# Attractivity for a k-dimensional system of fractional functional differential equations and global attractivity for a k-dimensional system of nonlinear fractional differential equations

## Abstract

In this paper, we present some results for the attractivity of solutions for a k-dimensional system of fractional functional differential equations involving the Caputo fractional derivative by using the classical Schauder’s fixed-point theorem. Also, the global attractivity of solutions for a k-dimensional system of fractional differential equations involving Riemann-Liouville fractional derivative are obtained by using Krasnoselskii’s fixed-point theorem. We give two examples to illustrate our main results.

## 1 Introduction

In recent years, many researchers have been focused on investigation of fractional differential equations which has played an important role in different areas of science (see for example,  and the references therein). As you know, there are many practical applications of fractional differential equations in different fields of science such as economy, biology, and the study of forced van der Pol oscillators (see for example,  and the references therein). On the other hand, there are a few papers on the attractivity of solutions for fractional differential equations and fractional functional differential equations (see for example,  and ). For the details of basic notions of this paper such the standard Caputo fractional derivative, the standard Riemann-Liouville fractional derivative, and the fractional integral of order q for a function f see . In 2011, Chen and Zhou reviewed the attractivity of solutions for the fractional functional differential equation $D α c x(t)=f(t, x t )$ for $t∈( t 0 ,∞)$ via the boundary value condition $x(t)=ϕ(t)$ for $t 0 −r≤t≤ t 0$, where $t 0 ≥0$, $r>0$, $0<α<1$, cD is the standard Caputo fractional derivative, $ϕ∈C([ t 0 −r, t 0 ],R)$ and $f:( t 0 ,∞)×C([−r,0],R)→R$ a function with some properties . In 2012, Chen et al. reviewed the global attractivity of solutions for the nonlinear fractional differential equation $D α x(t)=g(t,x(t))$ for $t∈( t 0 ,∞)$ via the boundary value problem $[ D α − 1 x ( t ) ] t = t 0 = x 0$, where $t 0 ≥0$, $0<α<1$, $x 0$ is a constant, D is the standard Riemann-Liouville fractional derivative, and $g:( t 0 ,∞)×R→R$ a function with some properties . Also, they investigated the global attractivity of solutions for the nonlinear fractional differential equation $D α c x(t)=g(t,x(t))$ for $t∈( t 0 ,∞)$ via the boundary value problem $x( t 0 )= x 0$, where $t 0 ≥0$, $0<α<1$, $x 0$ is a constant, cD is the standard Caputo fractional derivative, and $g:( t 0 ,∞)×R→R$ a function with some properties .

In this paper, we investigate the attractivity of solutions for a k-dimensional system of fractional differential equations. Also, we investigate the global attractivity of solutions for another k-dimensional system of nonlinear fractional differential equations.

## 2 Preliminaries

In this paper, we investigate the attractivity of solutions for k-dimensional system of fractional functional differential equations

${ D α 1 c x 1 ( t ) = f 1 ( t , x 1 t , x 2 t , … , x k t ) , D α 2 c x 2 ( t ) = f 2 ( t , x 1 t , x 2 t , … , x k t ) , ⋮ D α k c x k ( t ) = f k ( t , x 1 t , x 2 t , … , x k t ) ,$
(2.1)

via the boundary value problems $x 1 (t)= ϕ 1 (t), x 2 (t)= ϕ 2 (t),…, x k (t)= ϕ k (t)$ for $t 0 −r≤t≤ t 0$, where $0< α i <1$ for $i=1,2,…,k$, $t 0 ≥0$, $t∈( t 0 ,∞)$, cD is the standard Caputo fractional derivative, $J=( t 0 ,∞)$, $r>0$, $ϕ i ∈C([ t 0 −r, t 0 ], R n )$ for $i=1,2,…,k$ and $f i :J×C([−r,0], R n )×C([−r,0], R n )×⋯×C([−r,0], R n )→ R n$ is a function satisfying some assumptions that will be specified later for $i=1,2,…,k$. If $x∈C([ t 0 −r,∞), R n )$, then $x t$ is defined by $x t (θ)=x(t+θ)$ for $−r≤θ≤0$ and $t∈[ t 0 ,∞)$. Also, we investigate the global attractivity of solutions for the k-dimensional system of nonlinear fractional differential equations

${ D α 1 x 1 ( t ) = g 1 ( t , x 1 ( t ) , x 2 ( t ) , … , x k ( t ) ) , D α 2 x 2 ( t ) = g 2 ( t , x 1 ( t ) , x 2 ( t ) , … , x k ( t ) ) , ⋮ D α k x k ( t ) = g k ( t , x 1 ( t ) , x 2 ( t ) , … , x k ( t ) ) ,$
(2.2)

via the boundary value problems $[ D α 1 − 1 x 1 ( t ) ] t = t 0 = x 1 0 , [ D α 2 − 1 x 2 ( t ) ] t = t 0 = x 2 0 ,…, [ D α k − 1 x k ( t ) ] t = t 0 = x k 0$, where $0< α i <1$ for $i=1,2,…,k$, $t 0 ≥0$, $t∈( t 0 ,∞)$, D is the Riemann-Liouville fractional derivative, $J=( t 0 ,∞)$, $x 1 0 ,…, x k 0$ are constants, and $g i :J× R n × R n ×⋯× R n → R n$ is an integrable function satisfying some assumptions that will be specified later for $i=1,2,…,k$. In fact, we say that the solution $( x 1 (t), x 2 (t),…, x k (t))$ of the problem (2.1) is attractive whenever if there exists a constant $b i 0 ( t 0 )>0$ such that $| ϕ i (s)|≤ b i 0$ for all $i=1,2,…,k$ and $s∈[ t 0 −r, t 0 ]$, then $lim t → ∞ x i (t, t 0 , ϕ i )→0$. Also, the zero solution $x(t)$ of the problem (2.2) is said to be globally attractive whenever each solution tends to zero as $t→∞$. Let $X=C(J, R n )$ be the Banach space of all continuous functions from J into $R n$ with the norm $∥x∥= sup t ∈ J |x(t)|$, where $|⋅|$ denotes a suitable complete norm on $R n$. It is clear that the product space is also a Banach space, where $∥ ( x 1 , x 2 , … , x k ) ∥ ∗ =∥ x 1 ∥+∥ x 2 ∥+⋯+∥ x k ∥$. We need the following Schauder fixed-point theorem and improvement of a fixed-point theorem of Krasnoselskii due to Burton, which one can find in [14, 17] and .

Theorem 2.1 If U is a nonempty, closed, bounded, and convex subset of the Banach space X and $T:U→U$ is completely continuous, then T has a fixed point.

Theorem 2.2 Let S be a nonempty, closed, convex, and bounded subset of the Banach space X, $A:X→X$ a contraction with constant $l<1$, $B:S→X$ a continuous map which $B(S)$ resides in a compact subset of X and $x=Ax+By$ and $y∈S$ implies $x∈S$. Then the operator equation $Ax+Bx=x$ has a solution in S.

## 3 Main results

First, we investigate attractive solutions of the problem (2.1). In this way, suppose that $∥ x t ∥= sup − r ≤ θ ≤ 0 |x(t+θ)|$ for $t∈J$. We assume that $f i (t, x 1 t , x 2 t ,…, x k t )$ is Lebesgue measurable with respect to t on $[ t 0 ,∞)$ and $f i (t, φ 1 , φ 2 ,…, φ k )$ is continuous with respect to $φ j$ on $C([−r,0], R n )$ for $i,j=1,2,…,k$. Note that the problem (2.1) is equivalent to the system of equations

$x i (t)= { ϕ i ( t 0 ) + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s , t > t 0 , ϕ i ( t ) , t ∈ [ t 0 − r , t 0 ]$

or

$x i (t)= { 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 [ ϕ i ( t 0 ) Γ ( 1 − α i ) ( s − t 0 ) − α i + f i ( s , x 1 s , x 2 s , … , x k s ) ] d s , t > t 0 , ϕ i ( t ) , t ∈ [ t 0 − r , t 0 ]$

for $i=1,2,…,k$. Define the operator $T: X k → X k$ by

$T( x 1 , x 2 ,…, x k )(t)= ( T 1 ( x 1 , x 2 , … , x k ) ( t ) T 2 ( x 1 , x 2 , … , x k ) ( t ) ⋮ T k ( x 1 , x 2 , … , x k ) ( t ) ) ,$

where

$T i ( x 1 , x 2 ,…, x k )(t)= { ϕ i ( t 0 ) + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s , t > t 0 , ϕ i ( t ) , t ∈ [ t 0 − r , t 0 ]$

for $i=1,2,…,k$. It is easy to check that $( x 1 (t), x 2 (t),…, x k (t))$ is a solution of the problem (2.1) if and only if $( x 1 (t), x 2 (t),…, x k (t))$ is a fixed point of the operator T.

Theorem 3.1 Suppose that for each $i∈{1,2,…,k}$ there exist $γ i 1 >0$ and $α i 1 ∈(0, α i )$ such that

$| ϕ i ( t 0 ) + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s | ≤ ( t − t 0 ) − γ i 1$

for all $t∈J$ and $f i ∈ L 1 α i 1 (J×C([−r,0], R n )×C([−r,0], R n )×⋯×C([−r,0], R n ), R n )$. Then the problem (2.1) has at least one attractive solution $( x 1 , x 2 ,…, x k )$ such that $x i ∈C([ t 0 −r,∞), R n )$ for all $i=1,2,…,k$.

Proof Consider the set

where $t ˜$ is a constant. It is easy to check that $S 1$ is a closed, bounded, and convex subset of . We show that the operator T has a fixed point in $S 1$. This implies that the problem (2.1) has a solution. Note that $| T i ( x 1 , x 2 ,…, x k )(t)|≤ ( t − t 0 ) − γ i 1$ for all $i=1,2,…,k$ and so $T( S 1 )⊂ S 1$. Now, we show that T is continuous. Let $( x 1 m , x 2 m ,…, x k m ),( x 1 , x 2 ,…, x k )∈ S 1$ for all $m≥1$ and $lim m → ∞ | x i m (t)− x i (t)|=0$ for all $i=1,2,…,k$. Then, we have $lim m → ∞ f i (t, x 1 t m , x 2 t m ,…, x k t m )= f i (t, x 1 t , x 2 t ,…, x k t )$ for all $i=1,2,…,k$ and $t> t 0$. Let $ϵ>0$ be given. Choose $T ˜ > t 0$ such that $t≥ T ˜$ implies that $( t − t 0 ) − γ i 1 < ϵ 2$. Let $ν i 1 = α i − 1 1 − α i 1$ and note that $1+ ν i 1 >0$ for $i=1,2,…,k$. Also, we have

$| T i ( x 1 m , x 2 m , … , x k m ) ( t ) − T i ( x 1 , x 2 , … , x k ) ( t ) | ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 | f i ( s , x 1 s m , x 2 s m , … , x k s m ) − f i ( s , x 1 s , x 2 s , … , x k s ) | d s ≤ 1 Γ ( α i ) { ∫ t 0 t [ ( t − s ) α i − 1 ] 1 1 − α i 1 d s } 1 − α i 1 × [ ∫ t 0 t | f i ( s , x 1 s m , x 2 s m , … , x k s m ) − f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ( t − t 0 ) 1 + ν i 1 ) 1 − α i 1 × [ ∫ t 0 T ˜ | f i ( s , x 1 s m , x 2 s m , … , x k s m ) − f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ( T ˜ − t 0 ) 1 + ν i 1 ) 1 − α i 1 ( T ˜ − t 0 ) α i 1 × sup t 0 ≤ s ≤ T ˜ | f i ( s , x 1 s m , x 2 s m , … , x k s m ) − f i ( s , x 1 s , x 2 s , … , x k s ) |$

for $t 0 . Thus, $lim m → ∞ | T i ( x 1 m , x 2 m ,…, x k m )(t)− T i ( x 1 , x 2 ,…, x k )(t)|=0$ for all $t 0 . Also, we have

$| T i ( x 1 m , x 2 m , … , x k m ) ( t ) − T i ( x 1 , x 2 , … , x k ) ( t ) | = | 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 f i ( s , x 1 s m , x 2 s m , … , x k s m ) d s − 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s | ≤ 2 ( t − t 0 ) − γ i 1 ≤ ϵ$

for $t> T ˜$. Hence, $lim m → ∞ | T i ( x 1 m , x 2 m ,…, x k m )(t)− T i ( x 1 , x 2 ,…, x k )(t)|=0$ for $t> t 0$. This implies that $T i$ is continuous for $i=1,2,…,k$ and so T is continuous. Now, we show that the set $T( S 1 )$ is equi-continuous. Let $ϵ>0$. Since $lim t → ∞ ( t − t 0 ) − γ i 1 =0$ for $i=1,2,…,k$, there is a $T ˜ ′ > t 0$ such that $( t − t 0 ) − γ i 1 < ϵ 2$ for all $t> T ˜ ′$ and $i=1,2,…,k$. Let $t 1 , t 2 > t 0$ and $t 2 > t 1$. If $t 1 , t 2 ∈( t 0 , T ˜ ′ ]$, then

$| T i ( x 1 , x 2 , … , x k ) ( t 2 ) − T i ( x 1 , x 2 , … , x k ) ( t 1 ) | ≤ | 1 Γ ( α i ) ∫ t 0 t 2 ( t 2 − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s − 1 Γ ( α i ) ∫ t 0 t 1 ( t 1 − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s | ≤ 1 Γ ( α i ) ∫ t 0 t 1 [ ( t 1 − s ) α i − 1 − ( t 2 − s ) α i − 1 ] | f i ( s , x 1 s , x 2 s , … , x k s ) | d s + 1 Γ ( α i ) ∫ t 1 t 2 ( t 2 − s ) α i − 1 | f i ( s , x 1 s , x 2 s , … , x k s ) | d s ≤ 1 Γ ( α i ) { ∫ t 0 t 1 [ ( t 1 − s ) α i − 1 − ( t 2 − s ) α i − 1 ] 1 1 − α i 1 d s } 1 − α i 1 × [ ∫ t 0 t 1 | f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 + 1 Γ ( α i ) [ ∫ t 1 t 2 ( t 2 − s ) α i − 1 1 − α i 1 d s ] 1 − α i 1 [ ∫ t 1 t 2 | f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ) 1 − α i 1 [ ( t 1 − t 0 ) α i − 1 1 − α i 1 + 1 + ( t 2 − t 1 ) α i − 1 1 − α i 1 + 1 − ( t 2 − t 0 ) α i − 1 1 − α i 1 + 1 ] 1 − α i 1 × [ ∫ t 0 T ˜ ′ | f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 + 1 Γ ( α i ) ( 1 1 + ν i 1 ) 1 − α i 1 [ ( t 2 − t 1 ) α i − 1 1 − α i 1 + 1 ] 1 − α i 1 [ ∫ t 0 T ˜ ′ | f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ) 1 − α i 1 [ ∫ t 0 T ˜ ′ | f i ( s , x 1 s , x 2 s , … , x k s ) | 1 α i 1 d s ] α i 1 ( t 2 − t 1 ) α i − α i 1$

and so $lim t 2 → t 1 | T i ( x 1 , x 2 ,…, x k )( t 2 )− T i ( x 1 , x 2 ,…, x k )( t 1 )|=0$. If $t 1 , t 2 > T ˜ ′$, then

$| T i ( x 1 , x 2 , … , x k ) ( t 2 ) − T i ( x 1 , x 2 , … , x k ) ( t 1 ) | = | 1 Γ ( α i ) ∫ t 0 t 2 ( t 2 − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s − 1 Γ ( α i ) ∫ t 0 t 1 ( t 1 − s ) α i − 1 f i ( s , x 1 s , x 2 s , … , x k s ) d s | ≤ ( t 2 − t 0 ) − γ i 1 + ( t 1 − t 0 ) − γ i 1 ≤ ϵ .$

Now, let $t 0 < t 1 < T ˜ ′ < t 2$. Since

$| T i ( x 1 , x 2 , … , x k ) ( t 2 ) − T i ( x 1 , x 2 , … , x k ) ( t 1 ) | ≤ | T i ( x 1 , x 2 , … , x k ) ( t 2 ) − T i ( x 1 , x 2 , … , x k ) ( T ˜ ′ ) | + | T i ( x 1 , x 2 , … , x k ) ( T ˜ ′ ) − T i ( x 1 , x 2 , … , x k ) ( t 1 ) | ,$

we get $lim t 2 → t 1 | T i ( x 1 , x 2 ,…, x k )( t 2 )− T i ( x 1 , x 2 ,…, x k )( t 1 )|=0$ in all cases. This implies that the set $T( S 1 )$ is equi-continuous. Since $T( S 1 )⊂ S 1$ is uniformly bounded, $T( S 1 )$ is relatively compact. Now by using Theorem 2.1, T has a fixed point in $S 1$ which is a solution of the problem (2.1). Since $x(t)=( x 1 (t), x 2 (t),…, x k (t))∈ S 1$, $lim t → ∞ x(t)=0$. Thus, $x(t)$ is an attractive solution for the problem (2.1). □

Theorem 3.2 Suppose that for each $i∈{1,2,…,k}$ there exist $γ i 2 >0$, $α i 2 ∈(0, α i )$ and $l i ∈ L 1 α i 2 (J, R + )$ such that $1 Γ ( α i ) ∫ t 0 t ( t 1 − s ) α i − 1 l i (s) ( s − t 0 ) − γ i 2 ds≤ ( t − t 0 ) − γ i 2$ and

$| ϕ i ( t 0 ) Γ ( 1 − α i ) ( t − t 0 ) − α i + f i ( t , x 1 t , x 2 t , … , x k t ) | ≤ l i (t)∥ x i t ∥$

for all $i=1,2,…,k$, $t∈J$ and $x i ∈C([ t 0 −r,∞), R n )$. Then the problem (2.1) has at least one attractive solution $( x 1 , x 2 ,…, x k )$ such that $x i ∈C([ t 0 −r,∞), R n )$ for all $i=1,2,…,k$.

Proof It is sufficient we consider the set

where $t ˜$ is a constant. By using a similar techniques and proof in Theorem 3.1, one can show that $T( S 2 )⊂ S 2$, T is continuous and $T( S 2 )$ is relatively compact. Now by using Theorem 2.1, T has a fixed point in $S 2$ which is a solution of the problem (2.1). Since $x(t)=( x 1 (t), x 2 (t),…, x k (t))∈ S 2$, $lim t → ∞ x(t)=0$. Thus, $x(t)$ is an attractive solution for the problem (2.1). □

Theorem 3.3 Suppose that for each $i∈{1,2,…,k}$ there exists $β i 1 ∈( α i ,1)$ such that

$| ϕ i ( t 0 ) Γ ( 1 − α i ) ( t − t 0 ) − α i + f i ( t , x 1 t , x 2 t , … , x k t ) | ≤ Γ ( 1 + α i − β i 1 ) Γ ( 1 − β i 1 ) ( t − t 0 ) − β i 1$

for all $t∈J$. Then the problem (2.1) has at least one attractive solution $( x 1 , x 2 ,…, x k )$ such that $x i ∈C([ t 0 −r,∞), R n )$ for all $i=1,2,…,k$.

Proof It is sufficient we consider the set

where $t ˜$ is a constant. By using a similar techniques and proof in Theorem 3.1, one can show that $T( S 3 )⊂ S 3$, T is continuous and $T( S 3 )$ is relatively compact. Now, by using Theorem 2.1, T has a fixed point in $S 3$ which is a solution of the problem (2.1). Since $x(t)=( x 1 (t), x 2 (t),…, x k (t))∈ S 3$, $lim t → ∞ x(t)=0$. Thus, $x(t)$ is an attractive solution for the problem (2.1). □

Here, we are going to investigate global attractivity of solutions of the problem (2.2). We assume that $g i (t, x 1 (t), x 2 (t),…, x k (t))$ is Lebesgue measurable with respect to t on $[ t 0 ,∞)$ and there exists a constant $α i 1 ′ ∈(0, α i )$ such that $g i ∈ L 1 α i 1 ′ (J× R n × R n ×⋯× R n , R n )$ and $g i (t, x 1 (t), x 2 (t),…, x k (t))$ is continuous with respect to $x j$ on $[ t 0 ,∞)$ for all $i,j=1,2,…,k$. Note that the problem (2.2) is equivalent to the system of equations

$x i (t)= x i 0 Γ ( α i ) ( t − t 0 ) α i − 1 + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 g i ( s , x 1 ( s ) , x 2 ( s ) , … , x k ( s ) ) ds$

for all $t> t 0$ and $i=1,2,…,k$. Define the operator $T: X k → X k$ by

$T( x 1 , x 2 ,…, x k )(t)= ( T 1 ( x 1 , x 2 , … , x k ) ( t ) T 2 ( x 1 , x 2 , … , x k ) ( t ) ⋮ T k ( x 1 , x 2 , … , x k ) ( t ) ) ,$

where

$T i ( x 1 , x 2 ,…, x k )(t)= x i 0 Γ ( α i ) ( t − t 0 ) α i − 1 + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 g i ( s , x 1 ( s ) , x 2 ( s ) , … , x k ( s ) ) ds$

for all $i=1,2,…,k$. Now, define

$A( x 1 , x 2 ,…, x k )(t)= ( A 1 ( x 1 , x 2 , … , x k ) ( t ) A 2 ( x 1 , x 2 , … , x k ) ( t ) ⋮ A k ( x 1 , x 2 , … , x k ) ( t ) ) ,$

where $A i ( x 1 , x 2 ,…, x k )(t)= x i 0 Γ ( α i ) ( t − t 0 ) α i − 1$ for all $i=1,2,…,k$. Finally, define

$B( x 1 , x 2 ,…, x k )(t)= ( B 1 ( x 1 , x 2 , … , x k ) ( t ) B 2 ( x 1 , x 2 , … , x k ) ( t ) ⋮ B k ( x 1 , x 2 , … , x k ) ( t ) ) ,$

where $B i ( x 1 , x 2 ,…, x k )(t)= 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 g i (s, x 1 (s), x 2 (s),…, x k (s))ds$ for all $i=1,2,…,k$. It is easy to check that $( x 1 (t), x 2 (t),…, x k (t))$ is a solution of the problem (2.2) if and only if it is a fixed point of the operator T. Note that A is a contraction with constant 0.

Theorem 3.4 Suppose that for each $i∈{1,2,…,k}$ there exist $α i < β i 1 ′ <1$ and $M i ≥0$ such that $| g i (t, x 1 (t), x 2 (t),…, x k (t))|≤ M i ( t − t 0 ) − β i 1 ′$ for all $t∈J$ and $x 1 ,…, x k ∈C(( t 0 ,∞), R n )$. Then the zero solution of the problem (2.2) is globally attractive.

Proof Consider the set

where $γ i 1 ′ = 1 2 ( β i 1 ′ − α i )$ and $T ˜ 1$ is chosen such that $| x i 0 | Γ ( α i ) T ˜ 1 1 2 ( α i − 1 ) + M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) T ˜ 1 − 1 2 ( β i 1 ′ − α i ) ≤1$ for all $i=1,2,…,k$. First, we show that B maps $S 1 ′$ into $S 1 ′$. It is easy to check that $S 1 ′$ is a closed, bounded, and convex subset of $R n × R n ×⋯× R n$. Note that

$| B i ( y 1 , y 2 , … , y k ) ( t ) | ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 | g i ( s , x 1 ( s ) , x 2 ( s ) , … , x k ( s ) ) | d s ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 M i ( s − t 0 ) − β i 1 ′ d s ≤ M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − ( β i 1 ′ − α i )$

and $M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − 1 2 ( β i 1 ′ − α i ) ≤ M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) T ˜ 1 − 1 2 ( β i 1 ′ − α i ) ≤1$ for all $i=1,2,…,k$ and $t≥ t 0 + T ˜ 1$. Thus,

$| B i ( y 1 , y 2 ,…, y k )(t)|≤ [ M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − 1 2 ( β i 1 ′ − α i ) ] ( t − t 0 ) − 1 2 ( β i 1 ′ − α i ) ≤ ( t − t 0 ) − γ i 1 ′$

for all $i=1,2,…,k$ and $t≥ t 0 + T ˜ 1$. Hence, $B( S 1 ′ )⊂ S 1 ′$. Now, we show that B is continuous on $[ t 0 + T ˜ 1 ,∞)$. Let $( y 1 m , y 2 m ,…, y k m ),( y 1 , y 2 ,…, y k )∈ S 1 ′$ for all $m≥1$ and $lim m → ∞ | y i m (t)− y i (t)|=0$. Then, one can get $lim m → ∞ g i (t, y 1 m (t), y 2 m (t),…, y k m (t))= g i (t, y 1 (t), y 2 (t),…, y k (t))$ for all $t≥ t 0 + T ˜ 1$. Let $ϵ>0$ be given. Choose $T ˜ > t 0 + T ˜ 1$ such that $M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( T ˜ − t 0 ) − ( β i 1 ′ − α i ) < ϵ 2$ for all $t> T ˜$. Let $ν i 1 ′ = α i − 1 1 − α i 1 ′$ for $i=1,2,…,k$. Then, we have

$| B i ( y 1 m , y 2 m , … , y k m ) ( t ) − B i ( y 1 , y 2 , … , y k ) ( t ) | ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 | g i ( s , y 1 m ( s ) , y 2 m ( s ) , … , y k m ( s ) ) − g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s ≤ 1 Γ ( α i ) { ∫ t 0 t [ ( t − s ) α i − 1 ] 1 1 − α i 1 ′ d s } 1 − α i 1 ′ × [ ∫ t 0 t | g i ( s , y 1 m ( s ) , y 2 m ( s ) , … , y k m ( s ) ) − g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ′ ( t − t 0 ) 1 + ν i 1 ′ ) 1 − α i 1 ′ × [ ∫ t 0 T ˜ | g i ( s , y 1 m ( s ) , y 2 m ( s ) , … , y k m ( s ) ) − g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ′ ( T ˜ − t 0 ) 1 + ν i 1 ′ ) 1 − α i 1 ′ ( T ˜ − t 0 ) α i 1 ′ × sup t 0 ≤ s ≤ T ˜ | g i ( s , y 1 m ( s ) , y 2 m ( s ) , … , y k m ( s ) ) − g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) |$

for all $t 0 + T ˜ 1 ≤t≤ T ˜$. Hence, $lim m → ∞ | B i ( y 1 m , y 2 m ,…, y k m )(t)− B i ( y 1 , y 2 ,…, y k )(t)|=0$ for all $t 0 + T ˜ 1 ≤t≤ T ˜$. Also,

$| B i ( y 1 m , y 2 m , … , y k m ) ( t ) − B i ( y 1 , y 2 , … , y k ) ( t ) | ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 | g i ( s , y 1 m ( s ) , y 2 m ( s ) , … , y k m ( s ) ) − g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 [ | g i ( s , y 1 m ( s ) , y 2 m ( s ) , … , y k m ( s ) ) | + | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | ] d s ≤ 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 [ 2 M i ( s − t 0 ) − β i 1 ′ ] d s ≤ 2 M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − ( β i 1 ′ − α i ) ≤ 2 M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( T ˜ − t 0 ) − ( β i 1 ′ − α i ) ≤ ϵ$

for all $t> T ˜$. Thus, $lim m → ∞ | B i ( y 1 m , y 2 m ,…, y k m )(t)− B i ( y 1 , y 2 ,…, y k )(t)|=0$ for all $t≥ t 0 + T ˜ 1$. This implies that $B i$ is continuous on $[ t 0 + T ˜ 1 ,∞)$ for $i=1,2,…,k$ and so B is continuous on $[ t 0 + T ˜ 1 ,∞)$. Now, we show that $B( S 1 ′ )$ is equi-continuous. Let $ϵ>0$ be given. Since $lim t → ∞ ( t − t 0 ) − γ i 1 ′ =0$, there exists $T ˜ ′ > t 0 + T ˜ 1$ such that $( t − t 0 ) − γ i 1 ′ < ϵ 2$ for $t> T ˜ ′$. Let $t 1 , t 2 ≥ t 0 + T ˜ 1$ and $t 2 > t 1$. If $t 1 , t 2 ∈[ t 0 + T ˜ 1 , T ˜ ′ ]$, then we have

$| B i ( y 1 , y 2 , … , y k ) ( t 2 ) − B i ( y 1 , y 2 , … , y k ) ( t 1 ) | = | 1 Γ ( α i ) ∫ t 0 t 2 ( t 2 − s ) α i − 1 g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) d s − 1 Γ ( α i ) ∫ t 0 t 1 ( t 1 − s ) α i − 1 g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) d s | ≤ 1 Γ ( α i ) ∫ t 0 t 1 [ ( t 1 − s ) α i − 1 − ( t 2 − s ) α i − 1 ] | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s + 1 Γ ( α i ) ∫ t 1 t 2 ( t 2 − s ) α i − 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s ≤ 1 Γ ( α i ) [ ∫ t 0 t 1 [ ( t 1 − s ) α i − 1 − ( t 2 − s ) α i − 1 ] 1 1 − α i 1 ′ d s ] 1 − α i 1 ′ × [ ∫ t 0 t 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ + 1 Γ ( α i ) [ ∫ t 1 t 2 ( t 2 − s ) α i − 1 1 − α i 1 ′ d s ] 1 − α i 1 ′ [ ∫ t 0 t 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ ≤ 1 Γ ( α i ) ( 1 1 + ν i 1 ′ ) 1 − α i 1 ′ ( ( t 1 − t 0 ) 1 + ν i 1 ′ − ( t 2 − t 0 ) 1 + ν i 1 ′ + ( t 2 − t 1 ) 1 + ν i 1 ′ ) 1 − α i 1 ′ × [ ∫ t 0 T ˜ ′ | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ + 1 Γ ( α i ) ( 1 1 + ν i 1 ′ ) 1 − α i 1 ′ ( ( t 2 − t 1 ) 1 + ν i 1 ′ ) 1 − α i 1 ′ [ ∫ t 0 T ˜ ′ | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ ≤ 2 Γ ( α i ) ( 1 1 + ν i 1 ′ ) 1 − α i 1 ′ [ ∫ t 0 T ˜ ′ | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | 1 α i 1 ′ d s ] α i 1 ′ ( t 2 − t 1 ) α i − α i 1 ′$

and so $lim t 2 → t 1 | B i ( y 1 , y 2 ,…, y k )( t 2 )− B i ( y 1 , y 2 ,…, y k )( t 1 )|=0$. If $t 1 , t 2 > T ˜ ′$, then

$| B i ( y 1 , y 2 , … , y k ) ( t 2 ) − B i ( y 1 , y 2 , … , y k ) ( t 1 ) | ≤ 1 Γ ( α i ) ∫ t 0 t 2 ( t 2 − s ) α i − 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s + 1 Γ ( α i ) ∫ t 0 t 1 ( t 1 − s ) α i − 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s ≤ ( t 2 − t 0 ) − γ i 1 ′ + ( t 1 − t 0 ) − γ i 1 ′ ≤ ϵ .$

If $t 0 + T ˜ 1 ≤ t 1 < T ˜ ′ < t 2$, then we have

$| B i ( y 1 , y 2 , … , y k ) ( t 2 ) − B i ( y 1 , y 2 , … , y k ) ( t 1 ) | ≤ | B i ( y 1 , y 2 , … , y k ) ( t 2 ) − B i ( y 1 , y 2 , … , y k ) ( T ˜ ′ ) | + | B i ( y 1 , y 2 , … , y k ) ( T ˜ ′ ) − B i ( y 1 , y 2 , … , y k ) ( t 1 ) |$

and so $lim t 2 → t 1 | B i ( y 1 , y 2 ,…, y k )( t 2 )− B i ( y 1 , y 2 ,…, y k )( t 1 )|=0$. Thus, $B( S 1 ′ )$ is equi-continuous. Since $B( S 1 ′ )⊂ S 1 ′$ is uniformly bounded, $B( S 1 ′ )$ is relatively compact. Now, suppose that $x=( x 1 , x 2 ,…, x k )∈C(( t 0 ,∞), R n )×C(( t 0 ,∞), R n )×⋯×C(( t 0 ,∞), R n )$, $y=( y 1 , y 2 ,…, y k )∈ S 1 ′$ and $x=Ax+By$. Then,

$| x i ( t ) | ≤ | A i ( x 1 , x 2 , … , x k ) ( t ) | + | B i ( y 1 , y 2 , … , y k ) ( t ) | ≤ | x i 0 | Γ ( α i ) ( t − t 0 ) α i − 1 + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s ≤ | x i 0 | Γ ( α i ) ( t − t 0 ) α i − 1 + M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − ( β i 1 ′ − α i )$

for all $i=1,2,…,k$. Since $0< α i < β i 1 ′ <1$ for $i=1,2,…,k$, we get

$| x i 0 | Γ ( α i ) ( t − t 0 ) 1 2 ( α i − 1 ) + M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − 1 2 ( β i 1 ′ − α i ) ≤ | x i 0 | Γ ( α i ) T ˜ 1 1 2 ( α i − 1 ) + M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) T ˜ 1 − 1 2 ( β i 1 ′ − α i ) ≤ 1 .$

Thus, $| x i (t)|≤[ | x i 0 | Γ ( α i ) ( t − t 0 ) 1 2 ( α i − 1 ) + M i Γ ( 1 − β i 1 ′ ) Γ ( 1 + α i − β i 1 ′ ) ( t − t 0 ) − 1 2 ( β i 1 ′ − α i ) ] ( t − t 0 ) − γ i 1 ′ ≤ ( t − t 0 ) − γ i 1 ′$ for all $t≥ t 0 + T ˜ 1$ and $i=1,2,…,k$. This implies that $x(t)=( x 1 (t), x 2 (t),…, x k (t))∈ S 1 ′$ for all $t≥ t 0 + T ˜ 1$. Therefore, by using Theorem 2.2 T has a fixed point in $S 1 ′$ which is a solution of the problem (2.2). Since all elements of the set $S 1 ′$ tend to 0 as $t→∞$, the zero solution of the problem (2.2) is globally attractive. □

Theorem 3.5 Suppose that for each $i∈{1,2,…,k}$ there exist $α i < β i 2 ′ < 1 2 (1+ α i )$ and $l i ≥0$ such that $| g i (t, x 1 (t), x 2 (t),…, x k (t))|≤ l i ( t − t 0 ) − β i 2 ′ | x i (t)|$ for all $t∈J$ and $x 1 ,…, x k ∈C(( t 0 ,∞), R n )$. Then the zero solution of the problem (2.2) is globally attractive.

Proof It is sufficient to consider the set

where $γ i 2 ′ = 1 2 (1− α i )$ and $T ˜ 2$ is chosen such that $| x i 0 | Γ ( α i ) T ˜ 2 1 2 ( α i − 1 ) + l i Γ ( 1 − β i 2 ′ − γ i 2 ′ ) Γ ( 1 + α i − β i 2 ′ − γ i 2 ′ ) T ˜ 2 − ( β i 2 ′ − α i ) ≤1$ for all $i=1,2,…,k$. Similar to the proof of Theorem 3.4, one can show that $S 2 ′$ is a closed, bounded, and convex set, B maps $S 2 ′$ into $S 2 ′$, $B( S 2 ′ )$ is relatively compact, and B is continuous on $[ t 0 + T ˜ 2 ,∞)$. Now, suppose that $x=( x 1 , x 2 ,…, x k )∈C(( t 0 ,∞), R n )×C(( t 0 ,∞), R n )×⋯×C(( t 0 ,∞), R n )$, $y=( y 1 , y 2 ,…, y k )∈ S 2$ and $x=Ax+By$. Then,

$| x i ( t ) | ≤ | A i ( x 1 , x 2 , … , x k ) ( t ) | + | B i ( y 1 , y 2 , … , y k ) ( t ) | ≤ | x i 0 | Γ ( α i ) ( t − t 0 ) α i − 1 + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 | g i ( s , y 1 ( s ) , y 2 ( s ) , … , y k ( s ) ) | d s ≤ | x i 0 | Γ ( α i ) ( t − t 0 ) α i − 1 + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 l i ( s − t 0 ) − β i 2 ′ | y i ( s ) | d s ≤ | x i 0 | Γ ( α i ) ( t − t 0 ) α i − 1 + 1 Γ ( α i ) ∫ t 0 t ( t − s ) α i − 1 l i ( s − t 0 ) − β i 2 ′ − γ i 2 ′ d s ≤ | x i 0 | Γ ( α i ) ( t − t 0 ) α i − 1 + l i Γ ( 1 − β i 2 ′ − γ i 2 ′ ) Γ ( 1 + α i − β i 2 ′ − γ i 2 ′ ) ( t − t 0 ) − ( β i 2 ′ − γ i 2 ′ − α i )$

for all $i=1,2,…,k$. Since $0< α i < β i 2 ′ < 1 2 (1+ α i )<1$ for $i=1,2,…,k$, we get

$| x i 0 | Γ ( α i ) ( t − t 0 ) 1 2 ( α i − 1 ) + l i Γ ( 1 − β i 2 ′ − γ i 2 ′ ) Γ ( 1 + α i − β i 2 ′ − γ i 2 ′ ) ( t − t 0 ) − ( β i 2 ′ − α i ) ≤ | x i 0 | Γ ( α i ) T ˜ 2 1 2 ( α i − 1 ) + l i Γ ( 1 − β i 2 ′ − γ i 2 ′ ) Γ ( 1 + α i − β i 2 ′ − γ i 2 ′ ) T ˜ 2 − ( β i 2 ′ − α i ) ≤ 1 .$

Thus, $| x i (t)|≤[ | x i 0 | Γ ( α i ) ( t − t 0 ) 1 2 ( α i − 1 ) + l i Γ ( 1 − β i 2 ′ − γ i 2 ′ ) Γ ( 1 + α i − β i 2 ′ − γ i 2 ′ ) ( t − t 0 ) − ( β i 2 ′ − α i ) ] ( t − t 0 ) − γ i 2 ′ ≤ ( t − t 0 ) − γ i 2 ′$ for all $t≥ t 0 + T ˜ 2$ and $i=1,2,…,k$. This implies that $x(t)=( x 1 (t), x 2 (t),…, x k (t))∈ S 2$, for $t≥ t 0 + T ˜ 2$. Since all elements of the set $S 2 ′$ tend to 0 as $t→∞$, the zero solution of the problem (2.2) is globally attractive. □

## 4 Examples

Here, we give an example to illustrate our results.

Example 4.1 Consider the 3-dimensional system of fractional functional differential equations

${ D 1 2 c x 1 ( t ) = Γ ( 3 4 ) Γ ( 1 4 ) ( t + 3 ) − 3 4 sin 2 ( x 1 ( t − 1 ) ) 1 + ( x 2 ( t − 1 ) ) 2 × x 3 ( t − 1 ) 1 + | x 3 ( t − 1 ) | , t > 0 , D 1 4 c x 2 ( t ) = Γ ( 3 8 ) Γ ( 1 8 ) ( t + 3 2 ) − 7 8 cos 4 ( x 1 ( t − 1 ) ) 1 + sin 2 ( x 3 ( t − 1 ) ) + | x 2 ( t − 1 ) | , t > 0 , D 1 3 c x 3 ( t ) = Γ ( 5 6 ) π ( t + 1 ) − 1 2 ( x 1 ( t − 1 ) ) 4 1 + ( x 1 ( t − 1 ) ) 4 + 6 | x 2 ( t − 1 ) | 3 , t > 0 , x i ( t ) = t , i = 1 , 2 , 3 , t ∈ [ − 1 , 0 ] .$

Define the maps

$f 1 ( t , x 1 t , x 2 t , x 3 t ) = Γ ( 3 4 ) Γ ( 1 4 ) ( t + 3 ) − 3 4 sin 2 ( x 1 ( t − 1 ) ) 1 + ( x 2 ( t − 1 ) ) 2 × x 3 ( t − 1 ) 1 + | x 3 ( t − 1 ) | , f 2 ( t , x 1 t , x 2 t , x 3 t ) = Γ ( 3 8 ) Γ ( 1 8 ) ( t + 3 2 ) − 7 8 cos 4 ( x 1 ( t − 1 ) ) 1 + sin 2 ( x 3 ( t − 1 ) ) + | x 2 ( t − 1 ) | , f 3 ( t , x 1 t , x 2 t , x 3 t ) = Γ ( 5 6 ) π ( t + 1 ) − 1 2 ( x 1 ( t − 1 ) ) 4 1 + ( x 1 ( t − 1 ) ) 4 + 6 | x 2 ( t − 1 ) | 3$

and put $m 1 (t)= Γ ( 3 4 ) Γ ( 1 4 ) ( t + 3 ) − 3 4$, $m 2 (t)= Γ ( 3 8 ) Γ ( 1 8 ) ( t + 3 2 ) − 7 8$ and $m 3 (t)= Γ ( 5 6 ) π ( t + 1 ) − 1 2$. It is easy to check that $| f 1 (t, x 1 t , x 2 t , x 3 t )|≤ m 1 (t)$, $| f 2 (t, x 1 t , x 2 t , x 3 t )|≤ m 2 (t)$ and $| f 3 (t, x 1 t , x 2 t , x 3 t )|≤ m 3 (t)$. Since

$1 Γ ( α 1 ) ∫ t 0 t ( t − s ) α 1 − 1 m 1 ( s ) d s = 1 Γ ( 1 2 ) ∫ 0 t ( t − s ) − 1 2 × Γ ( 3 4 ) Γ ( 1 4 ) ( s + 3 ) − 3 4 d s 1 Γ ( α 1 ) ∫ t 0 t ( t − s ) α 1 − 1 m 1 ( s ) d s ≤ Γ ( 3 4 ) Γ ( 1 2 ) Γ ( 1 4 ) ∫ 0 t ( t − s ) − 1 2 s − 3 4 d s = t − 1 4 , 1 Γ ( α 2 ) ∫ t 0 t ( t − s ) α 2 − 1 m 2 ( s ) d s = 1 Γ ( 1 4 ) ∫ 0 t ( t − s ) − 3 4 × Γ ( 3 8 ) Γ ( 1 8 ) ( s + 3 2 ) − 7 8 d s 1 Γ ( α 2 ) ∫ t 0 t ( t − s ) α 2 − 1 m 2 ( s ) d s ≤ Γ ( 3 8 ) Γ ( 1 4 ) Γ ( 1 8 ) ∫ 0 t ( t − s ) − 3 4 s − 7 8 d s = t − 5 8$

and

$1 Γ ( α 3 ) ∫ t 0 t ( t − s ) α 3 − 1 m 3 ( s ) d s = 1 Γ ( 1 3 ) ∫ 0 t ( t − s ) − 2 3 × Γ ( 5 6 ) π ( s + 1 ) − 1 2 d s ≤ Γ ( 5 6 ) Γ ( 1 3 ) π ∫ 0 t ( t − s ) − 2 3 s − 1 2 d s = t − 1 6 ,$

we get $| 1 Γ ( α 1 ) ∫ 0 t ( t − s ) − 1 2 f 1 (s, x 1 s , x 2 s , x 3 s )ds|≤ t − 1 4$, $| 1 Γ ( α 2 ) ∫ 0 t ( t − s ) − 3 4 f 2 (s, x 1 s , x 2 s , x 3 s )ds|≤ t − 5 8$ and $| 1 Γ ( α 3 ) ∫ 0 t ( t − s ) − 2 3 f 1 (s, x 1 s , x 2 s , x 3 s )ds|≤ t − 1 6$. Now, let $α 11 = 1 4$, $α 21 = 1 8$ and $α 31 = 1 6$. Then,

$∫ t 0 ∞ | f 1 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 11 d t ≤ ∫ t 0 ∞ ( m 1 ( t ) ) 1 α 11 d t ∫ t 0 ∞ | f 1 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 11 d t = ∫ 0 ∞ [ Γ ( 3 4 ) Γ ( 1 4 ) ( t + 3 ) − 3 4 ] 4 d t ∫ t 0 ∞ | f 1 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 11 d t = 1 18 ( Γ ( 3 4 ) Γ ( 1 4 ) ) 4 , ∫ t 0 ∞ | f 2 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 21 d t ≤ ∫ t 0 ∞ ( m 2 ( t ) ) 1 α 21 d t ∫ t 0 ∞ | f 2 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 21 d t = ∫ 0 ∞ [ Γ ( 3 8 ) Γ ( 1 8 ) ( t + 3 2 ) − 7 8 ] 8 d t ∫ t 0 ∞ | f 2 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 21 d t = ( 2 5 3 7 ) ( Γ ( 3 8 ) Γ ( 1 8 ) ) 8$

and

$∫ t 0 ∞ | f 3 ( t , x 1 t , x 2 t , x 3 t ) | 1 α 31 d t ≤ ∫ t 0 ∞ ( m 3 ( t ) ) 1 α 31 d t = ∫ 0 ∞ [ Γ ( 5 6 ) π ( t + 1 ) − 1 2 ] 6 d t = 1 2 ( Γ ( 5 6 ) π ) 6 .$

Thus, all conditions of Theorem 3.1 hold and so this system of fractional functional differential equations has an attractive solution.

Example 4.2 Let $0< α i <1$, $M i >0$, $α i < β i 1 ′ <1$ and $x i 0$ be a constant for $i=1,2,3$. Consider the 3-dimensional system of fractional differential equations

${ D α 1 x 1 ( t ) = M 1 x 2 ( t ) sin 2 ( x 3 ( t ) ) 3 2 + | x 2 ( t ) | + | x 3 ( t ) | ( t − a ) − β 11 ′ , t > a , D α 2 x 2 ( t ) = M 2 t 2 ( x 1 ( t ) ) 2 ( 7 + 5 t 2 ) ( 1 + 2 ( x 1 ( t ) ) 2 + ( x 3 ( t ) ) 2 ) ( t − a ) − β 21 ′ , t > a , D α 3 x 3 ( t ) = M 3 cos 3 ( x 2 ( t ) ) ( x 3 ( t ) ) 3 8 + 3 ( x 2 ( t ) ) 2 + | x 3 ( t ) | 3 ( t − a ) − β 31 ′ , t > a , [ D α i − 1 x i ( t ) ] t = a = x i 0 , i = 1 , 2 , 3 .$

Define the maps

$g 1 ( t , x 1 ( t ) , x 2 ( t ) , … , x k ( t ) ) = M 1 x 2 ( t ) sin 2 ( x 3 ( t ) ) 3 2 + | x 2 ( t ) | + | x 3 ( t ) | ( t − a ) − β 11 ′ , g 2 ( t , x 1 ( t ) , x 2 ( t ) , … , x k ( t ) ) = M 2 t 2 ( x 1 ( t ) ) 2 ( 7 + 5 t 2 ) ( 1 + 2 ( x 1 ( t ) ) 2 + ( x 3 ( t ) ) 2 ) ( t − a ) − β 21 ′$

and

$g 3 ( t , x 1 ( t ) , x 2 ( t ) , … , x k ( t ) ) = M 3 cos 3 ( x 2 ( t ) ) ( x 3 ( t ) ) 3 8 + 3 ( x 2 ( t ) ) 2 + | x 3 ( t ) | 3 ( t − a ) − β 31 ′ .$

Thus, one can check that all conditions of Theorem 3.4 hold and so this system of fractional differential equations has a globally attractive solution.

## 5 Conclusions

Investigating the attractive solutions of the problems is an interesting topic within the fractional calculus. In this manuscript, we focus on the attractivity of solutions for two k-dimensional systems of fractional differential equations. Two illustrative examples show the applicability of the proposed methods. The techniques of the reported results can be applied for investigating the attractivity and global attractivity of solutions of different systems of (singular) fractional differential equations. Also, it is an interesting issue to investigate the attractivity and global attractivity of solutions of some systems of fractional differential inclusions.

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