International Journal of Analysis and Applications
ISSN 2291-8639
Volume 6, Number 1 (2014), 63-81
http://www.etamaths.com

EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS FOR

p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE

PROBLEMS

K. R. PRASAD1 AND B. M. B. KRUSHNA2,∗

Abstract. This paper deals with the existence of at least one and multiple

positive solutions for p-Laplacian fractional order two-point boundary value
problems,

D
q2
0+

(
φp

(
D

q1
0+
y(t)

))
= f(t,y(t)), t ∈ (0, 1),

y(j)(0) = 0, j = 0, 1, 2, · · ·,n− 2, y(q4)(1) = 0,

φp
(
D

q1
0+
y(0)

)
= 0 = D

q3
0+

(
φp

(
D

q1
0+
y(1)

))
,

where q2 ∈ (1, 2],q1 ∈ (n − 1,n],n ≥ 2,q3 ∈ (0, 1], q4 ∈ [1,q1 − 1] is a fixed
integer, φp(y) = |y|p−2y, p > 1, φ−1p = φq, 1/p + 1/q = 1, by applying
Krasnosel’skii and five functionals fixed point theorems.

1. Introduction

The goal of differential equations is to understand the phenomena of nature by
developing mathematical models. Fractional calculus is the field of mathematical
analysis, which deals with investigation and applications of derivatives and inte-
grals of an arbitrary order. Among all, a class of differential equations governed
by nonlinear differential operators appears frequently and generated great deal of
interest in studying such problems. In this theory, the most applicable operator is
the classical p-Laplacian, given by φp(y) = |y|p−2y, p > 1. These types of problems
arise in applied fields such as turbulent flow of a gas in a porous medium, modeling
of viscoelastic flows, biophysics, plasma physics and material science.

The existence of positive solutions of boundary value problems (BVPs) as-
sociated with integer order differential equations were studied by many authors
[11, 1, 9, 2, 19] and extended to p-Laplacian boundary value problems [17, 14, 4, 23].
Later, these results are further extended to fractional order boundary value prob-
lems [6, 10, 5, 7, 8, 20, 21, 22] by utilizing various fixed point theorems on cones. In
recent years, researchers are concentrating on the theory of fractional order bound-
ary value problems related with p-Laplacian operator.

2010 Mathematics Subject Classification. 26A33, 34B18, 35J05.
Key words and phrases. Fractional derivative, p-Laplacian, boundary value problem, two-

point, Green’s function, positive solution.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

63



64 PRASAD AND KRUSHNA

In 2012, Chai [7] investigated the existence and multiplicity of positive solutions
for a class of boundary value problem of fractional differential equation with p-
Laplacian operator,

D
β
0+(φp(D

α
0+u(t))) + f(t,u(t)) = 0, 0 < t < 1,

u(0) = 0,u(1) + σD
γ
0+u(1) = 0,D

α
0+u(0) = 0,

where 1 < α ≤ 2, 0 < β,γ ≤ 1, 0 ≤ α−γ−1,σ is a positive number, Dα
0+
,D

β
0+
,D

γ
0+

are the standard Riemann-Liouville fractional order derivatives.
The purpose of this paper is to establish the existence of at least one and multiple

positive solutions for p-Laplacian fractional order boundary value problems,

(1.1) D
q2
0+

(
φp

(
D
q1
0+
y(t)

))
= f(t,y(t)), t ∈ (0, 1),

(1.2)
y(j)(0) = 0, j = 0, 1, 2, · · ·,n− 2, y(q4)(1) = 0,

φp

(
D
q1
0+
y(0)

)
= 0 = D

q3
0+

(
φp

(
D
q1
0+
y(1)

))
,




where q2 ∈ (1, 2],q1 ∈ (n− 1,n],n ≥ 2,q3 ∈ (0, 1], q4 ∈ [1,q1 − 1] is a fixed integer,
φp(y) = |y|p−2y, p > 1, φ−1p = φq, 1/p + 1/q = 1, f : [0, 1] × R

+ → R+ is
continuous and D

qi
0+
, for i = 1, 2, 3 are the standard Riemann-Liouville fractional

order derivatives.
Define the nonnegative extended real numbers f0,f

0,f∞ and f
∞ by

f0 = lim
y→0+

min
t∈[0,1]

f(t,y)

φp(y)
, f0 = lim

y→0+
max
t∈[0,1]

f(t,y)

φp(y)
,

f∞ = lim
y→∞

min
t∈[0,1]

f(t,y)

φp(y)
and f∞ = lim

y→∞
max
t∈[0,1]

f(t,y)

φp(y)
,

and also assume that they will exist.

The rest of the paper is organized as follows. In Section 2, we construct the Green
functions for the homogeneous BVPs corresponding to (1.1)-(1.2) and estimate the
bounds for the Green functions. In Section 3, we establish criteria for the existence
of at least one positive solution for p-Laplacian fractional order BVP (1.1)-(1.2),
by using Krasnosel’skii fixed point theorem. In Section 4, we derive sufficient
conditions for the existence of at least three positive solutions for the p-Laplacian
fractional order BVP (1.1)-(1.2), by applying five functionals fixed point theorem.
We also establish the existence of at least 2k−1 positive solutions for an arbitrary
positive integer k. In Section 5, as an application, the results are demonstrated
with examples.

2. Green Functions and Bounds

In this section, we construct the Green functions for the homogeneous BVPs
and estimate the bounds for the Green functions, which are essential to establish
the main results.

Let G(t,s) be the Green’s function for the homogeneous BVP,

(2.1) −Dq1
0+
y(t) = 0, t ∈ (0, 1),

(2.2) y(j)(0) = 0, j = 0, 1, · · ·,n− 2, y(q4)(1) = 0.



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 65

Lemma 2.1. Let h(t) ∈ C[0, 1]. Then the fractional order differential equation,
(2.3) D

q1
0+
y(t) + h(t) = 0, t ∈ (0, 1),

satisfying (2.2) has a unique solution,

y(t) =

∫ 1
0

G(t,s)h(s)ds,

where

(2.4) G(t,s) =

{
G1(t,s), 0 ≤ t ≤ s ≤ 1,
G2(t,s), 0 ≤ s ≤ t ≤ 1,

G1(t,s) =
tq1−1(1 −s)q1−q4−1

Γ(q1)
,

G2(t,s) =
tq1−1(1 −s)q1−q4−1 − (t−s)q1−1

Γ(q1)
.

Proof. Let y(t) ∈ C[q1]+1[0, 1] be the solution of fractional order differential equa-
tion (2.3) satisfying (2.2). Then

y(t) =
−1

Γ(q1)

∫ t
0

(t−s)q1−1h(s)ds + c1tq1−1 + c2tq1−2 + · · · + cntq1−n.

From (2.2), ci = 0, i = 2, 3, · · ·,n and c1 =
∫ 1

0
(1−s)q1−q4−1

Γ(q1)
h(s)ds. Thus, the unique

solution of (2.3) with (2.2) is

y(t) =
−1

Γ(q1)

∫ t
0

(t−s)q1−1h(s)ds +
tq1−1

Γ(q1)

∫ 1
0

(1 −s)q1−q4−1h(s)ds

=

∫ 1
0

G(t,s)h(s)ds,

where G(t,s) is the Green’s function and given in (2.4). �

Lemma 2.2. Let z(t) ∈ C[0, 1]. Then the fractional order differential equation,

(2.5) D
q2
0+

(
φp

(
D
q1
0+
y(t)

))
= z(t), t ∈ (0, 1),

satisfying

(2.6) φp

(
D
q1
0+
y(0)

)
= 0,D

q3
0+

(
φp

(
D
q1
0+
y(1)

))
= 0,

has a unique solution,

y(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)z(τ)dτ
)
ds,

where

(2.7) H(t,s) =

{
tq2−1(1−s)q2−q3−1

Γ(q2)
, 0 ≤ t ≤ s ≤ 1,

tq2−1(1−s)q2−q3−1−(t−s)q2−1
Γ(q2)

, 0 ≤ s ≤ t ≤ 1.

Here H(t,s) is the Green’s function for

−Dq2
0+

(
φp(x(t))

)
= 0, t ∈ (0, 1),

φp(x(0)) = 0, D
q3
0+

(
φp(x(1))

)
= 0.



66 PRASAD AND KRUSHNA

Proof. An equivalent integral equation for (2.5) is given by

φp

(
D
q1
0+
y(t)

)
=

1

Γ(q2)

∫ t
0

(t− τ)q2−1z(τ)dτ + c1tq2−1 + c2tq2−2.

By (2.6), one can determine c2 = 0 and c1 =
−1

Γ(q2)

∫ 1
0

(1 − τ)q2−q3−1z(τ)dτ. Then,

φp

(
D
q1
0+
y(t)

)
=

1

Γ(q2)

∫ t
0

(t− τ)q2−1z(τ)dτ −
tq2−1

Γ(q2)

∫ 1
0

(1 − τ)q2−q3−1z(τ)dτ

= −
∫ 1

0

H(t,τ)z(τ)dτ.

Therefore, φ−1p

(
φp

(
D
q1
0+
y(t)

))
= −φ−1p

(∫ 1
0
H(t,τ)z(τ)dτ

)
. Consequently,

D
q1
0+
y(t) + φq

(∫ 1
0
H(t,τ)z(τ)dτ

)
= 0.

Hence,

y(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)z(τ)dτ
)
ds

is the solution of fractional order BVP (2.5), (1.2). �

Lemma 2.3. For t ∈ I =
[

1
4
, 3

4

]
, the Green’s function G(t,s) given in (2.4) satisfies

the following inequalities

(i) G(t,s) ≥ 0, for all (t,s) ∈ [0, 1] × [0, 1],
(ii) G(t,s) ≤ G(1,s), for all (t,s) ∈ [0, 1] × [0, 1],

(iii) G(t,s) ≥ ξG(1,s), for all (t,s) ∈ I × [0, 1],

where ξ =
(

1
4

)q1−1
.

Proof. The Green’s function G(t,s) of (2.1), (2.2) is given in (2.4).
For 0 ≤ t ≤ s ≤ 1,

G1(t,s) =
1

Γ(q1)

[
tq1−1(1 −s)q1−q4−1

]
≥ 0.

For 0 ≤ s ≤ t ≤ 1,

G2(t,s) =
1

Γ(q1)

[
tq1−1(1 −s)q1−q4−1 − (t−s)q1−1

]
≥

1

Γ(q1)

[
tq1−1(1 −s)q1−q4−1 − (t− ts)q1−1

]
=
tq1−1

Γ(q1)

[(
1 + δs +

q4(q4 + 1)

2
s2 + · · ·

)
− 1
]
(1 −s)q1−1

=
tq1−1

Γ(q1)

[
q4s + O(s

2)
]
(1 −s)q1−1 ≥ 0.

Hence the Green’s function G(t,s) is nonnegative.
For 0 ≤ t ≤ s ≤ 1,

∂G1(t,s)

∂t
=

1

Γ(q1)

[
(q1 − 1)tq1−2(1 −s)q1−q4−1

]
≥ 0.



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 67

Therefore, G1(t,s) is increasing in t, which implies G1(t,s) ≤ G1(1,s). Now, for
0 ≤ s ≤ t ≤ 1,

∂G2(t,s)

∂t

=
1

Γ(q1)

[
(q1 − 1)tq1−2(1 −s)q1−q4−1 − (q1 − 1)(t−s)q1−2

]
≥

1

Γ(q1)

[
(q1 − 1)tq1−2(1 −s)q1−q4−1 − (q1 − 1)(t− ts)q1−2

]
=

(q1 − 1)tq1−2

Γ(q1)

[
1 −

(
1 − (q4 − 1)s +

(q4 − 1)(q4 − 2)
2

s2 + · · ·
)]

(1 −s)q1−q4−1

=
(q1 − 1)tq1−2

Γ(q1)

[
(q4 − 1)s + O(s2)

]
(1 −s)q1−q4−1 ≥ 0.

Therefore, G2(t,s) is increasing in t, which implies G2(t,s) ≤ G2(1,s).
Let 0 ≤ t ≤ s ≤ 1 and t ∈ I. Then

G1(t,s) =
1

Γ(q1)

[
tq1−1(1 −s)q1−q4−1

]
=tq1−1

1

Γ(q1)

[
(1 −s)q1−q4−1

]
=tq1−1G1(1,s)

≥
(1

4

)q1−1
G1(1,s).

Let 0 ≤ s ≤ t ≤ 1 and t ∈ I. Then

G2(t,s) =
1

Γ(q1)

[
tq1−1(1 −s)q1−q4−1 − (t−s)q1−1

]
≥

1

Γ(q1)

[
tq1−1(1 −s)q1−q4−1 − (t− ts)q1−1

]
=tq1−1G2(1,s)

≥
(1

4

)q1−1
G2(1,s).

Hence the result. �

Lemma 2.4. For t,s ∈ [0, 1], the Green’s function H(t,s) given in (2.7) satisfies
the following inequalities

(i) H(t,s) ≥ 0,
(ii) H(t,s) ≤ H(s,s).

Proof. The Green’s function H(t,s) is given in (2.7). For 0 ≤ t ≤ s ≤ 1,

H(t,s) =
1

Γ(q2)

[
tq2−1(1 −s)q2−q3−1

]
≥ 0.



68 PRASAD AND KRUSHNA

For 0 ≤ s ≤ t ≤ 1,

H(t,s) =
1

Γ(q2)

[
tq2−1(1 −s)q2−q3−1 − (t−s)q2−1

]
≥

1

Γ(q2)

[
tq2−1(1 −s)q2−q3−1 − (t− ts)q2−1

]
=
tq2−1

Γ(q2)

[
(1 −s)−q3 − 1

]
(1 −s)q2−1

=
tq2−1

Γ(q2)

[(
1 + q3s +

q3(q3 + 1)

2
s2 + · · ·

)
− 1
]
(1 −s)q2−1

=
tq2−1

Γ(q2)

[
q3s + O(s

2)
]
(1 −s)q2−1 ≥ 0.

For 0 ≤ t ≤ s ≤ 1,

∂H(t,s)

∂t
=

1

Γ(q2)
[(q2 − 1)tq2−2(1 −s)q2−q3−1] ≥ 0.

Therefore, H(t,s) is increasing in t, for s ∈ [0, 1], which implies H(t,s) ≤ H(s,s).
Now, for 0 ≤ s ≤ t ≤ 1,

∂H(t,s)

∂t
=

1

Γ(q2)

[
(q2 − 1)tq2−2(1 −s)q2−q3−1 − (q2 − 1)(t−s)q2−2

]
≤

(q2 − 1)
Γ(q2)

[
(1 −s)q2−q3−1 − (1 −s)q2−2

]
=

(q2 − 1)
Γ(q2)

[
(1 −s)−q3+1 − 1

]
(1 −s)q2−2

=
(q2 − 1)
Γ(q2)

[(
1 − (1 −q3)s +

(1 −q3)(−q3)
2

s2 + · · ·
)
− 1
]
(1 −s)q2−2

=
(q2 − 1)
Γ(q2)

[
(q3 − 1)s + O(s2)

]
(1 −s)q2−2 ≤ 0.

Therefore, H(t,s) is decreasing in t, for s ∈ [0, 1] which implies H(t,s) ≤ H(s,s).
Hence the result. �

Lemma 2.5. Let µ ∈ ( 1
4
, 3

4
). Then the Green’s function H(t,s) holds the inequality,

(2.8) min
t∈I

H(t,s) ≥ δ∗(s)H(s,s), for 0 < s < 1,

where

(2.9) δ∗(s) =

{
( 3
4

)q2−1(1−s)q2−q3−1−( 3
4
−s)q2−1

sq2−1(1−s)q2−q3−1 , s ∈ (0,µ],
1

(4s)q2−1
, s ∈ [µ, 1).

Proof. For s ∈ (0, 1), H(t,s) is increasing in t for t ≤ s and decreasing in t for
s ≤ t.



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 69

We define

h1(t,s) =
[tq2−1(1 −s)q2−q3−1 − (t−s)q2−1]

Γ(q2)

h2(t,s) =
[tq2−1(1 −s)q2−q3−1]

Γ(q2)
, and

H(s,s) =
1

Γ(q2)

[
sq2−1(1 −s)q2−q3−1

]
.

Then,

min
t∈I

H(t,s) =




h1(
3
4
,s), s ∈ (0, 1

4
],

min{h1( 34,s),h2(
1
4
,s)}, s ∈ [ 1

4
, 3

4
],

h2(
1
4
,s), s ∈ [ 3

4
, 1),

=

{
h1(

3
4
,s), s ∈ (0,µ],

h2(
1
4
,s), s ∈ [µ, 1),

=




( 3
4

)q2−1(1−s)q2−q3−1−( 3
4
−s)q2−1

Γ(q2)
, s ∈ (0,µ],

1
Γ(q2)

(1−s)q2−q3−1
4q2−1

, s ∈ [µ, 1),

≥

{
( 3
4

)q2−1(1−s)q2−q3−1−( 3
4
−s)q2−1

sq2−1(1−s)q2−q3−1 H(s,s), s ∈ (0,µ],
1

(4s)q2−1
H(s,s), s ∈ [µ, 1),

= δ∗(s)H(s,s).

�

Let B = {y : y ∈ C[0, 1]} be the real Banach space equipped with the norm

‖y‖ = max
t∈[0,1]

|y(t)|.

Define a cone P ⊂ B by

P =
{
y ∈ B : y(t) ≥ 0, t ∈ [0, 1] and min

t∈I
y(t) ≥ ξ‖y‖

}
.

Let

K =
1∫ 1

0
G(1,s)φq

(∫ 1
0
H(τ,τ)dτ

)
ds

and

L =
1∫

s∈I ξG(1,s)φq

(∫
τ∈I δ

∗(τ)H(τ,τ)dτ
)
ds
.

Let T : P → B be the operator defined by

(2.10) Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds.

If y ∈ P is a fixed point of T, then y satisfies (2.10) and hence y is a positive
solution of the p-Laplacian fractional order BVP (1.1)-(1.2).

Lemma 2.6. The operator T defined by (2.10) is a self map on P .



70 PRASAD AND KRUSHNA

Proof. Let y ∈ P . Clearly, Ty(t) ≥ 0, for all t ∈ [0, 1], and

Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

so that

‖Ty‖≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds.

Next, if y ∈ P , then by the above inequality we have

min
t∈I

Ty(t) = min
t∈I

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≥ ξ
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≥ ξ‖Ty‖.

Hence, Ty ∈ P and so T : P → P. Standard arguments involving the Arzela-Ascoli
theorem shows that T is completely continuous. �

3. Existence of at Least One Positive Solution

In this section, we establish criteria for the existence of at least one positive
solution of the p-Laplacian fractional order BVP (1.1)-(1.2) by using Krasnosel’skii
fixed point theorem.

To establish the existence of at least one positive solution for p-Laplacian frac-
tional order BVP (1.1)-(1.2) by employing the following Krasnosel’skii fixed point
theorem.

Theorem 3.1. [15] Let X be a Banach Space, P ⊆ X be a cone and suppose
that Ω1, Ω2 are open subsets of X with 0 ∈ Ω1 and Ω1 ⊂ Ω2. Suppose further that
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 3.2. If f0 = 0 and f∞ = ∞, then the p-Laplacian fractional order BVP
(1.1)-(1.2) has at least one positive solution that lies in P .

Proof. Let T be the cone preserving, completely continuous operator defined by
(2.10). Since f0 = 0, we may choose H1 > 0 so that

max
t∈[0,1]

f(t,y)

φp(y)
≤ η1, for 0 < y ≤ H1,

where η1 > 0 satisfies

η
q−1
1

∫ 1
0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
ds = η

q−1
1

1

K
≤ 1.



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 71

Thus, if y ∈ P and ‖y‖ = H1, then we have

Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)η1φp(y)dτ
)
ds

=

∫ 1
0

G(1,s)η
q−1
1 φq

(∫ 1
0

H(τ,τ)dτ
)
yds

≤ ηq−11
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
ds‖y‖

= η
q−1
1

1

K
‖y‖≤‖y‖.

Therefore,

‖Ty‖≤‖y‖.

Now, if we let

Ω1 = {y ∈ B : ‖y‖ < H1},

then

(3.1) ‖Ty‖≤‖y‖, for y ∈ P ∩∂Ω1.

Further, since f∞ = ∞, there exists H
2
> 0 such that

min
t∈[0,1]

f(t,y)

φp(y)
≥ η2, for y ≥ H

2
,

where η2 > 0 is chosen such that

η
q−1
2 ξ

2

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ds = η

q−1
2 ξ

2 1

L
≥ 1.

Let

H2 = max
{

2H1,
1

ξ
H

2
}
,

and define

Ω2 = {y ∈ B : ‖y‖ < H2}.

If y ∈ P ∩∂Ω2 and ‖y‖ = H2, then

min
t∈I

y(t) ≥ ξ‖y‖≥ H
2
,



72 PRASAD AND KRUSHNA

and so

Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≥
∫
s∈I

ξG(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≥ ξ
∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)η2φp(y)dτ
)
ds

= ξη
q−1
2

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
yds

≥ ξηq−12
∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ξ‖y‖ds

= ξ2η
q−1
2

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ds‖y‖

= η
q−1
2 ξ

2 1

L
‖y‖

≥‖y‖.

Hence,

(3.2) ‖Ty‖≥‖y‖, for y ∈ P ∩∂Ω2.

An application of Theorem 3.1 to (3.1) and (3.2) yields a fixed point of T that lies
in P ∩ (Ω2\Ω1). This fixed point is the solution of the p-Laplacian fractional order
BVP (1.1)-(1.2). �

Theorem 3.3. If f0 = ∞ and f∞ = 0, then the p-Laplacian fractional order BVP
(1.1)-(1.2) has at least one positive solution that lies in P .

Proof. Let T be the cone preserving, completely continuous operator defined by
(2.10). Since f0 = ∞, we choose J1 > 0 such that

min
t∈[0,1]

f(t,y)

φp(y)
≥ η1, for 0 < y ≤ J

1,

where η1 > 0 satisfies

(η1)
q−1ξ2

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ds = (η1)

q−1ξ2
1

L
≥ 1.

In this case, we define

Ω1 = {y ∈ B : ‖y‖ < J1},



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 73

we have f(τ,y) ≥ η1φp(y),τ ∈ I, and moreover y(t) ≥ ξ‖y‖, t ∈ I
and so

Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≥
∫
s∈I

ξG(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≥ ξ
∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)η1φp(y)dτ
)
ds

= ξ(η1)
q−1

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
yds

≥ ξ(η1)
q−1

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ξ‖y‖ds

≥ ξ2(η1)
q−1

∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ds‖y‖

= (η1)
q−1ξ2

1

L
‖y‖

≥‖y‖.

Thus,

(3.3) ‖Ty‖≥‖y‖, for y ∈ P ∩∂Ω1.

Now, since f∞ = 0, there exists J
2
> 0 such that

max
t∈[0,1]

f(t,y)

φp(y)
≤ η2, for y ≥ J

2
,

where η2 > 0 satisfies

(η2)
q−1

∫ 1
0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
ds = (η2)

q−1 1

K
≤ 1.

It follows that

f(t,y) ≤ η2φp(y), for y ≥ J
2
.

We establish the result in two subcases.
Case(i) : f is bounded. Suppose N > 0 is such that maxt∈[0,1] f(t,y) ≤ N, for all 0 <
y < ∞. In this case, we may choose

J2 = max
{

2J1,Nq−1
1

K

}
,

and let

Ω2 = {y ∈ B : ‖y‖ < J2}.



74 PRASAD AND KRUSHNA

Then for y ∈ P ∩∂Ω2 and ‖y‖ = J2, we have

Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)Ndτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
Nq−1ds

≤ Nq−1
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
ds

= Nq−1
1

K
= J2 = ‖y‖,

and therefore

(3.4) ‖Ty‖≤‖y‖, for y ∈ P ∩∂Ω2.

Case(ii) : f is unbounded. Let J2 > max{2J1,J
2
} be such that f(t,y) ≤ f(t,J2),

for 0 < y ≤ J2, and let Ω2 = {y ∈ B : ‖y‖ < J2}. Choosing y ∈ P ∩ ∂Ω2 with
‖y‖ = J2, we have

Ty(t) =

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y)dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)f(τ,J2)dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)η2φp(J
2)dτ

)
ds

= (η2)
q−1

∫ 1
0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
dsJ2

= (η2)
q−1 1

K
J2

≤ J2 = ‖y‖.

And so

(3.5) ‖Ty‖≤‖y‖, for y ∈ P ∩∂Ω2.

An application of Theorem 3.1 to (3.3), (3.4) and (3.5) yields a fixed point of T that
lies in P ∩ (Ω2\Ω1). This fixed point is the solution of the p-Laplacian fractional
order BVP (1.1)-(1.2). �

4. Existence of Multiple Positive Solutions

In this section, we derive sufficient conditions for the existence of at least three
positive solutions for the p-Laplacian fractional order BVP (1.1)-(1.2), by applying
five functionals fixed point theorem. We also establish the existence of at least



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 75

2k − 1 positive solutions for an arbitrary positive integer k.

Let γ,β,θ be nonnegative continuous convex functionals on P and α,ψ be
nonnegative continuous concave functionals on P , then for nonnegative numbers
h′,a′,b′,d′ and c′, convex sets are defined.

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 for the p-Laplacian fractional order B-
VP (1.1)-(1.2), the following so called Five Functionals Fixed Point Theorem is
fundamental.

Theorem 4.1. [3] Let P be a cone in the real Banach space B. Suppose α and
ψ are nonnegative continuous concave functionals on P and γ,β,θ are nonnega-
tive 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.

(D1) {y ∈ P(γ,θ,α,a′,b′,c′) : α(y) > a′} 6= ∅ and
α(Ty) > a′ for y ∈ P(γ,θ,α,a′,b′,c′),

(D2) {y ∈ Q(γ,β,ψ,h′,d′,c′) : β(y) > d′} 6= ∅ and
β(Ty) > d′ for y ∈ Q(γ,β,ψ,h′,d′,c′),

(D3) α(Ty) > a′ provided y ∈ P(γ,α,a′,c′) with θ(Ty) > b′,
(D4) β(Ty) < d′ provided y ∈ Q(γ,β,ψ,h′,d′,c′) with ψ(Ty) < h′.

Then T has at least three fixed points y1,y2,y3 ∈ P(γ,c′) such that β(y1) < d′,a′ <
α(y2) and d

′ < β(y3) with α(y3) < a
′.

Define the nonnegative continuous concave functionals α,ψ and the nonnegative
continuous convex functionals β,θ,γ on P by

α(y) = min
t∈I

y(t),ψ(y) = min
t∈I1

y(t),

γ(y) = max
t∈[0,1]

y(t),β(y) = max
t∈I1

y(t),θ(y) = max
t∈I

y(t),

where I1 =
[

1
3
, 2

3

]
. For any y ∈ P ,

(4.1) α(y) = min
t∈I

y(t) ≤ max
t∈I1

y(t) = β(y)

and

(4.2) ‖y‖≤
1

ξ
min
t∈I

y(t) ≤
1

ξ
max
t∈[0,1]

y(t) =
1

ξ
γ(y).



76 PRASAD AND KRUSHNA

Theorem 4.2. Suppose there exist 0 < a′ < b′ < b
′

ξ
≤ c′ such that f satisfies the

following conditions:

(C1) f(t,y(t)) < φp

(
a′K

)
, t ∈ [0, 1] and y ∈ [ξa′,a′],

(C2) f(t,y(t)) > φp

(
b′L
)
, t ∈ I and y ∈

[
b′,
b′

ξ

]
,

(C3) f(t,y(t)) < φp

(
c′K
)
, t ∈ [0, 1] and y ∈ [0,c′].

Then the p-Laplacian fractional order BVP (1.1)-(1.2) has at least three positive
solutions y1, y2 and y3 such that β(y1) < a

′, b′ < α(y2) and a
′ < β(y3) with

α(y3) < b
′.

Proof. We seek three fixed points y1,y2,y3 ∈ P of T defined by (2.10). From
Lemma 2.6, (4.1) and (4.2), for each y ∈ P , α(y) ≤ β(y) and ‖y‖ ≤ 1

ξ
γ(y). We

show that T : P(γ,c′) → P(γ,c′). Let y ∈ P(γ,c′). Then 0 ≤ y ≤ c′. We may use
the condition (C3) to obtain

γ(Ty) = max
t∈[0,1]

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)f(τ,y(τ))dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)φp

(
c′K
)
dτ
)
ds

< c′K
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
ds = c′.

Therefore T : P(γ,c′) → P(γ,c′). Now we verify the conditions (D1) and (D2) of
Theorem 4.1 are satisfied. It is obvious that

b′ + b
′

ξ

2
∈
{
y ∈ P

(
γ,θ,α,b′,

b′

ξ
,c′
)

: α(y) > b′
}
6= ∅

and

ξa′ + a′

2
∈
{
y ∈ Q

(
γ,β,ψ,ξa′,a′,c′

)
: β(y) < a′

}
6= ∅.

Next, let y ∈ P(γ,θ,α,b′, b
′

ξ
,c′) or y ∈ Q(γ,β,ψ,ξa′,a′,c′). Then, b′ ≤ y ≤ b

′

ξ
and

ηa′ ≤ y ≤ a′. Now, we may apply the condition (C2) to get

α(Ty) = min
t∈I

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≥ ξ
∫
s∈I

G(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)φp

(
b′L
)
dτ
)
ds

> b′L
∫
s∈I

ξG(1,s)φq

(∫
τ∈I

δ∗(τ)H(τ,τ)dτ
)
ds = b′.



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 77

Clearly, by the condition (C1), we have

β(Ty) = max
t∈I1

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)φp

(
a′K

)
dτ
)
ds

< a′K
∫ 1

0

G(1,s)φq

(∫ 1
0

H(τ,τ)dτ
)
ds

= a′.

To see that (D3) is satisfied, let y ∈ P(γ,α,b′,c′) with θ(Ty) > b
′

ξ
. Then

α(Ty) = min
t∈I

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≥ ξ
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≥ ξ max
t∈[0,1]

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≥ ξ max
t∈I

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

= ξθ(Ty)

> b′.

Finally, we show that (D4) holds. Let y ∈ Q(γ,β,a′,c′) with ψ(Ty) < ξa′. Then,
we have

β(Ty) = max
t∈I1

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≤ max
t∈[0,1]

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≤
∫ 1

0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

=
1

ξ

[
ξ

∫ 1
0

G(1,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)]

≤
1

ξ
min
t∈I

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

≤
1

ξ
min
t∈I1

∫ 1
0

G(t,s)φq

(∫ 1
0

H(s,τ)f(τ,y(τ))dτ
)
ds

=
1

ξ
ψ(Ty) < a′.

We have proved that all the conditions of Theorem 4.1 are satisfied. Therefore, the
p-Laplacian fractional order BVP (1.1)-(1.2) has at least three positive solutions
y1,y2 and y3 such that β(y1) < a

′, b′ < α(y2) and a
′ < β(y3) with α(y3) < b

′. This
completes the proof. �



78 PRASAD AND KRUSHNA

Theorem 4.3. Let k be an arbitrary positive integer. Assume that there exist
numbers ar(r = 1, 2, · · ·,k) and bs(s = 1, 2, · · ·,k − 1) with 0 < a1 < b1 < b1ξ <
a2 < b2 <

b2
ξ
< · · · < ak−1 < bk−1 <

bk−1
ξ

< ak such that f satisfies the following

conditions:

(C4) f(t,y(t)) < φp

(
arK

)
, t ∈ [0, 1] and y ∈ [ξar,ar],r = 1, 2, · · ·,k,

(C5) f(t,y(t)) > φp

(
bsL
)
, t ∈ I and y ∈

[
bs,

bs
ξ

]
,s = 1, 2, · · ·,k − 1.

Then the p-Laplacian fractional order BVP (1.1)-(1.2) has at least 2k − 1 positive
solutions in Pak.

Proof. We use induction on k. First, for k = 1, we know from the condition (C4)
that T : Pa1 → Pa1, then it follows from the Schauder fixed point theorem that
the p-Laplacian fractional order BVP (1.1)-(1.2) has at least one positive solution
in Pa1. Next, we assume that this conclusion holds for k = l. In order to prove
that this conclusion holds for k = l + 1. We suppose that there exist numbers
ar(r = 1, 2, · · ·, l + 1) and bs(s = 1, 2, · · ·, l) with 0 < a1 < b1 < b1ξ < a2 < b2 <
b2
ξ
< · · · < al < bl < blξ < al+1 such that f satisfies the following conditions:

(4.3) f(t,y(t)) < φp

(
arK

)
, t ∈ [0, 1] and y ∈ [ξar,ar],r = 1, 2, · · ·, l + 1,

(4.4) f(t,y(t)) > φp

(
bsL
)
, t ∈ I and y ∈

[
bs,

bs
ξ

]
,s = 1, 2, · · ·, l.

By assumption, the fractional order BVP (1.1)-(1.2) has at least 2l − 1 positive
solutions y∗i (i = 1, 2, · · ·, 2l− 1) in Pal. At the same time, it follows from Theorem
4.2, (4.3) and (4.4) that the p-Laplacian fractional order BVP (1.1)-(1.2) has at
least three positive solutions y1, y2 and y3 in Pal+1 such that β(y1) < al, bl <
α(y2) and al < β(y3) with α(y3) < bl. Obviously y2 and y3 are distinct from y

∗
i (i =

1, 2, · · ·, 2l− 1) in Pal. Therefore, the p-Laplacian fractional order BVP (1.1)-(1.2)
has at least 2l + 1 positive solutions in Pal+1, which shows that this conclusion also
holds for k = l + 1. This completes the proof. �

5. Examples

In this section, as an application, the results of the previous sections are demon-
strated with examples.

Example 5.1 Consider the p-Laplacian fractional order BVP,

(5.1) D1.70+
(
φp

(
D2.60+ y(t)

))
= f(t,y(t)), t ∈ (0, 1),

(5.2) y(0) = y′(0) = y′′(1) = 0,φp

(
D2.60+ y(0)

)
= 0 = D0.60+

(
φp

(
D2.60+ y(1)

))
,

where

f(t,y) =
y2(650 − 649e−3y)

5
.



p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 79

Then the Green’s function G(t,s) for the homogeneous BVP,

−D2.60+ y(t) = 0, t ∈ (0, 1),
y(0) = y′(0) = y′′(1) = 0,

and is given by

G(t,s) =

{
t1.6(1−s)−0.6

Γ(2.6)
, t ≤ s,

t1.6(1−s)−0.6−(t−s)1.6
Γ(2.6)

, s ≤ t.
The Green’s function H(t,s) for the BVP,

−D1.70+
(
φp

(
D2.60+ y(t)

))
= 0, t ∈ (0, 1),

φp

(
D2.60+ y(0)

)
= 0 = D0.60+

(
φp

(
D2.60+ y(1)

))
,

and is given by

H(t,s) =

{
t0.7(1−s)0.1

Γ(1.7)
, t ≤ s,

t0.7(1−s)0.1−(t−s)0.7
Γ(1.7)

, s ≤ t.
Clearly, the Green functions G(t,s) and H(t,s) are positive. Let p = 2. By direct
calculations, ξ = 0.1088, K = 1.0077, L = 27.1209, f0 = 0 and f∞ = ∞. Then, all
the conditions of Theorem 3.2 are satisfied. Thus by Theorem 3.2, the p-Laplacian
fractional order BVP (5.1)-(5.2) has at least one positive solution.

Example 5.2 Consider the p-Laplacian fractional order BVP,

(5.3) D1.80+
(
φp

(
D2.70+ y(t)

))
= f(t,y(t)), t ∈ (0, 1),

(5.4) y(0) = y′(0) = y′′(1) = 0,φp

(
D2.70+ y(0)

)
= 0 = D0.70+

(
φp

(
D2.70+ y(1)

))
,

where

f(t,y) =

{
t

100
+ 15

32
y7, 0 ≤ y ≤ 2,

y + t
100

+ 116
2
, y > 2.

Then the Green’s function G(t,s) for the homogeneous BVP,

−D2.70+ y(t) = 0, t ∈ (0, 1),
y(0) = y′(0) = y′′(1) = 0,

and is given by

G(t,s) =

{
t1.7(1−s)−0.7

Γ(2.7)
, t ≤ s,

t1.7(1−s)−0.7−(t−s)1.7
Γ(2.7)

, s ≤ t,
The Green’s function H(t,s) for the BVP,

−D1.80+
(
φp

(
D2.70+ y(t)

))
= 0, t ∈ (0, 1),

φp

(
D2.70+ y(0)

)
= 0 = D0.70+

(
φp

(
D2.70+ y(1)

))
,

and is given by

H(t,s) =

{
t0.8(1−s)0.1

Γ(1.8)
, t ≤ s,

t0.8(1−s)0.1−(t−s)0.8
Γ(1.8)

, s ≤ t.



80 PRASAD AND KRUSHNA

Clearly, the Green functions G(t,s) and H(t,s) are positive and f is continuous
and increasing on [0,∞). Let p = 2. By direct calculations, ξ = 0.0947, K = 1.8901
and L = 29.6238. Choosing a′ = 1,b′ = 2 and c′ = 100, then 0 < a′ < b′ < b

′

ξ
≤ c′

and f satisfies

(i) f(t,y) < 1.8901 = φp

(
a′K

)
, t ∈ [0, 1] and y ∈ [0.0947, 1],

(ii) f(t,y) > 59.25 = φp

(
b′L
)
, t ∈

[1
4
,

3

4

]
and y ∈ [2, 21.12],

(iii) f(t,y) < 189.01 = φp

(
c′K
)
, t ∈ [0, 1] and y ∈ [0, 100].

Then, all the conditions of Theorem 4.2 are satisfied. Thus by Theorem 4.2, the
p-Laplacian fractional order BVP (5.3)-(5.4) has at least three positive solutions
y1,y2 and y3 satisfying

max
t∈[ 1

3
, 2
3

]
y1(t) < 1, 2 < min

t∈[ 1
4
, 3
4

]
y2(t),

1 < max
t∈[ 1

3
, 2
3

]
y3(t), min

t∈[ 1
4
, 3
4

]
y3(t) < 2.

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p-LAPLACIAN FRACTIONAL ORDER BOUNDARY VALUE PROBLEMS 81

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1Department of Applied Mathematics, Andhra University, Visakhapatnam, 530 003,

India,

2Department of Mathematics, MVGR College of Engineering, Vizianagaram, 535 005,

India,

∗Corresponding author