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

Weihua Jiang

doi:10.1186/1687-1847-2013-324

Research
Open Access

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

Weihua Jiang¹Email author

Advances in Difference Equations20132013:324

https://doi.org/10.1186/1687-1847-2013-324

Received: 5 June 2013
Accepted: 10 September 2013
Published: 19 November 2013

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.

Keywords

fractional differential equation
integral boundary conditions
resonance
Fredholm operator
coincidence degree theory

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 [3–24]). The existence of solutions for differential equations at resonance has been studied by many authors (see [19–23, 25–29] and references cited therein). In papers [19–22], 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

{\begin{matrix} 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) = \int_{0}^{1} h_{1} (t) D_{0^{+}}^{α - 1} x (t) d t, \\ D_{0^{+}}^{α - 1} x (1) = \int_{0}^{1} h_{2} (t) D_{0^{+}}^{α - 1} x (t) d t, \\ y (0) = 0, D_{0^{+}}^{β - 1} y (0) = \int_{0}^{1} g_{1} (t) D_{0^{+}}^{β - 1} y (t) d t, \\ D_{0^{+}}^{β - 1} y (1) = \int_{0}^{1} g_{2} (t) D_{0^{+}}^{β - 1} y (t) d t, \end{matrix}

(1.1)

where $2 < α, β \leq 3$ , $D_{0^{+}}^{α}$ 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 $dim Ker L = 4$ .

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

( $H_{1}$ ) $2 < α, β \leq 3$ , $h_{i}, g_{i} \in L [0, 1]$ , $\int_{0}^{1} h_{i} (t) d t = 1$ , $\int_{0}^{1} g_{i} (t) d t = 1$ , $i = 1, 2$ .

(

H_{2}

)

\begin{matrix} Δ_{1} = | \begin{array}{c} \int_{0}^{1} t h_{1} (t) d t & 1 - \int_{0}^{1} t h_{2} (t) d t \\ \frac{1}{2} \int_{0}^{1} t^{2} h_{1} (t) d t & \frac{1}{2} (1 - \int_{0}^{1} t^{2} h_{2} (t) d t) \end{array} | : = | \begin{array}{c} Δ_{11} & Δ_{12} \\ Δ_{21} & Δ_{22} \end{array} | \neq 0, \\ Δ_{2} = | \begin{array}{c} \int_{0}^{1} t g_{1} (t) d t & 1 - \int_{0}^{1} t g_{2} (t) d t \\ \frac{1}{2} \int_{0}^{1} t^{2} g_{1} (t) d t & \frac{1}{2} (1 - \int_{0}^{1} t^{2} g_{2} (t) d t) \end{array} | : = | \begin{array}{c} δ_{11} & δ_{12} \\ δ_{21} & δ_{22} \end{array} | \neq 0 . \end{matrix}

(

H_{3}

)

f_{i} : [0, 1] \times R^{3} \to 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) \in L [0, 1]

and constants

η_{1}, η_{2} \in (0, 1)

with

c_{0} < 1

,

c_{0}^{'} < 1

,

\frac{1}{Γ (α) (1 - c_{0})} (2 + \frac{1}{η_{1}^{α - 2}}) a_{0} < 1

,

\frac{1}{Γ (β) (1 - c_{0}^{'})} (2 + \frac{1}{η_{2}^{β - 2}}) b_{0}^{'} < 1

,

A_{1} A_{2} a_{0}^{'} b_{0} < 1

such that

\begin{matrix} | f_{1} (t, x, y, z) | \leq 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) | \leq a_{02} (t) | x | + b_{02} (t) | y | + c_{02} (t) | z | + d_{02} (t) {| y |}^{θ_{2}} + r_{2} (t), \end{matrix}

where $a_{0} = \int_{0}^{1} a_{01} (t) d t$ , $b_{0} = \int_{0}^{1} b_{01} (t) d t$ , $c_{0} = \int_{0}^{1} c_{01} (t) d t$ , $d_{0} = \int_{0}^{1} d_{01} (t) d t$ , $r_{0} = \int_{0}^{1} r_{1} (t) d t$ , $a_{0}^{'} = \int_{0}^{1} a_{02} (t) d t$ , $b_{0}^{'} = \int_{0}^{1} b_{02} (t) d t$ , $c_{0}^{'} = \int_{0}^{1} c_{02} (t) d t$ , $d_{0}^{'} = \int_{0}^{1} d_{02} (t) d t$ , $r_{0}^{'} = \int_{0}^{1} r_{2} (t) d t$ , $0 \leq θ_{1}$ , $θ_{2} < 1$ , $A_{1} = \frac{2 η_{1}^{α - 2} + 1}{Γ (α) (1 - c_{0}) η_{1}^{α - 2} - a_{0} (2 η_{1}^{α - 2} + 1)}$ , $A_{2} = \frac{2 η_{2}^{β - 2} + 1}{Γ (β) (1 - c_{0}^{'}) η_{2}^{β - 2} - b_{0}^{'} (2 η_{2}^{β - 2} + 1)}$ .

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) \subset X \to Y

be a Fredholm operator with index zero, let

P : X \to X

,

Q : Y \to Y

be projectors such that

Im P = Ker L, Ker Q = Im L, X = Ker L \oplus Ker P, Y = Im L \oplus Im Q .

It follows that

L |_{dom L \cap Ker P} : dom L \cap Ker P \to Im L

is invertible. We denote the inverse by $K_{P}$ .

Assume that Ω is an open bounded subset of X, $dom L \cap \bar{Ω} \neq \emptyset$ . The map $N : X \to Y$ will be called L-compact on $\bar{Ω}$ if $Q N (\bar{Ω})$ is bounded and $K_{P} (I - Q) N : \bar{Ω} \to X$ is compact.

Theorem 2.1 [30]

Let

L : dom L \subset X \to Y

be a Fredholm operator of index zero and

N : X \to Y

L-compact on

\bar{Ω}

. Assume that the following conditions are satisfied:

(1)

$L x \neq λ N x$ for every $(x, λ) \in [(dom L ∖ Ker L) \cap \partial Ω] \times (0, 1)$ ;
(2)

$N x \notin Im L$ for every $x \in Ker L \cap \partial Ω$ ;
(3)

$deg (Q N |_{Ker L}, Ω \cap Ker L, 0) \neq 0$ , where $Q : Y \to Y$ is a projection such that $Im L = Ker Q$ .

Then the equation $L x = N x$ has at least one solution in $dom L \cap \bar{Ω}$ .

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, \infty) \to R

is given by

I_{0^{+}}^{α} y (t) = \frac{1}{Γ (α)} \int_{0}^{t} {(t - s)}^{α - 1} y (s) d s,

(2.1)

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

Definition 2.2 The fractional derivative of order

α > 0

of a function

y : (0, \infty) \to R

is given by

D_{0^{+}}^{α} y (t) = \frac{1}{Γ (n - α)} \frac{d^{n}}{d t^{n}} \int_{0}^{t} {(t - s)}^{n - α - 1} y (s) d s,

(2.2)

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

Lemma 2.1 Assume

f \in L [0, 1]

,

q \geq p \geq 0

,

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} = \frac{Γ (λ + 1)}{Γ (n + λ - α + 1)} \frac{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} + \dots + c_{n} t^{α - n},

where n is the smallest integer greater than or equal to α, $c_{i} \in R$ , $i = 1, 2, \dots, n$ .

Take

X = C^{α - 1} [0, 1] \times C^{β - 1} [0, 1]

with the norm

∥ (x, y) ∥ = max {{∥ x ∥}_{\infty}, {∥ y ∥}_{\infty}, {∥ D_{0^{+}}^{α - 1} x ∥}_{\infty}, {∥ D_{0^{+}}^{β - 1} y ∥}_{\infty}},

where

C^{α - 1} [0, 1] = {x ∣ x, D_{0^{+}}^{α - 1} x \in C [0, 1]}

,

{∥ x ∥}_{\infty} = {max}_{t \in [0, 1]} | x (t) |

. Set

Y = L [0.1] \times L [0.1]

with the norm

∥ (f, g) ∥ = max {\int_{0}^{1} | f (x) | d x, \int_{0}^{1} | g (x) | d x} .

Define operators

L : dom L \subset X \to Y

,

N : X \to Y

as follows:

L (x, y) = (D_{0^{+}}^{α} x, D_{0^{+}}^{β} y), (x, y) \in dom L, N (x, y) = (N_{1} (x, y), N_{2} (x, y)), (x, y) \in X,

where

\begin{array}{rcl} dom L & = & {(x, y) | (x, y) \in X, (D_{0^{+}}^{α} x, D_{0^{+}}^{β} y) \in Y, x (0) = y (0) = 0, \\ D_{0^{+}}^{α - 1} x (0) = \int_{0}^{1} h_{1} (t) D_{0^{+}}^{α - 1} x (t) d t, D_{0^{+}}^{α - 1} x (1) = \int_{0}^{1} h_{2} (t) D_{0^{+}}^{α - 1} x (t) d t, \\ D_{0^{+}}^{β - 1} y (0) = \int_{0}^{1} g_{1} (t) D_{0^{+}}^{β - 1} y (t) d t, D_{0^{+}}^{β - 1} y (1) = \int_{0}^{1} g_{2} (t) D_{0^{+}}^{β - 1} y (t) d t}, \end{array}

$N_{1} (x, y) = f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t))$ , $N_{2} (x, y) = f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t))$ . 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 $(x, y) \in dom L$ 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] \to R

,

i = 1, 2, 3, 4

, and

Q_{j} : L [0, 1] \to L [0, 1]

,

j = 1, 2

as follows:

\begin{matrix} T_{1} u = \int_{0}^{1} u (t) \int_{t}^{1} h_{1} (s) d s d t, T_{2} u = \int_{0}^{1} u (t) \int_{0}^{t} h_{2} (s) d s d t, \\ T_{3} u = \int_{0}^{1} u (t) \int_{t}^{1} g_{1} (s) d s d t, T_{4} u = \int_{0}^{1} u (t) \int_{0}^{t} g_{2} (s) d s d t, \\ Q_{1} u = \frac{1}{Δ_{1}} (Δ_{22} T_{1} u - Δ_{21} T_{2} u) + \frac{1}{Δ_{1}} (Δ_{11} T_{2} u - Δ_{12} T_{1} u) t, \\ Q_{2} u = \frac{1}{Δ_{2}} (δ_{22} T_{3} u - δ_{21} T_{4} u) + \frac{1}{Δ_{2}} (δ_{11} T_{4} u - δ_{12} T_{3} u) t . \end{matrix}

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 : dom L \subset X \to Y

is a Fredholm operator of index zero, the linear continuous projectors

P : X \to X

and

Q : Y \to Y

can be defined as

\begin{aligned} P (x, y) = (\frac{D_{0^{+}}^{α - 1} x (0)}{Γ (α)} t^{α - 1} + \frac{D_{0^{+}}^{α - 2} x (0)}{Γ (α - 1)} t^{α - 2}, \frac{D_{0^{+}}^{β - 1} y (0)}{Γ (β)} t^{β - 1} + \frac{D_{0^{+}}^{β - 2} y (0)}{Γ (β - 1)} t^{β - 2}), \\ Q (u, v) = (Q_{1} u, Q_{2} v), \end{aligned}

respectively, and the linear operator

K_{P} : Im L \to dom L \cap Ker P

can be written by

K_{P} (u, v) = (I_{0^{+}}^{α} u, I_{0^{+}}^{β} v) .

Proof We can easily get that

Ker L = {(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} \in R} .

Obviously, $Im P = Ker L$ , $P^{2} (u, v) = P (u, v)$ .

By a simple calculation, we obtain that

Im L = {(u, v) \in 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

Im L = Ker Q

. It is clear that

X = Ker P \oplus Ker L, Y = Im L \oplus Im Q .

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

For

(u, v) \in Im L

, we can easily get that

K_{P} (u, v) = (I_{0^{+}}^{α} u, I_{0^{+}}^{β} v) \in dom L \cap Ker P

. Obviously,

L K_{P} (u, v) = (u, v)

,

(u, v) \in Im L

. For

(x, y) \in dom L \cap Ker P

, by Lemma 2.3 and

K_{P} L (x, y) \in dom L

, 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 $(x, y) \in Ker P$ 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 $K_{P} L (x, y) \in Ker P$ , means that $c_{1} = c_{2} = d_{1} = d_{2} = 0$ . So, $K_{P} L (x, y) = (x, y)$ . Therefore, $K_{P} = {(L |_{dom L \cap Ker P})}^{- 1}$ . The proof is completed. □

Lemma 3.2 Suppose that ( $H_{1}$ ), ( $H_{2}$ ) and ( $H_{3}$ ) hold. If $Ω \subset X$ is an open bounded subset and $dom L \cap \bar{Ω} \neq \emptyset$ , then N is L-compact on $\bar{Ω}$ .

Proof Since Ω is bounded, there exists a constant

r > 0

such that

∥ (x, y) ∥ < r

,

(x, y) \in \bar{Ω}

. It follows from condition (

H_{3}

) that there exist functions

Φ_{i} \in L [0, 1]

such that

| f_{i} (t, x, y, z) | \leq Φ_{i} (t)

for all

| x |, | y |, | z | \in [0, r]

, a.e.

t \in [0, 1]

,

i = 1, 2

. Thus,

\begin{matrix} | T_{1} N_{1} (x, y) | = | \int_{0}^{1} N_{1} (x, y) \int_{t}^{1} h_{1} (s) d s d t | \\ \leq \int_{0}^{1} Φ_{1} (t) d t \int_{0}^{1} | h_{1} (s) | d s < + \infty, (x, y) \in \bar{Ω}, \\ | T_{2} N_{1} (x, y) | = | \int_{0}^{1} N_{1} (x, y) \int_{0}^{t} h_{2} (s) d s d t | \\ \leq \int_{0}^{1} Φ_{1} (t) d t \int_{0}^{1} | h_{2} (s) | d s < + \infty, (x, y) \in \bar{Ω}, \\ | T_{3} N_{2} (x, y) | = | \int_{0}^{1} N_{2} (x, y) \int_{t}^{1} g_{1} (s) d s d t | \\ \leq \int_{0}^{1} Φ_{2} (t) d t \int_{0}^{1} | g_{1} (s) | d s < + \infty, (x, y) \in \bar{Ω}, \\ | T_{4} N_{2} (x, y) | = | \int_{0}^{1} N_{2} (x, y) \int_{0}^{t} g_{2} (s) d s d t | \\ \leq \int_{0}^{1} Φ_{2} (t) d t \int_{0}^{1} | g_{2} (s) | d s < + \infty, (x, y) \in \bar{Ω} . \end{matrix}

These mean that there exist constants

a_{i} > 0

,

b_{i} > 0

,

i = 1, 2

, such that

| Q_{1} N_{1} (x, y) | \leq a_{1} + b_{1} t, | Q_{2} N_{2} (x, y) | \leq a_{2} + b_{2} t, (x, y) \in \bar{Ω}, t \in [0, 1],

i.e., $Q N (\bar{Ω}) \subset Y$ is bounded. Now we will prove that $K_{P} (I - Q) N : \bar{Ω} \to X$ is compact.

Obviously,

K_{P} (I - Q) N (\bar{Ω})

is bounded. For

0 \leq t_{1} < t_{2} \leq 1

,

(x, y) \in \bar{Ω}

, we have

\begin{matrix} 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})), \end{matrix}

where $I_{0} : L [0, 1] \to L [0, 1]$ is an identical mapping.

It follows from

\begin{matrix} | 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}) | \\ = \frac{1}{Γ (α)} | \int_{0}^{t_{2}} {(t_{2} - s)}^{α - 1} (I_{0} - Q_{1}) N_{1} (x (s), y (s)) d s \\ - \int_{0}^{t_{1}} {(t_{1} - s)}^{α - 1} (I_{0} - Q_{1}) N_{1} (x (s), y (s)) d s | \\ \leq \frac{1}{Γ (α)} [\int_{0}^{t_{1}} ({(t_{2} - s)}^{α - 1} - {(t_{1} - s)}^{α - 1}) (Φ_{1} (s) + a_{1} + b_{1} s) d s \\ + \int_{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}) | \\ = | \int_{0}^{t_{2}} (I_{0} - Q_{1}) N_{1} (x (s), y (s)) d s - \int_{0}^{t_{1}} (I_{0} - Q_{1}) N_{1} (x (s), y (s)) d s | \\ \leq \int_{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}) | \\ = \frac{1}{Γ (β)} | \int_{0}^{t_{2}} {(t_{2} - s)}^{β - 1} (I_{0} - Q_{2}) N_{2} (x (s), y (s)) d s \\ - \int_{0}^{t_{1}} {(t_{1} - s)}^{β - 1} (I_{0} - Q_{2}) N_{2} (x (s), y (s)) d s | \\ \leq \frac{1}{Γ (β)} [\int_{0}^{t_{1}} ({(t_{2} - s)}^{β - 1} - {(t_{1} - s)}^{β - 1}) (Φ_{2} (s) + a_{2} + b_{2} s) d s \\ + \int_{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}) | \\ = | \int_{0}^{t_{2}} (I_{0} - Q_{2}) N_{2} (x (s), y (s)) d s \\ - \int_{0}^{t_{1}} (I_{0} - Q_{2}) N_{2} (x (s), y (s)) d s | \\ \leq \int_{t_{1}}^{t_{2}} (Φ_{2} (s) + a_{2} + b_{2} s) d s, \end{matrix}

the uniform continuity of ${(t - s)}^{α - 1}$ and ${(t - s)}^{β - 1}$ on $[0, 1] \times [0, 1]$ , 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 $K_{P} (I - Q) N : \bar{Ω} \to X$ 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 \in [η_{1}, 1]} | x (t) | > M_{1} or min_{t \in [η_{1}, 1]} | D_{0^{+}}^{α - 1} x (t) | > L_{1},

then either

\int_{0}^{1} f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t)) \int_{t}^{1} h_{1} (s) d s d t \neq 0

or

\int_{0}^{1} f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t)) \int_{0}^{t} h_{2} (s) d s d t \neq 0,

and if either

min_{t \in [η_{2}, 1]} | y (t) | > M_{2} or min_{t \in [η_{2}, 1]} | D_{0^{+}}^{β - 1} y (t) | > L_{2},

then either

\int_{0}^{1} f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t)) \int_{t}^{1} g_{1} (s) d s d t \neq 0

or

\int_{0}^{1} f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t)) \int_{0}^{t} g_{2} (s) d s d t \neq 0,

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}) \in Ker L

, there exist constants

k_{1}

,

k_{2}

,

l_{1}

,

l_{2}

such that either (1) or (2) holds, where

\begin{aligned} (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} | \leq 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} | \leq 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} | \leq 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} | \leq l_{1}, | d_{2} | > l_{2} . \end{aligned}

Lemma 3.3 Suppose that (

H_{1}

)-(

H_{4}

) hold, then the set

Ω_{1} = {(x, y) \in dom L ∖ Ker L ∣ L (x, y) = λ N (x, y), λ \in (0, 1)}

is bounded in X.

Proof Take

(x, y) \in Ω_{1}

. By

L (x, y) = λ N (x, y)

, we get

{\begin{matrix} x (t) = \frac{λ}{Γ (α)} \int_{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) = \frac{λ}{Γ (β)} \int_{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} . \end{matrix}

(3.1)

By Lemmas 2.1, 2.2 and (3.1), we have

{\begin{matrix} D_{0^{+}}^{α - 1} x (t) = λ \int_{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 \int_{0}^{t} f_{2} (s, x (s), y (s), D_{0^{+}}^{β - 1} y (s)) d s + b_{1} Γ (β) . \end{matrix}

(3.2)

It follows from

N (x, y) \in Im L

that

\begin{matrix} \int_{0}^{1} f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t)) \int_{t}^{1} h_{1} (s) d s d t = 0, \\ \int_{0}^{1} f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t)) \int_{0}^{t} h_{2} (s) d s d t = 0, \\ \int_{0}^{1} f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t)) \int_{t}^{1} g_{1} (s) d s d t = 0, \\ \int_{0}^{1} f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t)) \int_{0}^{t} g_{2} (s) d s d t = 0 . \end{matrix}

These, together with (

H_{4}

), mean that there exist constants

t_{0}, t_{1} \in [η_{1}, 1]

and

t_{0}^{'}, t_{1}^{'} \in [η_{2}, 1]

such that

| x (t_{0}) | \leq M_{1}, | D_{0^{+}}^{α - 1} x (t_{1}) | \leq L_{1}, | y (t_{0}^{'}) | \leq M_{2}, | D_{0^{+}}^{β - 1} y (t_{1}^{'}) | \leq L_{2} .

(3.3)

By (3.2), we get

\begin{matrix} D_{0^{+}}^{α - 1} x (t) = λ \int_{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 \int_{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}^{'}) . \end{matrix}

By (3.3) and (

H_{3}

), we obtain that

{\begin{matrix} {∥ D_{0^{+}}^{α - 1} x ∥}_{\infty} \leq \frac{1}{1 - c_{0}} (a_{0} {∥ x ∥}_{\infty} + b_{0} {∥ y ∥}_{\infty} + d_{0} {∥ x ∥}_{\infty}^{θ_{1}} + r_{0} + L_{1}), \\ {∥ D_{0^{+}}^{β - 1} y ∥}_{\infty} \leq \frac{1}{1 - c_{0}^{'}} (a_{0}^{'} {∥ x ∥}_{\infty} + b_{0}^{'} {∥ y ∥}_{\infty} + d_{0}^{'} {∥ y ∥}_{\infty}^{θ_{2}} + r_{0}^{'} + L_{2}) . \end{matrix}

(3.4)

Instead of t by

t_{0}

,

t_{0}^{'}

in (3.1) and

t_{1}

,

t_{1}^{'}

in (3.2), respectively, we get

{\begin{matrix} x (t) = \frac{λ}{Γ (α)} [\int_{0}^{t} {(t - s)}^{α - 1} f_{1} (s, x (s), y (s), D_{0^{+}}^{α - 1} x (s)) d s \\ + t^{α - 2} (t_{0} - t) \int_{0}^{t_{1}} f_{1} (s, x (s), y (s), D_{0^{+}}^{α - 1} x (s)) d s \\ - \frac{t^{α - 2}}{t_{0}^{α - 2}} \int_{0}^{t_{0}} {(t_{0} - s)}^{α - 1} f_{1} (s, x (s), y (s), D_{0^{+}}^{α - 1} x (s)) d s] \\ + \frac{t^{α - 2}}{Γ (α)} D_{0^{+}}^{α - 1} x (t_{1}) (t - t_{0}) + \frac{t^{α - 2}}{t_{0}^{α - 2}} x (t_{0}), \\ y (t) = \frac{λ}{Γ (β)} [\int_{0}^{t} {(t - s)}^{β - 1} f_{2} (s, x (s), y (s), D_{0^{+}}^{β - 1} y (s)) d s \\ + t^{β - 2} (t_{0}^{'} - t) \int_{0}^{t_{1}^{'}} f_{2} (s, x (s), y (s), D_{0^{+}}^{β - 1} y (s)) d s \\ - \frac{t^{β - 2}}{t_{0}^{' β - 2}} \int_{0}^{t_{0}^{'}} {(t_{0}^{'} - s)}^{β - 1} f_{2} (s, x (s), y (s), D_{0^{+}}^{β - 1} y (s)) d s] \\ + \frac{t^{β - 2}}{Γ (β)} D_{0^{+}}^{β - 1} y (t_{1}^{'}) (t - t_{0}^{'}) + \frac{t^{β - 2}}{t_{0}^{' β - 2}} y (t_{0}^{'}) . \end{matrix}

(3.5)

It follows from (3.4), (3.5) and (

H_{3}

) that

\begin{aligned} | x (t) | \leq & \frac{1}{Γ (α)} (2 + \frac{1}{η_{1}^{α - 2}}) \int_{0}^{1} | f_{1} (s, x (s), y (s), D_{0^{+}}^{α - 1} x (s)) | d s + (\frac{L_{1}}{Γ (α)} + \frac{M_{1}}{η_{1}^{α - 2}}) \\ \leq & \frac{1}{Γ (α)} (2 + \frac{1}{η_{1}^{α - 2}}) (a_{0} {∥ x ∥}_{\infty} + b_{0} {∥ y ∥}_{\infty} + c_{0} {∥ D_{0^{+}}^{α - 1} x ∥}_{\infty} + d_{0} {∥ x ∥}_{\infty}^{θ_{1}} + r_{0}) \\ + (\frac{L_{1}}{Γ (α)} + \frac{M_{1}}{η_{1}^{α - 2}}) \\ \leq & \frac{1}{Γ (α) (1 - c_{0})} (2 + \frac{1}{η_{1}^{α - 2}}) (a_{0} {∥ x ∥}_{\infty} + b_{0} {∥ y ∥}_{\infty} + d_{0} {∥ x ∥}_{\infty}^{θ_{1}} + r_{0} + c_{0} L_{1}) \\ + (\frac{L_{1}}{Γ (α)} + \frac{M_{1}}{η_{1}^{α - 2}}) \end{aligned}

and

\begin{aligned} | y (t) | \leq & \frac{1}{Γ (β)} (2 + \frac{1}{η_{2}^{β - 2}}) \int_{0}^{1} | f_{2} (s, x (s), y (s), D_{0^{+}}^{β - 1} y (s)) | d s + (\frac{L_{2}}{Γ (β)} + \frac{M_{2}}{η_{2}^{β - 2}}) \\ \leq & \frac{1}{Γ (β)} (2 + \frac{1}{η_{2}^{β - 2}}) (a_{0}^{'} {∥ x ∥}_{\infty} + b_{0}^{'} {∥ y ∥}_{\infty} + c_{0}^{'} {∥ D_{0^{+}}^{β - 1} y ∥}_{\infty} + d_{0}^{'} {∥ y ∥}_{\infty}^{θ_{2}} + r_{0}^{'}) \\ + (\frac{L_{2}}{Γ (β)} + \frac{M_{2}}{η_{2}^{β - 2}}) \\ \leq & \frac{1}{Γ (β) (1 - c_{0}^{'})} (2 + \frac{1}{η_{2}^{β - 2}}) (a_{0}^{'} {∥ x ∥}_{\infty} + b_{0}^{'} {∥ y ∥}_{\infty} + d_{0}^{'} {∥ y ∥}_{\infty}^{θ_{2}} + r_{0}^{'} + c_{0}^{'} L_{2}) \\ + (\frac{L_{2}}{Γ (β)} + \frac{M_{2}}{η_{2}^{β - 2}}) . \end{aligned}

Thus,

{∥ x ∥}_{\infty} \leq A_{1} [b_{0} {∥ y ∥}_{\infty} + d_{0} {∥ x ∥}_{\infty}^{θ_{1}}] + M_{0},

(3.6)

{∥ y ∥}_{\infty} \leq A_{2} [a_{0}^{'} {∥ x ∥}_{\infty} + d_{0}^{'} {∥ y ∥}_{\infty}^{θ_{2}}] + M_{0}^{'},

(3.7)

where $M_{0} = A_{1} [(r_{0} + c_{0} L_{1}) + (\frac{L_{1}}{Γ (α)} + \frac{M_{1}}{η_{1}^{α - 2}}) / \frac{1}{Γ (α) (1 - c_{0})} (2 + \frac{1}{η_{1}^{α - 2}})]$ , $M_{0}^{'} = A_{2} [(r_{0}^{'} + c_{0}^{'} L_{2}) + (\frac{L_{2}}{Γ (β)} + \frac{M_{2}}{η_{2}^{β - 2}}) / \frac{1}{Γ (β) (1 - c_{0}^{'})} (2 + \frac{1}{η_{2}^{β - 2}})]$ .

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) \in Ker L, N (x, y) \in Im L}

is bounded in X.

Proof For $(x, y) \in Ω_{2}$ , we have $(x, y) = (c_{1} t^{α - 1} + c_{2} t^{α - 2}, d_{1} t^{β - 1} + d_{2} t^{β - 2})$ , $c_{1}, c_{2}, d_{1}, d_{2} \in R$ 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 $| c_{1} | \leq k_{1}$ , $| c_{2} | \leq k_{2}$ , $| d_{1} | \leq l_{1}$ , $| d_{2} | \leq l_{2}$ . 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) \in Ker L ∣ λ J (x, y) + (1 - λ) θ Q N (x, y) = (0, 0), λ \in [0, 1]}

is bounded in X, where

J : Ker L \to Im Q

is a linear isomorphism given by

\begin{matrix} J (c_{1} t^{α - 1} + c_{2} t^{α - 2}, d_{1} t^{β - 1} + d_{2} t^{β - 2}) \\ = (\frac{1}{Δ_{1}} (Δ_{22} c_{1} - Δ_{21} c_{2}) + \frac{1}{Δ_{1}} (Δ_{11} c_{2} - Δ_{12} c_{1}) t, \\ \frac{1}{Δ_{2}} (δ_{22} d_{1} - δ_{21} d_{2}) + \frac{1}{Δ_{2}} (δ_{11} d_{2} - δ_{12} d_{1}) t), \\ θ = {\begin{matrix} 1, & if (H_{5}) (1) holds, \\ - 1, & if (H_{5}) (2) holds . \end{matrix} \end{matrix}

Proof For

(c_{1} t^{α - 1} + c_{2} t^{α - 2}, d_{1} t^{β - 1} + d_{2} t^{β - 2}) \in Ω_{3}

, there exists

λ \in [0, 1]

such that

λ J (c_{1} t^{α - 1} + c_{2} t^{α - 2}, d_{1} t^{β - 1} + d_{2} t^{β - 2}) = - (1 - λ) θ Q N (c_{1} t^{α - 1} + c_{2} t^{α - 2}, d_{1} t^{β - 1} + d_{2} t^{β - 2}) .

This means that

\begin{matrix} 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}) . \end{matrix}

If

λ = 0

, by (

H_{5}

), we get

| c_{1} | \leq k_{1}

,

| c_{2} | \leq k_{2}

,

| d_{1} | \leq l_{1}

,

| d_{2} | \leq l_{2}

. If

λ = 1

, then

c_{1} = c_{2} = d_{1} = d_{2} = 0

. For

λ \in (0, 1)

, if

| c_{1} | > k_{1}

, we can get

λ c_{1}^{2} = - (1 - λ) 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} | \leq k_{1}

and

| c_{2} | > k_{2}

, we can get

λ c_{2}^{2} = - (1 - λ) 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, $| c_{i} | \leq k_{i}$ , $i = 1, 2$ . By the same methods, we can obtain that $| d_{i} | \leq l_{i}$ , $i = 1, 2$ . 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 \supset ⋃_{i = 1}^{3} \bar{Ω_{i}} \cup {(0, 0)}

be a bounded open subset of X. It follows from Lemma 3.2 that N is L-compact on

\bar{Ω}

. By Lemmas 3.3 and 3.4, we get

(1)

$L (x, y) \neq λ N (x, y)$ for every $(x, y, λ) \in [(dom L ∖ Ker L) \cap \partial Ω] \times (0, 1)$ ,
(2)

$N (x, y) \notin Im L$ for every $(x, y) \in Ker L \cap \partial Ω$ .

We need only to prove
(3)

$deg (Q N |_{Ker L}, Ω \cap Ker L, (0, 0)) \neq 0$ .

Take

H (x, y, λ) = λ J (x, y) + θ (1 - λ) Q N (x, y) .

According to Lemma 3.5, we know

H (x, y, λ) \neq (0, 0)

for

(x, y) \in \partial Ω \cap Ker L

. By the homotopy of degree, we get that

\begin{array}{rcl} deg (Q N |_{Ker L}, Ω \cap Ker L, (0, 0)) & = & deg (θ H (\cdot, 0), Ω \cap Ker L, (0, 0)) \\ = & deg (θ H (\cdot, 1), Ω \cap Ker L, (0, 0)) \\ = & deg (θ J, Ω \cap Ker L, (0, 0)) \neq 0 . \end{array}

By Theorem 2.1, we can get that $L (x, y) = N (x, y)$ has at least one solution in $dom L \cap \bar{Ω}$ , 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:

{\begin{matrix} D_{0^{+}}^{\frac{5}{2}} x (t) = f_{1} (t, x (t), y (t), D_{0^{+}}^{\frac{3}{2}} x (t)), 0 < t < 1, \\ D_{0^{+}}^{\frac{5}{2}} y (t) = f_{2} (t, x (t), y (t), D_{0^{+}}^{\frac{3}{2}} y (t)), 0 < t < 1, \\ x (0) = 0, D_{0^{+}}^{\frac{3}{2}} x (0) = \int_{0}^{1} h_{1} (t) D_{0^{+}}^{\frac{3}{2}} x (t) d t, \\ D_{0^{+}}^{\frac{3}{2}} x (1) = \int_{0}^{1} h_{2} (t) D_{0^{+}}^{\frac{3}{2}} x (t) d t, \\ y (0) = 0, D_{0^{+}}^{\frac{3}{2}} y (0) = \int_{0}^{1} g_{1} (t) D_{0^{+}}^{\frac{3}{2}} y (t) d t, \\ D_{0^{+}}^{\frac{3}{2}} y (1) = \int_{0}^{1} g_{2} (t) D_{0^{+}}^{\frac{3}{2}} y (t) d t, \end{matrix}

(4.1)

where

\begin{matrix} f_{1} (t, x, y, z) = {\begin{matrix} \frac{1}{4} t sin x + \frac{1}{8} t^{3} sin y, & t \in [0, \frac{1}{4}), \\ \frac{1}{4} t sin x + \frac{1}{8} t^{3} sin y + t z, & t \in [\frac{1}{4}, \frac{1}{2}), \\ \frac{1}{4} t x + \frac{1}{8} t^{3} sin y + t sin z, & t \in [\frac{1}{2}, 1], \end{matrix} \\ f_{2} (t, x, y, z) = {\begin{matrix} \frac{1}{8} t^{3} sin x + \frac{1}{10} sin y, & t \in [0, \frac{1}{9}), \\ \frac{1}{8} t^{3} sin x + \frac{1}{10} sin y + t z, & t \in [\frac{1}{9}, \frac{1}{4}), \\ \frac{1}{8} t^{3} sin x + \frac{1}{10} y + t sin z, & t \in [\frac{1}{4}, 1], \end{matrix} \\ h_{1} (t) = {\begin{matrix} 2, & t \in [0, \frac{1}{2}), \\ 0, & t \in [\frac{1}{2}, 1], \end{matrix} h_{2} (t) = {\begin{matrix} 0, & t \in [0, \frac{1}{2}), \\ 2, & t \in [\frac{1}{2}, 1], \end{matrix} \\ g_{1} (t) = {\begin{matrix} 4, & t \in [0, \frac{1}{4}), \\ 0, & t \in [\frac{1}{4}, 1], \end{matrix} g_{2} (t) = {\begin{matrix} 0, & t \in [0, \frac{1}{4}), \\ \frac{4}{3}, & t \in [\frac{1}{4}, 1] . \end{matrix} \end{matrix}

Corresponding to problem (1.1), we have

α = β = \frac{5}{2}

,

Δ_{1} = | \begin{array}{c} \frac{1}{4} & \frac{1}{4} \\ \frac{1}{24} & \frac{5}{24} \end{array} | = \frac{1}{24} \neq 0, Δ_{2} = | \begin{array}{c} \frac{1}{8} & \frac{3}{8} \\ \frac{1}{96} & \frac{9}{32} \end{array} | = \frac{1}{32} \neq 0 .

Obviously,

\int_{0}^{1} h_{i} (t) d t = 1

,

\int_{0}^{1} g_{i} (t) d t = 1

,

i = 1, 2

. Thus, conditions (

H_{1}

) and (

H_{2}

) are satisfied. It is easy to get that

a_{0} = \frac{1}{8}

,

b_{0} = \frac{1}{32}

,

c_{0} = \frac{15}{32}

,

a_{0}^{'} = \frac{1}{32}

,

b_{0}^{'} = \frac{1}{10}

,

c_{0}^{'} = \frac{40}{81}

. Take

M_{1} = 8

,

L_{1} = 1

,

η_{1} = \frac{1}{4}

,

M_{2} = 20

,

L_{2} = 4

,

η_{2} = \frac{1}{9}

. By a simple calculation, we can get that (

H_{3}

) is satisfied and the following inequations hold

\begin{matrix} \int_{0}^{1} f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t)) \int_{t}^{1} h_{1} (s) d s d t \neq 0, if min_{t \in [η_{1}, 1]} | D_{0^{+}}^{α - 1} x (t) | > L_{1}, \\ \int_{0}^{1} f_{1} (t, x (t), y (t), D_{0^{+}}^{α - 1} x (t)) \int_{0}^{t} h_{2} (s) d s d t \neq 0, if min_{t \in [η_{1}, 1]} | x (t) | > M_{1}, \\ \int_{0}^{1} f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t)) \int_{t}^{1} g_{1} (s) d s d t \neq 0, if min_{t \in [η_{2}, 1]} | D_{0^{+}}^{β - 1} y (t) | > L_{2}, \end{matrix}

and

\int_{0}^{1} f_{2} (t, x (t), y (t), D_{0^{+}}^{β - 1} y (t)) \int_{0}^{t} g_{2} (s) d s d t \neq 0, if min_{t \in [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.

Declarations

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.

Authors’ Affiliations

(1)

College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei, 050018, P.R. China

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