Research Article Multiplicity of Positive Solutions for ... - Hindawi

Hindawi Publishing Corporation International Journal of Differential Equations Volume 2016, Article ID 6906049, 10 pages http://dx.doi.org/10.1155/2016/6906049

Research Article Multiplicity of Positive Solutions for Fractional Differential Equation with 𝑝-Laplacian Boundary Value Problems Sabbavarapu Nageswara Rao Department of Mathematics, Jazan University, P.O. Box 114, Jazan, Saudi Arabia Correspondence should be addressed to Sabbavarapu Nageswara Rao; sabbavarapu [email protected] Received 15 October 2015; Accepted 6 April 2016 Academic Editor: Kanishka Perera Copyright © 2016 Sabbavarapu Nageswara Rao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We investigate the existence of multiple positive solutions of fractional differential equations with 𝑝-Laplacian operator 𝛽 𝐷𝑎+ (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢(𝑡))) = 𝑓(𝑡, 𝑢(𝑡)), 𝑎 < 𝑡 < 𝑏, 𝑢(𝑗) (𝑎) = 0, 𝑗 = 0, 1, 2, . . . , 𝑛 − 2, 𝑢(𝛼1 ) (𝑏) = 𝜉𝑢(𝛼1 ) (𝜂), 𝜙𝑝 (𝐷𝑎𝛼+ 𝑢(𝑎)) = 0 = 𝛽

𝐷𝑎+1 (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢(𝑏))), where 𝛽 ∈ (1, 2], 𝛼 ∈ (𝑛 − 1, 𝑛], 𝑛 ≥ 3, 𝜉 ∈ (0, ∞), 𝜂 ∈ (𝑎, 𝑏), 𝛽1 ∈ (0, 1], 𝛼1 ∈ {1, 2, . . . , 𝛼 − 2} is a fixed integer, and 𝜙𝑝 (𝑠) = |𝑠|𝑝−2 𝑠, 𝑝 > 1, 𝜙𝑝−1 = 𝜙𝑞 , (1/𝑝) + (1/𝑞) = 1, by applying Leggett–Williams fixed point theorems and fixed point index theory.

1. Introduction

with the boundary conditions

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 integrals of an arbitrary order. Among all, a class of differential equations governed by nonlinear differential operators appears frequently and generated a great deal of interest in studying such problems. In this theory, the most applicable operator is the classical 𝑝-Laplacian, given by 𝜙𝑝 (𝑢) = |𝑢|𝑝−2 , 𝑝 > 1. The positive solutions of boundary value problems associated with ordinary differential equations were studied by many authors [1–4] and extended to 𝑝-Laplacian boundary value problems [5–8]. Later, these results are further extended to fractional order boundary value problems [9–12] 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 [13–19]. The above few papers motivated this work. In this paper, we are concerned with the existence of multiple positive solutions for the fractional differential equation with 𝑝-Laplacian operator 𝛽

𝐷𝑎+ (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑡))) = 𝑓 (𝑡, 𝑢 (𝑡)) , 𝑎 < 𝑡 < 𝑏,

(1)

𝑢(𝑗) (𝑎) = 0,

𝑗 = 0, 1, 2, . . . , 𝑛 − 2,

𝑢(𝛼1 ) (𝑏) = 𝜉𝑢(𝛼1 ) (𝜂) ,

(2)

𝛽

𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑎)) = 𝐷𝑎+1 (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑏))) = 0, where 𝜙𝑝 (𝑠) = |𝑠|𝑝−2 𝑠, 𝑝 > 1, 𝜙𝑝−1 = 𝜙𝑞 , (1/𝑝) + (1/𝑞) = 1, 𝛽 ∈ (1, 2], 𝛼 ∈ (𝑛 − 1, 𝑛], 𝑛 ≥ 3, 𝛽1 ∈ (0, 1], 𝛼1 ∈ [1, 𝛼 − 2] is a fixed integer, and 𝜉 ∈ (0, ∞), 𝜂 ∈ (𝑎, 𝑏) are constants. The function 𝑓 : [𝑎, 𝑏] × 𝑅+ → 𝑅+ is continuous and 𝐷𝑎𝛼+ , 𝛽 𝛽 𝐷𝑎+ , 𝐷𝑎+1 are the standard Riemann-Liouville fractional order derivatives. The rest of this paper is organized as follows. In Section 2, the Green functions for the homogeneous BVPs corresponding to (1)-(2) are constructed and the bounds for the Green functions are estimated. In Section 3, sufficient conditions for the existence of at least two or at least three positive solutions are established, by using fixed point index theory and LeggettWilliams fixed point theorems. In Section 4, as an application, an example is presented to illustrate our main result.

2

International Journal of Differential Equations

2. Green’s Function and Bounds In this section, we construct Green’s function for the homogeneous boundary value problem and estimate bounds for Green’s function that will be used to prove our main theorems. Let 𝐺(𝑡, 𝑠) be Green’s function for the homogeneous BVP

From 𝑢(𝑗) (𝑎) = 0, 0 ≤ 𝑗 ≤ 𝑛 − 2, we have 𝑐𝑛 = 𝑐𝑛−1 = 𝑐𝑛−2 = ⋅ ⋅ ⋅ = 𝑐2 = 0. Then 𝑢 (𝑡) =

−1 𝑡 ∫ (𝑡 − 𝑠)𝛼−1 𝑦 (𝑠) 𝑑𝑠 + 𝑐1 (𝑡 − 𝑎)𝛼−1 , Γ (𝛼) 𝑎 𝛼1

𝑢(𝛼1 ) (𝑡) = 𝑐1 ∏ (𝛼 − 𝑖) (𝑡 − 𝑎)𝛼−𝛼1 −1

(9)

𝑖=1

−𝐷𝑎𝛼+ 𝑢 (𝑡) = 0, 𝑎 < 𝑡 < 𝑏,

𝛼1

𝑢(𝑗) (𝑎) = 0, 0 ≤ 𝑗 ≤ 𝑛 − 2,

(3)

𝛼−𝛼1 −1

Lemma 1. Let 𝑑 = (𝑏 − 𝑎) − 𝜉(𝜂 − 𝑎) 𝐶[𝑎, 𝑏], then the fractional order BVP

> 0. If 𝑦 ∈

𝑢

𝛼1

𝛼1

𝑖=1

𝑖=1

𝑐1 ∏ (𝛼 − 𝑖) (𝑏 − 𝑎)𝛼−𝛼1 −1 − ∏ (𝛼 − 𝑖)

1 Γ (𝛼)

𝑏

⋅ ∫ (𝑏 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠

𝐷𝑎𝛼+ 𝑢 (𝑡) + 𝑦 (𝑡) = 0, 𝑎 < 𝑡 < 𝑏, (𝑗)

𝑖=1

𝑡 1 ∫ (𝑡 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠. Γ (𝛼) 𝑎

From 𝑢(𝛼1 ) (𝑏) = 𝜉𝑢(𝛼1 ) (𝜂), we have

𝑢(𝛼1 ) (𝑏) = 𝜉𝑢(𝛼1 ) (𝜂) . 𝛼−𝛼1 −1

− ∏ (𝛼 − 𝑖)

𝑎

(4)

(𝑎) = 0, 0 ≤ 𝑗 ≤ 𝑛 − 2,

𝑢(𝛼1 ) (𝑏) = 𝜉𝑢(𝛼1 ) (𝜂)

(10)

𝛼1

= 𝜉 [𝑐1 ∏ (𝛼 − 𝑖) (𝜂 − 𝑎) 𝑖=1

(5)

𝛼1

− ∏ (𝛼 − 𝑖) 𝑖=1

𝑏

has a unique solution, 𝑢(𝑡) = ∫𝑎 𝐺(𝑡, 𝑠)𝑦(𝑠)𝑑𝑠, where 𝜉 (𝑡 − 𝑎)𝛼−1 𝐺 (𝑡, 𝑠) = 𝐺1 (𝑡, 𝑠) + 𝐺2 (𝜂, 𝑠) ; 𝑑

𝛼−𝛼1 −1

𝜂 1 𝛼−𝛼 −1 ∫ (𝜂 − 𝑠) 1 𝑦 (𝑠) 𝑑𝑠] . Γ (𝛼) 𝑎

Therefore (6)

𝑐1 =

1 𝛼−𝛼1 −1

Γ (𝛼) [(𝑏 − 𝑎)𝛼−𝛼1 −1 − 𝜉 (𝜂 − 𝑎)

]

𝑏

× [∫ (𝑏 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠

here

𝑎

𝜂

𝑎

(𝑡 − 𝑎)𝛼−1 (𝑏 − 𝑠)𝛼−𝛼1 −1 { , 𝑎 ≤ 𝑡 ≤ 𝑠 ≤ 𝑏, { { { (𝑏 − 𝑎)𝛼−𝛼1 −1 ⋅{ 𝛼−𝛼 −1 𝛼−1 { { (𝑡 − 𝑎) (𝑏 − 𝑠) 1 { − (𝑡 − 𝑠)𝛼−1 , 𝑎 ≤ 𝑠 ≤ 𝑡 ≤ 𝑏, { (𝑏 − 𝑎)𝛼−𝛼1 −1 1 𝐺2 (𝜂, 𝑠) = Γ (𝛼) 𝛼−𝛼1 −1

𝛼−𝛼1 −1

− 𝜉 ∫ (𝜂 − 𝑠)

1 𝐺1 (𝑡, 𝑠) = Γ (𝛼)

𝑦 (𝑠) 𝑑𝑠] =

1 Γ (𝛼) [(𝑏 − 𝑎)𝛼−𝛼1 −1 ]

𝑏

⋅ ∫ (𝑏 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠 𝑎

𝛼−𝛼1 −1

(7)

+

𝜉 (𝜂 − 𝑎)

Γ (𝛼) (𝑏 − 𝑎)𝛼−𝛼1 −1 [(𝑏 − 𝑎)𝛼−𝛼1 −1 − 𝜉 (𝜂 − 𝑎) 𝑏

(𝜂 − 𝑎) (𝑏 − 𝑠) { { , 𝜂 ≤ 𝑠 ≤ 𝑏, { { 𝛼−𝛼1 −1 { (𝑏 − 𝑎) ⋅{ 𝛼−𝛼 −1 { { (𝜂 − 𝑎) 1 (𝑏 − 𝑠)𝛼−𝛼1 −1 𝛼−𝛼 −1 { { − (𝜂 − 𝑠) 1 , 𝑎 ≤ 𝑠 ≤ 𝜂. (𝑏 − 𝑎)𝛼−𝛼1 −1 {

𝑎

−

𝜂

Proof. Assume that 𝑢 ∈ 𝐶[𝛼]+1 [𝑎, 𝑏] is a solution of fractional order BVP (4)-(5) and is uniquely expressed as 𝐼𝑎𝛼+ 𝐷𝑎𝛼+ 𝑢(𝑡) = −𝐼𝑎𝛼+ 𝑦(𝑡), so that −1 𝑡 𝑢 (𝑡) = ∫ (𝑡 − 𝑠)𝛼−1 𝑦 (𝑠) 𝑑𝑠 + 𝑐1 (𝑡 − 𝑎)𝛼−1 Γ (𝛼) 𝑎 + 𝑐2 (𝑡 − 𝑎)

𝛼−2

+ ⋅ ⋅ ⋅ + 𝑐𝑛 (𝑡 − 𝑎)

𝛼−𝑛

.

𝜉 Γ (𝛼) [(𝑏 − 𝑎)𝛼−𝛼1 −1 − 𝜉 (𝜂 − 𝑎) 𝛼−𝛼1 −1

⋅ ∫ (𝜂 − 𝑠) 𝑎

(8)

𝛼−𝛼1 −1

𝑏

𝑎

𝜉 [(𝑏 − 𝑎) 𝑏

𝛼−𝛼1 −1

𝑎

𝛼−𝛼1 −1

− 𝜉 (𝜂 − 𝑎)

⋅ ∫ 𝐺2 (𝜂, 𝑠) 𝑦 (𝑠) 𝑑𝑠.

]

1 Γ (𝛼) (𝑏 − 𝑎)𝛼−𝛼1 −1

𝑦 (𝑠) 𝑑𝑠 =

⋅ ∫ (𝑏 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠 +

]

(11)

⋅ ∫ (𝑏 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠

𝛼−𝛼1 −1

𝛼−𝛼1 −1

]

International Journal of Differential Equations

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

Thus, the unique solution of (4)-(5) is

𝛽

−𝐷𝑎+ (𝜙𝑝 (𝑥 (𝑡))) = 0, 𝑎 < 𝑡 < 𝑏,

−1 𝑡 𝑢 (𝑡) = ∫ (𝑡 − 𝑠)𝛼−1 𝑦 (𝑠) 𝑑𝑠 Γ (𝛼) 𝑎 +

+

𝜙𝑝 (𝑥 (𝑎)) = 0, 𝛽

𝑏 (𝑡 − 𝑎)𝛼−1 ∫ (𝑏 − 𝑠)𝛼−𝛼1 −1 𝑦 (𝑠) 𝑑𝑠 𝛼−𝛼1 −1 𝑎 Γ (𝛼) (𝑏 − 𝑎)

𝐷𝑎+1 (𝜙𝑝 (𝑥 (𝑏))) = 0. Proof. An equivalent integral equation for (13) is given by

𝜉 (𝑡 − 𝑎)𝛼−1 [(𝑏 − 𝑎)𝛼−𝛼1 −1 − 𝜉 (𝜂 − 𝑎)

𝛼−𝛼1 −1

] (12)

𝑏

𝑏

𝑎

𝑎

𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑡)) =

𝑡 1 ∫ (𝑡 − 𝜏)𝛽−1 ℎ (𝜏) 𝑑𝜏 Γ (𝛽) 𝑎

+ 𝑐1 (𝑡 − 𝑎)

⋅ ∫ 𝐺2 (𝜂, 𝑠) 𝑦 (𝑠) 𝑑𝑠 = ∫ 𝐺1 (𝑡, 𝑠) 𝑦 (𝑠) 𝑑𝑠 +

[(𝑏 − 𝑎)

𝛽−1

+ 𝑐2 (𝑡 − 𝑎)

𝛽−2

(18) .

By (14), one can determine 𝑐2 = 0 and 𝑐1 = (−1/(𝑏 −

𝜉 (𝑡 − 𝑎)𝛼−1 𝛼−𝛼1 −1

(17)

𝑏

− 𝜉 (𝜂 − 𝑎)

𝑏

𝑏

𝑎

𝑎

𝛼−𝛼1 −1

𝑎)𝛽−𝛽1 −1 )(1/Γ(𝛽)) ∫𝑎 (𝑏 − 𝜏)𝛽−𝛽1 −1 ℎ(𝜏)𝑑𝜏. Thus, the unique solution of (13), (2) is

]

𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑡))

⋅ ∫ 𝐺2 (𝜂, 𝑠) 𝑦 (𝑠) 𝑑𝑠 = ∫ 𝐺 (𝑡, 𝑠) 𝑦 (𝑠) 𝑑𝑠,

= where 𝐺(𝑡, 𝑠) is given in (6).

𝑡 1 ∫ (𝑡 − 𝜏)𝛽−1 ℎ (𝜏) 𝑑𝜏 Γ (𝛽) 𝑎

−

Lemma 2. If ℎ ∈ 𝐶[𝑎, 𝑏], then the fractional order differential equation

(𝑡 − 𝑎)𝛽−1

𝑏 1 ∫ (𝑏 − 𝜏)𝛽−𝛽1 −1 ℎ (𝜏) 𝑑𝜏 𝛽−𝛽1 −1 Γ (𝛽) 𝑎 (𝑏 − 𝑎)

(19)

𝑏

= − ∫ 𝐻 (𝑡, 𝜏) ℎ (𝜏) 𝑑𝜏.

𝛽

𝐷𝑎+ (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑡))) = ℎ (𝑡) ,

𝑎 < 𝑡 < 𝑏,

𝑎

(13)

𝑏

Therefore, 𝜙𝑝−1 (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢(𝑡))) = −𝜙𝑝−1 (∫𝑎 𝐻(𝑡, 𝜏)ℎ(𝜏)𝑑𝜏). Con𝑏

sequently, 𝐷𝑎𝛼+ 𝑢(𝑡) + 𝜙𝑞 (∫𝑎 𝐻(𝑡, 𝜏)ℎ(𝜏)𝑑𝜏) = 0. Hence,

satisfying (5) and 𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑎)) = 0, (14)

𝛽

𝐷𝑎+1 (𝜙𝑝 (𝐷𝑎𝛼+ 𝑢 (𝑏))) = 0

𝑏

𝑢 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) ℎ (𝜏) 𝑑𝜏) 𝑑𝑠, 𝑎

𝑎

(15)

where 𝐻 (𝑡, 𝑠) =

𝑏

𝑎

𝑎

(20)

Lemma 3. Green’s function 𝐺(𝑡, 𝑠) satisfies the following inequalities:

has a unique solution, 𝑏

𝑏

𝑢 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) ℎ (𝜏) 𝑑𝜏) 𝑑𝑠.

(ii) 𝐺(𝑡, 𝑠) ≥ ((𝜂 − 𝑎)/(𝑏 − 𝑎))𝛼−1 𝐺(𝑏, 𝑠), 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑡, 𝑠) ∈ [𝜂, 𝑏] × [𝑎, 𝑏]. Proof. Consider Green’s function given by (6). Let 𝑎 ≤ 𝑡 ≤ 𝑠 ≤ 𝑏. Then, we have

1 Γ (𝛽)

(𝑡 − 𝑎)𝛽−1 (𝑏 − 𝑠)𝛽−𝛽1 −1 { { , { { { (𝑏 − 𝑎)𝛽−𝛽1 −1 ⋅{ 𝛽−𝛽 −1 { { (𝑡 − 𝑎)𝛽−1 (𝑏 − 𝑠) 1 { { − (𝑡 − 𝑠)𝛽−1 , 𝛽−𝛽1 −1 − 𝑎) (𝑏 {

(i) 𝐺(𝑡, 𝑠) ≤ 𝐺(𝑏, 𝑠), 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑡, 𝑠) ∈ [𝑎, 𝑏] × [𝑎, 𝑏],

1 𝜕 (𝑡 − 𝑎)𝛼−2 (𝑏 − 𝑠)𝛼−𝛼1 −1 [ ] 𝐺1 (𝑡, 𝑠) = 𝜕𝑡 Γ (𝛼 − 1) (𝑏 − 𝑎)𝛼−𝛼1 −1

𝑎 ≤ 𝑡 ≤ 𝑠 ≤ 𝑏, (16) 𝑎 ≤ 𝑠 ≤ 𝑡 ≤ 𝑏.

≥

(1 − 𝑎) 𝑡𝛼−2 [ Γ (𝛼 − 1)

⋅ (1 − 𝑠)𝛼−2 ≥ 0.

𝛼−2 𝛼−𝛼1 +1

𝑏

(1 − (1 − 𝛼1 ) 𝑠 + 𝑂 (𝑠2 ))

(𝑏 − 𝑎)𝛼−𝛼1 −1

]

(21)

4

International Journal of Differential Equations

Let 𝑎 ≤ 𝑠 ≤ 𝑡 ≤ 𝑏. Then, we have 𝜕 1 (𝑡 − 𝑎)𝛼−2 (𝑏 − 𝑠)𝛼−𝛼1 −1 − (𝑡 − 𝑠)𝛼−2 ] 𝐺1 (𝑡, 𝑠) = [ 𝜕𝑡 Γ (𝛼 − 1) (𝑏 − 𝑎)𝛼−𝛼1 −1 ≥

𝛼−2

𝑡 [ Γ (𝛼 − 1)

(1 − 𝑎)𝛼−2 𝑏𝛼−𝛼1 +1 (1 − (1 − 𝛼1 ) 𝑠 + 𝑂 (𝑠2 )) − (𝑏 − 𝑎)𝛼−𝛼1 −1 (𝑏 − 𝑎)𝛼−𝛼1 −1

Therefore 𝐺1 (𝑡, 𝑠) is increasing in 𝑡, which implies 𝐺1 (𝑡, 𝑠) ≤ 𝐺1 (𝑏, 𝑠). Now we prove 𝐺2 (𝜂, 𝑠) ≥ 0, 𝑠 ∈ [𝑎, 𝑏] .

≥

(𝜂 − 𝑎) 1 (𝑏 − 𝑠) 1 [ Γ (𝛼) (𝑏 − 𝑎)𝛼−𝛼1 −1 (𝜂 − 𝑎) 1 [ Γ (𝛼)

] (1 − 𝑠)𝛼−2 ≥ 0.

In fact, if 𝑠 ≥ 𝜂, obviously, (23) holds. If 𝑠 ≤ 𝜂, one has

(23)

𝛼−𝛼 −1

𝐺2 (𝜂, 𝑠) =

(22)

𝛼−𝛼1 −1

(𝑏 − 𝑠)

𝛼−𝛼1 −1

𝛼−𝛼1 −1

− (𝜂 − 𝑠)

𝛼−𝛼1 −1

− (𝜂 − 𝜂𝑠)

]

𝛼−𝛼1 −1

(24) 𝛼−𝛼1 −1

(𝑏 − 𝑎)

𝛼−𝛼1 −1

(𝑏 − 𝑎)

That implies that (23) is also true. Therefore, by (6), (21), and (23), we find

If 𝑎 ≤ 𝑡 ≤ 𝑠 ≤ 𝑏, we have 𝐺1 (𝑡, 𝑠) =

𝜕 𝜕 (𝛼 − 1) 𝜉 (𝑡 − 𝑎)𝛼−2 𝐺 (𝑡, 𝑠) = 𝐺1 (𝑡, 𝑠) + 𝐺2 (𝜂, 𝑠) 𝜕𝑡 𝜕𝑡 𝑑 (25) ≥ 0. Therefore 𝐺(𝑡, 𝑠) is increasing with respect to 𝑡 ∈ [𝑎, 𝑏]. Hence the inequality (i) is proved. Now, we establish the inequality (ii). On the other hand, if 𝑎 ≤ 𝑠 ≤ 𝑡 ≤ 𝑏, we have

] ≥ 0.

=

1 (𝑡 − 𝑎)𝛼−1 (𝑏 − 𝑠)𝛼−𝛼1 −1 [ ] Γ (𝛼) (𝑏 − 𝑎)𝛼−𝛼1 −1

𝑡 − 𝑎 𝛼−1 (𝑏 − 𝑎)𝛼−1 (𝑏 − 𝑠)𝛼−𝛼1 −1 1 ( ] ) [ Γ (𝛼) 𝑏 − 𝑎 (𝑏 − 𝑎)𝛼−𝛼1 −1

≥(

(27)

𝜂 − 𝑎 𝛼−1 ) 𝐺1 (𝑏, 𝑠) . 𝑏−𝑎

Therefore 𝐺1 (𝑡, 𝑠) ≥ (

𝜂 − 𝑎 𝛼−1 ) 𝐺1 (𝑏, 𝑠) . 𝑏−𝑎

(28)

From (6) and (28) we have 𝐺1 (𝑡, 𝑠)

𝐺 (𝑡, 𝑠) = 𝐺1 (𝑡, 𝑠) +

=

1 (𝑡 − 𝑎)𝛼−1 (𝑏 − 𝑠)𝛼−𝛼1 −1 [ − (𝑡 − 𝑠)𝛼−1 ] Γ (𝛼) (𝑏 − 𝑎)𝛼−𝛼1 −1

≥

1 𝑡 − 𝑎 𝛼−1 ( ) Γ (𝛼) 𝑏 − 𝑎

(𝑏 − 𝑎)𝛼−1 (𝑏 − 𝑠)𝛼−𝛼1 −1 ⋅[ − (𝑏 − 𝑠)𝛼−1 ] (𝑏 − 𝑎)𝛼−𝛼1 −1 ≥(

𝜂−𝑎 ) 𝑏−𝑎

𝛼−1

𝐺1 (𝑏, 𝑠) .

(26)

𝜉 (𝑡 − 𝑎)𝛼−1 𝐺2 (𝜂, 𝑠) 𝑑

≥(

𝜂 − 𝑎 𝛼−1 𝜉 (𝑡 − 𝑎)𝛼−1 ) 𝐺1 (𝑡, 𝑠) + 𝐺2 (𝜂, 𝑠) (29) 𝑏−𝑎 𝑑

≥(

𝜂 − 𝑎 𝛼−1 ) 𝐺 (𝑏, 𝑠) . 𝑏−𝑎

Therefore 𝐺 (𝑡, 𝑠) ≥ (

𝜂 − 𝑎 𝛼−1 ) 𝐺 (𝑏, 𝑠) 𝑏−𝑎

(30) ∀ (𝑡, 𝑠) ∈ [𝜂, 𝑏] × [𝑎, 𝑏] .

Hence the inequality (ii) is proved.

International Journal of Differential Equations Lemma 4. Green’s function 𝐻(𝑡, 𝑠) satisfies the following inequalities: (i) 𝐻(𝑡, 𝑠) ≤ 𝐻(𝑠, 𝑠), 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑡, 𝑠) ∈ [𝑎, 𝑏] × [𝑎, 𝑏], (ii) 𝐻(𝑡, 𝑠) ≥ 𝛾𝐻(𝑠, 𝑠), 𝑓𝑜𝑟 𝑎𝑙𝑙 (𝑡, 𝑠) ∈ 𝐼 × [𝑎, 𝑏],

5 is a compact map such that 𝑥 ≠ 𝑇𝑥 for 𝑥 ∈ 𝜕Ω𝑞 . Thus, one has the following conclusions: (𝐷1) If ‖𝑥‖ < ‖𝑇𝑥‖ for 𝑥 ∈ 𝜕Ω𝑞 , then 𝑖(𝑇, Ω𝑞 , 𝑃) = 0. (𝐷2) If ‖𝑥‖ ≥ ‖𝑇𝑥‖ for 𝑥 ∈ 𝜕Ω𝑞 , then 𝑖(𝑇, Ω𝑞 , 𝑃) = 1.

where 𝐼 = [(3𝑎 + 𝑏)/4, (𝑎 + 3𝑏)/4] and 𝛾 = (1/4)𝛽−1 .

3. Main Results

The method of proof is similar to that [20], and we omit it here.

In this section, the existence of at least two or at least three positive solutions for fractional differential equation with 𝑝-Laplacian operator BVP (1)-(2) is established by using fixed point index theory and Leggett-Williams fixed point theorems. Let 𝐸 = {𝑢 : 𝑢 ∈ 𝐶[𝑎, 𝑏]} be the real Banach space equipped with the norm ‖𝑢‖ = max𝑡∈[𝑎,𝑏] |𝑢(𝑡)|. Define the cone 𝑃 ⊂ 𝐸 by

Theorem 5 (Leggett-Williams [3]). Let 𝑇 : 𝑃𝑐 → 𝑃𝑐 be completely continuous and let 𝜙 be a nonnegative continuous concave functional on 𝑃 such that 𝜙(𝑦) ≤ ‖𝑦‖ for all 𝑦 ∈ 𝑃𝑐 . Suppose that there exist 𝑎, 𝑏, 𝑐, and 𝑑 with 0 < 𝑎 < 𝑏 < 𝑑 ≤ 𝑐 such that (𝐴1) {𝑦 ∈ 𝑃(𝜙, 𝑏, 𝑑) : 𝜙(𝑦) > 𝑏} ≠ 0 and 𝜙(𝑇𝑦) > 𝑏 for 𝑦 ∈ 𝑃(𝜙, 𝑏, 𝑑),

𝑃 = {𝑢 ∈ 𝐸 : 𝑢 (𝑡) ≥ 0, ∀𝑡 ∈ [𝑎, 𝑏] , min 𝑢 (𝑡) 𝑡∈[𝜂,𝑏]

(𝐴3) 𝜙(𝑇𝑦) > 𝑏 for 𝑦 ∈ 𝑃(𝜙, 𝑏, 𝑐) with ‖𝑇𝑦‖ > 𝑑. Then 𝑇 has at least three fixed points 𝑦1 , 𝑦2 , and 𝑦3 in 𝑃𝑐 satisfying ‖𝑦1 ‖ < 𝑎, 𝑏 < 𝜙(𝑦2 ), ‖𝑦3 ‖ > 𝑎, and 𝜙(𝑦3 ) < 𝑏. Theorem 6 (see [3]). Let 𝑇 : 𝑃𝑐 → 𝑃 be a completely continuous operator and let 𝜙 be a nonnegative continuous concave functional on 𝑃 such that 𝜙(𝑦) ≤ ‖𝑦‖ for all 𝑦 ∈ 𝑃𝑐 . Suppose that there exist 𝑎, 𝑏, and 𝑐 with 0 < 𝑎 < 𝑏 < 𝑐 such that (𝐵1) {𝑦 ∈ 𝑃(𝜙, 𝑏, 𝑐) : 𝜙(𝑦) > 𝑏} ≠ 0 and 𝜙(𝑇𝑦) > 𝑏 for 𝑦 ∈ 𝑃(𝜙, 𝑏, 𝑐), (𝐵2) ‖𝑇𝑦‖ < 𝑎 for ‖𝑦‖ ≤ 𝑎, (𝐵3) 𝜙(𝑇𝑦) > (𝑏/𝑐)‖𝑇𝑦‖ for 𝑦 ∈ 𝑃𝑐 with ‖𝑇𝑦‖ > 𝑐. Then 𝑇 has at least two fixed points 𝑦1 and 𝑦2 in 𝑃𝑐 satisfying ‖𝑦1 ‖ < 𝑎, ‖𝑦2 ‖ > 𝑎 and 𝜙(𝑦2 ) < 𝑏. Theorem 7 (see [21]). Let 𝑃 be a closed convex set in a Banach space 𝐸 and let Ω be a bounded open set such that Ω𝑝 fl Ω∩𝑃 ≠ 0. Let 𝑇 : Ω𝑝 → 𝑃 be a compact map. Suppose that 𝑥 ≠ 𝑇𝑥 for all 𝑥 ∈ 𝜕𝑝 :

(31)

𝜂 − 𝑎 𝛼−1 ≥( ) ‖𝑢‖} . 𝑏−𝑎

(𝐴2) ‖𝑇𝑦‖ < 𝑎 for ‖𝑦‖ ≤ 𝑎,

Let 𝑇 : 𝑃 → 𝐸 be the operator defined by 𝑇𝑢 (𝑡) 𝑏

𝑏

𝑎

𝑎

= ∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠.

If 𝑢 ∈ 𝑃 is a fixed point of 𝑇, then 𝑢 satisfies (32) and hence 𝑢 is a positive solution of 𝑝-Laplacian fractional order BVP (1)-(2). Lemma 9. The operator 𝑇 defined by (32) is a self-map on 𝑃. Proof. Let 𝑢 ∈ 𝑃. Clearly, 𝑇𝑢(𝑡) ≥ 0, for all 𝑡 ∈ [𝑎, 𝑏] and 𝑏

𝑏

𝑇𝑢 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

(33)

≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 so that 𝑏

𝑏

𝑎

𝑎

‖𝑇𝑢‖ ≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠. (34)

(𝐶1) Existence: if 𝑖(𝑇, Ω𝑝 , 𝑃) ≠ 0, then 𝑇 has a fixed point in Ω𝑝 .

On the other hand, by Lemma 3, we have

(𝐶2) Normalization: if 𝑢 ∈ Ω𝑝 , then 𝑖(̂ 𝑢, Ω𝑝 , 𝑃) = 1, where ̂ (𝑥) = 𝑢 for 𝑥 ∈ Ω𝑝 . 𝑢

min 𝑇𝑢 (𝑡) = min ∫ 𝐺 (𝑡, 𝑠)

(𝐶3) Homotopy: let V : [0, 1] × Ω𝑝 → 𝑃 be a compact map such that 𝑥 ≠ V(𝑡, 𝑥) for 𝑥 ∈ 𝜕Ω𝑝 and 𝑡 ∈ [0, 1]. Then 𝑖(V(0, ⋅), Ω𝑝 , 𝑃) = 𝑖(V(1, ⋅), Ω𝑝 , 𝑃). (𝐶4) Additivity: if 𝑈1, 𝑈2 are disjoint relatively open subsets of Ω𝑝 such that 𝑥 ≠ 𝑇𝑥 for 𝑥 ∈ Ω𝑝 \ (𝑈1 ∪ 𝑈2), then 𝑖(𝑇, Ω𝑝 , 𝑃) = 𝑖(𝑇, 𝑈1, 𝑃) + 𝑖(𝑇, 𝑈2, 𝑃), where 𝑖(𝑇, 𝑈𝑗, 𝑃) = 𝑖(𝑇 | 𝑢𝑗, 𝑈𝑗, 𝑃) (𝑗 = 1, 2). Theorem 8 (see [22]). Let 𝑃 be a cone in a Banach space 𝐸. For 𝑞 > 0, define Ω𝑞 = {𝑥 ∈ 𝑃 : ‖𝑥‖ < 𝑞}. Assume that 𝑇 : Ω𝑞 → 𝑃

(32)

𝑏

𝑡∈[𝜂,𝑏]

𝑡∈[𝜂,𝑏] 𝑎

𝑏

⋅ 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 𝑎

≥(

𝜂 − 𝑎 𝛼−1 𝑏 ) ∫ 𝐺 (𝑏, 𝑠) 𝑏−𝑎 𝑎 𝑏

⋅ 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 𝑎

≥(

𝜂 − 𝑎 𝛼−1 ) ‖𝑇𝑢‖ . 𝑏−𝑎

(35)

6

International Journal of Differential Equations

Hence 𝑇𝑢 ∈ 𝑃 and so 𝑇 : 𝑃 → 𝑃. Standard argument involving the Arzela-Ascoli theorem shows that 𝑇 is completely continuous.

𝑏

≥ 𝐴𝑏󸀠 ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝜂

= 2𝑏󸀠 (

For convenience of the reader, we denote

𝜏∈𝐼

𝛾𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠

𝑏 − 𝑎 𝛼−1 𝑏 − 𝑎 𝛼−1 ) ) . > 𝑏󸀠 ( 𝜂−𝑎 𝜂−𝑎 (38)

𝛼−1

𝐴=

𝐵=

2 ((𝑏 − 𝑎) / (𝜂 − 𝑎)) 𝑏

∫𝜂 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫𝜏∈𝐼 𝛾𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 1 𝑏

𝑏

∫𝑎 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫𝑎 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠

𝑓0 = lim+ min

𝑢→0 𝑡∈[𝑎,𝑏]

,

Consequently, min 𝑇𝑢 (𝑡) ≥ (

𝑡∈[𝜂,𝑏]

,

𝜃 (𝑇𝑢) > 𝑏󸀠 ,

Theorem 10. Let 𝑓(𝑡, 𝑢) be nonnegative continuous on [𝑎, 𝑏] × [0, ∞). Assume that there exist constants 𝑎󸀠 , 𝑏󸀠 with 𝑏󸀠 > 𝑎󸀠 > 0 such that the following conditions are satisfied: 󸀠

󸀠

󸀠

(𝐻1) 𝑓(𝑡, 𝑢(𝑡)) ≥ 𝜙𝑝 (𝐴𝑏 ) for all (𝑡, 𝑢) ∈ [𝜂, 𝑏] × [𝑏 , 𝑏 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) ]. (𝐻2) 𝑓(𝑡, 𝑢(𝑡)) < 𝜙𝑝 (𝐵𝑎󸀠 ) for all (𝑡, 𝑢) ∈ [𝑎, 𝑏] × [0, 𝑎󸀠 ].

Proof. Let 𝜃 : 𝑃 → [0, ∞) be the nonnegative continuous concave functional defined by 𝜃(𝑢) = min𝑡∈[𝜂,𝑏] 𝑢(𝑡), 𝑢 ∈ 𝑃. Evidently, for each 𝑢 ∈ 𝑃, we have 𝜃(𝑢) ≤ ‖𝑢‖. It is easy to see that 𝑇 : 𝑃𝑏󸀠 ((𝑏−𝑎)/(𝜂−𝑎))2(𝛼−1) → 𝑃 is completely continuous and 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) > 𝑏󸀠 > 𝑎󸀠 > 0. We choose 𝑢(𝑡) = 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) ; then

𝑏−𝑎 𝜃 (𝑢) = 𝑏 ( ) 𝜂−𝑎 󸀠

𝑏

𝑡∈[𝑎,𝑏]

𝑡∈[𝑎,𝑏]

𝑏

𝑏

𝑎

𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

< 𝐵𝑎󸀠 ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 = 𝑎󸀠 , which shows that 𝑇 : 𝑃𝑎󸀠 → 𝑃𝑎󸀠 , that is, ‖𝑇𝑢‖ < 𝑎󸀠 for 𝑢 ∈ 𝑃𝑎󸀠 . This shows that condition (𝐵2) of Theorem 6 is satisfied. Finally, we show that (𝐵3) of Theorem 6 also holds. Assume that 𝑢 ∈ 𝑃𝑏󸀠 ((𝑏−𝑎)/(𝜂−𝑎))2(𝛼−1) with ‖𝑇𝑢‖ > 𝑏󸀠 ((𝑏−𝑎)/(𝜂− 𝑎))2(𝛼−1) ; then by the definition of cone 𝑃, we have

>( =

) | 𝜃(𝑢) > 𝑏 } ≠ 0. So {𝑢 ∈ 𝑃(𝜃, 𝑏 , 𝑏 ((𝑏 − 𝑎)/(𝜂 − 𝑎)) 2(𝛼−1) 󸀠 󸀠 Hence, if 𝑢 ∈ 𝑃(𝜃, 𝑏 , 𝑏 ((𝑏−𝑎)/(𝜂−𝑎)) ), then 𝑏󸀠 ≤ 𝑢(𝑡) ≤ 2(𝛼−1) 󸀠 𝑏 ((𝑏 − 𝑎)/(𝜂 − 𝑎)) for 𝑡 ∈ [𝜂, 𝑏]. Thus for 𝑡 ∈ [𝜂, 𝑏], from assumption (𝐻1), we have

(41)

⋅ 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠

󸀠

󸀠

𝑎

⋅ (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠] ≤ ∫ 𝐺 (𝑏, 𝑠)

𝑡∈[𝜂,𝑏]

(37)

(40)

‖𝑇𝑢‖ = max |𝑇𝑢 (𝑡)| = max [∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞

>𝑏.

2(𝛼−1)

󸀠

𝑏 − 𝑎 2(𝛼−1) ). ) 𝜂−𝑎

Therefore, condition (𝐵1) of Theorem 6 is satisfied. Now if 𝑢 ∈ 𝑃𝑎󸀠 , then ‖𝑢‖ ≤ 𝑎󸀠 . By assumption (𝐻2), we have

𝜃 (𝑇𝑢) = min 𝑇𝑢 (𝑡) ≥ (

𝑏 − 𝑎 2(𝛼−1) ), ) 𝜂−𝑎

2(𝛼−1)

∀𝑢 ∈ 𝑃 (𝜃, 𝑏󸀠 , 𝑏󸀠 (

𝑏

Then fractional order BVP (1)-(2) has at least two positive solutions 𝑢1 and 𝑢2 satisfying ‖𝑢1 ‖ < 𝑎󸀠 , min𝑡∈[𝜂,𝑏] 𝑢2 (𝑡) < 𝑏󸀠 , and ‖𝑢2 ‖ > 𝑎󸀠 .

󸀠

(39)

for 𝜂 ≤ 𝑡 ≤ 𝑏, 𝑏󸀠 ≤ 𝑢(𝑡) ≤ 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) . That is,

𝑓 (𝑡, 𝑢) 𝑓∞ = lim max . 𝑢→∞ 𝑡∈[𝑎,𝑏] 𝜙 (𝑢) 𝑝

𝑢 ∈ 𝑃 (𝜃, 𝑏󸀠 , 𝑏󸀠 (

𝜂 − 𝑎 𝛼−1 𝑏 − 𝑎 𝛼−1 = 𝑏󸀠 >( ) × 𝑏󸀠 ( ) 𝑏−𝑎 𝜂−𝑎

(36)

𝑓 (𝑡, 𝑢) , 𝜙𝑝 (𝑢)

𝜂 − 𝑎 𝛼−1 ) ‖𝑇𝑢‖ 𝑏−𝑎

𝜂 − 𝑎 (𝛼−1) ) ‖𝑇𝑢‖ 𝑏−𝑎

𝜂 − 𝑎 2(𝛼−1) ) ‖𝑇𝑢‖ 𝑏−𝑎

(42)

𝑏󸀠 2(𝛼−1)

𝑏󸀠 ((𝑏 − 𝑎) / (𝜂 − 𝑎))

‖𝑇𝑢‖ .

So condition (𝐵3) of Theorem 6 is satisfied. Thus using Theorem 6, 𝑇 has at least two fixed points. Consequently, boundary value problem (1)-(2) has at least two positive solutions 𝑢1 and 𝑢2 in 𝑃𝑏󸀠 ((𝑏−𝑎)/(𝜂−𝑎))2(𝛼−1) satisfying 󵄩󵄩 󵄩󵄩 󸀠 󵄩󵄩𝑢1 󵄩󵄩 < 𝑎 ,

𝑇𝑢 (𝜂) 𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝜂

𝑎

= ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 ≥ ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠

min 𝑢2 (𝑡) < 𝑏󸀠 ,

𝑡∈[𝜂,𝑏]

󵄩󵄩 󵄩󵄩 󸀠 󵄩󵄩𝑢2 󵄩󵄩 > 𝑎 .

(43)

International Journal of Differential Equations

7

Theorem 11. Let 𝑓(𝑡, 𝑢) be nonnegative continuous on [𝑎, 𝑏] × [0, ∞). Assume that there exist constants 𝑎󸀠 , 𝑏󸀠 , 𝑐󸀠 with ((𝜂 − 𝑎)/(𝑏 − 𝑎))2(𝛼−1) 𝑐󸀠 > 𝑏󸀠 > 𝑎󸀠 > 0 such that 󸀠

󸀠

(𝐻3) 𝑓(𝑡, 𝑢(𝑡)) < 𝜙𝑝 (𝐵𝑎 ) for all (𝑡, 𝑢) ∈ [𝑎, 𝑏] × [0, 𝑎 ], (𝐻4) 𝑓(𝑡, 𝑢(𝑡)) ≥ 𝜙𝑝 (𝐴𝑏󸀠 ) for all (𝑡, 𝑢) ∈ [𝜂, 𝑏] × [𝑏󸀠 , 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) ], (𝐻5) 𝑓(𝑡, 𝑢(𝑡)) ≤ 𝜙𝑝 (𝐵𝑐󸀠 ) for all (𝑡, 𝑢) ∈ [𝑎, 𝑏] × [0, 𝑐󸀠 ]. Then fractional order BVP (1)-(2) has at least three positive solutions 𝑢1 , 𝑢2 , and 𝑢3 with ‖𝑢1 ‖ < 𝑎󸀠 , min𝑡∈[𝜂,𝑏] 𝑢2 (𝑡) > 𝑏󸀠 , ‖𝑢3 ‖ > 𝑎󸀠 , and min𝑡∈[𝜂,𝑏] 𝑢3 (𝑡) < 𝑏󸀠 . 󸀠

Proof. If 𝑢 ∈ 𝑃𝑐󸀠 , then ‖𝑢‖ ≤ 𝑐 . By assumption (𝐻5), we have

𝑡∈[𝑎,𝑏]

(𝐻6) 𝑓0 = 𝑓∞ = ∞; (𝐻7) there exists a constant 𝜇1 > 0 such that 𝑓(𝑡, 𝑢) < 𝜙𝑝 (𝐵𝜇1 ), for (𝑡, 𝑢) ∈ [𝑎, 𝑏] × [0, 𝜇1 ], then boundary value problem (1)-(2) has at least two positive solutions 𝑢1 and 𝑢2 such that 0 < ‖𝑢1 ‖ < 𝜇1 < ‖𝑢2 ‖. Proof. From Lemma 1, we obtain 𝑇 : 𝑃 → 𝑃 being completely continuous. In view of 𝑓0 = ∞, there exists 𝜎1 ∈ (0, 𝜇1 ) such that 𝑓(𝑡, 𝑢) ≥ 𝜙𝑝 (𝜂1 𝑢), for 𝑎 ≤ 𝑡 ≤ 𝑏, 0 < 𝑢 ≤ 𝜎1 , where 𝜂1 ∈ (𝐴/2, ∞). Let Ω𝜎1 = {𝑢 ∈ 𝑃 | ‖𝑢‖ < 𝜎1 }. Then, for any 𝑢 ∈ 𝜕Ω𝜎1 , we have 𝑏

𝑏

𝑎

𝑎

𝑇𝑢 (𝜂) = ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑏

𝑏

⋅ 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 ≥ ∫ 𝐺 (𝜂, 𝑠)

‖𝑇𝑢‖ = max |𝑇𝑢 (𝑡)| = max [∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞 𝑡∈[𝑎,𝑏]

Theorem 12. Let 𝑓(𝑡, 𝑢) be nonnegative continuous on [𝑎, 𝑏] × [0, ∞). If the following assumptions are satisfied:

𝜂

𝑎

𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝑎

𝜂

⋅ 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝜙𝑝 (𝜂1 𝑢) 𝑑𝜏) 𝑑𝑠 ≥ ∫ 𝐺 (𝜂,

⋅ (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠] ≤ ∫ 𝐺 (𝑏, 𝑠) (44)

𝑏

𝑠) 𝜙𝑞 (∫

⋅ 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠

𝜏∈𝐼

𝑎

󸀠

𝑏

𝑏

𝑎

𝑎

⋅ 𝜙𝑝 (𝜂1 (

󸀠

≤ 𝐵𝑐 ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 = 𝑐 . This shows that 𝑇 : 𝑃𝑐󸀠 → 𝑃𝑐󸀠 . Using the same arguments as in the proof of Theorem 10, we can show that 𝑇 : 𝑃𝑐󸀠 → 𝑃𝑐󸀠 is a completely continuous operator. It follows from conditions (𝐻3) and (𝐻4) in Theorem 11 that 𝑐󸀠 > 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) > 𝑏󸀠 > 𝑎󸀠 . Similarly to the proof of Theorem 10, we have 𝑇 : 𝑃𝑎󸀠 → 𝑃𝑎󸀠 and 𝑏 − 𝑎 2(𝛼−1) ) | 𝜃 (𝑢) > 𝑏󸀠 } ≠ 0, {𝑢 ∈ 𝑃 (𝜃, 𝑏 , 𝑏 ( ) 𝜂−𝑎 󸀠

(45)

𝜃 (𝑇𝑢) > 𝑏󸀠 , for all 𝑢 ∈ 𝑃(𝜃, 𝑏󸀠 , 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) ). Moreover, for 𝑢 ∈ 𝑃(𝜃, 𝑏󸀠 , 𝑐󸀠 ) and ‖𝑇𝑢‖ > 𝑏󸀠 ((𝑏 − 𝑎)/(𝜂 − 𝑎))2(𝛼−1) , we have

𝑏 − 𝑎 𝛼−1 > 𝑏󸀠 . >𝑏 ( ) 𝜂−𝑎

𝜂 − 𝑎 𝛼−1 ) ‖𝑢‖) 𝑑𝜏) 𝑑𝑠 𝑏−𝑎

𝑏 𝜂 − 𝑎 𝛼−1 ) ‖𝑢‖ ∫ 𝐺 (𝜂, 𝑠) 𝑏−𝑎 𝜂

⋅ 𝜙𝑞 (∫

𝜏∈𝐼

𝛾𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 =

2𝜂1 ‖𝑢‖ > ‖𝑢‖ , 𝐴

which implies ‖𝑇𝑢‖ > ‖𝑢‖ for 𝑢 ∈ 𝜕Ω𝜎1 . Hence, Theorem 8 implies 𝑖 (𝑇, Ω𝜎1 , 𝑃) = 0.

󸀠

𝜂 − 𝑎 𝛼−1 𝜃 (𝑇𝑢) = min 𝑇𝑢 (𝑡) ≥ ( ) ‖𝑇𝑢‖ 𝑡∈[𝜂,𝑏] 𝑏−𝑎

= 𝜂1 (

(47)

𝛾𝐻 (𝜏, 𝜏)

On the other hand, since 𝑓∞ = ∞, there exists 𝜎3 > 𝜇1 such that 𝑓(𝑡, 𝑢) ≥ 𝜙𝑝 (𝜂2 𝑢), for 𝑢 ≥ 𝜎3 , where 𝜂2 ∈ (𝐴/2, ∞). Let 𝜎2 > max{𝜎3 ((𝑏 − 𝑎)/(𝜂 − 𝑎))𝛼−1 , 𝜇1 } and Ω𝜎2 = {𝑢 ∈ 𝑃 | ‖𝑢‖ < 𝜎2 }. Then min𝑡∈[𝜂,𝑏] 𝑢(𝑡) ≥ ((𝑏 − 𝑎)/(𝜂 − 𝑎))𝛼−1 ‖𝑢‖ > 𝜎3 , for any 𝑢 ∈ 𝜕Ω𝜎2 . By using the method to get (48), we obtain 𝑇𝑢(𝜂) > (2𝜂1 /𝐴)‖𝑢‖ > ‖𝑢‖, which implies ‖𝑇𝑢‖ > ‖𝑢‖ for 𝑢 ∈ 𝜕Ω2 . Thus, from Theorem 8, we have 𝑖 (𝑇, Ω𝜎2 , 𝑃) = 0.

(46)

󸀠

So all the conditions of Theorem 5 are satisfied. Thus using Theorem 5, 𝑇 has at least three fixed points. So, th boundary value problem (1)-(2) has at least three positive solutions 𝑢1 , 𝑢2 , and 𝑢3 with ‖𝑢1 ‖ < 𝑎󸀠 , min𝑡∈[𝜂,𝑏] 𝑢2 (𝑡) > 𝑏󸀠 , ‖𝑢3 ‖(𝑡) > 𝑎󸀠 , and min𝑡∈[𝜂,𝑏] ‖𝑢3 ‖(𝑡) < 𝑏󸀠 .

(48)

(49)

Finally, let Ω𝜇1 = {𝑢 ∈ 𝑃 | ‖𝑢‖ < 𝜇1 }. Then, for any 𝑢 ∈ 𝜕Ω𝜇1 , by (𝐻7), we then get 𝑏

𝑏

𝑎

𝑎

𝑇𝑢 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 < ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝜙𝑝 (𝐵𝜇1 ) 𝑑𝜏) 𝑑𝑠

8

International Journal of Differential Equations 𝑏

𝑏

𝑎

𝑎

Next, since 𝑓∞ = 0, there exists 𝛿3 > 𝜇2 such that 𝑓(𝑡, 𝑢) ≤ 𝜙𝑝 (𝜂3 𝑢), for 𝑢 ≥ 𝛿3 , where 𝜂3 ∈ (0, 𝐵). We consider two cases.

= 𝐵𝜇1 ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 = 𝜇1 = ‖𝑢‖ , (50) which implies ‖𝑇𝑢‖ < ‖𝑢‖ for 𝑢 ∈ 𝜕Ω𝜇1 . Using Theorem 8 again, we get 𝑖 (𝑇, Ω𝜇1 , 𝑃) = 1.

(51)

Note that 𝜎1 < 𝜇1 < 𝜎2 , by the additivity of fixed point index and (48)–(51); we obtain 𝑖 (𝑇, Ω𝜇1 \ Ω𝜎1 , 𝑃) = 𝑖 (𝑇, Ω𝜇1 , 𝑃) − 𝑖 (𝑇, Ω𝜎1 , 𝑃) = 1, 𝑖 (𝑇, Ω𝜎2 \ Ω𝜇1 , 𝑃) = 𝑖 (𝑇, Ω𝜎2 , 𝑃) − 𝑖 (𝑇, Ω𝜇1 , 𝑃)

(52)

= −1.

𝑇𝑢 (𝑡) 𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 ≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝜙𝑝 (𝑁) 𝑑𝜏) 𝑑𝑠 𝑏

𝑏

𝑎

𝑎

= 𝑁 ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 =

(55)

𝑁 < 𝛿4 𝐵

= ‖𝑢‖ .

Hence, 𝑇 has a fixed point 𝑢1 in Ω𝜇1 \ Ω𝜎1 , and it has a fixed point 𝑢2 in Ω𝜎2 \ Ω𝜇1 . Clearly, 𝑢1 and 𝑢2 are positive solutions of boundary value problem (1)-(2) and 0 < ‖𝑢1 ‖ < 𝜇1 < ‖𝑢2 ‖. Theorem 13. Let 𝑓(𝑡, 𝑢) be nonnegative continuous on [𝑎, 𝑏] × [0, ∞). If the following assumptions are satisfied: (𝐻8) 𝑓0 = 𝑓∞ = 0; (𝐻9) there exists a constant 𝜇2 > 0 such that 𝑓(𝑡, 𝑢) > 𝜙𝑝 (𝐴𝜇2 ), for (𝑡, 𝑢) ∈ [𝜂, 𝑏]×[((𝜂−𝑎)/(𝑏−𝑎))𝛼−1 𝜇2 , 𝜇2 ], then boundary value problem (1)-(2) has at least two positive solutions 𝑢1 and 𝑢2 such that 0 < ‖𝑢1 ‖ < 𝜇2 < ‖𝑢2 ‖. Proof. From Lemma 9, we obtain 𝑇 : 𝑃 → 𝑃 being completely continuous. In view of 𝑓0 = 0, there exists 𝛿1 ∈ (0, 𝜇2 ) such that 𝑓(𝑡, 𝑢) ≤ 𝜙𝑝 (𝜂2 𝑢), for 𝑎 ≤ 𝑡 ≤ 𝑏, 0 < 𝑢 ≤ 𝛿1 , where 𝜂2 ∈ (0, 𝐵). Let Ω𝛿1 = {𝑢 ∈ 𝑃 | ‖𝑢‖ < 𝛿1 }. Then, for any 𝑢 ∈ 𝜕Ω𝛿1 , we have

Case (ii). Suppose that 𝑓 is unbounded. In view of 𝑓 : [𝑎, 𝑏] × [0, ∞) → [0, ∞) being continuous, there exist 𝑡⋆ ∈ [𝑎, 𝑏] and 𝛿5 > max{((𝑏 − 𝑎)/(𝜂 − 𝑎))𝛼−1 𝛿3 , 𝜇2 } such that 𝑓(𝑡, 𝑢) ≤ 𝑓(𝑡⋆ , 𝛿5 ), for 𝑎 ≤ 𝑡 ≤ 𝑏, 0 ≤ 𝑢 ≤ 𝛿5 . Then, for 𝑢 ∈ 𝑃 with ‖𝑢‖ = 𝛿5 , we obtain 𝑇𝑢 (𝑡) 𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 ≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝑡⋆ , 𝛿5 ) 𝑑𝜏) 𝑑𝑠 (56)

≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝜙𝑝 (𝜂3 ‖𝑢‖) 𝑑𝜏) 𝑑𝑠 𝑏

𝑏

𝑎

𝑎

≤ 𝜂3 𝛿5 ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 =

𝜂3 𝛿5 𝐵

< 𝛿5 = ‖𝑢‖ .

𝑇𝑢 (𝑡) 𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝑎

𝑎

So, in either case, if we always choose Ω𝛿2 = {𝑢 ∈ 𝐸 | ‖𝑢‖ < 𝛿2 = max{𝛿4 , 𝛿5 }}, then we have ‖𝑇𝑢‖ < ‖𝑢‖, for 𝑢 ∈ 𝜕Ω2 . Thus, from Theorem 8, we have

≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 ≤ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝜙𝑝 (𝜂2 𝑢) 𝑑𝜏) 𝑑𝑠 𝑏

𝑏

𝑎

𝑎

(57)

Finally, let Ω𝜇2 = {𝑢 ∈ 𝑃 | ‖𝑢‖ < 𝜇2 }. Then, for any 𝑢 ∈ 𝜕Ω𝜇2 , min𝑡∈[𝜂,𝑏] 𝑢(𝑡) ≥ ((𝜂−𝑎)/(𝑏−𝑎))𝛼−1 ‖𝑢‖ = ((𝜂−𝑎)/(𝑏−𝑎))𝛼−1 𝜇2 , by (𝐻9), and we then obtain

𝜂2 ‖𝑢‖ < ‖𝑢‖ , 𝐵

𝑇𝑢 (𝜂)

which implies ‖𝑇𝑢‖ < ‖𝑢‖ for 𝑢 ∈ 𝜕Ω𝛿1 . Hence, Theorem 8 implies 𝑖 (𝑇, Ω𝛿1 , 𝑃) = 0.

𝑖 (𝑇, Ω𝛿2 , 𝑃) = 1.

(53)

≤ 𝜂2 ‖𝑢‖ ∫ 𝐺 (𝑏, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠 =

Case (i). Suppose that 𝑓 is bounded, which implies that there exists 𝑁 > 0 such that 𝑓(𝑡, 𝑢) ≤ 𝜙𝑝 (𝑁) for all 𝑡 ∈ [𝑎, 𝑏] and 𝑢 ∈ [0, ∞). Take 𝛿4 > max{𝑁/𝐵, 𝛿3 }. Then, for 𝑢 ∈ 𝑃 with ‖𝑢‖ = 𝛿4 , we get

(54)

𝑏

𝑏

𝑎

𝑎

𝑏

𝑏

𝜂

𝑎

= ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝑓 (𝜏, 𝑢 (𝜏)) 𝑑𝜏) 𝑑𝑠 ≥ ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝐻 (𝑠, 𝜏) 𝜙𝑝 (𝐴𝜇2 ) 𝑑𝜏) 𝑑𝑠

International Journal of Differential Equations 𝑏

≥ 𝐴𝜇2 ∫ 𝐺 (𝜂, 𝑠) 𝜙𝑞 (∫ 𝜂

𝜏∈𝐼

9 (ii) 𝑓(𝑡, 𝑢(𝑡)) ≥ 68.364 = 𝜙𝑝 (𝐴𝑏󸀠 ) for all (𝑡, 𝑢) ∈ [0.5, 2] × [2, 16.0049],

𝛾𝐻 (𝜏, 𝜏) 𝑑𝜏) 𝑑𝑠

(iii) 𝑓(𝑡, 𝑢(𝑡)) ≤ 104.52 = 𝜙𝑝 (𝐵𝑐󸀠 ) for all (𝑡, 𝑢) ∈ [0, 1] × [0, 100].

𝑏 − 𝑎 𝛼−1 ) 𝜇2 > 𝜇2 = ‖𝑢‖ , = 2( 𝜂−𝑎 (58) which implies ‖𝑇𝑢‖ > ‖𝑢‖ for 𝑢 ∈ 𝜕Ω𝜇2 . An application of Theorem 8 again shows that 𝑖 (𝑇, Ω𝜇2 , 𝑃) = 0.

Consequently, all of the conditions of Theorem 11 are satisfied. With the use of Theorem 11, boundary value problem (61) has at least three positive solutions 𝑢1 , 𝑢2 , and 𝑢3 with 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑢1 󵄩󵄩 < 1,

(59)

min 𝑢2 (𝑡) > 2,

𝑡∈[1/2,1]

Note that 𝛿1 < 𝜇2 < 𝛿2 by the additivity of fixed point index and (54)–(59); we obtain

󵄩󵄩 󵄩󵄩 󵄩󵄩𝑢3 󵄩󵄩 > 1,

𝑖 (𝑇, Ω𝜇2 \ Ω𝛿1 , 𝑃) = 𝑖 (𝑇, Ω𝜇2 , 𝑃) − 𝑖 (𝑇, Ω𝛿1 , 𝑃) = −1,

min 𝑢3 (𝑡) < 2.

𝑡∈[1/2,1]

(60)

Competing Interests

𝑖 (𝑇, Ω𝛿2 \ Ω𝜇2 , 𝑃) = 𝑖 (𝑇, Ω𝛿2 , 𝑃) − 𝑖 (𝑇, Ω𝜇2 , 𝑃) = 1. Hence, 𝑇 has a fixed point 𝑢1 in Ω𝜇2 \ Ω𝛿1 , and it has a fixed point 𝑢2 in Ω𝛿2 \ Ω𝜇2 . Consequently, 𝑢1 and 𝑢2 are positive solutions of boundary value problem (1)-(2) and 0 < ‖𝑢1 ‖ < 𝜇2 < ‖𝑢2 ‖.

4. Example In this section, we consider boundary value problem of the fractional differential equation 2.5 𝐷01.5 + (𝜙𝑝 (𝐷0+ 𝑢 (𝑡))) = 𝑓 (𝑡, 𝑢 (𝑡)) ,

󸀠

𝑢 (0) = 0, 1 1 𝑢󸀠 (1) = 𝑢󸀠 ( ) , 3 2

(61)

The author expresses his gratitude to his guide Professor K. Rajendra Prsasd.

References

[2] D. R. Anderson and J. M. Davis, “Multiple solutions and eigenvalues for third-order right focal boundary value problems,” Journal of Mathematical Analysis and Applications, vol. 267, no. 1, pp. 135–157, 2002.

[4] Y. Sun, Q. Sun, and X. Zhang, “Existence and nonexistence of positive solutions for a higher-order three-point boundary value problem,” Abstract and Applied Analysis, vol. 2014, Article ID 513051, 7 pages, 2014.

= 0,

where 𝑡 { 0 ≤ 𝑢 ≤ 2, + 12𝑢5 , { { 100 𝑓 (𝑡, 𝑢) = { { { 𝑡 + 𝑢 + 58, 𝑢 ≥ 2. { 100

Acknowledgments

[3] R. W. Leggett and L. R. Williams, “Multiple positive fixed points of nonlinear operators on ordered Banach spaces,” Indiana University Mathematics Journal, vol. 28, no. 4, pp. 673–688, 1979.

𝜙𝑝 (𝐷02.5 + 𝑢 (0)) = 0, (𝜙𝑝 (𝐷02.5 + 𝑢 (1)))

The author declares that he has no competing interests regarding the publication of this paper.

[1] R. P. Agarwal, D. O’Regan, and P. J. Wong, Positive Solutions of Differential, Difference and Integral Equations, Kluwer Academic, Dordrecht, The Netherlands, 1999.

0 < 𝑡 < 1,

𝑢 (0) = 0,

𝐷00.5 +

(63)

(62)

Let 𝑝 = 1/2. We note that 𝑎 = 0, 𝑏 = 1, 𝜂 = 1/2, 𝜉 = 1/3, 𝛽 = 3/2, 𝛽1 = 1/2, 𝑛 = 3, 𝛼 = 5/2, and 𝛼1 = 1. By a simple calculation, we obtain ((𝜂 − 𝑎)/(𝑏 − 𝑎))𝛼−1 = 0.3535, 𝛾 = 0.5, 𝐵 = 1.0452, and 𝐴 = 34.1482. Choosing 𝑎󸀠 = 1, 𝑏󸀠 = 2, and 𝑐󸀠 = 100, then 0 < 𝑎󸀠 < 𝑏󸀠 < ((𝜂 − 𝑎)/(𝑏 − 𝑎))2(𝛼−1) 𝑐󸀠 and 𝑓 satisfies (i) 𝑓(𝑡, 𝑢(𝑡)) < 1.0452 = 𝜙𝑝 (𝐵𝑎󸀠 ) for all (𝑡, 𝑢) ∈ [0, 1] × [0, 1],

[5] S. Nageswararao, “Solvability of second order delta-nabla pLaplacian m-point eigenvalue problem on time scales,” TWMS Journal of Applied and Engineering Mathematics, vol. 5, no. 1, pp. 98–109, 2015. [6] M. Ramaswamy and R. Shivaji, “Multiple positive solutions for classes of p-Laplacian equations,” Differential and Integral Equations, vol. 17, no. 11-12, pp. 1255–1261, 2004. [7] C. Yang and J. Yan, “Positive solutions for third-order SturmLiouville boundary value problems with p-Laplacian,” Computers & Mathematics with Applications, vol. 59, no. 6, pp. 2059– 2066, 2010. [8] C. Zhai and C. Guo, “Positive solutions for third-order SturmLiouville boundary-value problems with p-Laplacian,” Electronic Journal of Differential Equations, vol. 2009, no. 154, pp. 1–9, 2009.

10 [9] S. Liang and J. Zhang, “Positive solutions for boundary value problems of nonlinear fractional differential equation,” Nonlinear Analysis: Theory, Methods & Applications, vol. 71, no. 11, pp. 5545–5550, 2009. [10] S. Nageswararao, “Multiple positive solutions for a system of Riemann-Liouville fractional order two-point boundary value problems,” Panamerican Mathematical Journal, vol. 25, no. 1, pp. 66–81, 2015. [11] S. Nageswararao, “Existence and multiplicity for a system of fractional higher-order two-pointboundary value problem,” Journal of Applied Mathematics and Computing, vol. 53, pp. 93– 107, 2016. [12] Y. Zhao, S. Sun, Z. Han, and M. Zhang, “Positive solutions for boundary value problems of nonlinear fractional differential equations,” Applied Mathematics and Computation, vol. 217, no. 16, pp. 6950–6958, 2011. [13] G. Chai, “Positive solutions for boundary value problem of fractional Differential equation with p-Laplacian operator,” Boundary Value Problems, vol. 2012, article 18, 2012. [14] Z. Han, H. Lu, S. Sun, and D. Yang, “Positive solutions to boundary-value problems of p-Laplacian fractional differential equations with a parameter in the boundary conditions,” Electronic Journal of Differential Equations, vol. 2012, no. 213, pp. 1– 14, 2012. [15] H. Lu, Z. Han, and S. Sun, “Multiplicity of positive solutions for Sturm-Liouville boundary value problems of fractional differential equations with p-Laplacian,” Boundary Value Problems, vol. 2014, article 26, 2014. [16] C. T. Ledesma, “Boundary value problem with fractional pLaplacian operator,” Advances in Nonlinear Analysis, 2015. [17] Y. Li and A. Qi, “Positive solutions for multi-point boundary value problems of fractional differential equations with pLaplacian,” Mathematical Methods in the Applied Sciences, vol. 39, no. 6, pp. 1425–1434, 2016. [18] K. R. Prasad and B. M. B. Krushna, “Existence of multiple positive solutions for p-Laplacian fractional order boundary value problems,” International Journal of Analysis and Applications, vol. 6, no. 1, pp. 63–81, 2014. [19] B. Zhang, “Solvability of fractional boundary value problems with p-Laplacian operator,” Advances in Difference Equations, vol. 2015, article 352, 2015. [20] A. Kameswararao and S. Nageswararao, “Eigenvalue problems for systems of nonlinear higher order boundary value problems,” Journal of Applied Mathematics and Informatics, vol. 28, no. 3-4, pp. 711–721, 2010. [21] D. J. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, Orlando, Fla, USA, 1988. [22] K. Deimling, Nonlinear Functional Analysis, Springer, New York, NY, USA, 1985.

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