# Multiple Positive Solutions for a Coupled System ofp-Laplacian Fractional Order Two-Point Boundary Value Problems

Authors: K. R. Prasad; B. M. B. Krushna
Journal: International Journal of Differential Equations
 (2014)

---

## Abstract

This paper establishes the existence of at least three positive solutions for a coupled system ofp-Laplacian fractional order two-point boundary value problems, D
0
+
β
1
(
ϕ
p
(
D
0
+
α
1
u
(
t
)
)
)
=
f
1
(
t
,
u
(
t
)
,
v
(
t
)
), t
∈
(
0,1
), D
0
+
β
2
(
ϕ
p
(
D
0
+
α
2
v
(
t
)
)
)
=
f
2
(
t
,
u
(
t
)
,
v
(
t
)
), t
∈
(
0,1
), u
(
0
)
=
D
0
+
q
1
u
(
0
)
=
0, γ
u
(
1
)
+
δ
D
0
+
q
2
u
(
1
)
=
0, D
0
+
α
1
u
(
0
)
=
D
0
+
α
1
u
(
1
)
=
0, v
(
0
)
=
D
0
+
q
1
v
(
0
)
=
0, γ
v
(
1
)
+
δ
D
0
+
q
2
v
(
1
)
=
0, D
0
+
α
2
v
(
0
)
=
D
0
+
α
2
v
(
1
)
=
0, by applying five functionals fixed point theorem.


---

## Body

## 1. Introduction

The theory of differential equations offers a broad mathematical basis to understand the problems of modern society which are complex and interdisciplinary by nature. Fractional order differential equations have gained importance due to their applications to almost all areas of science, engineering, and technology. Among all the theories, the most applicable operator is the classicalp-Laplacian, given by ϕ
p
(
s
)
=
|
s
|
p
-
2
s, p
>
1. These types of problems have a wide range of applications in physics and related sciences such as biophysics, plasma physics, and chemical reaction design.The positive solutions of boundary value problems associated with ordinary differential equations were studied by many authors [1–3] and extended to p-Laplacian boundary value problems [4–6]. Later, these results are further extended to fractional order boundary value problems [7–15] by applying various fixed point theorems on cones. Recently, researchers are concentrating on the theory of fractional order boundary value problems associated with p-Laplacian operator.In 2012, Chai [16] investigated the existence and multiplicity of positive solutions for a class of boundary value problem of fractional differential equation with p-Laplacian operator,
(1)
D
0
+
β
(
ϕ
p
(
D
0
+
α
u
(
t
)
)
)
+
f
(
t
,
u
(
t
)
)
=
0
,
0
<
t
<
1
,
u
(
0
)
=
0
,
u
(
1
)
+
σ
D
0
+
γ
u
(
1
)
=
0
,
D
0
+
α
u
(
0
)
=
0
,
by means of the fixed point theorem on cones.This paper is concerned with the existence of positive solutions for a coupled system ofp-Laplacian fractional order boundary value problems:
(2)
D
0
+
β
1
(
ϕ
p
(
D
0
+
α
1
u
(
t
)
)
)
=
f
1
(
t
,
u
(
t
)
,
v
(
t
)
)
,
t
∈
(
0,1
)
,
(3)
D
0
+
β
2
(
ϕ
p
(
D
0
+
α
2
v
(
t
)
)
)
=
f
2
(
t
,
u
(
t
)
,
v
(
t
)
)
,
t
∈
(
0,1
)
,
(4)
u
(
0
)
=
D
0
+
q
1
u
(
0
)
=
0
,
γ
u
(
1
)
+
δ
D
0
+
q
2
u
(
1
)
=
0
,
D
0
+
α
1
u
(
0
)
=
D
0
+
α
1
u
(
1
)
=
0
,
(5)
v
(
0
)
=
D
0
+
q
1
v
(
0
)
=
0
,
γ
v
(
1
)
+
δ
D
0
+
q
2
v
(
1
)
=
0
,
D
0
+
α
2
v
(
0
)
=
D
0
+
α
2
v
(
1
)
=
0
,
where ϕ
p
(
s
)
=
|
s
|
p
-
2
s,  p
>
1,  ϕ
p
-
1
=
ϕ
q,  1
/
p
+
1
/
q
=
1, γ
,
δ are positive real numbers, 2
<
α
i
≤
3
,
1
<
β
i,  q
i
≤
2,  f
i
:
[
0,1
]
×
ℝ
2
→
ℝ
+ are continuous functions, and D
0
+
α
i,  D
0
+
β
i,  D
0
+
q
i,  for i
=
1,2 are the standard Riemann-Liouville fractional order derivatives.The rest of the paper is organized as follows. In Section2, the Green functions for the homogeneous BVPs corresponding to (2), (4) are constructed and the bounds for the Green functions are estimated. In Section 3, sufficient conditions for the existence of at least three positive solutions for a coupled system of p-Laplacian fractional order BVP (2)–(5) are established, by using five functionals fixed point theorem. In Section 4, as an application, the results are demonstrated with an example.


## 2. Green Functions and Bounds

In this section, the Green functions for the homogeneous BVPs are constructed and the bounds for the Green functions are estimated, which are essential to establish the main results.LetG
1
(
t
,
s
) be Green’s function for the homogeneous BVP:
(6)
-
D
0
+
α
1
u
(
t
)
=
0
,
t
∈
(
0,1
)
,
(7)
u
(
0
)
=
0
,
D
0
+
q
1
u
(
0
)
=
0
,
γ
u
(
1
)
+
δ
D
0
+
q
2
u
(
1
)
=
0
.Lemma 1.
Letd
=
δ
Γ
(
α
1
)
+
γ
Γ
(
α
1
-
q
2
)
≠
0. If h
∈
C
[
0,1
], then the fractional order BVP
(8)
D
0
+
α
1
u
(
t
)
+
h
(
t
)
=
0
,
t
∈
(
0,1
)
,
with (7) has a unique solution
(9)
u
(
t
)
=
∫
0
1
‍
G
1
(
t
,
s
)
h
(
s
)
d
s
,
where
(10)
G
1
(
t
,
s
)
=
{
G
11
(
t
,
s
)
,
0
≤
t
≤
s
≤
1
,
G
12
(
t
,
s
)
,
0
≤
s
≤
t
≤
1
,
(11)
G
11
(
t
,
s
)
=
1
d
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
(
1
-
s
)
]
α
1
-
1
,
G
12
(
t
,
s
)
=
G
11
(
t
,
s
)
-
1
d
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
t
-
s
)
α
1
-
1
.Proof.
Letu
∈
C
[
α
1
]
+
1
[
0,1
] be the solution of fractional order BVP (8), (7). Then
(12)
I
0
+
α
1
D
0
+
α
1
u
(
t
)
=
-
I
0
+
α
1
h
(
t
)
,
and hence
(13)
u
(
t
)
=
-
1
Γ
(
α
1
)
∫
0
t
‍
(
t
-
s
)
α
1
-
1
h
(
s
)
d
s
+
c
1
t
α
1
-
1
+
c
2
t
α
1
-
2
+
c
3
t
α
1
-
3
.
Using the boundary conditions (7), c
1, c
2, and c
3 are determined as
(14)
c
1
=
1
d
∫
0
1
‍
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
(
1
-
s
)
α
1
-
1
h
(
s
)
d
s
,
c
2
=
0
,
c
3
=
0
.
Hence, the unique solution of (8), (7) is
(15)
u
(
t
)
=
∫
0
t
‍
{
1
d
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
(
1
-
s
)
]
α
1
-
1
-
(
t
-
s
)
α
1
-
1
Γ
(
α
1
)
}
h
(
s
)
d
s
+
∫
t
1
‍
1
d
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
×
[
t
(
1
-
s
)
]
α
1
-
1
h
(
s
)
d
s
=
∫
0
1
‍
G
1
(
t
,
s
)
h
(
s
)
d
s
.Lemma 2.
Lety
(
t
)
∈
C
[
0,1
] and 2
<
α
1
≤
3, 1
<
β
1
≤
2. Then the fractional order BVP
(16)
D
0
+
β
1
(
ϕ
p
(
D
0
+
α
1
u
(
t
)
)
)
=
y
(
t
)
,
t
∈
(
0,1
)
,
with (4) has a unique solution
(17)
u
(
t
)
=
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
s
,
τ
)
y
(
τ
)
d
τ
)
d
s
,
where
(18)
H
1
(
t
,
s
)
=
{
[
t
(
1
-
s
)
]
β
1
-
1
Γ
(
β
1
)
,
0
≤
t
≤
s
≤
1
,
[
t
(
1
-
s
)
]
β
1
-
1
-
(
t
-
s
)
β
1
-
1
Γ
(
β
1
)
,
0
≤
s
≤
t
≤
1
.Proof.
An equivalent integral equation for (16) is given by
(19)
ϕ
p
(
D
0
+
α
1
u
(
t
)
)
=
1
Γ
(
β
1
)
∫
0
t
‍
(
t
-
τ
)
β
1
-
1
y
(
τ
)
d
τ
+
c
1
t
β
1
-
1
+
c
2
t
β
1
-
2
.
Using the conditions D
0
+
α
1
u
(
0
)
=
0,  D
0
+
α
1
u
(
1
)
=
0,  c
1, and c
2 are determined as c
1
=
(
-
1
/
Γ
(
β
1
)
)
∫
0
1
‍
(
1
-
τ
)
β
1
-
1
y
(
τ
)
d
τ and c
2
=
0. Then,
(20)
ϕ
p
(
D
0
+
α
1
u
(
t
)
)
=
1
Γ
(
β
1
)
∫
0
t
‍
(
t
-
τ
)
β
1
-
1
y
(
τ
)
d
τ
-
1
Γ
(
β
1
)
∫
0
1
‍
[
t
(
1
-
τ
)
]
β
1
-
1
y
(
τ
)
d
τ
=
-
∫
0
1
‍
H
1
(
t
,
τ
)
y
(
τ
)
d
τ
.
Therefore,
(21)
ϕ
p
-
1
(
ϕ
p
(
D
0
+
α
1
u
(
t
)
)
)
=
-
ϕ
p
-
1
(
∫
0
1
‍
H
1
(
t
,
τ
)
y
(
τ
)
d
τ
)
.
Consequently,
(22)
D
0
+
α
1
u
(
t
)
+
ϕ
q
(
∫
0
1
‍
H
1
(
t
,
τ
)
y
(
τ
)
d
τ
)
=
0
.
Hence, u
(
t
)
=
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
s
,
τ
)
y
(
τ
)
d
τ
)
d
s is the solution of fractional order BVP (16) and (4).Lemma 3.
Assume thatδ
(
q
2
-
1
)
>
γ
Γ
(
α
1
-
q
2
)
/
Γ
(
α
1
). Then Green’s function G
1
(
t
,
s
) satisfies the following inequalities:(i)
G
1
(
t
,
s
)
≥
0, for all (
t
,
s
)
∈
[
0,1
]
×
[
0,1
],
(ii)
G
1
(
t
,
s
)
≤
G
1
(
1
,
s
), for all (
t
,
s
)
∈
[
0,1
]
×
[
0,1
],
(iii)
G
1
(
t
,
s
)
≥
(
1
/
4
α
1
-
1
)
G
1
(
1
,
s
), for all (
t
,
s
)
∈
I
×
[
0,1
],where I
=
[
1
/
4,3
/
4
].Proof.
Green’s functionG
1
(
t
,
s
) is given in (10). For 0
≤
t
≤
s
≤
1,
(23)
G
11
(
t
,
s
)
=
1
d
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
(
1
-
s
)
]
α
1
-
1
≥
0
.
For 0
≤
s
≤
t
≤
1,
(24)
G
12
(
t
,
s
)
=
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
-
t
s
]
α
1
-
1
d
-
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
t
-
s
)
α
1
-
1
d
≥
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
-
t
s
]
α
1
-
1
d
-
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
t
-
t
s
)
α
1
-
1
d
≥
δ
d
[
(
1
-
s
)
-
q
2
-
1
]
[
t
-
t
s
]
α
1
-
1
≥
0
.
Hence, the inequality (i) is proved. For 0
≤
t
≤
s
≤
1,
(25)
∂
G
11
(
t
,
s
)
∂
t
=
1
d
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
(
1
-
s
)
α
1
-
1
×
(
α
1
-
1
)
t
α
1
-
2
≥
0
.
Therefore, G
11
(
t
,
s
) is increasing with respect to t, which implies that G
11
(
t
,
s
)
≤
G
11
(
1
,
s
). Now, for 0
≤
s
≤
t
≤
1,
(26)
∂
G
12
(
t
,
s
)
∂
t
=
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
×
(
α
1
-
1
)
[
t
-
t
s
]
α
1
-
2
(
1
-
s
)
d
-
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
α
1
-
1
)
(
t
-
s
)
α
1
-
2
d
≥
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
×
(
α
1
-
1
)
[
t
-
t
s
]
α
1
-
2
(
1
-
s
)
d
-
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
α
1
-
1
)
(
t
-
t
s
)
α
1
-
2
d
=
(
α
1
-
1
)
(
t
-
t
s
)
α
1
-
2
d
×
[
(
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
)
(
1
-
s
)
-
(
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
)
]
=
(
α
1
-
1
)
(
t
-
t
s
)
α
1
-
2
d
×
[
δ
(
(
1
-
s
)
-
(
q
2
-
1
)
-
1
)
-
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
s
]
=
(
α
1
-
1
)
(
t
-
t
s
)
α
1
-
2
d
×
[
δ
(
(
q
2
-
1
)
s
+
(
q
2
-
1
)
q
2
2
s
2
+
·
·
·
)
-
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
s
(
q
2
-
1
)
q
2
2
]
=
(
α
1
-
1
)
(
t
-
t
s
)
α
1
-
2
d
×
[
(
δ
(
q
2
-
1
)
-
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
)
s
+
O
(
s
2
)
]
≥
0
.
Therefore, G
12
(
t
,
s
) is increasing with respect to t, which implies that G
12
(
t
,
s
)
≤
G
12
(
1
,
s
). Hence, the inequality (ii) is proved. Now, the inequality (iii) can be established.
Let0
≤
t
≤
s
≤
1 and t
∈
I. Then
(27)
G
11
(
t
,
s
)
=
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
(
1
-
s
)
]
α
1
-
1
=
t
α
1
-
1
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
(
1
-
s
)
α
1
-
1
=
t
α
1
-
1
G
11
(
1
,
s
)
≥
1
4
α
1
-
1
G
11
(
1
,
s
)
.
Let 0
≤
s
≤
t
≤
1 and t
∈
I. Then
(28)
G
12
(
t
,
s
)
=
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
(
1
-
s
)
]
α
1
-
1
-
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
t
-
s
)
α
1
-
1
≥
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
]
[
t
(
1
-
s
)
]
α
1
-
1
-
[
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
]
(
t
-
t
s
)
α
1
-
1
=
t
α
1
-
1
[
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
+
δ
(
1
-
s
)
-
q
2
-
(
δ
+
γ
Γ
(
α
1
-
q
2
)
Γ
(
α
1
)
)
]
(
1
-
s
)
α
1
-
1
=
t
α
1
-
1
G
12
(
1
,
s
)
≥
1
4
α
1
-
1
G
12
(
1
,
s
)
.
Hence the inequality (iii) is proved.Lemma 4.
Fort
,
s
∈
[
0,1
], Green’s function H
1
(
t
,
s
) satisfies the following inequalities:(i)
H
1
(
t
,
s
)
≥
0,
(ii)
H
1
(
t
,
s
)
≤
H
1
(
s
,
s
).Proof.
Green’s functionH
1
(
t
,
s
) is given in (18). Clearly, it is observed that, for 0
≤
t
≤
s
≤
1, H
1
(
t
,
s
)
≥
0.
For0
≤
s
≤
t
≤
1,
(29)
H
1
(
t
,
s
)
=
1
Γ
(
β
1
)
[
t
β
1
-
1
(
1
-
s
)
β
1
-
1
-
(
t
-
s
)
β
1
-
1
]
≥
1
Γ
(
β
1
)
[
t
β
1
-
1
(
1
-
s
)
β
1
-
1
-
(
t
-
t
s
)
β
1
-
1
]
=
1
Γ
(
β
1
)
[
t
β
1
-
1
(
1
-
s
)
β
1
-
1
-
t
β
1
-
1
(
1
-
s
)
β
1
-
1
]
=
0
.
Hence, the inequality (i) is proved. Now we establish the inequality (ii), for 0
≤
t
≤
s
≤
1,
(30)
∂
H
1
(
t
,
s
)
∂
t
=
1
Γ
(
β
1
)
[
(
β
1
-
1
)
t
β
1
-
2
(
1
-
s
)
β
1
-
1
]
>
0
.
Therefore, H
1
(
t
,
s
) is increasing with respect to t, for s
∈
[
0,1
), which implies that H
1
(
t
,
s
)
≤
H
1
(
s
,
s
). Similarly, it can be proved that H
1
(
t
,
s
)
≤
H
1
(
s
,
s
) for 0
≤
s
≤
t
≤
1. Hence the inequality (ii) is proved.Lemma 5.
Green’s functionH
1
(
t
,
s
) satisfies the following inequality: (
A
) there exists a positive function γ
1
*
(
s
)
∈
C
(
0,1
) such that
(31)
min
⁡
t
∈
I
⁡
H
1
(
t
,
s
)
≥
γ
1
*
(
s
)
H
1
(
s
,
s
)
,
f
o
r
0
<
s
<
1
.Proof.
SinceH
1
(
t
,
s
) is monotonic function, for all t
,
s
∈
[
0,1
], we have
(32)
max
⁡
0
≤
t
≤
1
⁡
H
1
(
t
,
s
)
=
H
1
(
s
,
s
)
=
1
Γ
(
β
1
)
[
s
(
1
-
s
)
]
β
1
-
1
.
From (i) of Lemma 4, H
1
(
t
,
s
)
≥
0, for t
,
s
∈
[
0,1
]. For s
∈
(
0,1
/
4
),  H
1
(
t
,
s
) is increasing with respect to t for t
∈
(
0
,
s
/
(
1
-
(
1
-
s
)
(
β
1
-
1
)
/
(
β
1
-
2
)
)
) and decreasing with respect to t for t
∈
(
s
/
(
1
-
(
1
-
s
)
(
β
1
-
1
)
/
(
β
1
-
2
)
)
,
1
/
4
). For s
∈
(
1
/
4,1
),  H
1
(
t
,
s
) is decreasing with respect to t for s
≤
t and increasing with respect to t for s
≥
t. If we define
(33)
h
1
(
t
,
s
)
=
[
t
(
1
-
s
)
]
β
1
-
1
-
(
t
-
s
)
β
1
-
1
Γ
(
β
1
)
,
h
2
(
t
,
s
)
=
[
t
(
1
-
s
)
]
β
1
-
1
Γ
(
β
1
)
.
Then,
(34)
min
⁡
t
∈
I
⁡
H
1
(
t
,
s
)
=
{
h
1
(
3
4
,
s
)
,
s
∈
(
0
,
1
4
]
,
min
⁡
{
h
1
(
3
4
,
s
)
,
h
2
(
1
4
,
s
)
}
,
s
∈
[
1
4
,
3
4
]
,
h
2
(
1
4
,
s
)
,
s
∈
[
3
4
,
1
)
,
=
{
h
1
(
3
4
,
s
)
,
s
∈
(
0
,
ξ
]
,
h
2
(
1
4
,
s
)
,
s
∈
[
ξ
,
1
)
,
=
{
[
(
3
/
4
)
(
1
-
s
)
]
β
1
-
1
-
(
(
3
/
4
)
-
s
)
β
1
-
1
Γ
(
β
1
)
,
s
∈
(
0
,
ξ
]
,
1
Γ
(
β
1
)
(
1
-
s
)
β
1
-
1
4
β
1
-
1
,
s
∈
[
ξ
,
1
)
,
≥
{
[
(
3
/
4
)
(
1
-
s
)
]
β
1
-
1
-
(
(
3
/
4
)
-
s
)
β
1
-
1
[
s
(
1
-
s
)
]
β
1
-
1
×
H
1
(
s
,
s
)
,
s
∈
(
0
,
ξ
]
,
1
(
4
s
)
β
1
-
1
H
1
(
s
,
s
)
s
∈
[
ξ
,
1
)
,
=
γ
1
*
(
s
)
H
1
(
s
,
s
)
,
where
(35)
γ
1
*
(
s
)
=
{
[
(
3
/
4
)
(
1
-
s
)
]
β
1
-
1
-
(
(
3
/
4
)
-
s
)
β
1
-
1
[
s
(
1
-
s
)
]
β
1
-
1
,
s
∈
(
0
,
ξ
]
,
1
(
4
s
)
β
1
-
1
,
s
∈
[
ξ
,
1
)
,
and ξ
∈
(
1
/
4,3
/
4
) satisfy the equation [
(
3
/
4
)
(
1
-
ξ
)
]
β
1
-
1
-
(
(
3
/
4
)
-
ξ
)
β
1
-
1
=
(
(
1
-
ξ
)
/
4
)
β
1
-
1. In particular, ξ
=
0.5 if β
1
=
2; ξ
→
0.5 as β
1
→
2; and ξ
→
0.76 as β
1
→
1. Hence the inequality in (31) holds.In a similar manner, the results of the Green functionsG
2
(
t
,
s
) and H
2
(
t
,
s
) for the homogeneous BVPs corresponding to the fractional order BVP (3) and (5) are obtained.Remark 6.
Consider the following.
G
1
(
t
,
s
)
≥
η
G
1
(
1
,
s
) and G
2
(
t
,
s
)
≥
η
G
2
(
1
,
s
), for all (
t
,
s
)
∈
I
×
[
0,1
], where η
=
min
⁡
⁡
{
1
/
4
α
1
-
1
,
1
/
4
α
2
-
1
}.Remark 7.
Consider the following.
H
1
(
t
,
s
)
≥
γ
*
(
s
)
H
1
(
s
,
s
) and H
2
(
t
,
s
)
≥
γ
*
(
s
)
H
2
(
s
,
s
), for all (
t
,
s
)
∈
I
×
[
0,1
], where γ
*
(
s
)
=
min
⁡
{
γ
1
*
(
s
)
,
γ
2
*
(
s
)
}.


## 3. Existence of Multiple Positive Solutions

In this section, the existence of at least three positive solutions for a coupled system ofp-Laplacian fractional order BVP (2)–(5) is established by using five functionals fixed point theorem.Letγ, β, θ be nonnegative continuous convex functionals on P and let α, ψ be nonnegative continuous concave functionals on P; then for nonnegative numbers h
′, a
′, b
′, d
′, and c
′, convex sets are defined:
(36)
P
(
γ
,
c
′
)
=
{
y
∈
P
:
γ
(
y
)
<
c
′
}
,
P
(
γ
,
α
,
a
′
,
c
′
)
=
{
y
∈
P
:
a
′
≤
α
(
y
)
;
γ
(
y
)
≤
c
′
}
,
Q
(
γ
,
β
,
d
′
,
c
′
)
=
{
y
∈
P
:
β
(
y
)
≤
d
′
;
γ
(
y
)
≤
c
′
}
,
P
(
γ
,
θ
,
α
,
a
′
,
b
′
,
c
′
)
=
{
y
∈
P
:
a
′
≤
α
(
y
)
;
θ
(
y
)
≤
b
′
;
γ
(
y
)
≤
c
′
}
,
Q
(
γ
,
β
,
ψ
,
h
′
,
d
′
,
c
′
)
=
{
y
∈
P
:
h
′
≤
ψ
(
y
)
;
β
(
y
)
≤
d
′
;
γ
(
y
)
≤
c
′
}
.In obtaining multiple positive solutions of thep-Laplacian fractional order BVP (2)-(5), the following so-called five functionals fixed point theorem is fundamental.Theorem 8 (see [17]).
LetP be a cone in the real Banach space B. Suppose that α and ψ are nonnegative continuous concave functionals on P and γ, β, θ are nonnegative continuous convex functionals on P, such that, for some positive numbers c
′ and e
′, α
(
y
)
≤
β
(
y
) and ∥
y
∥
≤
e
′
γ
(
y
), for all y
∈
P
(
γ
,
c
′
)
¯. Suppose further that T
:
P
(
γ
,
c
′
)
¯
→
P
(
γ
,
c
′
)
¯ is completely continuous and there exist constants h
′, d
′, a
′, and b
′
≥
0 with 0
<
d
′
<
a
′ such that each of the following is satisfied:(B1)
{
y
∈
P
(
γ
,
θ
,
α
,
a
′
,
b
′
,
c
′
)
:
α
(
y
)
>
a
′
}
≠
∅ and α
(
T
y
)
>
a
′ for y
∈
P
(
γ
,
θ
,
α
,
a
′
,
b
′
,
c
′
),
(B2)
{
y
∈
Q
(
γ
,
β
,
ψ
,
h
′
,
d
′
,
c
′
)
:
β
(
y
)
>
d
′
}
≠
∅ and β
(
T
y
)
>
d
′ for y
∈
Q
(
γ
,
β
,
ψ
,
h
′
,
d
′
,
c
′
),
(B3)
α
(
T
y
)
>
a
′ provided that y
∈
P
(
γ
,
α
,
a
′
,
c
′
) with θ
(
T
y
)
>
b
′,
(B4)
β
(
T
y
)
<
d
′ provided that y
∈
Q
(
γ
,
β
,
ψ
,
h
′
,
d
′
,
c
′
) with ψ
(
T
y
)
<
h
′.Then, T has at least three fixed points y
1
,
y
2
,
y
3
∈
P
(
γ
,
c
′
)
¯ such that β
(
y
1
)
<
d
′
,
a
′
<
α
(
y
2
) and d
′
<
β
(
y
3
) with α
(
y
3
)
<
a
′.Consider the Banach spaceB
=
E
×
E, where E
=
{
u
:
u
∈
C
[
0,1
]
} equipped with the norm ∥
(
u
,
v
)
∥
=
∥
u
∥
0
+
∥
v
∥
0, for (
u
,
v
)
∈
B and the norm, is defined as
(37)
∥
u
∥
0
=
max
⁡
0
≤
t
≤
1
|
u
(
t
)
|
.
Define a cone P
⊂
B by
(38)
P
=
{
min
⁡
t
∈
I
[
u
(
t
)
+
v
(
t
)
]
≥
η
∥
(
u
,
v
)
∥
(
u
,
v
)
∈
B
∣
u
(
t
)
≥
0
,
v
(
t
)
≥
0
,
t
∈
[
0,1
]
,
min
⁡
t
∈
I
[
u
(
t
)
+
v
(
t
)
]
≥
η
∥
(
u
,
v
)
∥
}
.
Define the nonnegative continuous concave functionals α
,
ψ and the nonnegative continuous convex functionals β, θ, γ on P by
(39)
α
(
u
,
v
)
=
min
⁡
t
∈
I
{
|
u
|
+
|
v
|
}
,
ψ
(
u
,
v
)
=
min
⁡
t
∈
I
1
{
|
u
|
+
|
v
|
}
,
γ
(
u
,
v
)
=
max
⁡
t
∈
[
0,1
]
{
|
u
|
+
|
v
|
}
,
β
(
u
,
v
)
=
max
⁡
t
∈
I
1
{
|
u
|
+
|
v
|
}
,
θ
(
u
,
v
)
=
max
⁡
t
∈
I
{
|
u
|
+
|
v
|
}
,
where I
1
=
[
1
/
3,2
/
3
]. For any (
u
,
v
)
∈
P,
(40)
α
(
u
,
v
)
=
min
⁡
t
∈
I
{
|
u
|
+
|
v
|
}
≤
max
⁡
t
∈
I
1
{
|
u
|
+
|
v
|
}
=
β
(
u
,
v
)
,
∥
(
u
,
v
)
∥
≤
1
η
min
⁡
t
∈
I
{
|
u
|
+
|
v
|
}
≤
1
η
max
⁡
t
∈
[
0,1
]
{
|
u
|
+
|
v
|
}
=
1
η
γ
(
u
,
v
)
.
Let
(41)
L
=
min
⁡
{
(
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
τ
,
τ
)
d
τ
)
d
s
)
-
1
,
(
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
2
(
τ
,
τ
)
d
τ
)
d
s
)
-
1
}
,
M
=
max
⁡
{
(
∫
s
∈
I
‍
η
G
1
(
1
,
s
)
ϕ
q

(
∫
τ
∈
I
‍
γ
*

(
τ
)
H
1

(
τ
,
τ
)
d
τ

)
d
s

)
-
1

,
(
∫
s
∈
I
‍
η
G
2
(
1
,
s
)
ϕ
q
×
(
∫
τ
∈
I
‍
γ
*
(
τ
)
H
2
(
τ
,
τ
)
d
τ
)
d
s
)
-
1
}
.Theorem 9.
Suppose that there exist0
<
a
′
<
b
′
<
b
′
/
η
<
c
′ such that f
i, for i
=
1,2 satisfies the following conditions:(A1)
f
i
(
t
,
u
(
t
)
,
v
(
t
)
)
<
ϕ
p
(
a
′
L
/
2
),  t
∈
[
0,1
] and u
,
v
∈
[
η
a
′
,
a
′
],
(A2)
f
i
(
t
,
u
(
t
)
,
v
(
t
)
)
>
ϕ
p
(
b
′
M
/
2
),  t
∈
I and u
,
v
∈
[
b
′
,
b
′
/
η
],
(A3)
f
i
(
t
,
u
(
t
)
,
v
(
t
)
)
<
ϕ
p
(
c
′
L
/
2
),  t
∈
[
0,1
] and u
,
v
∈
[
0
,
c
′
].Then, the fractional order BVP (2)–(5) has at least three positive solutions, (
x
1
,
x
2
),  (
y
1
,
y
2
), and (
z
1
,
z
2
) such that β
(
x
1
,
x
2
)
<
a
′,  b
′
<
α
(
y
1
,
y
2
) and a
′
<
β
(
z
1
,
z
2
) with α
(
z
1
,
z
2
)
<
b
′.Proof.
LetT
1
,
T
2
:
P
→
E and T
:
P
→
B be the operators defined by
(42)
T
1
(
u
,
v
)
(
t
)
=
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
,
T
2
(
u
,
v
)
(
t
)
=
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
,
T
(
u
,
v
)
(
t
)
=
(
T
1
(
u
,
v
)
(
t
)
,
T
2
(
u
,
v
)
(
t
)
)
,
for
(
u
,
v
)
∈
B
.
It is obvious that a fixed point of T is the solution of the fractional order BVP (2)–(5). Three fixed points of T are sought. First, it is shown that T
:
P
→
P. Let (
u
,
v
)
∈
P. Clearly, T
1
(
u
,
v
)
(
t
)
≥
0 and T
2
(
u
,
v
)
(
t
)
≥
0, for t
∈
[
0,1
]. Also, for (
u
,
v
)
∈
P,
(43)
∥
T
1
(
u
,
v
)
∥
0
≤
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
×
(
∫
0
1
‍

H
1
(
s
,
τ
)

f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)

d
s
,
∥
T
2
(
u
,
v
)
∥
0
≤
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
×
(
∫
0
1
‍

H
2
(
s
,
τ
)

f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)

d
s
,
min
⁡
t
∈
I
⁡
T
1
(
u
,
v
)
(
t
)
=
min
⁡
t
∈
I
∫
0
1
‍

G
1
(
t
,
s
)
ϕ
q
×

(

∫
0
1
‍

H
1

(
s
,
τ
)

f
1

(
τ
,
u
(
τ
)

,
v
(
τ
)
)
d
τ


)

d
s
≥
η
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
×
(
∫
0
1
‍

H
1
(
s
,
τ
)

f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)

d
s
≥
η
∥
T
1
(
u
,
v
)
∥
0
.
Similarly, min
⁡
⁡
t
∈
I
T
2
(
u
,
v
)
(
t
)
≥
η
∥
T
2
(
u
,
v
)
∥
0. Therefore,
(44)
min
⁡
⁡
t
∈
I
{
T
1
(
u
,
v
)
(
t
)
+
T
2
(
u
,
v
)
(
t
)
}
≥
η
∥
T
1
(
u
,
v
)
∥
0
+
η
∥
T
2
(
u
,
v
)
∥
0
=
η
(
∥
T
1
(
u
,
v
)
∥
0
+
∥
T
2
(
u
,
v
)
∥
0
)
=
η
∥
(
T
1
(
u
,
v
)
,
T
2
(
u
,
v
)
)
∥
=
η
∥
T
(
u
,
v
)
∥
.
Hence, T
(
u
,
v
)
∈
P and so T
:
P
→
P. Moreover, T is completely continuous operator. From (40), for each (
u
,
v
)
∈
P, α
(
u
,
v
)
≤
β
(
u
,
v
), and ∥
(
u
,
v
)
∥
≤
(
1
/
η
)
γ
(
u
,
v
). It is shown that T
:
P
(
γ
,
c
′
)
¯
→
P
(
γ
,
c
′
)
¯. Let (
u
,
v
)
∈
P
(
γ
,
c
′
)
¯. Then 0
≤
|
u
|
+
|
v
|
≤
c
′. Condition (A3) is used to obtain
(45)
γ
(
T
(
u
,
v
)
(
t
)
)
=
max
⁡
t
∈
[
0,1
]
[
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
∫
0
1
‍
]
≤
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
s
,
τ
)
ϕ
p
(
c
′
L
2
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
(
∫
0
1
‍
H
2
(
s
,
τ
)
ϕ
p
(
c
′
L
2
)
d
τ
)
d
s
<
c
′
L
2
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
τ
,
τ
)
d
τ
)
d
s
+
c
′
L
2
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
2
(
τ
,
τ
)
d
τ
)
d
s
<
c
′
2
+
c
′
2
=
c
′
.
Therefore T
:
P
(
γ
,
c
′
)
¯
→
P
(
γ
,
c
′
)
¯. Now conditions (B1) and (B2) of Theorem 8 are to be verified. It is obvious that
(46)
b
′

+

(
b
′
/
η
)
2

∈

{

(
u
,
v
)
∈
P

(

γ
,
θ
,
α
,
b
′
,
b
′
η
,
c
′

)
:
α
(
u
,
v
)

>
b
′

}


≠
∅
,
η
a
′
+
a
′
2
∈

{
(
u
,
v
)
∈
Q
(
γ
,
β
,
ψ
,
η
a
′
,
a
′
,
c
′
)
:
β
(
u
,
v
)

<
a
′
}

≠
∅
.
Next, let (
u
,
v
)
∈
P
(
γ
,
θ
,
α
,
b
′
,
b
′
/
η
,
c
′
) or (
u
,
v
)
∈
Q
(
γ
,
β
,
ψ
,
η
a
′
,
a
′
,
c
′
). Then, b
′
≤
|
u
(
t
)
|
+
|
v
(
t
)
|
≤
b
′
/
η and η
a
′
≤
|
u
(
t
)
|
+
|
v
(
t
)
|
≤
a
′. Now, condition (A2) is applied to get
(47)
α
(
T
(
u
,
v
)
(
t
)
)
=
min
⁡
⁡
t
∈
I
[
∫
0
1
‍

G
1
(
t
,
s
)
ϕ
q

(
∫
0
1
‍

H
1
(
s
,
τ
)

f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≥
η
[
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
γ
*
(
τ
)
H
1
(
τ
,
τ
)
ϕ
p
(
b
′
M
2
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
×
(
∫
0
1
‍
γ
*
(
τ
)
H
2
(
τ
,
τ
)
ϕ
p
(
b
′
M
2
)
d
τ
)
d
s
]
>
b
′
M
2
∫
s
∈
I
‍
η
G
1
(
1
,
s
)
ϕ
q
(
∫
τ
∈
I
‍
γ
*
(
τ
)
H
1
(
τ
,
τ
)
d
τ
)
d
s
+
b
′
M
2
∫
s
∈
I
‍
η
G
2
(
1
,
s
)
ϕ
q
(
∫
τ
∈
I
‍
γ
*
(
τ
)
H
2
(
τ
,
τ
)
d
τ
)
d
s
≥
b
′
2
+
b
′
2
=
b
′
.
Clearly, condition (A1) leads to
(48)
β
(
T
(
u
,
v
)
(
t
)
)
=
max
⁡
⁡
t
∈
I
1
[
∫
0
1
‍

G
1

(
t
,
s
)
ϕ
q

(

∫
0
1
‍

H
1
(
s
,
τ
)

f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ


)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≤
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
s
,
τ
)
ϕ
p
(
a
′
L
2
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
2
(
s
,
τ
)
ϕ
p
(
a
′
L
2
)
d
τ
)
d
s
<
a
′
L
2
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
τ
,
τ
)
d
τ
)
d
s
+
a
′
L
2
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
2
(
τ
,
τ
)
d
τ
)
d
s
≤
a
′
2
+
a
′
2
=
a
′
.
To see that (B3) is satisfied, let (
u
,
v
)
∈
P
(
γ
,
α
,
b
′
,
c
′
) with θ
(
T
(
u
,
v
)
(
t
)
)
>
b
′
/
η. Then
(49)
α
(
T
(
u
,
v
)
(
t
)
)
=
min
⁡
⁡
t
∈
I
[
∫
0
1
‍

G
1
(
t
,
s
)
ϕ
q

(
∫
0
1
‍

H
1
(
s
,
τ
)

f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≥
η
[
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≥
η
max
⁡
⁡
t
∈
[
0,1
]
[
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≥
η
max
⁡
⁡
t
∈
I
[
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
=
η
θ
(
T
(
u
,
v
)
(
t
)
)
>
b
′
.
Finally, it is shown that (B4) holds. Let (
u
,
v
)
∈
Q
(
γ
,
β
,
a
′
,
c
′
) with ψ
(
T
(
u
,
v
)
)
<
η
a
′. Then, we have
(50)
β
(
T
(
u
,
v
)
(
t
)
)
=
max
⁡
⁡
t
∈
I
1
[
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q

(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1

(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≤
max
⁡
⁡
t
∈
[
0,1
]

[
∫
0
1
‍
G
1
(
t
,
s
)
ϕ
q

(
∫
0
1
‍
H
1

(
s
,
τ
)
f
1

(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≤
∫
0
1
‍
G
1
(
1
,
s
)
ϕ
q

(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1

(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
=
1
η
[
η
∫
0
1
‍

G
1
(
1
,
s
)
ϕ
q

(
∫
0
1
‍
H
1

(
s
,
τ
)
f
1

(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
η
∫
0
1
‍
G
2
(
1
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≤
1
η
min
⁡
⁡
t
∈
I

[
∫
0
1

‍
G
1
(
t
,
s
)
ϕ
q

(
∫
0
1
‍
H
1

(
s
,
τ
)
f
1

(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
≤
1
η
min
⁡
⁡

t
∈
I
1
[
∫
0
1
‍

G
1
(
t
,
s
)
ϕ
q

(
∫
0
1
‍
H
1
(
s
,
τ
)
f
1

(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ

)
d
s
+
∫
0
1
‍
G
2
(
t
,
s
)
ϕ
q
×
(
∫
0
1
‍
H
2
(
s
,
τ
)
f
2
(
τ
,
u
(
τ
)
,
v
(
τ
)
)
d
τ
)
d
s
]
=
1
η
ψ
(
T
(
u
,
v
)
(
t
)
)
<
a
′
.
It has been proved that all the conditions of Theorem 8 are satisfied. Therefore, the fractional order BVP (2)–(5) has at least three positive solutions, (
x
1
,
x
2
),  (
y
1
,
y
2
), and (
z
1
,
z
2
) such that β
(
x
1
,
x
2
)
<
a
′, b
′
<
α
(
y
1
,
y
2
), and a
′
<
β
(
z
1
,
z
2
) with α
(
z
1
,
z
2
)
<
b
′. This completes the proof of the theorem.


## 4. Example

In this section, as an application, the result is demonstrated with an example.Consider a coupled system ofp-Laplacian fractional order BVP:
(51)
D
0
+
1.7
(
ϕ
p
(
D
0
+
2.6
u
(
t
)
)
)
=
f
1
(
t
,
u
(
t
)
,
v
(
t
)
)
,
t
∈
(
0,1
)
,
D
0
+
1.9
(
ϕ
p
(
D
0
+
2.8
v
(
t
)
)
)
=
f
2
(
t
,
u
(
t
)
,
v
(
t
)
)
,
t
∈
(
0,1
)
,
u
(
0
)
=
D
0
+
0.5
u
(
0
)
=
0
,
5
u
(
1
)
+
8
D
0
+
0.7
u
(
1
)
=
0
,
D
0
+
2.6
u
(
0
)
=
D
0
+
2.6
u
(
1
)
=
0
,
v
(
0
)
=
D
0
+
0.5
v
(
0
)
=
0
,
5
v
(
1
)
+
8
D
0
+
0.7
v
(
1
)
=
0
,
D
0
+
2.8
v
(
0
)
=
D
0
+
2.8
v
(
1
)
=
0
,
where
(52)
f
1
(
t
,
u
,
v
)
=
{
e
t
99
+
sin
⁡
(
u
+
v
)
10
+
87
(
u
+
v
)
3
10
,
0
≤
u
+
v
≤
10
,
e
t
99
+
87000
10
+
sin
⁡
(
u
+
v
)
10
,
u
+
v
>
10
.
f
2
(
t
,
u
,
v
)
=
{
3
e
t
159
+
cos
⁡
⁡
(
u
+
v
)
10
+
93
(
u
+
v
)
3
10
,
0
≤
u
+
v
≤
10
,
3
e
t
159
+
93000
10
+
cos
⁡
⁡
(
u
+
v
)
10
,
u
+
v
>
10
,
Then the Green functions G
i
(
t
,
s
) and H
i
(
t
,
s
), for i
=
1,2, are given by
(53)
G
1
(
t
,
s
)
=
{
1
16.24
[
5
Γ
(
1.9
)
Γ
(
2.6
)
+
8
(
1
-
s
)
-
0.7
]
×
[
t
(
1
-
s
)
]
1.6
,
t
≤
s
,
1
16.24
[
5
Γ
(
1.9
)
Γ
(
2.6
)
+
8
(
1
-
s
)
-
0.7
]
×
[
t
(
1
-
s
)
]
1.6
-
(
t
-
s
)
1.6
Γ
(
2.6
)
,
s
≤
t
,
G
2
(
t
,
s
)
=
{
1
18.64
[
5
Γ
(
2.1
)
Γ
(
2.6
)
+
8
(
1
-
s
)
-
0.7
]
×
[
t
(
1
-
s
)
]
1.8
,
t
≤
s
,
1
18.64
[
5
Γ
(
2.1
)
Γ
(
2.6
)
+
8
(
1
-
s
)
-
0.7
]
×
[
t
(
1
-
s
)
]
1.8
-
(
t
-
s
)
1.8
Γ
(
2.8
)
,
s
≤
t
,
H
1
(
t
,
s
)
=
{
[
t
(
1
-
s
)
]
0.7
-
(
t
-
s
)
0.7
Γ
(
1.7
)
,
s
≤
t
,
[
t
(
1
-
s
)
]
0.7
Γ
(
1.7
)
,
t
≤
s
,
H
2
(
t
,
s
)
=
{
[
t
(
1
-
s
)
]
0.9
-
(
t
-
s
)
0.9
Γ
(
1.9
)
,
s
≤
t
,
[
t
(
1
-
s
)
]
0.9
Γ
(
1.9
)
,
t
≤
s
.
Clearly, the Green functions G
i
(
t
,
s
) and H
i
(
t
,
s
), for i
=
1,2, are positive and f
1,  f
2 are continuous and increasing on [
0
,
∞
). By direct calculations, η
=
0.08,  p
=
2,  L
=
33.16, and M
=
1677.73. Choosing a
′
=
1,  b
′
=
10 and c
′
=
900 and then 0
<
a
′
<
b
′
<
b
′
/
η
≤
c
′ and f
i, for i
=
1,2 satisfies(i)
f
i
(
t
,
u
,
v
)
<
16.5845
=
ϕ
p
(
a
′
L
/
2
),  t
∈
[
0,1
] and u
,
v
∈
[
0.08,1
],
(ii)
f
i
(
t
,
u
,
v
)
>
8388.65
=
ϕ
p
(
b
′
M
/
2
),  t
∈
[
1
/
4,3
/
4
] and u
,
v
∈
[
10,125
],
(iii)
f
i
(
t
,
u
,
v
)
<
14926.09
=
ϕ
p
(
c
′
L
/
2
),  t
∈
[
0,1
] and u
,
v
∈
[
0,900
].Then, all the conditions of Theorem 9 are satisfied. Therefore, it follows from Theorem 8 that the p-Laplacian fractional order BVP (51) has at least three positive solutions.



---

*Source: 485647-2014-05-07.xml*