International Journal of Analysis and Applications
ISSN 2291-8639
Volume 11, Number 2 (2016), 81-92
http://www.etamaths.com

EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS FOR THE SYSTEM OF

NONLINEAR FRACTIONAL ORDER BOUNDARY VALUE PROBLEM

SABBAVARAPU NAGESWARA RAO∗

Abstract. This paper is concerned with boundary value problems for system of nonlinear fractional

differential equations involving the Caputo fractional derivatives


cDq1u(t) + f1(t,u(t),v(t)) = 0, t ∈ [0, 1],
cDq2v(t) + f2(t,u(t),v(t)) = 0, t ∈ [0, 1],

u(0) −αu′(0) = u′(η) = βu(1) + γu′′(1) = 0,
v(0) −αv′(0) = v′(η) = βv(1) + γv′′(1) = 0,

where cDq1 and cDq2 are the standard Caputo fractional derivatives of orders q1 and q2 respectively,
with 2 < q1,q2 ≤ 3. The functions fi : [0, 1] × [0,∞) × [0,∞) → [0,∞) are continuous for i = 1, 2,
α > 0,β > 0,γ > 0,η ∈ (0, 1). Under the suitable conditions, the existence and multiplicity of
positive solutions are established by using abstract fixed point theorems.

1. Introduction

In recent years, the study of fractional order differential equations has emerged as an important
area of mathematics. It has wide range of applications in various fields of science and engineering such
as physics, mechanics, control systems, flow in porous media, electromagnetics and viscoelasticity.
There has been much attention paid in developing the theory of existence of positive solutions for
fractional order differential equations satisfying initial (or) boundary conditions to mention a few
references [15, 16, 18, 24]. To mention a few references much interest has been created in establishing
positive solutions and multiple positive solutions for two-point, multi-point fractional order boundary
value problems (BVPs). To mention the related papers along these lines − see, Bai and Sun [2], Bai,
Sun and Zhang [3], Bai and Lü [4], Chai [5], Goodrich [10], Liang and Zhang [17], Nageswararao [21],
Prasad and Krushna [23], and Tian and Liu [26].

Motivated by above papers, in this paper we are concerned with the existence of multiple positive
solutions to the couple system of nonlinear fractional order differential equations

(1)

cDq1u(t) + f1(t,u(t),v(t)) = 0, t ∈ [0, 1],
cDq2v(t) + f2(t,u(t),v(t)) = 0, t ∈ [0, 1],

with the three-point boundary conditions

(2)
u(0) −αu′(0) = u′(η) = βu(1) + γu′′(1) = 0,
v(0) −αv′(0) = v′(η) = βv(1) + γv′′(1) = 0,

where cDq1 and cDq2 are the Caputo fractional derivatives of orders q1 and q2 respectively, with
2 < q1,q2 ≤ 3. The functions fi : [0, 1] × [0,∞) × [0,∞) → [0,∞) are continuous for i = 1, 2,
α > 0,β > 0,γ > 0, and η ∈ (0, 1).

By a positive solution of the fractional order boundary value problem (1)-(2), we understand a pair
of functions (u,v) ∈ C([0, 1]) ×C([0, 1]) satisfying (1)-(2) with u(t) ≥ 0, v(t) ≥ 0 for all t ∈ [0, 1], and
supt∈[0,1] u(t) > 0, supt∈[0,1] v(t) > 0.

2010 Mathematics Subject Classification. 39A10, 34B15, 34A40.

Key words and phrases. Caputo fractional derivative; couple system; fixed point theorem; Green’s function; cone.

c©2016 Authors retain the copyrights of
their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

81



82 SABBAVARAPU NAGESWARA RAO

The rest of this paper is organized as follows, In Section 2, we present some definitions and back-
ground results. For sake of convenience, we also state the fixed point theorems. In Section 3, we
construct the Green’s function for the homogeneous BVP corresponding to (1)-(2), and estimate the
bounds for the Green’s function. In Section 4, we establish the existence and multiplicity positive
solutions of the BVP (1)-(2). In Section 5, some examples are given to illustrate our existence results.

We assume the following conditions hold throughout the paper:

(A1) The functions fi : [0, 1] × [0,∞) × [0,∞) → [0,∞) are continuous and
fi(t, 0, 0) ≡ 0, for 1 ≤ i ≤ 2;

(A2) α > 0, β > 0, 0 < γ < β(1 − 2η(1 + α)) and 1 − 2η(1 + α) > 0;
(A3) limu+v→0+ supt∈[0,1]

f1(t,u,v)
u+v

= 0, limu+v→0+ supt∈[0,1]
f2(t,u,v)
u+v

= 0;

(A4) limu+v→∞ inft∈[0,1]
f1(t,u,v)
u+v

= ∞, limu+v→∞ inft∈[0,1]
f2(t,u,v)
u+v

= ∞;
(A5) limu+v→0+ inft∈[0,1]

f1(t,u,v)
u+v

= ∞, limu+v→0+ inft∈[0,1]
f2(t,u,v)
u+v

= ∞;
(A6) limu+v→∞ supt∈[0,1]

f1(t,u,v)
u+v

= 0, limu+v→∞ supt∈[0,1]
f2(t,u,v)
u+v

= 0;

(A7) For each t ∈ [0, 1],fi(t,u,v) are nondecreasing with respect to u,v and there exists a constant
N > 0 such that fi(t,u,v) <

N
2N′

, for 1 ≤ i ≤ 2, where N′ = (1−2η(1+α))(β+2γ)
2d

.

2. Preliminaries

In this section, we recall some definitions and properties of the fractional calculus. We also state
a fixed point theorem of Krasnosel’skii [14] is yield the existence of positive and multiple positive
solutions.
Definition: For a continuous function f : [0,∞) → R, the Caputo derivative of fractional order q is
defined by

cDqf(t) =
1

Γ(n−q)

∫ t
0

(t−s)n−q−1f(n)(s)ds, n− 1 < q < n, n = [q] + 1

provided that f(n)(t) exists, where [q] denotes the integer part of the real number q
Definition: The Riemann-Liouville fractional integral of order q for a continuous function f(t) is
defined as

Iqf(t) =
1

Γ(q)

∫ t
0

(t−s)q−1f(s)ds, q > 0

provided that such integral exists.

Definition: The Riemann-Liouville fractional derivative of order q for a continuous function f(t)
is defined by

Dqf(t) =
1

Γ(n−q)

( d
dt

)n ∫ t
0

(t−s)n−q−1f(s)ds, n = [q] + 1

provided that the right-hand side is pointwise defined on (0,∞).

Furthermore, we note that the Riemann-Liouville fractional derivative of a constant is usually
nonzero which can cause serious problems in real world applications. Actually, the relationship between
the two-types of fractional derivative is as follows

cDqf(t) =
1

Γ(n−q)

∫ t
0

f(n)(s)

(t−s)q+1−n
ds

= Dqf(t) −
n−1∑
k=0

f(k)(s)

Γ(k −q + 1)
tk−q

= Dq
[
f(t) −

n−1∑
k=0

f(k)

k!
tk
]
, t > 0,n− 1 < q < n.

So, we prefer to use Caputos definition which gives better results than those of Riemann- Liouville.
Lemma 2.1[25] Let q > 0, then the fractional differential equation cDqu(t) = 0 has solution u(t) =



MULTIPLE POSITIVE SOLUTIONS 83

c0 + c1t + c2t
2 + · · · + cn−1tn−1, ci ∈ R, i = 0, 1, 2, · · ·n− 1 where n is the smallest integer greater

than or equal to q.
Lemma 2.2[25] Let q > 0, then IqcDqu(t) = u(t) + c0 + c1t + c2t

2 + · · · + cn−1tn−1, for some ci ∈
R, i = 0, 1, 2, · · ·n− 1 where n is the smallest integer greater than or equal to q.

Theorem 2.1([6, 9, 14]) Let (E,‖ · ‖) be a Banach space, and let P ⊂ E be a cone in E. As-
sume that Ω1 and Ω2 are open subsets of E with 0 ∈ Ω1 and Ω1 ⊂ Ω2. If T : P ∩ (Ω2\Ω1) → P is
completely continuous operator such that either

(i) ‖ Tu ‖≤‖ u ‖, u ∈ P ∩∂Ω1 and ‖ Tu ‖≥‖ u ‖, u ∈ P ∩∂Ω2,(or)
(ii) ‖ Tu ‖≥‖ u ‖, u ∈ P ∩∂Ω1 and ‖ Tu ‖≤‖ u ‖, u ∈ P ∩∂Ω2,

holds. Then T has a fixed point in P ∩ (Ω2\Ω1).

Theorem 2.2([6, 9, 14]) Let (E,‖ · ‖) be a Banach space, and let P ⊂ E be a cone in E. Assume
that Ω1, Ω2 and Ω3 are open bounded subsets of E such that 0 ∈ Ω1 and Ω1 ⊂ Ω2, Ω2 ⊂ Ω3. If
T : P ∩ (Ω3\Ω1) → P is completely continuous operator such that:

(i) ‖ Tu ‖≥‖ u ‖, ∀u ∈ P ∩∂Ω1;
(ii) ‖ Tu ‖≤‖ u ‖, Tu 6= u, ∀u ∈ P ∩∂Ω2;
(iii) ‖ Tu ‖≥‖ u ‖, ∀u ∈ P ∩∂Ω3,

then T has at least two fixed points x∗, x∗∗ in P ∩(Ω3\Ω1), and furthermore x∗ ∈ P ∩(Ω2\Ω1), x∗∗ ∈
P ∩ (Ω3\Ω2).

3. Green’s Function and Bounds

In this section, we construct the Green’s function and bounds for the homogeneous boundary value
problem corresponding (1)-(2) that will be used to prove our main theorems.
Lemma 3.1 Let d = β + 2γ−2ηβ(1 + α) > 0. If h ∈ C[0, 1], then the fractional order boundary value
problem

(3) cDq1u(t) + h(t) = 0, 0 < t < 1,

(4) u(0) −αu′(0) = u′(η) = βu(1) + γu′′(1) = 0

has a unique solution

u(t) =

∫ 1
0

G1(t,s)h(s)ds,

where G1(t,s) is the Green’s function for the problem (3)-(4) and is given by

(5) G1(t,s) =




G1(t,s)
t∈[0,η] =

{
G11(t,s), 0 ≤ t ≤ s ≤ η < 1,
G12(t,s), 0 ≤ s ≤ min{t,η} < 1,

G1(t,s)
t∈[η,1] =

{
G13(t,s), 0 ≤ max{t,η}≤ s ≤ 1,
G14(t,s), 0 < η ≤ s ≤ t ≤ 1,



84 SABBAVARAPU NAGESWARA RAO

G11(t,s) =
1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)
− (1 + α)

(
β
(
(t−η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)(η −s)q1−2
Γ(q1 − 1)

]
G12(t,s) =

1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)
− (1 + α)

(
β
(
(t−η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)(η −s)q1−2
Γ(q1 − 1)

]
−

(t−s)q1−1

Γ(q1)

G13(t,s) =
1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)]
G14(t,s) =

1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)]
−

(t−s)q1−1

Γ(q1)

Proof. Assume that u ∈ C[q1]+1[0, 1] is a solution of fractional order boundary value problem by
(3)-(4) and is uniquely expressed as Iq1cDq1u(t) = −Iq1h(t), so that

u(t) =
−1

Γ(q1)

∫ t
0

(t−s)q1−1h(s)ds + c1 + c2t + c3t2.

Using the boundary conditions (4), we obtain that

c1 =
−2αη
d

∫ 1
0

(
β(1 −s)q1−1

Γ(q1)
+
γ(1 −s)q1−3

Γ(q1 − 2)

)
h(s)ds

+
1

d

(
2ηβ(1 + α) −α(β + 2γ − 2ηβ(1 + α))

)∫ η
0

(η −s)q1−2

Γ(q1 − 1)
h(s)ds

c2 =
−2η
d

∫ 1
0

(
β(1 −s)q1−1

Γ(q1)
+
γ(1 −s)q1−3

Γ(q1 − 2)

)
h(s)ds

+
1

d

(
2ηβ(1 + α) − (β + 2γ − 2ηβ(1 + α))

)∫ η
0

(η −s)q1−2

Γ(q1 − 1)
h(s)ds

and

c3 =
1

d

∫ 1
0

(
β(1 −s)q1−1

Γ(q1)
+
γ(1 −s)q1−3

Γ(q1 − 2)

)
h(s)ds

−
1

d

(
β(1 + α)

)∫ η
0

(η −s)q1−2

Γ(q1 − 1)
h(s)ds.

Hence, the unique solution of (3) and (4) is

u(t) =
1

d

[(
(t−η)2 −η2 − 2αη

)∫ 1
0

(
β(1 −s)q1−1

Γ(q1)
+
γ(1 −s)q1−3

Γ(q1 − 2)

)
h(s)ds

]
−

1

d

[
(1 + α)

(
β
(
(t−η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)∫ η
0

(η −s)q1−2

Γ(q1 − 1)
h(s)ds

]
−

1

Γ(q1)

∫ t
0

(t−s)q1−1h(s)ds

=

∫ 1
0

G1(t,s)h(s)ds

where G1(t,s) is given in (5).

Lemma 3.2 Assume that the condition (A2) is satisfied. Then the Green’s function G1(t,s) given
in (5) is nonnegative, for all (t,s) ∈ [0, 1] × [0, 1].



MULTIPLE POSITIVE SOLUTIONS 85

Proof. Consider the Green’s function G1(t,s) given by (5)
Let 0 ≤ t ≤ s ≤ η < 1. Then

G11(t,s) =
1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)
− (1 + α)

(
β
(
(t−η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)(η −s)q1−2
Γ(q1 − 1)

]
≥

1

d

[(
(t− tη)2 −η2 − 2αη

)( β
Γ(q1)

+
γ(1 −s)−2

Γ(q1 − 2)

)
(1 −s)q1−1

− (1 + α)
(
β
(
(t− tη)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)(η −ηs)q1−2
Γ(q1 − 1)

]
=

1

d

[(
t2(1 −η)2 − (η + α)2 + α2

)( β
Γ(q1)

+
γ(1 + 2s + O(s2))

Γ(q1 − 2)

)
− (1 + α)

(
β
(
t2(1 −η)2 + (1 + η)2 − 1

)
+ d
)ηq1−2(1 + s + s2)

Γ(q1 − 1)

]
(1 −s)q1−1 ≥ 0

Let 0 ≤ s ≤ min{t,η} < 1. Then

G12(t,s) =
1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)
− (1 + α)

(
β
(
(t−η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)(η −s)q1−2
Γ(q1 − 1)

]
−

(t−s)q1−1

Γ(q1)

≥
1

d

[(
(t− tη)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)
− (1 + α)

(
β
(
(t− tη)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)(η −ηs)q1−2
Γ(q1 − 1)

]
−

(t− ts)q1−1

Γ(q1)

=
1

d

[(
t2(1 −η)2 −η2 − 2αη

)( β
Γ(q1)

+
γ(1 + 2s + O(s2))

Γ(q1 − 2)

)
− (1 + α)

(
β
(
t2(1 −η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

) ηq1−1
Γ(q1 − 1)

−
dtq1−1

Γ(q1)

]
×

× (1 −s)q1−1 ≥ 0
Let 0 ≤ max{t,η}≤ s ≤ 1. Then

G13(t,s) =
1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)]
≥

1

d

[(
t2(1 −η)2 − (η + α)2 + α2

)( β
Γ(q1)

+
γ(1 + 2s + O(s2))

Γ(q1 − 2)

)]
×

× (1 −s)q1−1 ≥ 0
Let 0 < η ≤ s ≤ t ≤ 1. Then

G14(t,s) =
1

d

[(
(t−η)2 −η2 − 2αη

)(β(1 −s)q1−1
Γ(q1)

+
γ(1 −s)q1−3

Γ(q1 − 2)

)]
−

(t−s)q1−1

Γ(q1)

≥
1

d

[(
t2(1 −η)2 − (η + α)2 + α2

)( β
Γ(q1)

+
γ(1 + 2s + O(s2))

Γ(q1 − 2)

)
−
dtq1−1

Γ(q1)

]
×

× (1 −s)q1−1 ≥ 0

Lemma 3.3 Assume that the condition (A2) is satisfied. Then the Green’s function satisfies the
following inequality,

(6) m1G1(1,s) ≤ G1(t,s) ≤ G1(1,s), for all (t,s) ∈ [0, 1] × [0, 1],

where 0 < m1 = min
{

η2

1−2η(1+α),
2αηγ

η2+ηαβ+2γ
, 2γη

2γ(1+η)+β
(

1−2η(1+α)
)} < 1.



86 SABBAVARAPU NAGESWARA RAO

Proof. Consider the Green’s function G1(t,s) is given in (5).
Case (i): For 0 ≤ max{t,η}≤ s ≤ 1

G13(t,s)

G13(1,s)
=

1
d

[(
(t−η)2 −η2 − 2αη

)(
β(1−s)q1−1

Γ(q1)
+

γ(1−s)q1−3
Γ(q1−2)

)]
1
d

[(
(1 −η)2 −η2 − 2αη

)(
β(1−s)q1−1

Γ(q1)
+

γ(1−s)q1−3
Γ(q1−2)

)]
we have G13(t,s) ≤ G13(1,s). And also from (A2), we have

G13(t,s)

G13(1,s)
=

1
d

[(
(t−η)2 −η2 − 2αη

)(
β(1−s)q1−1

Γ(q1)
+

γ(1−s)q1−3
Γ(q1−2)

)]
1
d

[(
(1 −η)2 −η2 − 2αη

)(
β(1−s)q1−1

Γ(q1)
+

γ(1−s)q1−3
Γ(q1−2)

)]
≥

η2

1 − 2η(1 + α)

Case (ii): For 0 ≤ η ≤ s ≤ t < 1
From (A2) and case (i),we have G14(t,s) ≤ G14(1,s). And also, we have

G14(t,s)

G14(1,s)
=

1
d

[(
(t−η)2 −η2 − 2αη

)(
β(1−s)q1−1

Γ(q1)
+

γ(1−s)q1−3
Γ(q1−2)

)]
− (t−s)

q1−1

Γ(q1)

1
d

[(
(t−η)2 −η2 − 2αη

)(
β(1−s)q1−1

Γ(q1)
+

γ(1−s)q1−3
Γ(q1−2)

)]
− (t−s)

q1−1

Γ(q1)

≥
2γη

2γ(1 + η) + β
(
1 − 2η(1 + α)

)
Case (iii): For 0 ≤ t ≤ s ≤ η < 1.
From (A2) and case (i), we have G11(t,s) ≤ G11(1,s). And also, from (A2), we have

G11(t,s)

G11(1,s)
=

G13(t,s) − 1d(1 + α)
(
β
(
(t−η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)
(η−s)q1−2
Γ(q1−1)

G13(1,s) − 1d(1 + α)
(
β
(
(1 −η)2 + (1 + η)2

)
+ 2γ − 2βη(1 + α)

)
(η−s)q1−2
Γ(q1−1)

≥
2αηγ

η2 + αβη + 2γ

Case (iv): For 0 ≤ s ≤ min{t,η} < 1
From (A2) and case (iii), we have G12(t,s) ≤ G12(1,s). And also, from (A2), we have

G12(t,s)

G12(1,s)
=
G11(t,s) − [β + 2γ − 2ηβ(1 + α)]

(t−s)q1−1
Γ(q1)

G11(1,s) − [β + 2γ − 2ηβ(1 + α)]
(1−s)q1−1

Γ(q1)

≥
2γη

2γ(1 + η) + β
(
1 − 2η(1 + α)

)
By above all cases, we get

m1G1(1,s) ≤ G1(t,s) ≤ G1(1,s), for all (t,s) ∈ [0, 1] × [0, 1],

where 0 < m1 = min
{

η2

1−2η(1+α),
2αηγ

η2+ηαβ+2γ
, 2γη

2γ(1+η)+β
(

1−2η(1+α)
)} < 1.

We can also formulate similar results as Lemma (3.1) - Lemma (3.3) above, for the fractional
boundary value problem

(7) cDq2v(t) + y(t) = 0, 0 < t < 1,

(8) v(0) −αv′(0) = v′(η) = βv(1) + γv′′(1) = 0



MULTIPLE POSITIVE SOLUTIONS 87

where cDq2 is the Caputo fractional derivative of order q2 with 2 < q2 ≤ 3, α > 0,β > 0,γ > 0,η ∈
(0, 1). We denote by G2 and m2 the corresponding Green’s function and constant for the problem
(7)-(8) defined in a similar manner as G1 and m1 respectively.

By using Green functions G1 and G2 our problem (1)-(2) can be written equivalently as the following
nonlinear system of integral equations


u(t) =

∫ 1
0

G1(t,s)f1(s,u(s),v(s))ds, t ∈ [0, 1],

v(t) =

∫ 1
0

G2(t,s)f2(s,u(s),v(s))ds, t ∈ [0, 1].

We consider the Banach space E = C([0, 1]) with supremum norm ‖ · ‖, and the Banach space
B = E ×E with the norm ‖ (u,v) ‖=‖ u ‖ + ‖ v ‖ . We define the cone P ⊂ B by

P =

{
(u,v) ∈ B; u(t) ≥ 0, v(t) ≥ 0,∀t ∈ [0, 1], and inf

t∈ [η,1]
(u(t) + v(t)) ≥ m ‖ (u,v) ‖

}
,

where m = min{m1,m2}.
We introduce the operators T1, T2 : P → B and T : P → B defined by

T1(u,v)(t) =

∫ 1
0

G1(t,s)f1(s,u(s),v(s))ds, t ∈ [0, 1],

T2(u,v)(t) =

∫ 1
0

G2(t,s)f2(s,u(s),v(s))ds, t ∈ [0, 1].

(9) T(u,v) =
(
T1(u,v),T2(u,v)

)
, (u,v) ∈ P.

The solutions of our problem (1)-(2) are the fixed points of the operator T .

Lemma 3.4 If (A1) − (A2) hold, then T : P → P is a completely continuous operator.

Proof. Let (u,v) ∈ P be an arbitrary element. Because T1(u,v) and T2(u,v) satisfy the problem
(3)-(4) for h(t) = f1(t,u(t),v(t)), t ∈ [0, 1], and the problem (7)-(8) for y(t) = f2(t,u(t),v(t)), t ∈ [0, 1]
respectively, then by Lemma 3, we obtain

inf
t∈[η,1]

T1(u,v)(t) ≥ m1 max
t∈[0,1]

T1(u,v)(t) = m1 ‖ T1(u,v) ‖,

inf
t∈[η,1]

T2(u,v)(t) ≥ m2 max
t∈[0,1]

T2(u,v)(t) = m2 ‖ T2(u,v) ‖ .

Hence, we conclude

inf
t∈[η,1]

[
T1(u,v)(t) + T2(u,v)(t)

]
≥ inf
t∈[η,1]

T1(u,v)(t) + inf
t∈[η,1]

T2(u,v)(t)

≥ m1 ‖ T1(u,v) ‖ +m2 ‖ T2(u,v) ‖
≥ m ‖ (T1(u,v),T2(u,v)) ‖= m ‖ T(u,v) ‖ .

Clearly, we obtain T1(u,v)(t) ≥ 0, T2(u,v)(t) ≥ 0 for all t ∈ [0, 1], and so, we deduce that T(u,v) ∈ P.
Hence, we get T(P) ⊂ P. By using standard arguments involving the Arzela-Ascoli theorem, we can
easily show that T1 and T2 are completely continuous, and then T is a completely continuous operator
from P to P .

4. Existence of Multiple Positive Solutions

In this section, we establish the existence of at least one and two positive solutions for the BVP
(1)-(2) by using abstract fixed point theorems [6, 9, 14].
Theorem 4.1 Assume that (A1) − (A4) are hold, then the BVP (1)-(2) has at least one positive
solution (u(t),v(t)), t ∈ [0, 1].



88 SABBAVARAPU NAGESWARA RAO

Proof. From assumption (A3) we deduce that there exists H1 > 0 such that for all t ∈ [0, 1],u,v ∈
R+ with 0 ≤ u + v ≤ H1, we have f1(t,u,v) ≤ η(u + v), f2(t,u,v) ≤ η′(u + v), where η and η′ are
satisfy

η

∫ 1
0

G1(1, t)dt ≤
1

2
and η′

∫ 1
0

G2(1, t)dt ≤
1

2
.

We define the set Ω1 = {(u,v) ∈ B :‖ (u,v) ‖< H1}. Now let (u,v) ∈ P ∩∂Ω1, that is (u,v) ∈ P with
‖ (u,v) ‖= H1 or equivalently ‖ u ‖ + ‖ v ‖= H1. Then u(t) + v(t) ≤ H1, thus we have

T1(u,v)(t) =

∫ 1
0

G1(t,s)f1(s,u(s),v(s))ds

≤ η
∫ 1

0

G1(1,s)(u(s) + v(s))ds

≤ η
∫ 1

0

G1(1,s)
[
‖ u ‖ + ‖ v ‖

]
ds

≤
1

2

[
‖ u ‖ + ‖ v ‖

]
=

1

2
‖ (u,v) ‖

and so, ‖ T1(u,v) ‖≤ 12 ‖ (u,v) ‖ . Similarly, we may take

T2(u,v)(t) =

∫ 1
0

G2(t,s)f2(s,u(s),v(s))ds

≤ η′
∫ 1

0

G2(1,s)(u(s) + v(s))ds

≤ η′
∫ 1

0

G2(1,s)
[
‖ u ‖ + ‖ v ‖

]
ds

≤
1

2

[
‖ u ‖ + ‖ v ‖

]
=

1

2
‖ (u,v) ‖

and so, ‖ T2(u,v) ‖≤ 12 ‖ (u,v) ‖ . Thus, for (u,v) ∈ P ∩∂Ω1 it follows that

‖ T(u,v) ‖ =‖
(
T1(u,v),T2(u,v)

)
‖=‖ T1(u,v) ‖ + ‖ T2(u,v) ‖

≤
1

2
‖ (u,v) ‖ +

1

2
‖ (u,v) ‖=‖ (u,v) ‖ .

Therefore,

(10) ‖ T(u,v) ‖≤‖ (u,v) ‖, for all (u,v) ∈ P ∩∂Ω1.

On the other hand, from (A4) there exist four positive constants µ,µ′,C1 and C2 such that f1(t,u,v) ≥
µ(u + v) −C1, f2(t,u,v) ≥ µ

′
(u + v) −C2, ∀(u,v) ∈ R+ ×R+, where µ and µ′ satisfy

µm2
∫ 1
η

G1(1,s)ds ≥ 1, µ′m2
∫ 1
η

G2(1,s)ds ≥ 1.

For (u,v) ∈ P, τ ∈ (0, 1), we have

T1(u,v)(τ) =

∫ 1
0

G1(τ,s)f1(s,u(s),v(s))ds

≥
∫ 1

0

G1(τ,s)
[
µ(u + v) −C1

]
ds

≥ µ
∫ 1
η

G1(τ,s)
(
u(s) + v(s)

)
ds−C1

∫ 1
η

G1(τ,s)ds

≥ µm2
∫ 1
η

G1(1,s)ds
(
‖ u ‖ + ‖ v ‖

)
−C1

∫ 1
η

G1(τ,s)ds

≥
(
‖ u ‖ + ‖ v ‖

)
−C1

∫ 1
η

G1(τ,s)ds.



MULTIPLE POSITIVE SOLUTIONS 89

In a similar manner, we deduce

T2(u,v)(τ) =

∫ 1
0

G2(τ,s)f2(s,u(s),v(s))ds

≥
∫ 1

0

G2(τ,s)
[
µ′(u + v) −C2

]
ds

≥ µ′
∫ 1
η

G1(τ,s)
(
u(s) + v(s)

)
ds−C2

∫ 1
η

G2(τ,s)ds

≥ µ′m2
∫ 1
η

G1(1,s)
(
‖ u ‖ + ‖ v ‖

)
ds−C2

∫ 1
η

G2(τ,s)ds

≥
(
‖ u ‖ + ‖ v ‖

)
−C2

∫ 1
η

G2(τ,s)ds.

Therefore T(u,v)(τ) ≥ 2 ‖ (u,v) ‖−C3, where C3 = C1
∫ 1
η
G1(τ,s)ds + C2

∫ 1
η
G2(τ,s)ds. From which

it follows that ‖ T(u,v) ‖≥ T(u,v)(τ) ≥‖ (u,v) ‖ as ‖ (u,v) ‖→∞.
Let Ω2 = {(u,v) ∈ B :‖ (u,v) ‖< H2}. Then for (u,v) ∈ P and ‖ (u,v) ‖= H2 > 0 sufficiently large,
we have

(11) ‖ T(u,v) ‖≥‖ (u,v) ‖, for all (u,v) ∈ P ∩∂Ω2.

Thus, from (10), (11) and Theorem (2.1), we know that the operator T has a fixed point in P∩(Ω2 \ Ω1).

Theorem 4.2 Assume that (A1), (A2), (A5) and (A6) are hold, then (1)-(2) has at least one
positive solution (u(t),v(t)), t ∈ [0, 1]

Proof. From (A5) there is a number Ĥ3 ∈ (0, 1) such that for each (t,u,v) ∈ [0, 1] × (0,Ĥ3) ×
(0,Ĥ3). One has f1(t,u,v) ≥ λ(u + v), where λ satisfy λm2

∫ 1
η
G1(1,s)ds ≥ 1. From (A1) that implies

f1(t, 0, 0) ≡ 0 and the continuity of f1(t,u,v), we know that there exists a number H3 ∈ (0,Ĥ3) small
enough such that

f1(t,u,v) ≤
Ĥ3∫ 1

0
G1(1, t)dt

whenever (t,u,v) ∈ [0, 1] × (0,H3) × (0,H3).

For every (u,v) ∈ P and ‖ (u,v) ‖= H3, note that∫ 1
0

G1(1,τ)f1(τ,u(τ),v(τ))dτ ≤
∫ 1

0

G1(1,τ)
Ĥ3∫ 1

0
G1(1, t)dt

dτ ≤ Ĥ3.

Thus

T1(u,v)(τ) =

∫ 1
0

G1(τ,s)f1(s,u,v)ds

≥ mλ
∫ 1
η

G1(1,s)
(
u(s) + v(s)

)
ds

≥ λm2
∫ 1
η

G1(1,s)ds(‖ u ‖ + ‖ v ‖)

≥
(
‖ u ‖ + ‖ v ‖

)
=‖ (u,v) ‖

that is

T1(u,v)(t) ≥‖ (u,v) ‖ for all t ≥ τ.
So, ‖ T(u,v) ‖≥‖ T1(u,v) ‖≥‖ (u,v) ‖. If set Ω3 = {(u,v) ∈ B :‖ (u,v) ‖< H3}, then

(12) ‖ T(u,v) ‖≥‖ (u,v) ‖, for all (u,v) ∈ P ∩∂Ω3.

On the other hand, we know from (A6) that there exist four positive numbers η,η′,C4 and C5 such
that for every (t,u,v) ∈ [0, 1]×R+×R+, we have f1(t,u,v) ≤ η(u+v)+C4, f2(t,u,v) ≤ η′(u+v)+C5,



90 SABBAVARAPU NAGESWARA RAO

where η and η
′

satisfy

η

∫ 1
0

G1(1,s)ds ≤
1

2
and η

′
∫ 1

0

G2(1,s)ds ≤
1

2
.

Thus we have

T1(u,v)(t) =

∫ 1
0

G1(t,s)f1(s,u,v)ds

≤
∫ 1

0

G1(1,s)(η(u + v) + C4)ds

≤ η
∫ 1

0

G1(1,s)
(
‖ u ‖ + ‖ v ‖

)
ds + C4

∫ 1
0

G1(1,s)ds

≤
1

2
‖ (u,v) ‖ +C4

∫ 1
0

G1(1,s)ds.

Similarly, we deduce

T2(u,v)(t) =

∫ 1
0

G2(t,s)f2(s,u,v)ds

≤
∫ 1

0

G2(1,s)(η
′(u + v) + C5)ds

≤ η′
∫ 1

0

G2(1,s)
(
‖ u ‖ + ‖ v ‖

)
ds + C5

∫ 1
0

G2(1,s)ds

≤
1

2
‖ (u,v) ‖ +C5

∫ 1
0

G2(1,s)ds.

Therefore T(u,v)(t) ≤‖ (u,v) ‖ +C6, where C6 = C4
∫ 1

0
G1(1,s)ds + C5

∫ 1
0
G2(1,s)ds, from which it

follows that T(u,v)(t) ≤‖ (u,v) ‖ as ‖ (u,v) ‖→ ∞. Let Ω4 = {(u,v) ∈ B :‖ (u,v) ‖< H4}. For each
(u,v) ∈ P and ‖ (u,v) ‖= H4 > 0 large enough, we have

(13) ‖ T(u,v) ‖≤‖ (u,v) ‖, for all (u,v) ∈ P ∩∂Ω4.

From (12),(13) and Theorem (2.1), we know that the operator T has a fixed point in P ∩ (Ω4 \ Ω3).

Theorem 4.3 Assume that (A1), (A2), (A4), (A5) and (A7) are satisfied, then (1)-(2) has at least
two positive solutions (u1(t),v1(t)), (u2(t),v2(t)), t ∈ [0, 1].

Proof. Note that we have Gi(t,s) ≤
(1−2η(1+α))(β+2γ)

2d
= N′ for i = 1, 2 for all (t,s) ∈ [0, 1]× [0, 1].

Let BN = {(u,v) ∈ P :‖ (u,v) ‖< N}. By using (A7), for any (u,v) ∈ ∂BN ∩P, we obtain

T1(u,v)(t) =

∫ 1
0

G1(t,s)f1(s,u,v)ds < N
′ N

2N′
=
N

2

which implies ‖ T1(u,v) ‖≤ N2 . In a similar manner, we may take ‖ T2(u,v) ‖≤
N
2

. Therefore

‖ T(u,v) ‖=‖ T1(u,v) ‖ + ‖ T2(u,v) ‖≤
N

2
+
N

2
= N.

Thus

(14) ‖ T(u,v) ‖≤‖ (u,v) ‖, for all (u,v) ∈ P ∩∂BN.

And from (A4) and (A5) we have

(15) ‖ T(u,v) ‖≥‖ (u,v) ‖, for all (u,v) ∈ P ∩∂Ω2,

(16) ‖ T(u,v) ‖≥‖ (u,v) ‖, for all (u,v) ∈ P ∩∂Ω3.

We have choose H2,H3 and N such that H3 ≤ N ≤ H2 and (14)-(16) are satisfied. From Theorem
(2.2), T has at least two fixed points in P ∩ (Ω2 \BN ) and P ∩ (BN \ Ω3), respectively.



MULTIPLE POSITIVE SOLUTIONS 91

5. Example

In this section, we demonstrate our results with some examples.
We consider the system of fractional order differential equations

(17)

cD2.5u(t) + f1(t,u(t),v(t)) = 0, t ∈ (0, 1)
cD2.5v(t) + f2(t,u(t),v(t)) = 0, t ∈ (0, 1)

with the three-point boundary conditions

(18)
u(0) −u′(0) = u′

(1
8

)
= 2u(1) +

1

2
u′′(1) = 0,

v(0) −v′(0) = v′
(1

8

)
= 2v(1) +

1

2
v′′(1) = 0.

Here q1 = q2 =
5
2
,α = 1, β = 2, η = 1

8
, γ = 1

2
and we deduce that m = min{m1,m2} = 0.03125

Example 5.1: Let f1(t,u,v) =
t
4
(u + v) + t2 + 4, f2(t,u,v) =

t4

2
(u + v) + e−(u+v), then conditions of

Theorem (4.1) are satisfied. From Theorem (4.1), the BVP (17)-(18) has at least one positive solution.

Example 5.2: Let f1(t,u,v) = (1 − t)
[
e−(u+v)(u + v)

]
, f2(t,u,v) =

4
1+t

(u2 + v2) then N′ = 9
8
.

We can choose N = 1 and conditions of Theorem (4.3) are satisfied. From Theorem (4.3), the BVP
(17)-(18) has at least two positive solutions.

Acknowledgement: The author is very grateful to his guide professor K. Rajendra Prasad and
to the referees for their valuable suggestions and comments on improving this paper.

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Department of Mathematics, Jazan University, Jazan, Kingdom of Saudi Arabia

∗Corresponding author: sabbavarapu−nag@yahoo.co.in; snrao@jazanu.edu.sa