International Journal of Analysis and Applications
ISSN 2291-8639
Volume 13, Number 1 (2017), 108-118
http://www.etamaths.com

POSITIVE SOLUTIONS FOR A SINGULAR SUM FRACTIONAL

DIFFERENTIAL SYSTEM

MEHDI SHABIBI1, MIHAI POSTOLACHE2,3,∗ AND SHAHRAM REZAPOUR2,4

Abstract. We investigate the existence of positive solutions for a singular sum fractional differential

system under some boundary conditions by providing different conditions. Also, we give an example

to illustrate one of our main results.

1. Preliminaries

It has been published many works on the existence of solutions for different singular fractional
differential systems (see for example, [2], [3], [6], [7], [10] and [12]). In 2010, the existence of positive
solutions for the singular Dirichlet problem

Dαu(t) + f(t,u(t),Dµu(t)) = 0

with boundary conditions u(0) = u(1) = 0 is investigated, where 1 < α < 2, 0 < µ ≤ α − 1, f is a
Caratheodory function on [0, 1] × (0,∞) ×R and Dα is Riemann-Liouville fractional derivative ( [1]).
In 2013, the singular problem

Dαu + f(t,u,Dγu,Dµu) + g(t,u,Dγu,Dµu) = 0

with boundary conditions u(0) = u′(0) = u′′(0) = u′′′(0) = 0 is reviewed, where 3 < α < 4, 0 < γ < 1,
1 < µ < 2, Dα is the Caputo fractional derivative and f is a Caratheodory function on [0, 1]×(0,∞)3
( [4]).

By using main idea of [1] and [4], we investigate positive solutions for the singular differential system
of equations


Dα1u1 + f1(t, u1, . . . , um, D

µ1u1, . . . , D
µmum) + g1(t, u1, . . . , um, D

µ1u1, . . . , D
µmum) = 0,

. . . . . .
Dαmum + fm(t, u1, . . . , um, D

µ1u1, . . . , D
µmum) + gm(t, u1, . . . , um, D

µ1u1, . . . , D
µmum) = 0,

(1.1)

with boundary conditions ui(0) = 0, u
′
i(1) = 0 and

dk

dtk
[ui(t)]t=0 = 0 for 1 ≤ i ≤ m and 2 ≤ k ≤ n−1,

where αi ≥ 2, [αi] = n − 1, 0 < µi < 1, D is the Caputo fractional derivative, fi is a Caratheodory
function, gi satisfies Lipschitz condition and fi(t,x1, . . . ,x2m) is singular at t = 0 of for all 1 ≤ i ≤ m.

We say that a map f : [0, 1] × D ⊆ [0, 1] × D → Rn is Caratheodory whenever the function
t 7→ f(t,x1, . . . ,xn) is measurable for all (x1, . . . ,xn) ∈ D and (x1, . . . ,xn) 7→ f(t,x1, . . . ,xn) is
continuous for almost all t ∈ [0, 1] and for each compact K ⊆ D there exists ϕK ∈ L1[0, 1] such that
|f(t,x1, . . . ,xn)| ≤ ϕK(t) for almost all t ∈ [0, 1] and (x1, . . . ,xn) ∈ K.

Put

‖x‖1 =
∫ 1

0

|x(t|dt, ‖x‖ = sup{|x(t)| : t ∈ [0, 1]}, ‖(x1, . . . ,xn)‖∗ = max{‖x1‖, . . . ,‖xn‖},

‖(x1, . . . ,xn)‖∗∗ = max{‖x1‖, . . . ,‖xn‖,‖x′1‖, . . . ,‖x
′
n‖}, Y = CR([0, 1]), X = C

1
R([0, 1]).

By considering the problem (1.1), we assume the following hypotheses:
H1: f1, . . . ,fm are Caratheodory functions on [0, 1] × (0,∞)2m and there exists positive constants

m1, . . . ,mm such that
fi(t,x1, . . . ,x2m) ≥ mi

Received 24th July, 2016; accepted 17th September, 2016; published 3rd January, 2017.

2010 Mathematics Subject Classification. 34A08; 34B16.

Key words and phrases. Caputo derivative; fixed point; singular system; positive solution.

c©2017 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

108



A SINGULAR SUM FRACTIONAL DIFFERENTIAL SYSTEM 109

for almost all t ∈ [0, 1] and all (x1, . . . ,x2m) ∈ D := (0,∞)2m.
H2: g1, . . . ,gm are nonnegative and

|gi(t,x1, . . . ,x2m) −gi(t,y1, . . . ,y2m)| ≤
2m∑
k=1

Lik|xk −yk|,

for almost all t ∈ [0, 1] and for all (x1, . . . ,x2m), (y1, . . . ,y2m) ∈ D, where L11, . . . ,Lm1 , . . . ,L12m, . . . ,Lm2m
in [0,∞) are constants such that

1

Γ(αi − 1)

( m∑
k=1

Lik +

m∑
k=1

Lim+k
Γ(2 −µi)

)
< 1.

H3: there exist some maps γ1, . . . ,γm ∈ L1([0, 1]), some non-increasing maps p1, . . . ,pm ∈ CR(D)
with ∫ 1

0

pi

(
M1t

α1, . . . ,Mmt
αm,

M1(1 −µ1)
2

t1−µ1, . . . ,
Mm(1 −µm)

2
t1−µm

)
dt < ∞

and some functions h1, . . . ,hm ∈ CR([0,∞)2m) such that limx→∞
hi(x,...,x)

x
= 0, hi is nondecreasing in

all components and

fi(t,x1, . . . ,x2m) + gi(t,x1, . . . ,x2m) ≤ pi(x1, . . . ,x2m) + γi(t)hi(x1, . . . ,x2m),

for almost all t ∈ [0, 1] and all (x1, . . . ,x2m) ∈ D, where Mi = mi αi−1Γ(αi+1) for all 1 ≤ i ≤ m.

Now for each 1 ≤ i ≤ m and n ≥ 1, put fi,n(t,x1, . . . ,x2m) = fi(t,χ1(x1), . . . ,χn(x2m)), where
χn(u) = u whenever u ≥ 1n and χn(u) = u whenever u <

1
n

.
It is easy to check that

fi,n(t,x1, . . . ,x2m) + gi(t,x1, . . . ,x2m) ≤ pi
( 1
n
,. . . ,

1

n

)
+ γi(t)hi

(
x1 +

1

n
,. . . ,x2m +

1

n

)
,

fi,n(t,x1, . . . ,x2m) ≥ mi
and

fi,n(t,x1, . . . ,x2m) + gi(t,x1, . . . ,x2m) ≤ pi(x1, . . . ,x2m) + γi(t)hi
(
x1 +

1

n
,. . . ,x2m +

1

n

)
,

for all (x1, . . . ,xn) ∈ D, 1 ≤ i ≤ m and almost all t ∈ [0, 1].
First, we investigate the regular fractional differential system


Dα1u1 + f1,n(t,u1, . . . ,um,D

µ1u1, . . . ,D
µmum) = 0

. . . . . .
Dαmum + fm,n(t,u1, . . . ,um,D

µ1u1, . . . ,D
µmum) = 0,

(1.2)

with same boundary conditions in (1.1).
Now, we present some necessary notions.
According to [5], the Riemann-Liouville integral of order p for a function f : (0,∞) → R is

Ipf(t) =
1

Γ(p)

∫ t
0

(t−s)p−1f(s)ds

if the right-hand side map is defined pointwise on (0,∞) .
The Caputo fractional derivative of order α > 0 for a function f : (a,∞) → R is defined by

cDαf(t) =
1

Γ(n−α)

∫ t
0

fn(s)

(t−s)α+1−n
ds,

where n = [α] + 1; please, see [5].

Lemma 1.1 ( [8]). If x ∈ CR[0, 1]∩L1[0, 1], then IαDαx(t) = x(t) +
∑n−1
i=0 cit

i for some real constants
c0,c1, . . . ,cn−1, where 0 < n− 1 ≤ α < n.



110 SHABIBI, POSTOLACHE AND REZAPOUR

It has been proved in [11] that∫ t
0

(t−s)α−1sβds = B(β + 1,α)tα+β, B(β,α) =
Γ(α)Γ(β)

Γ(α + β)
, β > 0, α > −1.

We need the next result.

Lemma 1.2 ( [9]). Let M be a closed, convex and nonempty subset of a Banach space X, A a compact
and continuous operator and B a contraction. Then, there exist z ∈ M such that z = Az + Bz.

Lemma 1.3. Let y ∈ L1[0, 1], α ≥ 2 and n = [α] + 1. Then, the unique solution of the equation
Dαu(t) + y(t) = 0 with boundary conditions u′(1) = u(0) = u′′(0) = · · · = un−1(0) = 0 is u(t) =∫ 1

0
Gα(t,s)y(s)ds, where t ∈ [0, 1] and

Gα(t,s) =




t(1 −s)α−2

Γ(α− 1)
, 0 ≤ t ≤ s ≤ 1

t(1 −s)α−2

Γ(α− 1)
−

(t−s)α−1

Γ(α)
, 0 ≤ s ≤ t ≤ 1.

Proof. By Using Lemma 1.1 and the boundary conditions, we get

u(t) = −
1

Γ(α)

∫ t
0

(t−s)α−1y(s)ds + c1t

and so u′(1) = − 1
Γ(α−1)

∫ 1
0

(1 − s)α−2y(s)ds + c1. Since u′(1) = 0, c1 = 1Γ(α−1)
∫ 1

0
(1 − s)α−2y(s)ds.

Thus, u(t) = − 1
Γ(α)

∫ t
0
(t−s)α−1y(s)ds + t

Γ(α−1)
∫ 1

0
(1 −s)α−2y(s)ds. Hence, we conclude that u(t) =∫ 1

0
Gα(t,s)y(s)ds, where Gα(t,s) =

t(1−s)α−2
Γ(α−1) −

(t−s)α−1
Γ(α)

whenever 0 ≤ s ≤ t ≤ 1 and Gα(t,s) =
t(1−s)α−2

Γ(α−1) whenever 0 ≤ t ≤ s ≤ 1. �

Consider the Green function Gα(t,s) as in Lemma 1.3.
If 0 < t ≤ s < 1, then it is clear that Gα(t,s) > 0.
If 0 < s < t < 1, then

t(1−s)α−2
Γ(α−1) −

(t−s)α−1
Γ(α)

> 0 if and only if α− 1 > (t−s)
α−1

t(1−s)α−2 and so Gα(t,s) > 0.

One can check that Gα(t,s) > 0, Gα(t,s) ≤ 1Γ(α−1) and
∫ 1

0
Gα(t,s)ds ≤ 1Γ(α) , for all t,s ∈ (0, 1).

Also,
∫ 1

0
Gα(t,s)ds ≥ tΓ(α) −

tα

Γ(α+1)
≥ t

α(α−1)
Γ(α+1)

and ∂
∂t
Gα(t,s) > 0 for all t,s ∈ (0, 1). Moreover,

Gα,
∂
∂t
Gα ∈ CR([0, 1] × [0, 1]), ∂∂tGα(t,s) ≤

1
Γ(α−1) , for all t,s ∈ [0, 1] and

∫ 1
0

∂
∂t
Gα(t,s) ≥ 1−t

α−1

Γ(α)
, for

all t ∈ [0, 1].
Suppose that x ∈ C1R[0, 1] and 0 ≤ µ ≤ 1. Since D

µx(t) = 1
Γ(2−µ)

∫ t
0
(t−s)−µx′(s)ds for all 0 ≤ t ≤ 1,

|Dµx| ≤ ‖
x′‖

Γ(1−µ)
∫ t

0
(t−s)−µds = ‖

x′‖
Γ(2−µ)t

1−µ and so |Dµx| ≤ ‖
x′‖

Γ(2−µ) and D
µx ∈ CR[0, 1].

Now, put

P = {(x1, . . . ,xm) ∈ Xm : xi(t) ≥ 0 and x′i(t) ≥ 0 for all t ∈ [0, 1] and 0 ≤ i ≤ m}.
For each n ≥ 1 and 0 ≤ i ≤ m, define the maps

Φi,n(x1, . . . ,xm)(t) =

∫ 1
0

Gαi(t,s)fi,n(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))ds,

and

Ψi(x1, . . . ,xm)(t) =

∫ 1
0

Gαi(t,s)gi(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))ds,

Tn(x1, . . . ,xm)(t) =


Φ1,n(x1, . . . ,xm)(t)...

Φm,n(x1, . . . ,xm)(t)




and Ψ(x1, . . . ,xm)(t) =


Ψ1(x1, . . . ,xm)(t)...

Ψm(x1, . . . ,xm)(t)


 for all (x1, . . . ,xm) ∈ P .



A SINGULAR SUM FRACTIONAL DIFFERENTIAL SYSTEM 111

Lemma 1.4. The map Ψ : P → P is a contraction.

Proof. It is easy to check that Ψi(x1, . . . ,xm)(t) ≥ 0 and

Ψ′i(x1, . . . ,xm)(t) =

∫ 1
0

∂

∂t
Gαi(t,s)gi(s,x1(s), . . . ,xm(s),D

µ1x1(s), . . . ,D
µmxm(s))ds ≥ 0

for all t ∈ [0, 1], (x1, . . . ,xm) ∈ P and 0 ≤ i ≤ m.
Note that,

‖Ψi(x1, . . . ,xm) − Ψi(y1, . . . ,ym)‖

= sup
t∈[0,1]

∣∣∣∫ 1
0

Gαi(t,s)[gi(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))

−gi(s,y1(s), . . . ,ym(s),Dµ1y1(s), . . . ,Dµmym(s))]ds
∣∣∣

≤
∣∣∣∫ 1

0

Gαi(t,s)ds
∣∣∣(Li1‖x1 −y1‖ + · · · + Lim‖xm −ym‖

+Lim+1‖D
µ1x1 −Dµ1y1‖ + · · · + Li2m‖D

µmxm −Dµmym‖)

≤
1

Γ(αi)
(Li1‖x1 −y1‖ + · · · + L

i
m‖xm −ym‖

+
Lim+1

Γ(2 −µ1)
‖x′1 −y

′
1‖ + · · · +

Li2m
Γ(2 −µm)

‖x′m −y
′
m‖)

≤
1

Γ(αi)

( m∑
k=1

Lik +

m∑
k=1

Lim+k
Γ(2 −µi)

)
‖(x1, . . . ,xm) − (y1, . . . ,ym)‖∗∗

≤
1

Γ(αi − 1)

(
m∑
k=1

Lik +

m∑
k=1

Lim+k
Γ(2 −µi)

)
‖(x1, . . . ,xm) − (y1, . . . ,ym)‖∗∗

for all 0 ≤ i ≤ m and so

‖Ψ(x1, . . . ,xm) − Ψ(y1, . . . ,ym)‖∗ = max
1≤i≤m

‖Ψi(x1, . . . ,xm) − Ψi(y1, . . . ,ym)‖

≤ max
1≤i≤m

{
1

Γ(αi − 1)

(
m∑
k=1

Lik +

m∑
k=1

Lim+k
Γ(2 −µi)

)}
‖(x1, . . . ,xm) − (y1, . . . ,ym)‖∗∗ .

Similarly, one can show that

‖Ψ′(x1, . . . ,xm) − Ψ′(y1, . . . ,ym) ‖∗

≤ max
1≤i≤m

{
1

Γ(αi − 1)

(
m∑
k=1

Lik +

m∑
k=1

Lim+k
Γ(2 −µi)

)}
‖(x1, . . . ,xm) − (y1, . . . ,ym) ‖∗∗.

Thus, we get

‖Ψ(x1, . . . ,xm) − Ψ(y1, . . . ,ym) ‖∗∗

≤ max
1≤i≤m

{
1

Γ(αi − 1)

(
m∑
k=1

Lik +

m∑
k=1

Lim+k
Γ(2 −µi)

)}
‖(x1, . . . ,xm) − (y1, . . . ,ym)‖∗∗ .

Since max1≤i≤m

{
1

Γ(αi−1)

(∑m
k=1 L

i
k +

∑m
k=1

Lim+k
Γ(2−µi)

)}
< 1, Ψ is a contraction mapping. �

Lemma 1.5. For each n ≥ 1, Tn is a completely continuous operator on P.



112 SHABIBI, POSTOLACHE AND REZAPOUR

Proof. Let n ≥ 1, (x1, . . . ,xm) ∈ P and 1 ≤ i ≤ m. Choose a positive constant mi such that

fi,n(t,x1(t), . . . ,xm(t),D
µ1x1(t), . . . ,D

µmxm(t)) ≥ mi
for almost all t ∈ [0, 1].

Since Gαi and
∂
∂t
Gαi are nonnegative and continuous on [0, 1]× [0, 1] for all 1 ≤ i ≤ m, we conclude

that Φi,n(x1, . . . ,xm)(t) ≥ 0 and (Φi,n(x1, . . . ,xm))
′
(t) ≥ 0 for all t ∈ [0, 1] and 1 ≤ i ≤ m. Hence, Tn

maps P into P .
Now, we prove that Tn is continuous.
Let {(x1,k, . . . ,xm,k)} be a convergent sequence in P with limk→∞(x1,k, . . . ,xm,k) = (x1, . . . ,xm).

In this case, we get limk→∞xi,k = xi and limk→∞x
′
i,k = x

′
i uniformly on [0, 1] (i = 1, 2, . . . ,m).

But, |Dµixi,k(t) − Dµixi(t)| ≤
‖x′i,k−x′i‖
Γ(2−µi)

for all t ∈ [0, 1] and 1 ≤ i ≤ m. Thus, we conclude that
limk→∞D

µixi,k(t) = D
µixi(t) uniformly on [0, 1]. Hence,

lim
k→∞

fi,n(t,x1,k(t), . . . ,xm,k(t),D
µ1x1,k(t), . . . ,D

µmxm,k(t))

= fi,n(t,x1(t), . . . ,xm(t),D
µ1x1(t), . . . ,D

µmxm(t)).

Since fi,n ∈ Car([0, 1] × R2m), {(x1,k, . . . ,xm,k)} is bounded in Xm there exist a map ϕi ∈ L1[0, 1]
such that

mi ≤ fi,n(t,x1,k(t), . . . ,xm,k(t),Dµ1x1,k(t), . . . ,Dµmxm,k(t)) ≤ ϕi(t) (1.3)
for almost all t ∈ [0, 1], 1 ≤ i ≤ m and k ≥ 1.

By using the Lebesgue dominated convergence theorem, we conclude that

|Φi,n(x1,k, . . . ,xm,k)(t) − Φi,n(x1, . . . ,xm)(t)|

≤
1

Γ(αi)

∫ 1
0

|fi,n(s,x1,k(s), . . . ,xm,k(s),Dµ1x1,k(s), . . . ,Dµmxm,k(s))

−fi,n(s,x1(s), . . . ,xm(s),Dµ1x1(s), . . . ,Dµmxm(s))|ds,
and

|(Φi,n(x1,k, . . . ,xm,k) )′(t) − (Φi,n(x1, . . . ,xm) )′(t) |

≤
1

Γ(αi − 1)

∫ 1
0

|fi,n(s,x1,k(s), . . . ,xm,k(s),Dµ1x1,k(s), . . . ,Dµmxm,k(s))

−fi,n(s,x1(s), . . . ,xm(s),Dµ1x1(s), . . . ,Dµmxm(s))|ds.

Hence, limk→∞

∣∣∣(Φi,n(x1,k, . . . ,xm,k))j (t) − (Φi,n(x1, . . . ,xm))j (t)∣∣∣ = 0 uniformly on [0, 1] for j = 0, 1.
Thus, ‖Tn(x1,k, . . . ,xm,k)(t) −Tn(x1, . . . ,xm)(t)‖∗∗ → 0 and so Tn is continuous.

Now, we prove that Tn maps bounded sets to relatively compact subsets.
Let {(x1,k, . . . ,xm,k)} be a bounded sequence in P. Choose a positive number S such that ‖xi,k‖≤ S

and
∥∥∥x′i,k∥∥∥ ≤ S for all 1 ≤ i ≤ m and for k ≥ 1. Since ‖Dµixi,k‖ ≤ 1Γ(2−µi) for all 1 ≤ i ≤ m, there

exist a map ϕi ∈ L1[0, 1] such that (1.3) holds for almost all t ∈ [0, 1], 1 ≤ i ≤ m and k ≥ 1.
Note that

0 ≤ Φi,n(x1,k, . . . ,xm,k)(t)

=

∫ 1
0

Gαi(t,s)fi,n(s,x1,k(s), . . . ,xm,k(s),D
µ1x1,k(s), . . . ,D

µmxm,k(s))ds

≤
1

Γ(αi)

∫ 1
0

ϕi(s)ds =
‖ϕi‖1
Γ(αi)

and

0 ≤ (Φi,n(x1,k, . . . ,xm,k))′(t)

=

∫ 1
0

∂

∂t
Gαi(t,s)fi,n(s,x1,k(s), . . . ,xm,k(s),D

µ1x1,k(s), . . . ,D
µmxm,k(s))ds

≤
1

Γ(αi − 1)

∫ 1
0

ϕi(s)ds =
‖ϕi‖1

Γ(αi − 1)



A SINGULAR SUM FRACTIONAL DIFFERENTIAL SYSTEM 113

for all 1 ≤ i ≤ m. Thus, ‖Tn(x1,k, . . . ,xm,k)(t)‖∗∗ ≤ B, where B = max1≤i≤m
‖ϕi‖1

Γ(αi−1)
. This implies

that {Tn(x1,k, . . . ,xm,k)} is bounded in Xm. Let 0 ≤ t1 ≤ t2 ≤ 1 and 1 ≤ i ≤ m. Then, we have∣∣(Φi,n(x1,k, . . . ,xm,k))′ (t2) − (Φi,n(x1,k, . . . ,xm,k))′ (t1)∣∣
≤

t2 − t1
Γ(αi − 1)

∫ 1
0

(1 −s)αi−2fi,n(s,x1,k(s), . . . ,xm,k(s),Dµ1x1,k(s), . . . ,Dµmxm,k(s))ds

+
1

Γ(αi)

∣∣∣∫ t2
0

(t2 −s)αi−1fi,n(s,x1,k(s), . . . ,xm,k(s),Dµ1x1,k(s), . . . ,Dµmxm,k(s))ds

−
∫ t1

0

(t1 −s)αi−1fi,n(s,x1,k(s), . . . ,xm,k(s),Dµ1x1,k(s), . . . ,Dµmxm,k(s))ds
∣∣∣

‖fi,n‖1
Γ(αi − 1)

(t2 − t1) +
1

Γ(αi)
[

∫ t1
0

((t2 −s)αi−1 − (t1 −s)αi−1)×

fi,n(s,x1,k(s), . . . ,xm,k(s),D
µ1x1,k(s), . . . ,D

µmxm,k(s))ds

+

∫ t2
t1

(t2 −s)αi−2fi,n(s,x1,k(s), . . . ,xm,k(s),Dµ1x1,k(s), . . . ,Dµmxm,k(s))ds]

≤
‖ϕi‖1

Γ(αi − 1)
(t2 − t1) +

1

Γ(αi)

[∫ t1
0

(
(t2 −s)αi−1 − (t1 −s)αi−1

)
ϕi(s)ds + (t2 − t1)αi−1 ‖ϕi‖1

]
.

Let ε > 0 be given. Since the function |t − s|αi−1 is uniformly continuous on [0, 1] × [0, 1], there
exist δ > 0 such that (t2 − s)αi−1 − (t1 − s)αi−1 < ε for all 0 ≤ t1 ≤ t2 ≤ 1 with t2 − t1 < δ and
0 ≤ s ≤ t1. If 0 ≤ t1 ≤ t2 ≤ 1 with t2 − t1 < min{δ,ε}, then we have∣∣(Φi,n(x1,k, . . . ,xm,k))′ (t2) − (Φi,n(x1,k, . . . ,xm,k))′ (t1)∣∣ < 3ε‖ϕi‖1

Γ(αi)
.

Thus,

‖T ′n(x1,k, . . . ,xm,k)(t2) −T
′
n(x1,k, . . . ,xm,k)(t1)‖∗ < max1≤i≤m

3ε‖ϕi‖1
Γ(αi)

.

This implies that {T ′n(x1,k, . . . ,xm,k)} is equi-continuous on [0, 1]. Now by using the Arzela-Ascoli
theorem, {Tn(x1,k, . . . ,xm,k)} is relatively compact and so Tn is completely continuous. �

2. Main Results

Now, we are ready to provide our main results about the problem (1.1).

Theorem 2.1. Assume that hypotheses H1 and H2 hold. Then, the problem (1.2) with the boundary

conditions in (1.1) has a solution (x1,n, . . . ,xm,n) in P such that xi,n(t) ≥
mit

αi(αi−1)
Γ(αi+1)

, for all t ∈ [0, 1]
and 1 ≤ i ≤ m.

Proof. By using Lemma 1.4, the mapping Ψ : P → P is a contraction. Also by using Lemma 1.5,
the operator Tn : P → P is a completely continuous one. Now by using Lemma 1.2, there exists
(x1,n, . . . ,xm,n) ∈ P such that (x1,n, . . . ,xm,n) = Tn(x1,n, . . . ,xm,n) + Ψ(x1,n, . . . ,xm,n). Thus, xi,n =
Φi,n(x1,n, . . . ,xm,n) + Ψi(x1,n, . . . ,xm,n) for all 1 ≤ i ≤ m. Hence,

xi,n(t) =

∫ 1
0

Gαi(t,s)fi,n(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))ds

+

∫ 1
0

Gαi(t,s)gi(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))ds

for all 1 ≤ i ≤ m. By using the assumptions, we get xi,n(t) ≥
mit

αi(αi−1)
Γ(αi+1)

for all t ∈ [0, 1] and
1 ≤ i ≤ m. One can check that the element (x1,n, . . . ,xm,n) ∈ P is a solution for the problem (1.2)
with the boundary conditions in (1.1). �

Lemma 2.1. Assume that hypotheses H1, H2 and H3 hold. If (x1,n, . . . ,xm,n) is a solution for the
problem (1.2) with the boundary conditions in (1.1), then {(x1,n, . . . ,xm,n)}n≥1 is relatively compact
in P.



114 SHABIBI, POSTOLACHE AND REZAPOUR

Proof. As we found in the last result,

xi,n(t) =

∫ 1
0

Gαi(t,s)fi,n(s,x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))ds

+

∫ 1
0

Gαi(t,s)gi(s,x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))ds

for all n ≥ 1, t ∈ [0, 1] and 1 ≤ i ≤ m. Thus,

x′i,n(t) ≥ mi
∫ 1

0

∂

∂t
Gαi(t,s)ds ≥

mi(1 − tαi−1)
Γ(αi)

,

for all t ∈ [0, 1]. Hence,

Dµixi,n(t) =
1

Γ(1 −µi)

∫ t
0

(t−s)−µix′i,n(s)ds

≥
mi

Γ(αi)Γ(1 −µi)

∫ t
0

(t−s)−µi(1 −sαi−1)ds >
mi

Γ(αi)Γ(1 −µi)

∫ t
0

(t−s)−µi(1 −s)ds

for all t ∈ [0, 1]. Thus,

Dµixi,n(t) >
mit

1−µi

Γ(αi)Γ(2 −µi)
−

mit
2−µi

Γ(αi)Γ(3 −µi)

=
mit

1−µi

Γ(αi)

(
Γ(3 −µi) − tΓ(2 −µi)

Γ(2 −µi)Γ(3 −µi)

)
=
mit

1−µi

Γ(αi)

(
2 −µi − t
Γ(3 −µi)

)
≥
mit

1−µi(1 −µi)
Γ(αi)Γ(3 −µi)

for all t ∈ [0, 1]. Since Γ(3 −µi) ≤ 2, we get Dµixi,n(t) ≥
mit

1−µi(1−µi)
2Γ(αi)

.

Now, put

Mi = mi min
{ 1

Γ(αi)
,
αi − 1

Γ(αi + 1)

}
.

Then, xi,n(t) ≥ Mitαi and Dµixi,n(t) ≥
Mi(1−µi)

2
t1−µi for all n ≥ 1, t ∈ [0, 1] and 1 ≤ i ≤ m. Hence,

pi(x1,n(t), . . . ,xm,n(t),D
µ1x1,n(t), . . . ,D

µmxm,n(t))

≤ pi
(
M1t

α1, . . . ,Mmt
αm,

M1(1 −µ1)
2

t1−µ1 . . . ,
Mm(1 −µm)

2
t1−µm

)
for all n ≥ 1, t ∈ [0, 1] and 1 ≤ i ≤ m. This implies that

0 ≤ x′i,n(t) =
∫ 1

0

∂

∂t
Gαi(t,s)fi,n(s,x1,n(s), . . . ,xm,n(s),D

µ1x1,n(s), . . . ,D
µmxm,n(s))ds

+

∫ 1
0

∂

∂t
Gαi(t,s)gi(s,x1,n(s), . . . ,xm,n(s),D

µ1x1,n(s), . . . ,D
µmxm,n(s))ds

≤
1

Γ(αi − 1)

∫ 1
0

pi

(
M1s

α1, . . . ,Mms
αm,

M1(1 −µ1)
2

s1−µ1 . . . ,
Mm(1 −µm)

2
s1−µm

)
ds

+
1

Γ(αi − 1)

∫ 1
0

γi(s)hi(x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))ds

for all n ≥ 1, t ∈ [0, 1] and 1 ≤ i ≤ m.
Also, we have∫ 1

0

pi

(
M1s

α1, . . . ,Mms
αm,

M1(1 −µ1)
2

s1−µ1, . . . ,
Mm(1 −µm)

2
s1−µm

)
ds := Λi < ∞

for all 1 ≤ i ≤ m. If ηn = ‖(x1,n, . . . ,xm,n)‖∗∗, then ‖xi,n‖ ≤ ηn and
∥∥x′i,n∥∥ ≤ ηn for all i and n.

Thus, |Dµixi,n(t)| ≤ ηnΓ(2−µi) for all n ≥ 1, t ∈ [0, 1] and 1 ≤ i ≤ m and so

0 ≤ x′i,n(t)

≤
1

Γ(αi − 1)

(
Λi + hi(1 + ηn, . . . , 1 + ηn, 1 +

ηn
Γ(2 −µ1)

, . . . , 1 +
ηn

Γ(2 −µm)
)

)
‖γi‖1

and so 0 ≤ xi,n(t) =
∫ t

0
x′i,n(s)ds for all n ≥ 1, t ∈ [0, 1] and 1 ≤ i ≤ m.



A SINGULAR SUM FRACTIONAL DIFFERENTIAL SYSTEM 115

Similarly, we obtain

0 ≤ xi,n(t)

≤
1

Γ(αi − 1)

(
Λi + hi(1 + ηn, . . . , 1 + ηn, 1 +

ηn
Γ(2 −µ1)

, . . . , 1 +
ηn

Γ(2 −µm)
)

)
‖γi‖1

and ηn ≤ 1Γ(αi−1)
(

Λi + hi(1 + ηn, . . . , 1 + ηn, 1 +
ηn

Γ(2−µ1)
, . . . , 1 + ηn

Γ(2−µm)
)
)
‖γi‖1 for all i.

Since limx→∞
hi(x,...,x)

x
= 0 for all 1 ≤ i ≤ m, there exists Li > 0 such that

1

Γ(αi − 1)

(
Λi + hi(1 + νi, . . . , 1 + νi, 1 +

νi
Γ(2 −µ1)

, . . . , 1 +
νi

Γ(2 −µm)
)

)
‖γi‖1 < νi

for all νi > Li. If L = max{L1, . . . ,Lm}, then
1

Γ(αi − 1)

(
Λi + hi(1 + ν, . . . , 1 + ν, 1 +

ν

Γ(2 −µ1)
, . . . , 1 +

ν

Γ(2 −µm)
)

)
‖γi‖1 < ν

for all ν > L. Thus, ηn = ‖(x1,n, . . . ,xm,n)‖∗∗ = max1≤i≤m{‖xi,n‖ ,
∥∥x′i,n∥∥} < L which implies

{‖(x1,n, . . . ,xm,n)‖∗∗} is bounded in Xm.
Now, put

Bi := hi

(
1 + L,. . . , 1 + L, 1 +

L

Γ(2 −µ1)
, . . . , 1 +

L

Γ(2 −µm)

)
and

Fi(t) := pi

(
M1t

α1, . . . ,Mmt
αm,

M1(1 −µ1)
2

t1−µ1 . . . ,
Mm(1 −µm)

2
t1−µm

)
,

for all i and almost all t ∈ [0, 1].
Then, we have Λi =

∫ 1
0
Fi(t)dt and

fi,n(t,x1,n(t), . . . ,xm,n(t),D
µ1x1,n(t), . . . ,D

µmxm,n(t))

+gi(t,x1,n(t), . . . ,xm,n(t),D
µ1,nx1(t), . . . ,D

µmxm,n(t))

≤ Fi(t) + Biγi(t).
If 0 ≤ t1 ≤ t2 ≤ 1, then

|x′i,n(t2) −x
′
i,n(t1)| =

∣∣∣∫ 1
0

(
∂

∂t
Gαi(t2,s) −

∂

∂t
Gαi(t2,s))×

[fi,n(s,x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))

+gi(s,x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))]ds
∣∣∣

≤
1

Γ(αi − 1)
[(t2 − t1)

∫ 1
0

Fi(s) + Biγi(s)ds +

∫ t1
0

((t2 −s)αi−2 − (t1 −s)αi−2)(Fi(s) + Biγi(s) )ds

+

∫ t2
t1

(t2 −s)αi−2 (Fi(s) + Biγi(s)) ds]

≤
1

Γ(αi − 1)
[(t2 − t1)(Λi + Bi‖γi‖1) +

∫ t1
0

((t2 −s)αi−2 − (t1 −s)αi−2)(Fi(s) + Biγi(s))ds

+(t2 − t1)αi−2(Λi + Bi‖γi‖1)].
Let �i > 0 be given. Choose δ(�i) > 0 such that (t2−s)αi−2−(t1−s)αi−2 < �i for all 0 ≤ t1 < t2 ≤ 1

with t2 − t1 < δ(�i) and 0 ≤ s ≤ t. If we put
0 < δ < min{δ(�1), . . . ,δ(�m), α1−2

√
�1, . . . ,

αm−2
√
�m},

then |x′i,n(t2) − x
′
i,n(t1)| ≤

3 �i
Γ(αi−1)

(Λi + Bi‖γi‖1) for all 1 ≤ i ≤ m. Hence, {(x1,n, . . . ,xm,n)
′} is

equi-continuous and so {(x1,n, . . . ,xm,n)}n≥1 is relatively compact in Xm. �

Theorem 2.2. Assume that hypotheses H1, H2 and H3 hold. Then the system (1.1) has a solution

(x1, . . . ,xm) in P such that D
µixi(t) ≥

Mi(1−µi)
2

t1−µi and xi(t) ≥ Mitαi for all t ∈ [0, 1] and 1 ≤ i ≤
m.



116 SHABIBI, POSTOLACHE AND REZAPOUR

Proof. By Theorem 2.1, for each n ≥ 1 the system (1.2) with the boundary conditions in (1.1) has
a solution (x1,n, . . . ,xm,n) ∈ P . By Lemma 2.1, {(x1,n, . . . ,xm,n)}n≥1 is relatively compact in Xm.
By using the Arzela-Ascoli theorem, there exists (x1, . . . ,xm) such that limn→∞(x1,n, . . . ,xm,n) =
(x1, . . . ,xm). It is obvious that (x1, . . . ,xm) satisfies the boundary conditions of the problem (1.1),
Dµixi,n → Dµixi and

lim
n→∞

fi,n(t,x1,n(t), . . . ,xm,n(t),D
µ1x1,n(t), . . . ,D

µmxm,n(t))

+gi(t,x1,n(t), . . . ,xm,n(t),D
µ1x1,n(t), . . . ,D

µmxm,n(t))

= fi(t,x1(t), . . . ,xm(t),D
µ1x1(t), . . . ,D

µmxm(t))

+gi(t,x1(t), . . . ,xm(t),D
µ1x1(t), . . . ,D

µmxm(t))

for almost all t ∈ [0, 1] and 1 ≤ i ≤ m and so (x1, . . . ,xm) ∈ P . Now, suppose that K :=
supn≥1 ‖(x1,n, . . . ,xm,n)‖∗∗. Then, we have ‖D

µixi,n‖≤ KΓ(2−µi) for all n and 1 ≤ i ≤ m. Hence,

0 ≤ Gαi(t,s)[fi,n(s,x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))

+gi(s,x1,n(s), . . . ,xm,n(s),D
µ1x1,n(s), . . . ,D

µmxm,n(s))]

≤
1

Γ(αi − 1)

(
Fi(s) + hi(1 + K,.. . , 1 + K, 1 +

K

Γ(2 −µi)
, . . . , 1 +

K

Γ(2 −µi)
)γi(s)

)
for almost all (t,s) ∈ [0, 1] × [0, 1], n ≥ 1 and 1 ≤ i ≤ m. Now by using the Lebesgue dominated
theorem, we conclude that

xi(t) =

∫ 1
0

Gαi(t,s)fi(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))ds

+

∫ 1
0

Gαi(t,s)gi(s,x1(s), . . . ,xm(s),D
µ1x1(s), . . . ,D

µmxm(s))ds

for all 1 ≤ i ≤ m and t ∈ [0, 1], and this completes the proof. �

Next example illustrates our last result.

Example 2.1. Let us study the system


D
5
2 x1 +

1

t
2
3

(2 + a1x1 + a2x2 + a3D
1
3 x1 + a4D

1
2 x2)

+(0.1e
1

1+x1 + 0.2e
1

1+x2 + 0.1e
1

1+D13x1 + 0.2e
1

1+D12x2 ) = 0

D
7
3 x2 +

1

t
1
2

(1 + b1x1 + b2x2 + b3D
1
3 x1 + b4D

1
2 x2)

+(0.2e
1

1+x1 + 0.2e
1

1+x2 + 0.3e
1

1+D13x1 + 0.1e
1

1+D12x2 ) = 0

with boundary condition x1(0) = x2(0) = 0, x
′
1(1) = x

′
2(1) = 0 and x

′′
1 (0) = x

′′
2 (0) = 0, where

a1,a2,a3,a4,b1,b2,b3 and b4 are positive constants.
Consider the functions

f1(t,x1,x2,x3,x4) =
1

t
2
3

(2 + a1x1 + a2x2 + a3x3 + a4x4),

f2(t,x1,x2,x3,x4) =
1

t
1
2

(1 + b1x1 + b2x2 + b3x3 + b4x4),

g1(t,x1,x2,x3,x4) = p1(x1,x2,x3,x4) = 0.1e
1

1+x1 + 0.2e
1

1+x2 + 0.1e
1

1+x3 + 0.2e
1

1+x4 ,

g2(t,x1,x2,x3,x4) = p2(x1,x2,x3,x4) = 0.2e
1

1+x1 + 0.2e
1

1+x2 + 0.3e
1

1+x3 + 0.1e
1

1+x4 ,

h1(x1,x2,x3,x4) = 2 + a1x1 + a2x2 + a3x3 + a4x4,

h2(x1,x2,x3,x4) = 1 + b1x1 + b2x2 + b3x3 + b4x4,

λ1(t) =
1

t
2
3

and λ2(t) =
1

t
1
2

. Put m = 2, α1 =
5
2
, α2 =

7
3
, µ1 =

1
2
, µ2 =

1
3
, L11 = 0.1, L

1
2 = 0.2,

L13 = 0.1, L
1
4 = 0.2, L

2
1 = 0.2, L

2
2 = 0.2, L

2
3 = 0.3, L

2
4 = 0.1, m1 = 2 and m2 = 1.

One can check that f1 and f2 are Caratheodory functions,

f1(t,x1,x2,x3,x4) ≥ 2, f2(t,x1,x2,x3,x4) ≥ 1



A SINGULAR SUM FRACTIONAL DIFFERENTIAL SYSTEM 117

for all (x1,x2,x3,x4) ∈ (0,∞)4 and almost all t ∈ [0, 1], g1 and g2 are nonnegative,

|g1(t,x1,x2,x3,x4) −g1(t,y1,y2,y3,y4)| ≤
4∑
i=1

L1i |xi −yi|

and

|g2(t,x1,x2,x3,x4) −g2(t,y1,y2,y3,y4)| ≤
4∑
i=1

L2i |xi −yi|

for all (x1,x2,x3,x4), (y1,y2,y3,y4) ∈ (0,∞)4 and t ∈ [0, 1].
Also, we have

1

Γ(α1 − 1)

( 2∑
k=1

L1k +

2∑
k=1

L12+k
Γ(2 −µ1)

)
=

1

Γ( 3
2
− 1)

(0.1 + 0.2 +
0.1

Γ( 5
3
)
) +

0.2

Γ( 5
3
)
) < 1

and

1

Γ(α2 − 1)

( 2∑
k=1

L2k +

2∑
k=1

L22+k
Γ(2 −µ2)

)
=

1

Γ( 4
3
− 1)

(0.2 + 0.2 +
0.3

Γ( 3
2
)
) +

0.1

Γ( 3
2
)
) < 1.

Note that the maps p1 and p2 are non-increasing respect to all components.
If

M1 := m1
α1 − 1

Γ(α1 + 1)
= 2 ×

3
2

Γ( 7
2
)

=
3

Γ( 7
2
)
, M2 := m2

α2 − 1
Γ(α2 + 1)

= 1 ×
4
3

Γ( 10
3

)
=

4

3Γ( 10
3

)
,

then ∫ 1
0

p1

(
M1t

α1,M2t
α2,

M1(1 −µ1)
2

t1−µ1,
M2(1 −µ2)

2
t1−µ2

)
dt < ∞

and ∫ 1
0

p2

(
M1t

α1,M2t
α2,

M1(1 −µ1)
2

t1−µ1,
M2(1 −µ2)

2
t1−µ2

)
dt < ∞.

Also, the functions h1 and h2 are non-decreasing respect to all components,

lim
x→∞

h1(x,. . . ,x)

x
= lim
x→∞

2 + a1x + a2x + a3x + a4x

x
= 0

and

lim
x→∞

h2(x,. . . ,x)

x
= lim
x→∞

1 + b1x + b2x + b3x + b4x

x
= 0.

Now by using Theorem 2.2, the problem (2.1) has a positive solution.

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1Department of Mathematics, Science and Research Branch, Islamic Azad University, Tehran, Iran

2Department of Medical Research, China Medical University Hospital, China Medical University, Taichung,
Taiwan

3Department of Mathematics & Computer Science, University Politehnica of Bucharest, 313 Splaiul

Independenţei, 060042 Bucharest, Romania

4Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran

∗Corresponding author: mi.postolache@mail.cmuh.org.tw; mihai@mathem.pub.ro


	1. Preliminaries
	2. Main Results
	References