, we will always suppose that the following conditions hold:

(C1):
i = 1 m a i ξ i = i = 1 m a i = 1 , i = 1 m b i η i α - 1 = i = 1 m b i η i α - 2 = 1 , j = 1 m c j γ j = j = 1 m c j = 1 , j = 1 m d j δ j β - 1 = j = 1 m d j δ j β - 2 = 1 ;
(C2):
Λ = Λ 1 Λ 4 - Λ 2 Λ 3 0 , Δ = Δ 1 Δ 4 - Δ 2 Δ 3 0 .
The subject of fractional calculus has gained considerable popularity and importance because of its frequent appearance in various fields such as physics, chemistry, and engineering. In consequence, the subject of fractional differential equations has attracted much attention. For details, refer to [14] and the references therein. Some basic theory for the initial value problems of fractional differential equations(FDE) involving Riemann-Liouville differential operator has been discussed by Lakshmikantham [57], El-Sayed et al. [8, 9], Diethelm and Ford [10], Bai [11], and so on. Also, there are some articles which deal with the existence and multiplicity of solutions for nonlinear FDE BVPs using techniques of topological degree theory. For example, Su [12] considered the BVP of the coupled system
D α u ( t ) = f ( t , v ( t ) , D μ v ( t ) ) , D β v ( t ) = g ( t , u ( t ) , D ν u ( t ) ) .

By using the Schauder fixed point theorem, one existence result was given.

However, there are few articles which consider the BVP at resonance for nonlinear ordinary differential equations of fractional order. In [13], Zhang and Bai investigated the nonlinear nonlocal problem
D 0 + α u ( t ) = f ( t , u ( t ) ) , 0 < t < 1 , u ( 0 ) = 0 , β u ( η ) = u ( 1 ) ,

where 1 < α ≤ 2, we consider the case βη α-1= 1, i.e., the resonance case.

In [14], Bai investigated the BVP at resonance
D 0 + α u ( t ) = f ( t , u ( t ) , D 0 + α - 1 u ( t ) ) + e ( t ) , 0 < t < 1 , I 0 + 2 - α u ( t ) | t = 0 = 0 , D 0 + α - 1 u ( 1 ) = i = 0 m - 2 β i D 0 + α - 1 u ( η i )

is considered, where 1 < α ≤ 2 is a real number, D 0 + α and I 0 + α are the standard Riemann-Liouville fractional derivative and fractional integral, respectively, and f : [0, 1] × R 2R is continuous, and e(t) L 1[0, 1], m ≥ 2, 0 < ξ i < 1, β i R, i = 1, 2, ..., m - 2, are given constants such that i = 1 m - 2 β i = 1 .

The coupled system (1.1)-(1.3) happens to be at resonance in the sense that the associated linear homogeneous coupled system
D 0 + α u ( t ) = 0 , 0 < t < 1 , D 0 + β v ( t ) = 0 , 0 < t < 1 ,
I 0 + 3 - α u ( t ) | t = 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 ( t ) | t = 0 = 0 , D 0 + β - 2 v ( 1 ) = j = 1 m c j D 0 + β - 2 v ( γ j ) , v ( 1 ) = i = 1 m d j v ( δ j )

has (u(t), v(t)) = (at α-1+ bt α-2, ct β-1+ dt β-2), a, b, c, d R as a nontrivial solution.

The purpose of this article is to study the existence of solution for BVP (1.1)-(1.3) at resonance case, and establish an existence theorem under nonlinear growth restriction of f. Our method is based upon the coincidence degree theory of Mawhin.

Now, we will briefiy recall some notation and an abstract existence result.

Let Y, Z be real Banach spaces, L : domL YZ be a Fredholm map of index zero and P : YY, Q : Z → Z be continuous projectors such that
Y = K e r L K e r P , Z = I m Q I m L , I m P = K e r L , K e r Q = I m L .

It follows that L| domLKerP : domLKer PImL is invertible. We denote the inverse of the map by K p . If Ω is an open bounded subset of Y such that domL ∩ Ω ≠ , the map N : YZ will be called L-compact on Ω if Q N ( Ω ¯ ) is bounded, and K p ( I - Q ) N : Ω ¯ Y is compact.

The theorem that we used is Theorem 2.4 of [15].

Theorem 1.1. Let L be a Fredholm operator of index zero and N be L-compact on Ω ¯ . Assume that the following conditions are satisfied:
  1. (i)

    LxλNx (x, λ) [domL\KerL ∩ ∂Ω] × [0, 1];

     
(ii)) Nx ImL, x KerL ∩ ∂Ω;
  1. (iii)

    deg(JQN | KerL , KerL ∩ Ω, 0) ≠ 0;

     

where Q : ZZ is a projection as above with KerQ = ImL, and J : ImQKerL is any isomorphism. Then, the equation Lx = Nx has at least one solution in d o m L Ω ¯ .

The rest of this article is organized as follows. In Section 2, we give some notation and lemmas. In Section 3, we establish a theorem of existence of a solution for the problem (1.1)-(1.3).

2. Background materials and preliminaries

For the convenience of the reader, we present here some necessary basic knowledge and definitions about fractional calculus theory. These definitions can be found in the recent literature [114, 16].

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 ) d s ,

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

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 d t n 0 t y ( s ) ( t - s ) α - n + 1 d s ,

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

Definition 2.3. We say that the map f : [0, 1] × R n R satisfies Carathéodory conditions with respect to L 1[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 r > 0, there exists ρ r L 1 ([0, 1], R) such that, for a.e. t [0, 1] and every |z| ≤ r, we have f (t, z) ≤ ρ r (t).

     
Lemma 2.1. [13] Assume that u C(0, 1) ∩ L 1(0, 1) with a fractional derivative of order α > 0 that belongs to C(0, 1) ∩ L 1(0, 1). Then,
I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α - 1 + c 1 t α - 2 + + c N t α - N

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

We use the classical Banach space C[0, 1] with the norm
| | u | | = max t [ 0 , 1 ] | u ( t ) | ,

L[0, 1] with the norm | | u | | 1 = 0 1 | u ( t ) | d t . For n N, we denote by AC n [0, 1] the space of functions u(t) which have continuous derivatives up to order n - 1 on [0, 1] such that u (n-1)(t) is absolutely continuous:

AC n [0, 1] = {u|[0, 1] → R and D n-1 u(t) is absolutely continuous in [0, 1]}.

Definition 2.4. Given μ > 0 and N = [μ] + 1 we can define a linear 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 t μ - ( N - 1 ) , t ( 0 , 1 ) } ,

where x C[0, 1], c i R, i = 1, 2, ..., N - 1.

Remark 2.1. By means of the linear functional analysis theory, we can prove that with the
| | u | | C μ = | | D 0 + μ u | | + + | | D 0 + μ - ( N - 1 ) u | | + | | u | | ,

C μ [0, 1] is a Banach space.

Remark 2.2. If μ is a natural number, then C μ [0, 1] is in accordance with the classical Banach space C n [0, 1].

Lemma 2.2. [13] f C μ [0, 1] is a sequentially compact set if and only if f is uniformly bounded and equicontinuous. Here, uniformly bounded means there exists M > 0, such that for every u f
| | u | | C μ = | | D 0 + μ u | | + + | | 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 = 1 , 2 , , N - 1 )
Lemma 2.3. [14] Let α > 0, n = [α] + 1. Assume that u L 1 (0, 1) with a fractional integration of order n - α that belongs to AC n [0, 1]. Then, the equality
( I 0 + α D 0 + α u ) ( t ) = u ( t ) - i = 1 n ( ( I 0 + n - α u ) ( t ) ) n - i | t = 0 Γ ( α - i + 1 ) t α - i

holds almost everywhere on [0, 1].

Definition 2.5. [14] Let I 0 + α ( L 1 ( 0 , 1 ) ) , α > 0 denote the space of functions u(t), represented by fractional integral of order α of a summable function: u = I 0 + α v , v L 1 ( 0 , 1 ) .

In the following lemma, we use the unified notation of both for fractional integrals and fractional derivatives assuming that I 0 + α = D 0 + - α for α > 0.

Let Z 1 = L 1[0, 1], with the norm | | y | | = 0 1 | y ( s ) | d s , Y 1 = C α-1[0, 1], Y 2 = C β-1[0, 1], defined by Remark 2.1, with the norm
| | u | | Y 1 = | | D 0 + α - 1 u | | + | | D 0 + α - 2 u | | + | | u | | , | | v | | Y 2 = | | D 0 + β - 1 v | | + | | D 0 + β - 2 v | | + | | v | | ,
where Y = Y 1 × Y 2 is a Banach space, with the norm
| | ( u , v ) | | Y = max { | | u | | Y 1 , | | v | | Y 2 } ,
and Z = Z 1 × Z 1 is a Banach space, with the norm
| | ( x , y ) | | Z = max { | | x | | 1 , | | y | | 1 } .
Define L 1 to be the linear operator from domL 1Y 1 to Z 1 with
d o m L 1 = { u C α 1 [ 0,1 ] | D 0 + α u L 1 [ 0,1 ] , u satisfies (1 .2)},
and
L 1 u = D 0 + α u , u d o m L 1 .
Define L 2 to be the linear operator from domL 2Y 2 to Z 1 with
d o m L 2 = { v C β 1 [ 0,1 ] | D 0 + β v L 1 [ 0,1 ] , v satisfies (1 .3)},
and
L 2 v = D 0 + β v , v d o m L 2 .
Define L to be the linear operator from domLY to Z with
d o m L = { ( u , v ) Y | u d o m L 1 , v d o m L 2 } ,
and
L ( u , v ) = ( L 1 u , L 2 v ) ,
we define N : YZ by setting
N ( u , v ) = ( N 1 v , N 2 u ) ,
where N 1 : Y 2Z 1 is defined by
N 1 v ( t ) = f ( t , v ( t ) , D 0 + β 1 v ( t ) ) , D 0 + β 2 v ( t ) ) ,
and N 2 : Y 1Z 2 is defined by
N 2 u ( t ) = f ( t , u ( t ) , D 0 + α 1 u ( t ) ) , D 0 + α 2 u ( t ) ) .
Then, the coupled system of BVPs (1.1) can be written as
L ( u , v ) = N ( u , v ) .

3. Main results

Lemma 3.1. The mapping L : domL YZ is a Fredholm operator of index zero.

Proof. Let L 1 u = D 0 + α u , by Lemma 2.3, D 0 + α u ( t ) = 0 has solution
u ( t ) = i = 1 3 ( ( I 0 + 3 α u ) ( t ) ) 3 i | t = 0 Γ ( α i + 1 ) t α i = ( ( I 0 + 3 α u ) ( t ) ) | t = 0 Γ ( α ) t α 1 + ( ( I 0 + 3 α u ) ( t ) ) | t = 0 Γ ( α 1 ) t α 2 + ( ( I 0 + 3 α u ) ( t ) ) | t = 0 Γ ( α 2 ) t α 3 = 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 .
Combine with (1.2), so
K e r L 1 = { a t α - 1 + b t α - 2 | a , b R } R 2 .

Similarly, let L 2 v = D 0 + β v , by Lemmas 2.3, 2.4, D 0 + β v ( t ) = 0 , combine with (1.3),

so
K e r L 2 = { c t β - 1 + d t β - 2 | c , d R } R 2 .
It is clear that
K e r L = { ( a t α - 1 + b t α - 2 , c t β - 1 + d t β - 2 ) | a , b , c , d R } R 2 × R 2 .
Let (x, y) ImL, then there exists (u, v) domL, such that (x, y) = L(u, v), that is u Y 1, x = D 0 + α u and v Y 2, y = D 0 + β v . By Lemma 2.3, we have
I 0 + α x ( t ) = u ( t ) - c 1 t α - 1 - c 2 t α - 2 - c 3 t α - 3 , I 0 + β y ( t ) = v ( t ) - d 1 t β - 1 - d 2 t β - 2 - d 3 t β - 3 ,
where
c 1 = D 0 + α 1 u ( t ) | t = 0 Γ ( α ) , c 2 = D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) , c 3 = I 0 + 3 α u ( t ) | t = 0 Γ ( α 2 ) , d 1 = D 0 + β 1 v ( t ) | t = 0 Γ ( β ) , d 2 = D 0 + β 2 v ( t ) | t = 0 Γ ( β 1 ) , d 3 = I 0 + 3 β v ( t ) | t = 0 Γ ( β 2 ) ,
and by the boundary condition (1.2), we obtain c 3 = 0, c 1, c2 can be any constant, and x satisfies
0 1 ( 1 - s ) x ( s ) d s - i = 1 m a i 0 ξ i ( ξ i - s ) x ( s ) d s = 0 , 0 1 ( 1 - s ) α - 1 x ( s ) d s - i = 1 m b i 0 η i ( η i - s ) α - 1 x ( s ) d s = 0 .
(3.1)
Similarly, by the boundary condition (1.3), we obtain d 3 = 0, d 1, d 2 can be any constant, and y satisfies
0 1 ( 1 - s ) y ( s ) d s - j = 1 m c j 0 γ j ( γ j - s ) y ( s ) d s = 0 , 0 1 ( 1 - s ) β - 1 y ( s ) d s - j = 1 m d j 0 δ j ( δ j - s ) β - 1 y ( s ) d s = 0 .
(3.2)
On the other hand, suppose x, y Z 1 satisfy (3.1), (3.2), respectively, let u ( t ) = I 0 + α x ( t ) , v ( t ) = I 0 + β y ( t ) , then u domL 1, D 0 + α u ( t ) = x ( t ) and v domL 2, D 0 + β v ( t ) = y ( t ) . That is to say, (x, y) ImL. From the above argument, we obtain
I m L = { ( x , y ) Z | x satisfies (3 .1),  y satisfies (3 .2)} .
Consider the continuous linear mapping A i , B i , T i , R i , Q i : Z 1Z 1, i = 1, 2 and Q : ZZ defined by
A 1 x = 0 1 ( 1 - s ) x ( s ) d s - i = 1 m a i 0 ξ i ( ξ i - s ) x ( s ) d s , A 2 x = 0 1 ( 1 - s ) α - 1 x ( s ) d s - i = 1 m b i 0 η i ( η i - s ) α - 1 x ( s ) d s
and
B 1 y = 0 1 ( 1 - s ) y ( s ) d s - j = 1 m c j 0 γ j ( γ j - s ) y ( s ) d s , B 2 y = 0 1 ( 1 - s ) β - 1 y ( s ) d s - j = 1 m d j 0 δ j ( δ j - s ) β - 1 y ( s ) d s . T 1 x = 1 Λ ( Λ 4 A 1 x - Λ 2 A 2 x ) , T 2 x = 1 Λ ( Λ 3 A 1 x - Λ 1 A 2 x )
(3.3)
and
R 1 y = 1 Δ ( Δ 4 B 1 y - Δ 2 B 2 y ) , R 2 y = 1 Δ ( Δ 3 B 1 y - Δ 1 B 2 y ) .
(3.4)
Since the conditions (C1) and (C2) hold, the mapping defined by
Q 1 x ( t ) = ( T 1 x ( t ) ) t α - 1 + ( T 2 x ( t ) ) t α - 2 , Q 2 y ( t ) = ( R 1 y ( t ) ) t β - 1 + ( R 2 y ( t ) ) t β - 2
(3.5)

is well-defined. It is clear that dimImQ 1 = dimImQ 2 = 2.

Recall (C1) and (C2) and note that
T 1 ( T 1 x t α - 1 ) = 1 Λ ( Λ 4 A 1 ( T 1 x t α - 1 ) - Λ 2 A 2 ( T 1 x t α - 1 ) ) = 1 Λ Λ 4 Λ 4 Λ 1 Λ A 1 x - Λ 1 Λ 2 Λ A 2 x - Λ 2 Λ 4 Λ 3 Λ A 1 x - Λ 2 Λ 3 Λ A 2 x = T 1 x ,
and similarly we can derive that
T 1 ( T 2 x t α - 2 ) = 0 , T 2 ( T 1 x t α - 1 ) = 0 , T 2 ( T 2 x t α - 2 ) = T 2 x .
Hence, for x Z 1, it follows from the four relations above that
Q 1 2 x = Q 1 ( ( T 1 x ) t α - 1 + ( T 2 x ) t α - 2 ) = T 1 ( ( T 1 x ) t α - 1 + ( T 2 x ) t α - 2 ) t α - 1 + T 2 ( ( T 1 x ) t α - 1 + ( T 2 x ) t α - 2 ) t α - 2 = ( T 1 x ) t α - 1 + ( T 2 x ) t α - 2 = Q 1 x ,

that is, the map Q 1 is idempotent. In fact, Q 1 is a continuous linear projector.

Similarly, the map Q 2 is a continuous linear projector.

Therefore,
Q ( x , y ) = ( Q 1 x , Q 2 y ) .

It is clear that Q is a continuous linear projector.

Note (x, y) ImL implies Q(x, y) = (Q 1 x, Q 2 y) = (0, 0). Conversely, if Q(x, y) = (0, 0),

so
Λ 4 A 1 x - Λ 2 A 2 x = 0 , Λ 1 A 2 x - Λ 3 A 1 x = 0 , Δ 4 B 1 y - Δ 2 B 2 y = 0 , Δ 1 B 2 y - Δ 3 B 1 y = 0 ,
but
Λ 4 - Λ 2 - Λ 3 Λ 1 = Λ 4 Λ 1 - Λ 2 Λ 3 0 , Δ 4 - Δ 2 - Δ 3 Δ 1 = Δ 4 Δ 1 - Δ 2 Δ 3 0 ,

then we must have A i x = B i y = 0, i = 1, 2, that is, (x, y) ImL. In fact, KerQ = ImL.

Take (x, y) Z in the form (x, y) = ((x, y) - Q(x, y)) + Q(x, y) so that ((x, y) - Q(x, y)) KerQ = ImL, Q(x, y) ImQ. Thus, Z = ImL + ImQ. Let (x, y) ImLImQ and assume that (x, y) = (at α-1+ bt α-2, ct β-1+ dt β-2) is not identically zero on [0, 1]. Then, since (x, y) ImL, from (3.1) and (3.2) and the condition (C2), we have
A 1 x = 0 1 ( 1 - s ) ( a s α - 1 + b s α - 2 ) d s - i = 1 m a i 0 ξ i ( ξ i - s ) ( a s α - 1 + b s α - 2 ) d s = 0 , A 2 x = 0 1 ( 1 - s ) α - 1 ( a s α - 1 + b s α - 2 ) d s - i = 1 m b i 0 η i ( η i - s ) α - 1 ( a s α - 1 + b s α - 2 ) d s = 0 , B 1 y = 0 1 ( 1 - s ) ( c s β - 1 + d s β - 2 ) d s - j = 1 m c j 0 γ j ( γ j - s ) ( c s β - 1 + d s β - 2 ) d s = 0 , B 2 y = 0 1 ( 1 - s ) β - 1 ( c s β - 1 + d s β - 2 ) d s - j = 1 m d j 0 δ j ( δ j - s ) β - 1 ( c s β - 1 + d s β - 2 ) d s = 0 .
So,
a Λ 1 + b Λ 2 = 0 , a Λ 3 + b Λ 4 = 0 , c Δ 1 + d Δ 2 = 0 , c Δ 3 + d Δ 4 = 0 ,
but
Λ 1 Λ 2 Λ 3 Λ 4 = Λ 1 Λ 4 - Λ 2 Λ 3 0 , Δ 1 Δ 2 Δ 3 Δ 4 = Δ 1 Δ 4 - Δ 2 Δ 3 0 ,

we derive a = b = c = d = 0, which is a contradiction. Hence, ImLImQ = {0, 0}; thus, Z = ImL ImQ.

Now, IndL = dimKerL - codimImL = 0, and so L is a Fredholm operator of index zero.

Let P 1 : Y 1Y 1, P 2 : Y 2Y 2, P : YY be defined by
P 1 u ( t ) = 1 Γ ( α ) D 0 + α - 1 u ( t ) | t = 0 t α - 1 + 1 Γ ( α - 1 ) D 0 + α - 2 u ( t ) | t = 0 t α - 2 , t [ 0 , 1 ] , P 2 v ( t ) = 1 Γ ( β ) D 0 + β - 1 v ( t ) | t = 0 t β - 1 + 1 Γ ( β - 1 ) D 0 + β - 1 v ( t ) | t = 0 t β - 2 , t [ 0 , 1 ] ,
and
P ( u , v ) = ( P 1 u , P 2 v ) .
Note that P 1, P 2, P are continuous linear projectors and
K e r P = ( K e r P l , K e r P 2 ) = { ( u , v ) Y | D 0 + α - 1 u ( 0 ) = D 0 + α - 2 u ( 0 ) = 0 , D 0 + β - 1 v ( 0 ) = D 0 + β - 2 v ( 0 ) = 0 } .

It is clear that Y = KerL KerP.

Note that the projectors P and Q are exact. Define by K p : ImLdomLKerP by
K P ( x , y ) = ( I 0 + α x , I 0 + β y ) , K p x ( t ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 x ( s ) d s = I 0 + α x ( t ) , x I m L .
Hence, we have
D 0 + α - 1 ( K p x ) t = 0 t x ( s ) d s , D 0 + α - 2 ( K p x ) t = 0 t ( t - s ) x ( s ) d s .
Then,
| | 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 ,
and thus
| | K p x | | Y 1 1 Γ ( α ) + 2 | | x | | 1
and
K p y ( t ) = 1 Γ ( β ) 0 t ( t - s ) β - 1 y ( s ) d s = I 0 + β y ( t ) , y I m L .
Hence, we have
D 0 + β - 1 ( K p y ) t = 0 t y ( s ) d s , D 0 + β - 2 ( K p y ) t = 0 t ( t - s ) y ( s ) d s .
Then,
| | K p y | | 1 Γ ( β ) | | y | | 1 , | | D 0 + β - 1 ( K p y ) | | | | y | | 1 , | | D 0 + β - 2 ( K p y ) | | | | y | | 1 ,
and thus
| | K p y | | Y 2 1 Γ ( β ) + 2 | | y | | 1 ,
so
K p ( x , y ) Y = ( I 0 + α x , I 0 + β y ) Y = max { I 0 + α x Y 1 , I 0 + β y Y 2 } max 1 Γ ( α ) + 2 x 1 , 1 Γ ( β ) + 2 y 1 .
For (x, y) ImL, we have
L K P ( x , y ) = L ( I 0 + α x , I 0 + β y ) = ( D 0 + α I 0 + α x , D 0 + β I 0 + β y ) = ( x , y ) .
Also, if (u, v) domLKerP, we have u domL 1, D 0 + α - 1 u ( 0 ) = D 0 + α - 2 u ( 0 ) = 0 , v domL 2, D 0 + β - 1 v ( 0 ) = D 0 + β - 2 v ( 0 ) = 0 , so the coefficients c i , d i , i = 1, 2, 3 in the expressions then
( K p L 1 ) u ( t ) = I 0 + α D 0 + α u ( t ) = u ( t ) + c 1 t α - 1 + c 2 t α - 2 + c 3 t α - 3 ,
where
c 1 = D 0 + α 1 u ( t ) | t = 0 Γ ( α ) , c 2 = D 0 + α 2 u ( t ) | t = 0 Γ ( α 1 ) , c 3 = I 0 + 3 α u ( t ) | t = 0 Γ ( α 2 ) , ( K p L 2 ) v ( t ) = I 0 + β D 0 + β v ( t ) = v ( t ) + d 1 t β 1 + d 2 t β 2 + d 3 t β 3 ,
where
d 1 = D 0 + β - 1 v ( t ) | t = 0 Γ ( β ) , d 2 = D 0 + β - 2 v ( t ) | t = 0 Γ ( β - 1 ) , d 3 = I 0 + 3 - β v ( t ) | t = 0 Γ ( β - 2 ) ,
and from the boundary value conditions (1.2), (1.3) and the fact that (u, v) domLKerP, P(u, v) = 0, we have c i = d i = 0, thus
( K p L ) ( u , v ) = K p ( L 1 u , L 2 v ) = ( u , v ) .

This shows that K p = [L| domLKerP ]-1.

Using (3.3)-(3.5), we write
K p ( I - Q ) N ( x , y ) = 1 Γ ( α ) 0 t ( t - s ) α - 1 [ N 1 x ( s ) - Q 1 N 1 x ( s ) ] d s , 1 Γ ( β ) 0 t ( t - s ) β - 1 [ N 2 y ( s ) - Q 2 N 2 y 2 ( s ) ] d s .

By Lemma 2.2 and a standard method, we obtain the following lemma.

Lemma 3.2. [16] K p (I - Q)N : YY is completely continuous.

In this section, we shall prove existence results for (1.1)-(1.3).

First, let us set the following notations for convenience:
m = 3 + 1 Γ ( α ) + + 1 Γ ( α - 1 ) , n = 3 + 1 Γ ( β ) + 1 Γ ( β - 1 ) , j = 2 + 1 Γ ( α ) , k = 2 + 1 Γ ( β ) , q = 5 + 2 Γ ( α ) + + 1 Γ ( α - 1 ) , w = 5 + 2 Γ ( β ) + + 1 Γ ( β - 1 ) .

Assume that the following conditions on the function f(t, x, y, z), g(t, x, y, z) are satisfied:

(H1) There exist functions a i (t), b i (t), c i (t), d i (t), r i (t) L 1[0, 1], i = 1, 2 and a constant θ i [0, 1), i = 1, 2 such that for all (x, y, z) R 3, t [0, 1], one of the following inequalities is satisfied:
| f ( t , x , y , z ) | a 1 ( t ) | x | + b 1 ( t ) | y | + c 1 ( t ) | z | + d 1 ( t ) | x | θ 1 + r 1 ( t ) ,
(3.6)
| g ( t , x , y , z ) | a 2 ( t ) | x | + b 2 ( t ) | y | + c 2 ( t ) | z | + d 2 ( t ) | x | θ 2 + r 2 ( t ) .
(3.7)
(H2) There exists a constant A > 0, such that for (u, v) domL\KerL satisfying
min { | D 0 + α - 1 u ( t ) | + | D 0 + α - 2 u ( t ) | , | D 0 + β - 1 v ( t ) | + | D 0 + β - 2 v ( t ) | } > A
or for all t [0, 1], we have
A 1 N 1 v ( t ) 0 , B 1 N 2 u ( t ) 0 or A 2 N 1 v ( t ) 0 , B 2 N 2 u ( t ) 0 .
(H3) There exists a constant B > 0 such that for every a, b, c, d R satisfying min{a 2 + b 2, c 2 + d 2} > B then either
a T 1 N 1 ( a t α - 1 + b t α - 2 ) + b T 2 N 1 ( a t α - 1 + b t α - 2 ) > 0 ,
(3.8)
c R 1 N 2 ( c t β - 1 + d t β - 2 ) + d R 2 N 2 ( c t β - 1 + d t β - 2 ) > 0 ,
(3.9)
or
a T 1 N 1 ( a t α - 1 + b t α - 2 ) + b T 2 N 1 ( a t α - 1 + b t α - 2 ) < 0 ,
(3.10)
c R 1 N 2 ( c t β - 1 + d t β - 2 ) + d R 2 N 2 ( c t β - 1 + d t β - 2 ) < 0 ,
(3.11)
Theorem 3.1 If (C1)-(C2) and (H1)-(H3) hold, then the BVP (1.1)-1.3) has at least one solution provided that
max { q ( a 1 1 + b 1 1 + c 1 1 ) , w ( a 2 1 + b 2 1 + c 2 1 ) , j a 1 1 + n a 2 1 + j b 1 1 + n b 2 1 + j c 1 1 + n c 2 1 , k a 1 1 + m a 2 1 + k b 1 1 + m b 2 1 + k c 1 1 + m c 2 1 } < 1.
Proof. Set
Ω 1 = { ( u , v ) d o m L \ K e r L : L ( u , v ) = λ N ( u , v ) , λ [ 0 , 1 ] } .
Then, for (u, v) Ω1, L(u, v) = λN (u, v), thus λ ≠ 0, N(u, v) ImL = KerQ, and hence QN(u, v) = (0, 0) for all t [0, 1]. By the definition of Q, we have Q 1 N 1 v(t) = Q 2 N 2 u(t) = 0. It follows from (H2) that there exists t 0, t 1 [0, 1], such that
min { | D 0 + α - 1 u ( t 0 ) | + | D 0 + α - 2 u ( t 0 ) | , | D 0 + β - 1 v ( t 1 ) | + | D 0 + β - 2 v ( t 1 ) | } A .
Now
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 , D 0 + β - 1 v ( t ) = D 0 + β - 1 v ( t 1 ) + t 1 t D 0 + β v ( s ) d s , D 0 + β - 2 v ( t ) = D 0 + β - 2 v ( t 1 ) + t 1 t D 0 + β - 1 v ( s ) d s ,
and so
| D 0 + α - 1 u ( 0 ) | D 0 + α - 1 u ( t ) | D 0 + α - 1 u ( t 0 ) | + D 0 + α - 1 u 1 A + L 1 u 1 A + N 1 v 1 ,
and
| D 0 + α - 2 u ( 0 ) | D 0 + α - 2 u ( t ) | D 0 + α - 2 u ( t 0 ) | + D 0 + α - 1 u | D 0 + α - 2 u ( t 0 ) | + | D 0 + α - 1 u ( t 0 ) | + D 0 + α u 1 A + L 1 u 1 A + N 1 v 1 .
Similarly,
| D 0 + β - 1 v ( 0 ) | A + N 2 u 1 , | D 0 + β - 2 v ( 0 ) | A + N 2 u 1 .

Therefore, we have noted that (I - P)(u, v) domLKerP for (u, v) Ω1.

Then,
P ( u , v ) Y = ( P 1 u , P 2 v ) Y = max { P 1 u Y 1 , P 2 v Y 2 } = max 1 Γ ( α ) D 0 + α - 1 u ( 0 ) t α - 1 + 1 Γ ( α - 1 ) D 0 + α - 2 u ( 0 ) t α - 2 Y 1 , 1 Γ ( β ) D 0 + β - 1 v ( 0 ) t β - 1 + 1 Γ ( β - 1 ) D 0 + β - 2 v ( 0 ) t β - 2 Y 2 = max 1 Γ ( α ) D 0 + α - 1 u ( 0 ) t α - 1 + 1 Γ ( α - 1 ) D 0 + α - 2 u ( 0 ) t α - 2 + D 0 + α - 1 u ( 0 ) + D 0 + α - 1 u ( 0 ) t + D 0 + α - 2 u ( 0 ) , 1 Γ ( β ) D 0 + β - 1 v ( 0 ) t β - 1 + 1 Γ ( β - 1 ) D 0 + β - 2 v ( 0 ) t β - 2 + D 0 + β - 1 v ( 0 ) + D 0 + β - 1 v ( 0 ) t + D 0 + β - 2 v ( 0 ) = max 3 + 1 Γ ( α ) + 1 Γ ( α - 1 ) ( A + N 1 v 1 ) , 3 + 1 Γ ( β ) + 1 Γ ( β - 1 ) ( A + N 2 u 1 ) = max { m ( A + N 1 v 1 ) , n ( A + N 2 u 1 ) } .
( I - P ) ( u , v ) Y = K p L ( I - P ) ( u , v ) Y = K p ( L 1 u , L 2 v ) Y max { ( 2 + 1 Γ ( α ) ) L 1 u 1 , ( 2 + 1 Γ ( β ) ) L 2 v 1 } max { j N 1 v 1 , k N 2 u 1 } ,
so, we have
( u , v ) Y ( I P ) ( u , v ) Y + P ( u , v ) Y = max { m ( A + N 1 v 1 ) , n ( A + N 2 u 1 ) } + max { j N 1 v 1 , k N 2 u 1 } = max { q N 1 v 1 + m A , m ( A + N 1 v 1 ) + k N 2 u 1 , n ( A + N 2 u 1 ) + j N 1 v 1 , w N 2 u 1 + n A } .
(3.12)

If the first condition of (H1) is satisfied, then from (3.12), the proof can be divided into four cases:

Case 1. ||(u, v)|| Y q||N 1 v||1 + mA.

From (3.6), we have
( u , v ) Y q [ a 1 1 v + b 1 1 D 0 + β - 1 v + c 1 1 D 0 + β - 2 v + d 1 1 D 0 + β - 2 v θ 1 + D ] ,
where D = r 1 1 + m A q , and consequently, for
v , D 0 + β - 1 v , D 0 + β - 2 v ( u , v ) Y ,
so
v q 1 - q a 1 1 [ b 1 1 D 0 + β - 1 v + c 1 1 D 0 + β - 2 v + d 1 1 D 0 + β - 2 v θ 1 + D ] , D 0 + β - 1 v q 1 - q a 1 1 - q b 1 1 [ c 1 1 D 0 + β - 2 x + d 1 1 D 0 + β - 2 v θ 1 + D ] , D 0 + β - 2 v q 1 - q a 1 1 - q b 1 1 - q c 1 1 ( d 1 1 D 0 + β - 2 v θ 1 + D ) .

But θ 1 [0, 1) and a 1 1 + b 1 1 + c 1 1 1 q , so there exists A 1, A 2, A 3 > 0 such

that
D 0 + β - 2 v A 1 , D 0 + β - 1 v A 2 , v A 3 .
Therefore, for all (u, v) Ω1,
( u , v ) Y = max { v , D 0 + β - 1 v , D 0 + β - 2 v } max { A 1 , A 2 , A 3 } ,

we can prove that Ω1 is also bounded.

Case 2. ||(u, v)|| Y w||N 2 u||1 + nA.

The proof is similar to that of case 1. Here, we omit it, where
a 2 1 + b 2 1 + c 2 1 1 w .

Case 3. ||(u, v)|| Y n(A + ||N 2 u||1) + j||N 1 v||1.

From (3.6) and (3.7), we have
( u , v ) Y j [ a 1 1 v + b 1 1 D 0 + β 1 v + c 1 1 D 0 + β 2 v + d 1 1 D 0 + β 2 v θ 1 + r 1 1 ] + n [ a 2 1 u + b 2 1 D 0 + α 1 u + c 2 1 D 0 + α 2 u + d 2 1 D 0 + α 2 u θ 2 + A + r 2 1 ] ,
v 1 1 j a 1 1 [ j b 1 1 D 0 + β 1 v + j c 1 1 D 0 + β 2 v + j d 1 1 D 0 + β 2 v θ 1 + j r 1 1 + n a 2 1 u + n b 2 1 D 0 + α 1 u + n c 2 1 D 0 + α 2 u + n d 2 1 D 0 + α 2 u θ 2 + n A + n r 2 1 ] ,
u 1 1 j a 1 1 n a 2 1 [ j b 1 1 D 0 + β 1 v + j c 1 1 D 0 + β 2 v + j d 1 1 D 0 + β 2 v θ 1 + j r 1 1 + n b 2 1 D 0 + α 1 u + n c 2 1 D 0 + α 2 u + n d 2 1 D 0 + α 2 u θ 2 + n A + n r 2 1 ] ,
D 0 + β 1 v 1 1 j a 1 1 n a 2 1 j b 1 1 [ j c 1 1 D 0 + β 2 v + j d 1 1 D 0 + β 2 v θ 1 + j r 1 + n b 2 1 D 0 + α 1 u + n c 2 1 D 0 + α 2 u + n d 2 1 D 0 + α 2 u θ 2 + n A + n r 2 ] ,
D 0 + α 1 u 1 1 j a 1 1 n a 2 1 j b 1 1 n b 2 1 [ j c 1 1 D 0 + β 2 v + j d 1 1 D 0 + β 2 v θ 1 + j r 1 1 + n c 2 1 D 0 + α 2 u + n d 2 1 D 0 + α 2 u θ 2 + n A + n r 2 1 ] ,
D 0 + β - 2 v j d 1 1 D 0 + β - 2 v θ 1 + j r 1 | | 1 + n d 2 1 | | D 0 + α - 2 u θ 2 + n A + n | | r 2 1 1 - j a 1 | | 1 - n a 2 1 - j b 1 1 - n b 2 | | 1 - j | | c 1 1 - n c 2 | | 1 ,
D 0 + α - 2 u j d 1 1 D 0 + β - 2 v θ 1 + j r 1 | | 1 + n d 2 1 | | D 0 + α - 2 u θ 2 + n A + n | | r 2 1 1 - j a 1 | | 1 - n a 2 1 - j b 1 1 - n b 2 | | 1 - j | | c 1 1 - n c 2 | | 1 .
If n d 2 1 D 0 + α - 2 u θ 2 j d 1 1 D 0 + β - 2 v θ 1 , then we have
D 0 + α - 2 u j | | r 1 1 + 2 n d 2 | | 1 | | D 0 + α - 2 u θ 2 + n A + n | | r 2 1 1 - j a 1 | | 1 - n a 2 1 - j b 1 | | 1 - n | | b 2 1 - j c 1 | | 1 - n c 2 1 .
But θ 2 [0, 1) and j||a 1||1 + n||a 2||1 + j||b 1||1 + n||b 2||1 + j||c 1||1 + n||c 2||1 < 1, so there exists A i > 0, i = 1, ..., 6 such that
D 0 + β - 2 v A 1 , D 0 + β - 1 v A 2 , v A 3 , D 0 + α - 2 u A 4 , D 0 + α - 1 u A 5 , u A 6 .
Therefore, for all (u, v) Ω1,
( u , v ) Y = max { v , D 0 + β - 1 v , D 0 + β - 2 v u , D 0 + α - 1 u , D 0 + α - 2 u } max { A i } , i = 1 , , 6 .

If n d 2 1 D 0 + α - 2 u θ 2 j d 1 1 D 0 + β - 2 v θ 1 , similarly to the above argument, we can also prove that Ω1 is bounded.

Case 4. ||(u, v)|| Y m(A + ||N 1 v||1) + k||N 2 u||1.

The proof is similar to that of case 3. Here, we omit it, where
k