R E S E A R C H
Open Access
Attractivity for a
k
-dimensional system of
fractional functional differential equations
and global attractivity for a
k
-dimensional
system of nonlinear fractional differential
equations
Dumitru Baleanu
1,2,3*, Sayyedeh Zahra Nazemi
4and Shahram Rezapour
4*Correspondence:
1Department of Mathematics,
Cankaya University, Ogretmenler Cad. 14, Balgat, Ankara, 06530, Turkey
2Institute of Space Sciences,
Magurele, Bucharest, Romania Full list of author information is available at the end of the article
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 ak-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 dif-ferential 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 econ-omy, 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 attractiv-ity 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 orderqfor a functionf see []. In , Chen and Zhou reviewed the attractivity of solutions for the fractional functional differential equationcDαx(t) =f(t,x
t) fort∈(t,∞) via the boundary value conditionx(t) =φ(t) for
t–r≤t≤t, wheret≥,r> , <α< ,cDis the standard Caputo fractional derivative, φ∈C([t–r,t],R) andf : (t,∞)×C([–r, ],R)→Ra function with some properties []. In , Chenet al.reviewed the global attractivity of solutions for the nonlinear fractional differential equationDαx(t) =g(t,x(t)) fort∈(t,∞) via the boundary value
problem [Dα–x(t)]
t=t=x, where t≥, <α< ,xis a constant,Dis the standard
problemx(t) =x, wheret≥, <α< ,xis a constant,cDis the standard Caputo frac-tional derivative, andg: (t,∞)×R→Ra function with some properties [].
In this paper, we investigate the attractivity of solutions for ak-dimensional system of fractional differential equations. Also, we investigate the global attractivity of solutions for anotherk-dimensional system of nonlinear fractional differential equations.
2 Preliminaries
In this paper, we investigate the attractivity of solutions fork-dimensional system of frac-tional funcfrac-tional differential equations
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
cDαx(t) =f(t,x
t,xt, . . . ,xkt),
cDαx(t) =f(t,x
t,xt, . . . ,xkt), ..
.
cDαkx
k(t) =fk(t,xt,xt, . . . ,xkt),
(.)
via the boundary value problems x(t) =φ(t),x(t) =φ(t), . . . ,xk(t) =φk(t) for t–r≤
t≤t, where <αi< fori= , , . . . ,k,t≥,t∈(t,∞), cDis the standard Caputo fractional derivative,J= (t,∞),r> ,φi∈C([t–r,t],Rn) fori= , , . . . ,kandfi:J×
C([–r, ],Rn)×C([–r, ],Rn)× · · · ×C([–r, ],Rn)→Rn is a function satisfying some
assumptions that will be specified later fori= , , . . . ,k. Ifx∈C([t–r,∞),Rn), thenxt
is defined byxt(θ) =x(t+θ) for –r≤θ≤ andt∈[t,∞). Also, we investigate the global
attractivity of solutions for thek-dimensional system of nonlinear fractional differential equations
⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩
Dαx(t) =g(t,x(t),x(t), . . . ,x
k(t)),
Dαx(t) =g(t,x
(t),x(t), . . . ,xk(t)),
.. .
Dαkx
k(t) =gk(t,x(t),x(t), . . . ,xk(t)),
(.)
via the boundary value problems [Dα–x
(t)]t=t =x
, [Dα–x(t)]t=t=x
, . . . , [Dαk–×
xk(t)]t=t =x
k, where <αi< fori= , , . . . ,k,t≥,t∈(t,∞),Dis the Riemann-Liouville fractional derivative,J= (t,∞),x, . . . ,xk are constants, andgi:J×Rn×Rn×
· · · ×Rn→Rnis an integrable function satisfying some assumptions that will be specified
later fori= , , . . . ,k. In fact, we say that the solution (x(t),x(t), . . . ,xk(t)) of the problem
(.) is attractive whenever if there exists a constantb
i(t) > such that|φi(s)| ≤bi for all
i= , , . . . ,kands∈[t–r,t], thenlimt→∞xi(t,t,φi)→. Also, the zero solutionx(t) of
the problem (.) is said to be globally attractive whenever each solution tends to zero as
t→ ∞. LetX=C(J,Rn) be the Banach space of all continuous functions fromJintoRn
with the normx=supt∈J|x(t)|, where| · |denotes a suitable complete norm onRn. It is
clear that the product space (Xk=X×X× · · · ×X
k
, · ∗) is also a Banach space, where
(x,x, . . . ,xk)∗=x+x+· · ·+xk. 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 [, ] and [].
Theorem . Let S be a nonempty,closed,convex,and bounded subset of the Banach space X,A:X→X a contraction with constant l< ,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 (.). In this way, suppose that
xt=sup–r≤θ≤|x(t+θ)|fort∈J. We assume thatfi(t,xt,xt, . . . ,xkt) is Lebesgue mea-surable with respect toton [t,∞) andfi(t,ϕ,ϕ, . . . ,ϕk) is continuous with respect toϕj
onC([–r, ],Rn) fori,j= , , . . . ,k. Note that the problem (.) is equivalent to the system
of equations
xi(t) =
φi(t) +(α
i) t
t(t–s) αi–f
i(s,xs,xs, . . . ,xks)ds, t>t,
φi(t), t∈[t–r,t]
or
xi(t) =
(αi) t
t(t–s)
αi–[ φi(t)
(–αi)(s–t)–
αi+f
i(s,xs,xs, . . . ,xks)]ds, t>t,
φi(t), t∈[t–r,t]
fori= , , . . . ,k. Define the operatorT:Xk→Xkby
T(x,x, . . . ,xk)(t) =
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
T(x,x, . . . ,xk)(t)
T(x,x, . . . ,xk)(t)
.. .
Tk(x,x, . . . ,xk)(t)
⎞ ⎟ ⎟ ⎟ ⎟ ⎠,
where
Ti(x,x, . . . ,xk)(t) =
φi(t) +(αi) t
t(t–s) αi–f
i(s,xs,xs, . . . ,xks)ds, t>t,
φi(t), t∈[t–r,t]
fori= , , . . . ,k. It is easy to check that (x(t),x(t), . . . ,xk(t)) is a solution of the problem
(.) if and only if (x(t),x(t), . . . ,xk(t)) is a fixed point of the operatorT.
Theorem . Suppose that for each i∈ {, , . . . ,k}there existγi> andαi∈(,αi)such
that
φi(t) +
(αi)
t t
(t–s)αi–f
i(s,xs,xs, . . . ,xks)ds
≤(t–t)–γi
for all t∈J and fi∈L
αi(J×C([–r, ],Rn)×C([–r, ],Rn)× · · · ×C([–r, ],Rn),Rn).Then
the problem(.)has at least one attractive solution(x,x, . . . ,xk)such that xi∈C([t–
r,∞),Rn)for all i= , , . . . ,k.
Proof Consider the set
S=
(x,x, . . . ,xk) :xi∈C
[t–r,∞),Rn
,xi(t)≤(t–t)–γi
for alli= , , . . . ,kandt≥ ˜t>t
where t˜ is a constant. It is easy to check that S is a closed, bounded, and convex subset of Rn×Rn× · · · ×Rn
k
. We show that the operator T has a fixed point in S.
This implies that the problem (.) has a solution. Note that |Ti(x,x, . . . ,xk)(t)| ≤(t–
t)–γi for all i= , , . . . ,k and so T(S)⊂S. Now, we show that T is continuous. Let
(xm
,xm, . . . ,xmk), (x,x, . . . ,xk)∈ S for all m≥ and limm→∞|xim(t) –xi(t)|= for all
i= , , . . . ,k. Then, we have limm→∞fi(t,xmt,xmt, . . . ,xmkt) =fi(t,xt,xt, . . . ,xkt) for all i= , , . . . ,k and t>t. Let > be given. Choose T˜ >t such that t≥ ˜T implies that (t–t)–γi<
. Letνi=
αi–
–αi and note that +νi> fori= , , . . . ,k. Also, we have
Ti
xm,xm, . . . ,xmk(t) –Ti(x,x, . . . ,xk)(t)
≤
(αi)
t t
(t–s)αi–fi
s,xms,xms, . . . ,xmks–fi(s,xs,xs, . . . ,xks)ds
≤
(αi)
t
t
(t–s)αi––αids
–αi
× t
t
fi
s,xms,xms, . . . ,xmks–fi(s,xs,xs, . . . ,xks)
αids αi
≤
(αi)
+νi
(t–t)
+νi–αi
×
˜
T t
fi
s,xms,xms, . . . ,xmks–fi(s,xs,xs, . . . ,xks)
αids αi
≤
(αi)
+νi
(T˜ –t)
+νi–αi
(T˜ –t)αi
× sup
t≤s≤ ˜T
fi
s,xms,xms, . . . ,xmks–fi(s,xs,xs, . . . ,xks)
for t<t≤ ˜T. Thus,limm→∞|Ti(xm,xm, . . . ,xmk)(t) –Ti(x,x, . . . ,xk)(t)|= for all t<
t≤ ˜T. Also, we have Ti
xm,xm, . . . ,xmk(t) –Ti(x,x, . . . ,xk)(t)
=
(αi)
t t
(t–s)αi–f
i
s,xms,xms, . . . ,xmksds
–
(αi)
t t
(t–s)αi–fi(s,xs,xs, . . . ,xks)ds
≤(t–t)–γi≤
fort>T˜. Hence,limm→∞|Ti(xm,xm, . . . ,xkm)(t) –Ti(x,x, . . . ,xk)(t)|= fort>t. This
im-plies thatTiis continuous fori= , , . . . ,kand soTis continuous. Now, we show that the
setT(S) is equi-continuous. Let> . Sincelimt→∞(t–t)–γi= fori= , , . . . ,k, there is aT˜>tsuch that (t–t)–γi< for allt>T˜andi= , , . . . ,k. Lett,t>tandt>t. Ift,t∈(t,T˜], then
Ti(x,x, . . . ,xk)(t) –Ti(x,x, . . . ,xk)(t)
≤
(αi)
t
t
– (αi)
t
t
(t–s)αi–fi(s,xs,xs, . . . ,xks)ds
≤
(αi)
t
t
(t–s)αi–– (t–s)αi–fi(s,xs,xs, . . . ,xks)ds
+
(αi)
t
t
(t–s)αi–fi(s,xs,xs, . . . ,xks)ds
≤
(αi)
t
t
(t–s)αi–– (t–s)αi–
–αids
–αi
×
t
t
fi(s,xs,xs, . . . ,xks)
αids
αi
+
(αi)
t
t
(t–s) αi– –αids
–αi t
t
fi(s,xs,xs, . . . ,xks)
αids αi
≤
(αi)
+νi
–αi
(t–t) αi–
–αi++ (t –t)
αi–
–αi+– (t –t)
αi–
–αi+–αi
×
T˜
t
fi(s,xs,xs, . . . ,xks)
αids
αi
+
(αi)
+νi
–αi
(t–t) αi– –αi+–αi
T˜
t
fi(s,xs,xs, . . . ,xks)
αids
αi
≤
(αi)
+νi
–αi T˜ t
fi(s,xs,xs, . . . ,xks)
αi ds
αi
(t–t)αi–αi
and solimt→t|Ti(x,x, . . . ,xk)(t) –Ti(x,x, . . . ,xk)(t)|= . Ift,t>T˜, then
Ti(x,x, . . . ,xk)(t) –Ti(x,x, . . . ,xk)(t)
=
(αi)
t
t
(t–s)αi–fi(s,xs,xs, . . . ,xks)ds
–
(αi)
t
t
(t–s)αi–fi(s,xs,xs, . . . ,xks)ds
≤(t–t)–γi+ (t–t)–γi≤.
Now, lett<t<T˜<t. Since
Ti(x,x, . . . ,xk)(t) –Ti(x,x, . . . ,xk)(t)
≤Ti(x,x, . . . ,xk)(t) –Ti(x,x, . . . ,xk)
˜
T
+Ti(x,x, . . . ,xk)
˜
T–Ti(x,x, . . . ,xk)(t),
we getlimt→t|Ti(x,x, . . . ,xk)(t) –Ti(x,x, . . . ,xk)(t)|= in all cases. This implies that
the setT(S) is equi-continuous. SinceT(S)⊂Sis uniformly bounded,T(S) is relatively compact. Now by using Theorem .,T has a fixed point inSwhich is a solution of the problem (.). Sincex(t) = (x(t),x(t), . . . ,xk(t))∈S,limt→∞x(t) = . Thus,x(t) is an
Theorem . Suppose that for each i∈ {, , . . . ,k} there existγi> ,αi∈(,αi)and
li∈L
αi(J,R+)such that (αi)
t t(t–s)
αi–l
i(s)(s–t)–γids≤(t–t)–γi and
φi(t)
( –αi)
(t–t)–αi+f
i(t,xt,xt, . . . ,xkt)
≤li(t)xit
for all i= , , . . . ,k,t∈J and xi∈C([t–r,∞),Rn).Then the problem(.)has at least one
attractive solution(x,x, . . . ,xk)such that xi∈C([t–r,∞),Rn)for all i= , , . . . ,k.
Proof It is sufficient we consider the set
S=
(x,x, . . . ,xk) :xi∈C
[t–r,∞),Rn
,xit ≤(t–t) –γi
for alli= , , . . . ,kandt≥ ˜t>t
,
where˜tis a constant. By using a similar techniques and proof in Theorem ., one can show that T(S)⊂S, T is continuous andT(S) is relatively compact. Now by using Theorem ., T has a fixed point inS which is a solution of the problem (.). Since
x(t) = (x(t),x(t), . . . ,xk(t))∈S,limt→∞x(t) = . Thus,x(t) is an attractive solution for
the problem (.).
Theorem . Suppose that for each i∈ {, , . . . ,k}there existsβi∈(αi, )such that
φi(t)
( –αi)
(t–t)–αi+f
i(t,xt,xt, . . . ,xkt)
≤( +αi–βi)
( –βi)
(t–t)–βi
for all t∈J.Then the problem(.)has at least one attractive solution(x,x, . . . ,xk)such
that xi∈C([t–r,∞),Rn)for all i= , , . . . ,k.
Proof It is sufficient we consider the set
S=
(x,x, . . . ,xk) :xi∈C
[t–r,∞),Rn
,xi(t)≤(t–t)βi–αi
for alli= , , . . . ,kandt≥ ˜t>t
,
where˜tis a constant. By using a similar techniques and proof in Theorem ., one can show that T(S)⊂S,T is continuous and T(S) is relatively compact. Now, by using Theorem ., T has a fixed point inS which is a solution of the problem (.). Since
x(t) = (x(t),x(t), . . . ,xk(t))∈S,limt→∞x(t) = . Thus,x(t) is an attractive solution for
the problem (.).
Here, we are going to investigate global attractivity of solutions of the problem (.). We assume thatgi(t,x(t),x(t), . . . ,xk(t)) is Lebesgue measurable with respect toton [t,∞) and there exists a constantαi∈(,αi) such thatgi∈L
αi(J×Rn×Rn× · · · ×Rn,Rn) and
gi(t,x(t),x(t), . . . ,xk(t)) is continuous with respect toxjon [t,∞) for alli,j= , , . . . ,k.
Note that the problem (.) is equivalent to the system of equations
xi(t) =
xi
(αi)
(t–t)αi–+ (αi)
t t
(t–s)αi–gi
s,x(s),x(s), . . . ,xk(s)
for allt>tandi= , , . . . ,k. Define the operatorT:Xk→Xkby
T(x,x, . . . ,xk)(t) =
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
T(x,x, . . . ,xk)(t)
T(x,x, . . . ,xk)(t)
.. .
Tk(x,x, . . . ,xk)(t)
⎞ ⎟ ⎟ ⎟ ⎟ ⎠,
where
Ti(x,x, . . . ,xk)(t) =
x
i
(αi)
(t–t)αi–+ (αi)
t t
(t–s)αi–gi
s,x(s),x(s), . . . ,xk(s)
ds
for alli= , , . . . ,k. Now, define
A(x,x, . . . ,xk)(t) =
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
A(x,x, . . . ,xk)(t)
A(x,x, . . . ,xk)(t)
.. .
Ak(x,x, . . . ,xk)(t)
⎞ ⎟ ⎟ ⎟ ⎟ ⎠,
whereAi(x,x, . . . ,xk)(t) = xi
(αi)(t–t)
αi–for alli= , , . . . ,k. Finally, define
B(x,x, . . . ,xk)(t) =
⎛ ⎜ ⎜ ⎜ ⎜ ⎝
B(x,x, . . . ,xk)(t)
B(x,x, . . . ,xk)(t)
.. .
Bk(x,x, . . . ,xk)(t)
⎞ ⎟ ⎟ ⎟ ⎟ ⎠,
whereBi(x,x, . . . ,xk)(t) =(αi) t
t(t–s) αi–g
i(s,x(s),x(s), . . . ,xk(s))dsfor alli= , , . . . ,k.
It is easy to check that (x(t),x(t), . . . ,xk(t)) is a solution of the problem (.) if and only if
it is a fixed point of the operatorT. Note thatAis a contraction with constant .
Theorem . Suppose that for each i∈ {, , . . . ,k}there existαi<βi< and Mi≥such
that|gi(t,x(t),x(t), . . . ,xk(t))| ≤Mi(t–t)–β
ifor all t∈J and x, . . . ,xk∈C((t,∞),Rn). Then the zero solution of the problem(.)is globally attractive.
Proof Consider the set
S=(x,x, . . . ,xk) :xi∈C
(t,∞),Rn,xi(t)≤(t–t)–γ
i
for alli= , , . . . ,kandt≥t+T˜
,
whereγi=(βi–αi) andT˜is chosen such that
|xi|
(αi)T˜
(αi–)
+
Mi(–βi) (+αi–βi)T˜
–(βi–αi)
≤ for
alli= , , . . . ,k. First, we show thatBmapsSintoS. It is easy to check thatSis a closed, bounded, and convex subset ofRn×Rn× · · · ×Rn. Note that
Bi(y,y, . . . ,yk)(t)
≤
(αi)
t t
(t–s)αi–gi
≤
(αi)
t t
(t–s)αi–Mi(s–t)–β
ids
≤ Mi( –βi) ( +αi–βi)
(t–t)–(βi–αi)
and Mi(–βi) (+αi–βi)(t–t)
–(βi–αi)≤ Mi(–βi) (+αi–βi) ˜ T–
(βi–αi)
≤ for alli= , , . . . ,kandt≥t+T˜. Thus,
Bi(y,y, . . . ,yk)(t)≤
Mi( –βi) ( +αi–βi)
(t–t)–(βi–αi)
(t–t)–(βi–αi)≤(t–t)–γi
for alli= , , . . . ,kandt≥t+T˜. Hence,B(S)⊂S. Now, we show thatBis continuous on [t+T˜,∞). Let (ym,ym, . . . ,ymk), (y,y, . . . ,yk)∈S for allm≥ andlimm→∞|ymi (t) –
yi(t)|= . Then, one can getlimm→∞gi(t,ym(t),ym(t), . . . ,ymk(t)) =gi(t,y(t),y(t), . . . ,yk(t))
for allt≥t+T˜. Let> be given. ChooseT˜ >t+T˜such that Mi
(–βi)
(+αi–βi)(T˜–t)
–(βi–αi)<
for allt>T˜. Letνi=
αi–
–αi fori= , , . . . ,k. Then, we have
Bi
ym,ym, . . . ,ymk(t) –Bi(y,y, . . . ,yk)(t)
≤
(αi)
t t
(t–s)αi–gi
s,ym(s),ym(s), . . . ,ymk(s)–gi
s,y(s),y(s), . . . ,yk(s)ds
≤
(αi)
t
t
(t–s)αi–
–αi
ds
–αi
×
t
t
gi
s,ym(s),ym(s), . . . ,ymk(s)–gi
s,y(s),y(s), . . . ,yk(s)
αids αi
≤
(αi)
+νi(t–t) +νi
–αi
×
T˜
t
gi
s,ym(s),ym(s), . . . ,ymk(s)–gi
s,y(s),y(s), . . . ,yk(s)
αids αi
≤
(αi)
+νi(T˜ –t) +νi
–αi
(T˜ –t)αi
× sup
t≤s≤ ˜T
gi
s,ym(s),ym(s), . . . ,ymk(s)–gi
s,y(s),y(s), . . . ,yk(s)
for allt+T˜≤t≤ ˜T. Hence,limm→∞|Bi(ym,ym, . . . ,ymk)(t) –Bi(y,y, . . . ,yk)(t)|= for all
t+T˜≤t≤ ˜T. Also,
Bi
ym,ym, . . . ,ymk(t) –Bi(y,y, . . . ,yk)(t)
≤
(αi)
t t
(t–s)αi–gi
s,ym(s),ym(s), . . . ,ymk(s)–gi
s,y(s),y(s), . . . ,yk(s)ds
≤
(αi)
t t
(t–s)αi–gi
s,ym(s),ym(s), . . . ,ymk(s)+gi
s,y(s),y(s), . . . ,yk(s)ds
≤
(αi)
t t
(t–s)αi–Mi(s–t)–β
≤ Mi( –βi) ( +αi–βi)
(t–t)–(βi–αi)
≤ Mi( –βi) ( +αi–βi)
(T˜ –t)–(β
i–αi)≤
for allt>T˜. Thus,limm→∞|Bi(ym,ym, . . . ,ymk)(t) –Bi(y,y, . . . ,yk)(t)|= for allt≥t+T˜. This implies thatBiis continuous on [t+T˜,∞) fori= , , . . . ,kand soBis continuous on [t+T˜,∞). Now, we show thatB(S) is equi-continuous. Let > be given. Since limt→∞(t–t)–γi= , there existsT˜>t+T˜ such that (t–t)–γi <
fort>T˜. Let
t,t≥t+T˜andt>t. Ift,t∈[t+T˜,T˜], then we have Bi(y,y, . . . ,yk)(t) –Bi(y,y, . . . ,yk)(t)
=
(αi)
t
t
(t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)
ds
–
(αi)
t
t
(t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)
ds
≤
(αi)
t
t
(t–s)αi–– (t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)ds
+
(αi)
t
t
(t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)ds
≤
(αi)
t
t
(t–s)αi–– (t–s)αi–
–αi
ds
–αi
× t
t
gi
s,y(s),y(s), . . . ,yk(s)
αi
ds
αi
+
(αi)
t
t
(t–s) αi– –αi
ds
–αi t
t
gi
s,y(s),y(s), . . . ,yk(s)
αi
ds
αi
≤
(αi)
+νi
–αi
(t–t)+ν
i– (t–t)+νi+ (t–t)+νi–α
i
×
T˜
t
gi
s,y(s),y(s), . . . ,yk(s)
αi
ds
αi
+
(αi)
+νi
–αi
(t–t)+ν
i–α
i
T˜
t
gi
s,y(s),y(s), . . . ,yk(s)
αi
ds
αi
≤
(αi)
+νi
–αi T˜ t
gi
s,y(s),y(s), . . . ,yk(s)
αi
ds
αi
(t–t)αi–α
i
and solimt→t|Bi(y,y, . . . ,yk)(t) –Bi(y,y, . . . ,yk)(t)|= . Ift,t>T˜, then Bi(y,y, . . . ,yk)(t) –Bi(y,y, . . . ,yk)(t)
≤
(αi)
t
t
(t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)ds
+
(αi)
t
t
(t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)ds
≤(t–t)–γ
Ift+T˜≤t<T˜<t, then we have
Bi(y,y, . . . ,yk)(t) –Bi(y,y, . . . ,yk)(t)
≤Bi(y,y, . . . ,yk)(t) –Bi(y,y, . . . ,yk)
˜
T
+Bi(y,y, . . . ,yk)
˜
T–Bi(y,y, . . . ,yk)(t)
and solimt→t|Bi(y,y, . . . ,yk)(t) –Bi(y,y, . . . ,yk)(t)|= . Thus,B(S) is equi-contin-uous. Since B(S)⊂S is uniformly bounded, B(S) is relatively compact. Now, sup-pose that x= (x,x, . . . ,xk)∈C((t,∞),Rn)×C((t,∞),Rn)× · · · ×C((t,∞),Rn),y=
(y,y, . . . ,yk)∈Sandx=Ax+By. Then,
xi(t)≤Ai(x,x, . . . ,xk)(t)+Bi(y,y, . . . ,yk)(t)
≤ |xi|
(αi)
(t–t)αi–+ (αi)
t t
(t–s)αi–gi
s,y(s),y(s), . . . ,yk(s)ds
≤ |xi|
(αi)
(t–t)αi–+ Mi( –β
i) ( +αi–βi)
(t–t)–(βi–αi)
for alli= , , . . . ,k. Since <αi<βi< fori= , , . . . ,k, we get
|x
i|
(αi)
(t–t)(αi–)+ Mi( –β
i) ( +αi–βi)
(t–t)–(βi–αi)
≤ |xi|
(αi)
˜ T (αi–)
+
Mi( –βi) ( +αi–βi)
˜ T–
(βi–αi)
≤.
Thus, |xi(t)| ≤[
|xi|
(αi)(t–t)
(αi–)+ Mi(–β
i) (+αi–βi)(t–t)
–(βi–αi)](t–t )–γ
i ≤(t–t)–γi for
allt≥t+T˜ andi= , , . . . ,k. This implies thatx(t) = (x(t),x(t), . . . ,xk(t))∈S for all
t≥t+T˜. Therefore, by using Theorem .Thas a fixed point inSwhich is a solution of the problem (.). Since all elements of the setStend to ast→ ∞, the zero solution
of the problem (.) is globally attractive.
Theorem . Suppose that for each i∈ {, , . . . ,k} there existαi<βi < ( +αi)and
li≥such that|gi(t,x(t),x(t), . . . ,xk(t))| ≤li(t–t)–β
i|xi(t)|for all t∈J and x, . . . ,xk∈ C((t,∞),Rn).Then the zero solution of the problem(.)is globally attractive.
Proof It is sufficient to consider the set
S=(x,x, . . . ,xk) :xi∈C
(t,∞),Rn,xi(t)≤(t–t)–γ
i
for alli= , , . . . ,kandt≥t+T˜
,
whereγi=( –αi) andT˜is chosen such that |
xi|
(αi)T˜
(αi–)
+
li(–βi–γi) (+αi–βi–γi)T˜
· · · ×C((t,∞),Rn),y= (y,y, . . . ,y
k)∈Sandx=Ax+By. Then, xi(t)≤Ai(x,x, . . . ,xk)(t)+Bi(y,y, . . . ,yk)(t)
≤ |xi|
(αi)
(t–t)αi–+ (αi)
t t
(t–s)αi–g
i
s,y(s),y(s), . . . ,yk(s)ds
≤ |xi|
(αi)
(t–t)αi–+ (αi)
t t
(t–s)αi–l
i(s–t)–β
iy
i(s)ds
≤ |xi|
(αi)
(t–t)αi–+ (αi)
t t
(t–s)αi–li(s–t)–β
i–γids
≤ |xi|
(αi)
(t–t)αi–+ li( –β
i–γi) ( +αi–βi–γi)
(t–t)–(βi–γi–αi)
for alli= , , . . . ,k. Since <αi<βi<( +αi) < fori= , , . . . ,k, we get
|x
i|
(αi)
(t–t)
(αi–)+ li( –β
i–γi) ( +αi–βi–γi)
(t–t)–(β
i–αi)
≤ |xi|
(αi)
˜ T (αi–)
+
li( –βi–γi) ( +αi–βi–γi)
˜ T–(βi–αi)
≤.
Thus,|xi(t)| ≤[
|xi|
(αi)(t–t)
(αi–)+ li(–β
i–γi) (+αi–βi–γi)(t–t)
–(βi–αi)](t–t )–γ
i ≤(t–t)–γi for
allt≥t+T˜andi= , , . . . ,k. This implies thatx(t) = (x(t),x(t), . . . ,xk(t))∈S, fort≥
t+T˜. Since all elements of the setStend to ast→ ∞, the zero solution of the problem
(.) is globally attractive.
4 Examples
Here, we give an example to illustrate our results.
Example . Consider the -dimensional system of fractional functional differential
equations
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
cDx(t) = () ()(t+ )
– sin
(x (t–))
+(x(t–)) ×
x(t–)
+|x(t–)|, t> ,
cDx
(t) =
()
()
(t+ )
– cos
(x (t–))
+sin(x(t–))+|x(t–)|, t> ,
cDx(t) =( )
√
π (t+ ) –
(x(t–))
+(x(t–))+|x(t–)|, t> , xi(t) =t, i= , , ,t∈[–, ].
Define the maps
f(t,xt,xt,xt) = ()
( )
(t+ )– sin
(x(t– ))
+ (x(t– )) ×
x(t– ) +|x(t– )|,
f(t,xt,xt,xt) = ()
()
t+
–
cos(x(t– ))
+sin(x(t– )) +|x(t– )| ,
f(t,xt,xt,xt) = ()
√
π (t+ )
–
(x(t– ))
and put m(t) = (
) ()(t+ )
–
,m(t) = ( ) ()(t+
)
–
andm(t) = ( )
√
π (t+ ) –
. It is easy
to check that|f(t,xt,xt,xt)| ≤m(t),|f(t,xt,xt,xt)| ≤m(t) and|f(t,xt,xt,xt)| ≤
m(t). Since
(α)
t t
(t–s)α–m(s)ds=
() t
(t–s)– ×(
) ()(s+ )
–ds
≤ ( ) ()() t
(t–s)–s–ds=t–,
(α)
t t
(t–s)α–m
(s)ds= (
) t
(t–s)–×(
) (
)
s+ – ds ≤ ( ) ()() t
(t–s)–s–ds=t–
and
(α)
t t
(t–s)α–m(s)ds=
() t
(t–s)– ×(
)
√
π (s+ ) –ds
≤ () ()√π
t
(t–s)–s–ds=t–,
we get|
(α)
t
(t–s) –f
(s,xs,xs,xs)ds| ≤t–
,|
(α)
t
(t–s) –f
(s,xs,xs,xs)ds| ≤t–
and|(α
)
t
(t–s)
–f(s,xs,x
s,xs)ds| ≤t–
. Now, letα=
,α=
andα= . Then,
∞
t
f(t,xt,xt,xt)
α dt≤
∞
t
m(t)
αdt
= ∞ () ( )
(t+ )–
dt = () () , ∞ t
f(t,xt,xt,xt)
α dt≤
∞
t
m(t)
α dt
= ∞ () ( )
t+ – dt = () () and ∞ t
f(t,xt,xt,xt)
α dt≤
∞
t
m(t)
α dt
= ∞ () √
Thus, all conditions of Theorem . hold and so this system of fractional functional dif-ferential equations has an attractive solution.
Example . Let <αi< ,Mi> ,αi<βi< andxi be a constant fori= , , . Consider
the -dimensional system of fractional differential equations
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
Dαx(t) =Mx(t)sin(x(t))
+|x(t)|+|x(t)|(t–a)
–β , t>a,
Dαx(t) = Mt(x(t))
(+t)(+(x
(t))+(x(t)))(t–a)
–β , t>a,
Dαx(t) = Mcos(x(t))(x(t))
+(x(t))+|x(t)| (t–a)
–β , t>a,
[Dαi–x
i(t)]t=a=xi, i= , , .
Define the maps
g
t,x(t),x(t), . . . ,xk(t)
=Mx(t)sin (x(t))
+|x(t)|+|x(t)|
(t–a)–β ,
g
t,x(t),x(t), . . . ,xk(t)
= Mt
(x(t))
( + t)( + (x
(t))+ (x(t)))
(t–a)–β
and
g
t,x(t),x(t), . . . ,xk(t)
=Mcos
(x(t))(x(t))
+ (x(t))+|x(t)|(t–a) –β .
Thus, one can check that all conditions of Theorem . hold and so this system of frac-tional 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.
Competing interests
The authors declare that they have no competing interests regarding the publication of this article.
Authors’ contributions
All authors read and approved the final version of this manuscript.
Author details
1Department of Mathematics, Cankaya University, Ogretmenler Cad. 14, Balgat, Ankara, 06530, Turkey.2Institute of Space
Sciences, Magurele, Bucharest, Romania. 3Department of Chemical and Materials Engineering, Faculty of Engineering,
King Abdulaziz University, P.O. Box 80204, Jeddah, 21589, Saudi Arabia. 4Department of Mathematics, Azarbaijan Shahid
Madani University, Azarshahr, Tabriz, Iran.
Acknowledgements
Research of the second and third authors was supported by Azarbaijan Shahid Madani University. Also, the authors express their gratitude to the referees for their helpful suggestions which improved the final version of this paper.
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