Positive Solutions for Singular Semipositone Fractional Differential

Hindawi Journal of Function Spaces Volume 2017, Article ID 5892616, 7 pages https://doi.org/10.1155/2017/5892616

Research Article Positive Solutions for Singular Semipositone Fractional Differential Equation Subject to Multipoint Boundary Conditions Rui Pu,1 Xingqiu Zhang,1 Yujun Cui,2 Peilong Li,1 and Weiwei Wang1 1

School of Medical Information Engineering, Jining Medical College, Rizhao, Shandong 276826, China Department of Applied Mathematics, Shandong University of Science and Technology, Qingdao 266590, China

2

Correspondence should be addressed to Xingqiu Zhang; [email protected] Received 5 April 2017; Accepted 1 June 2017; Published 2 July 2017 Academic Editor: Xinguang Zhang Copyright © 2017 Rui Pu et al. 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. Existence result together with multiplicity result of positive solutions of higher-order fractional multipoint boundary value problems is given by considering the integrations of height functions on some special bounded sets. The nonlinearity may change its sign and may possess singularities on the time and the space variables at the same time.

1. Introduction Very recently, Henderson and Luca [1] obtained the existence result of positive solution for the following singular differential equation with fractional derivative: 𝛼 𝑢 (𝑡) + 𝑓 (𝑡, 𝑢 (𝑡)) = 0, 𝐷0+

0 < 𝑡 < 1,

(1)

subject to multipoint BCs 𝑢 (0) = 𝑢󸀠 (0) = ⋅ ⋅ ⋅ = 𝑢(𝑛−2) (0) = 0, 𝑝

𝑚

𝑞

𝐷0+ 𝑢 (1) = ∑𝑎𝑖 𝐷0+ 𝑢 (𝜉𝑖 ) .

For example, people have realized that electrical conductance of cell membranes of biological organism took on fractionalorder style [2]. As a consequence, fractional-order ordinary differential equations can give a more accurate description on spread process of some infectious diseases such as HIV, handfoot-mouth disease [3], malaria, tuberculosis, and measles. Arafa et al. [4] formulated a model with orders 𝛼1 , 𝛼2 , 𝛼3 > 0 to describe the infection of CD4+ T cells. In 2015, Wang et al. [5] concentrated on the positive solutions for a class of modified HIV-1 population dynamics model presented in [6]

(2)

𝑖=1

𝛼 is the traditional Riemann-Liouville derivative Here 𝐷0+ with order 𝛼, 𝑛 − 1 < 𝛼 ≤ 𝑛 (𝑛 ≥ 3), 𝑎𝑖 ≥ 0, 𝑖 = 1, 2, . . . , 𝑚 (𝑚 ∈ N+ ), 0 < 𝜉1 < 𝜉2 < ⋅ ⋅ ⋅ < 𝜉𝑚 < 1, 𝑝, 𝑞 ∈ R, 1 ≤ 𝑝 ≤ 𝑛 − 2, and 0 ≤ 𝑞 ≤ 𝑝 with Δ = Γ(𝛼)/Γ(𝛼 − 𝛼−𝑞−1 > 0. The nonlinear term 𝑝) − (Γ(𝛼)/Γ(𝛼 − 𝑞)) ∑𝑚 𝑖=1 𝑎𝑖 𝜉𝑖 𝑓 permits sign-changing and singularities at 𝑡 = 0, 1. Fractional derivative gives a more precise exhibition on long-memory behavior owing to its nonlocal characteristic. Nowadays, researchers have reached a consensus that fractional calculus is one of the effective tools to describe phenomena in almost every field of science and technology.

𝛽

𝐷𝑡𝛼 𝑢 (𝑡) + 𝜆𝑓 (𝑡, 𝑢 (𝑡) , 𝐷𝑡 𝑢 (𝑡) , V (𝑡)) = 0, 𝛾

𝐷𝑡 V (𝑡) + 𝜆𝑔 (𝑡, 𝑢 (𝑡)) = 0, 𝛽

0 < 𝑡 < 1,

𝛽+1

𝐷𝑡 𝑢 (0) = 𝐷𝑡 𝑢 (0) = 0, 1

𝛽

𝛽

𝐷𝑡 𝑢 (1) = ∫ 𝐷𝑡 𝑢 (𝑠) d𝐴 (𝑠) , 0

󸀠

V (0) = V (0) = 0, 1

V (1) = ∫ V (𝑠) d𝐵 (𝑠) , 0

(3)

2

Journal of Function Spaces

where 2 < 𝛼, 𝛾 ≤ 3, 0 < 𝛽 < 1, 𝜆 > 0 is a parameter, 𝛽 𝛾 𝛼 − 𝛽 > 2, and 𝐷𝑡𝛼 , 𝐷𝑡 , and 𝐷𝑡 are traditional R-L derivatives. 1 𝛽 1 ∫0 𝐷𝑡 𝑢(𝑠)d𝐴(𝑠) and ∫0 V(𝑠)d𝐵(𝑠) represent R-S integrals and 𝐴, 𝐵 are bounded variations, and the nonlinearities permit sign-changing and singularities at 𝑡 = 0, 1. Other relative papers on nonlocal boundary value problems can be found in [7–20]. Motivated by the above papers, we devote ourselves to the existence result as well as multiplicity result of positive solutions for BVP (1)-(2). This article admits some new features. First, the nonlinear term 𝑓 may take negative infinity and change its sign. Compared with [1], 𝑓 permits singularities on the time and the space variables at the same time. Second, the method exploited in this paper is different from that in [1] in essence. More precisely, height functions with their integrations on some special bounded set are utilized to get the existence result for positive solutions of BVP (1)-(2). Thirdly, a result of multiple positive solutions is also given in this paper. Conditions employed in this paper are easy to be verified.

2. Preliminaries and Several Lemmas Two traditional Banach spaces 𝐸 = 𝐶[0, 1] and 𝐿1 (0, 1) are involved in this article, where 𝐸 = 𝐶[0, 1] and 𝐿1 (0, 1) represent the spaces of the continuous functions and Lebesgue integrable functions equipped with the norms 1 ‖𝑢‖ = max0≤𝑡≤1 |𝑢(𝑡)| and ‖𝑢‖1 = ∫0 |𝑢(𝑡)|d𝑡, respectively. 1

𝐺2 (𝑡, 𝑠) =

1 Γ (𝛼 − 𝑞)

𝛼−𝑞−1 (1 − 𝑠)𝛼−𝑝−1 − (𝑡 − 𝑠)𝛼−𝑞−1 , {𝑡 ⋅{ 𝑡𝛼−𝑞−1 (1 − 𝑠)𝛼−𝑝−1 , {

𝑚

𝑝

Lemma 2 (see [1]). Assume that 𝑎𝑖 > 0 (𝑖 = 1, 2, . . . , 𝑚) and Δ > 0. Then the Green function 𝐺 of (4) given by (6) is continuous on [0, 1] × [0, 1] and meets (a) 𝐺(𝑡, 𝑠) ≤ 𝐽(𝑠), ∀𝑡, 𝑠 ∈ [0, 1], and here 𝐽(𝑠) = ℎ1 (𝑠) + 𝛼−𝑝−1 (1 − (1 − (1/Δ) ∑𝑚 𝑖=1 𝑎𝑖 𝐺2 (𝜉𝑖 , 𝑠), ℎ1 (𝑠) = (1 − 𝑠) 𝑝 𝑠) )/Γ(𝛼), and 𝑠 ∈ [0, 1]; (b) 𝐺(𝑡, 𝑠) ≥ 𝑡𝛼−1 𝐽(𝑠), ∀𝑡, 𝑠 ∈ [0, 1]; (c) 𝐺(𝑡, 𝑠) ≤ 𝜎𝑡𝛼−1 ∀𝑡, 𝑠 ∈ [0, 1], and here 𝜎 = 1/Γ(𝛼) + 𝛼−𝑞−1 /(ΔΓ(𝛼 − 𝑞)). ∑𝑚 𝑖=1 𝑎𝑖 𝜉𝑖 Lemma 3. Let 𝑢 ∈ 𝐶[0, 1] satisfy (4), where 𝑦 ∈ 𝐿1 (0, 1), 𝑦(𝑡) ≥ 0, and 0 ≤ 𝑡 ≤ 1. Then, 𝑢(𝑡) ≥ 𝑡𝛼−1 ‖𝑢‖, and 0 ≤ 𝑡 ≤ 1. Proof. It can be easily seen from Lemma 5 in [1]; we omit the details. Lemma 4. Suppose that 𝑤(𝑡) ∈ 𝐶[0, 1] be the solution of 𝛼 𝑢 (𝑡) + 𝑘 (𝑡) = 0, 𝐷0+

0 < 𝑡 < 1,

𝑢 (0) = 𝑢󸀠 (0) = ⋅ ⋅ ⋅ = 𝑢(𝑛−2) (0) = 0, 𝑚

𝑝

(0) = 0,

(4)

𝑞

𝑞

where 𝑘 ∈ 𝐿1 (0, 1) and 𝑘(𝑡) > 0. Then, 𝑤(𝑡) ≤ 𝜎‖𝑘‖1 𝑡𝛼−1 and 0 ≤ 𝑡 ≤ 1. Proof. For any 0 ≤ 𝑡 ≤ 1, we get by Lemma 2

𝑖=1

1

1

𝑤 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝑘 (𝑠) d𝑠 ≤ 𝜎𝑡𝛼−1 ∫ 𝑘 (𝑠) d𝑠

satisfies

0

1

𝑢 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) 𝑦 (𝑠) d𝑠, 0

0

𝛼−1

𝑡 ∈ [0, 1] ,

(5)

and here 𝐺 (𝑡, 𝑠) = 𝐺1 (𝑡, 𝑠) +

(9)

𝑖=1

(𝑛−2)

𝐷0+ 𝑢 (1) = ∑𝑎𝑖 𝐷0+ 𝑢 (𝜉𝑖 )

𝐺1 (𝑡, 𝑠) =

0 ≤ 𝑡 ≤ 𝑠 ≤ 1,

𝐷0+ 𝑢 (1) = ∑𝑎𝑖 𝐷0+ 𝑢 (𝜉𝑖 ) ,

𝛼 𝐷0+ 𝑢 (𝑡) + 𝑦 (𝑡) = 0, 0 < 𝑡 < 1,

𝑢 (0) = 𝑢 (0) = ⋅ ⋅ ⋅ = 𝑢

(8)

∀(𝑡, 𝑠) ∈ [0, 1] × [0, 1].

Lemma 1 (see [1]). If Δ ≠ 0. Given 𝑦 ∈ 𝐶(0, 1) ∩ 𝐿 (0, 1), the solution of the following differential equation:

󸀠

0 ≤ 𝑠 ≤ 𝑡 ≤ 1,

𝑡𝛼−1 𝑚 ∑𝑎 𝐺 (𝜉 , 𝑠) , Δ 𝑖=1 𝑖 2 𝑖

(6)

1 Γ (𝛼)

𝛼−𝑝−1 𝛼−1 − (𝑡 − 𝑠)𝛼−1 , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1, {𝑡 (1 − 𝑠) ⋅{ 𝑡𝛼−1 (1 − 𝑠)𝛼−𝑝−1 , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1, {

= 𝜎 ‖𝑘‖1 𝑡

(10)

.

Lemma 5 (see [21]). Let Ω1 and Ω2 be two bounded open sets in Banach space 𝐸 such that 𝜃 ∈ Ω1 and Ω1 ⊂ Ω2 and 𝐴 : 𝑃 ∩ (Ω2 \ Ω1 ) → 𝑃 a completely continuous operator, where 𝜃 denotes the zero element of 𝐸 and 𝑃 a cone of 𝐸. Suppose that one of the two conditions holds: (i) ‖𝐴𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃∩𝜕Ω1 ; ‖𝐴𝑢‖ ≥ ‖𝑢‖, ∀𝑢 ∈ 𝑃∩𝜕Ω2 .

(7)

(ii) ‖𝐴𝑢‖ ≥ ‖𝑢‖, ∀𝑢 ∈ 𝑃∩𝜕Ω1 ; ‖𝐴𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃∩𝜕Ω2 . Then 𝐴 has a fixed point in 𝑃 ∩ (Ω2 \ Ω1 ).


3 The operator 𝑇𝑛 is given by

3. Main Results

(𝑇𝑛 𝑢) (𝑡)

Let 𝐾 = {𝑢 : 𝑢 (𝑡) ≥ 𝑡𝛼−1 ‖𝑢‖} .

(11)

1

= ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠, 0

Obviously, 𝐾 is a cone in 𝐸 and (𝐸, 𝐾) is a partial ordering Banach space. (A0 ) 𝑓 ∈ 𝐶((0, 1) × (0, +∞), (−∞, +∞)), and there exists a function 𝑘 ∈ 𝐿1 (0, 1) and 𝑘(𝑡) > 0, such that 𝑓(𝑡, 𝑢) ≥ −𝑘(𝑡), ∀𝑡 ∈ (0, 1), 𝑢 > 0. (A1 ) For any positive numbers 𝑟1 < 𝑟2 , there exists a nonnegative function 𝛾𝑟1 ,𝑟2 ∈ 𝐿1 (0, 1) such that 󵄨󵄨 󵄨 𝛼−1 ≤ 𝑢 ≤ 𝑟2 󵄨󵄨𝑓 (𝑡, 𝑢)󵄨󵄨󵄨 ≤ 𝛾𝑟1 ,𝑟2 (𝑡) , 0 < 𝑡 < 1, 𝑟1 𝑡

(12)

0 ≤ 𝑡 ≤ 1. Next, we will give the proof from three steps. (I) For Any 𝜎‖𝑘‖1 < 𝑟1 < 𝑟2 , 𝑇𝑛 : 𝐾∩(Ω𝑟2 \Ω𝑟1 ) → 𝐾, 𝑛 > 1/𝑟1 , Is Completely Continuous. For any 0 < 𝑡 < 1, considering that 𝑢(𝑡) − 𝑤(𝑡) ≥ (𝑟1 − 𝜎‖𝑘‖1 )𝑡𝛼−1 > 0, one has 1 (𝑟1 − 𝜎 ‖𝑘‖1 ) 𝑡𝛼−1 ≤ max {𝑢 (𝑡) − 𝑤 (𝑡) , } ≤ 𝑟2 , 𝑛 1 𝑛> , 𝑟1

with 1

∫ (1 − 𝑠)𝛼−𝑝−1 𝛾𝑟1 ,𝑟2 (𝑠) d𝑠 < +∞. 0

(13)

∫ 𝐽 (𝑠) 𝜑 (𝑠, 𝑟̃1 ) d𝑠 < 𝑟̃1 ,

(14)

0

1

∫ 𝐽 (𝑠) 𝜓 (𝑠, 𝑟̃2 ) d𝑠 > 𝑟̃2 ,

(15)

0

where 𝜓(𝑡, 𝑟̃2 ) = min{𝑓(𝑡, 𝑢) : (̃𝑟2 − 𝜎‖𝑘‖1 )𝑡𝛼−1 ≤ 𝑢 ≤ 𝑟̃2 } + 𝑘(𝑡). Theorem 6. Suppose that conditions (A0 ), (A1 ), (A2 ), and (A3 ) hold. Then BVP (1)-(2) admits at least one positive solution. Proof. First of all, we concentrate on the following modified approximating BVP (MABVP for short) to overcome difficulties caused by singularities: 𝛼 𝑢 (𝑡) + 𝑓 (𝑡, 𝜒𝑛 (𝑢 − 𝑤) (𝑡)) + 𝑘 (𝑡) = 0, 𝐷0+

1

≤ ∫ 𝐽 (𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 0

1

≤∫ [ 0

+

(1 − 𝑠)𝛼−𝑝−1 (1 − (1 − 𝑠)𝑝 ) Γ (𝛼)

𝑚 1 𝛼−𝑞−1 ∑𝑎𝑖 𝜉𝑖 (1 − 𝑠)𝛼−𝑝−1 ] ΔΓ (𝛼 − 𝑞) 𝑖=1

(21)

⋅ [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 1

≤∫ [ 0

+

(1 − 𝑠)𝛼−𝑝−1 (1 − (1 − 𝑠)𝑝 ) Γ (𝛼)

𝑚 1 𝛼−𝑞−1 ∑𝑎𝑖 𝜉𝑖 (1 − 𝑠)𝛼−𝑝−1 ] ΔΓ (𝛼 − 𝑞) 𝑖=1

⋅ (𝛾𝑟1 −𝜎‖𝑘‖1 ,𝑟2 (𝑠) + 𝑘 (𝑠)) d𝑠 < +∞. 0 < 𝑡 < 1,

𝑚

1

(𝑇𝑛 𝑢) (𝑡) = ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 0

where 𝜑(𝑡, 𝑟̃1 ) = max{𝑓(𝑡, 𝑢) : (̃𝑟1 − 𝜎‖𝑘‖1 )𝑡𝛼−1 ≤ 𝑢 ≤ 𝑟̃1 } + 𝑘(𝑡). (A3 ) There exists a 𝑟̃2 > 𝑟̃1 such that

𝛽

(20)

Thus, one gets from (A0 ), (A1 ), and Lemma 2 that

1

𝑢 (0) = 𝑢󸀠 (0) = ⋅ ⋅ ⋅ = 𝑢(𝑛−2) (0) = 0,

(19)

which means (𝑟1 − 𝜎 ‖𝑘‖1 ) 𝑡𝛼−1 ≤ 𝜒𝑛 (𝑢 − 𝑤) (𝑡) ≤ 𝑟2 , 0 < 𝑡 < 1.

(A2 ) There exists a 𝑟̃1 > 𝜎‖𝑘‖1 such that

(18)

(16)

So, 𝑇𝑛 : 𝐾 ∩ (Ω𝑟2 \ Ω𝑟1 ) → 𝐸 is well defined. At the same time, for given 𝑢 ∈ 𝐾 ∩ (Ω𝑟2 \ Ω𝑟1 ), 𝑡 ∈ [0, 1], one gets from (21) and Lemma 2 that 1

𝑞

(𝑇𝑛 𝑢) (𝑡) = ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠

𝐷0+ (1) = ∑𝑎𝑖 𝐷0+ 𝑢 (𝜉𝑖 ) ,

0

𝑖=1

1

(22)

≤ ∫ 𝐽 (𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠.

and, here, 1 { {𝑢, 𝑢 ≥ 𝑛 , 𝜒𝑛 (𝑢) = { 1 { , 𝑢 < 1. {𝑛 𝑛

0

Hence, (17)

1

󵄩󵄩 󵄩󵄩 󵄩󵄩𝑇𝑛 𝑢󵄩󵄩 ≤ ∫ 𝐽 (𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠. 0

(23)

4


Given 𝑢 ∈ 𝐾, it follows from Lemma 2 that

Equicontinuity of 𝑇𝑛 (𝐷) can be derived from the absolute continuity of Lebesgue integral and (25). Thus, the fact that 𝑇𝑛 : 𝐾 ∩ (Ω𝑟2 \ Ω𝑟1 ) → 𝐾, 𝑛 > 1/𝑟1 , is completely continuous has been proved.

1

(𝑇𝑛 𝑢) (𝑡) = ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 0

1

≥ 𝑡𝛼−1 ∫ 𝐽 (𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠

(24)

0

󵄩 󵄩 ≥ 𝑡𝛼−1 󵄩󵄩󵄩𝑇𝑛 𝑢󵄩󵄩󵄩 ,

which means that 𝑇𝑛 : 𝐾 ∩ (Ω𝑟2 \ Ω𝑟1 ) → 𝐾. Given bounded set 𝐷 ⊂ 𝐾 ∩ (Ω𝑟2 \ Ω𝑟1 ), we can see from (21) that 𝑇𝑚 (𝐷) is uniformly bounded. By Ascoli-Arzela theorem, in order to show the complete continuity, we need only to prove that 𝑇𝑛 (𝐷) is equicontinuous. It follows from (6), (7), (8), and (A1 ) that 󵄨󵄨 1 󵄨󵄨󵄨(𝑇 𝑢)󸀠 (𝑡)󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ 𝜕 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒 (𝑢 − 𝑤) (𝑠)) 󵄨󵄨 𝑛 󵄨󵄨 󵄨󵄨 𝑛 󵄨󵄨 0 𝜕𝑡 󵄨󵄨 󵄨󵄨 1 𝜕 󵄨 󵄨 + 𝑘 (𝑠)] d𝑠󵄨󵄨󵄨󵄨 ≤ 󵄨󵄨󵄨󵄨∫ 𝐺 (𝑡, 𝑠) 󵄨󵄨 󵄨󵄨 0 𝜕𝑡 1 󵄨󵄨 󵄨 ⋅ [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 (𝛼 − 1) 𝑡𝛼−2 𝑚 + 󵄨󵄨󵄨∫ ∑𝑎𝑖 𝐺2 (𝜉𝑖 , 𝑠) 󵄨󵄨 0 Δ 𝑖=1 󵄨 󵄨󵄨 󵄨󵄨 ⋅ [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 𝑡 1 󵄨󵄨 󵄨󵄨∫ [(𝛼 − 1) 𝑡𝛼−2 (1 − 𝑠)𝛼−𝑝−1 − (𝛼 − 1) ≤ Γ (𝛼) 󵄨󵄨󵄨 0 ⋅ (𝑡 − 𝑠)

𝛼−2

] [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠

𝑡

󵄨󵄨 󵄨 + 𝑘 (𝑠)] d𝑠󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 (𝛼 − 1) 𝑚 𝛼−𝑞−1 + 󵄨󵄨󵄨∫ ∑𝑎 𝜉 (1 − 𝑠)𝛼−𝑝−1 󵄨󵄨 0 ΔΓ (𝛼 − 𝑞) 𝑖=1 𝑖 𝑖 󵄨 󵄨󵄨 󵄨󵄨 ⋅ [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠󵄨󵄨󵄨 󵄨󵄨 󵄨 ≤

1

1 [∫ [(𝛼 − 1) (1 − 𝑠)𝛼−𝑝−1 + (𝛼 − 1) Γ (𝛼) 0

⋅ (1 − 𝑠)𝛼−2 ] (𝛾𝑟1 −𝜎‖𝑘‖1 ,𝑟2 (𝑠) + 𝑘 (𝑠)) d𝑠] 󵄨󵄨 1 󵄨󵄨 (𝛼 − 1) 𝑚 𝛼−𝑞−1 + 󵄨󵄨󵄨∫ ∑𝑎 𝜉 (1 − 𝑠)𝛼−𝑝−1 󵄨󵄨 0 ΔΓ (𝛼 − 𝑞) 𝑖=1 𝑖 𝑖 󵄨 󵄨󵄨 󵄨󵄨 ⋅ (𝛾𝑟1 −𝜎‖𝑘‖1 ,𝑟2 (𝑠) + 𝑘 (𝑠)) d𝑠󵄨󵄨󵄨 < +∞, 𝑡 ∈ [0, 1] . 󵄨󵄨 󵄨

(̃𝑟1 − 𝜎 ‖𝑘‖1 ) 𝑡𝛼−1 ≤ 𝜒𝑛 (𝑢 − 𝑤) (𝑡) ≤ 𝑟̃1 , 0 < 𝑡 < 1.

(26)

From the definition of 𝜑(𝑡, 𝑟̃1 ), we have 𝑓 (𝑡, 𝜒𝑛 (𝑢 − 𝑤) (𝑡)) + 𝑘 (𝑡) ≤ 𝜑 (𝑡, 𝑟̃1 ) .

(27)

By (A2 ) and Lemma 2, one gets 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑇𝑛 𝑢󵄩󵄩 1

= max ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 0≤𝑡≤1 0

(28)

1

≤ ∫ 𝐽 (𝑠) 𝜑 (𝑠, 𝑟̃1 ) d𝑠 < 𝑟̃1 ; 0

that is, 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑇𝑛 𝑢󵄩󵄩 ≤ ‖𝑢‖ ,

∀𝑢 ∈ 𝐾 ∩ 𝜕Ω (̃𝑟1 ) .

(29)

Given 𝑢 ∈ 𝐾 ∩ 𝜕Ω(̃𝑟2 ), one has 𝑢(𝑡) ≥ 𝑟̃2 𝑡𝛼−1 , 0 ≤ 𝑡 ≤ 1. Thus, for 𝑛 > 1/̃𝑟2 , we get 𝑟̃2 ≥ 𝜒𝑛 (𝑢 − 𝑤) (𝑡) ≥ (̃𝑟2 − 𝜎 ‖𝑘‖1 ) 𝑡𝛼−1 .

(30)

According to the definition of 𝜓(𝑡, 𝑟̃2 ), one gets

1

+ ∫ (𝛼 − 1) 𝑡𝛼−2 (1 − 𝑠)𝛼−𝑝−1 [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠))

(II) For Sufficiently Large 𝑛, MABVP (16) Admits at Least One Positive Solution. Given 𝑢 ∈ 𝐾∩𝜕Ω(̃𝑟1 ), one has 𝑢(𝑡) ≥ 𝑟̃1 𝑡𝛼−1 , 0 ≤ 𝑡 ≤ 1. Thus, for 𝑛 > 1/̃𝑟1 , similar to (20), we get

𝑓 (𝑡, 𝜒𝑛 (𝑢 − 𝑤) (𝑡)) + 𝑘 (𝑡) ≥ 𝜓 (𝑡, 𝑟̃2 ) .

(25)

(31)

By Lemma 2, (A3 ), and (31), we have 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑇𝑛 𝑢󵄩󵄩 1

= max ∫ 𝐺 (𝑡, 𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 0≤𝑡≤1 0

1

≥ max 𝑡𝛼−1 ∫ 𝐽 (𝑠) [𝑓 (𝑠, 𝜒𝑛 (𝑢 − 𝑤) (𝑠)) + 𝑘 (𝑠)] d𝑠 0≤𝑡≤1

(32)

0

1

≥ ∫ 𝐽 (𝑠) 𝜓 (𝑠, 𝑟̃2 ) d𝑠 ≥ 𝑟̃2 ; 0

that is, 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑇𝑛 𝑢󵄩󵄩 ≥ ‖𝑢‖ ,

∀𝑢 ∈ 𝐾 ∩ 𝜕Ω (̃𝑟2 ) .

(33)

Take 𝑚0 = max{1/̃𝑟1 , 1/̃𝑟2 }. Let 𝑁 = {𝑚0 , 𝑚0 + 1, . . .}. Then, for 𝑛 ∈ 𝑁, both (29) and (33) hold. This together with Lemma 5 shows that 𝑇𝑛 (𝑛 ∈ 𝑁) has at least one fixed point 𝑢̃𝑛 ∈ 𝐾 ∩ Ω𝑟̃2 \ Ω𝑟̃1 .


5

(III) BVP (1)-(2) Admits at Least One Positive Solution. By Lemma 4, one has 󵄩 󵄩 𝑢̃𝑛 (𝑡) ≥ 󵄩󵄩󵄩𝑢̃𝑛 󵄩󵄩󵄩 𝑡𝛼−1 ≥ 𝑟̃1 𝑡𝛼−1 > 𝜎 ‖𝑘‖1 𝑡𝛼−1 ≥ 𝑤 (𝑡) , (34) 𝑚 > 𝑁, 𝑡 ∈ [0, 1] , 1

𝑢𝑛 − 𝑤) (𝑠)) + 𝑘 (𝑠)) d𝑠, 𝑢̃𝑛 (𝑡) = ∫ 𝐺 (𝑡, 𝑠) (𝑓 (𝑠, 𝜒𝑛 (̃ 0

(35)

0 < 𝑡 < 1. By (A1 ), we know {̃ 𝑢𝑛 : 𝑛 > 𝑁} are bounded and equicontinuous sets [0, 1]. It follows from Arzela-Ascoli theorem that there exist a subsequence 𝑁0 of 𝑁 and a function 𝑢̃ ∈ 𝐶[0, 1] such that 𝑢̃𝑛 converges to 𝑢̃ uniformly on [0, 1] as 𝑛 → ∞ through 𝑁0 . Taking limit as 𝑛 → ∞ on both sides of (35) together with the fact 𝑢̃𝑛 (𝑡) ≥ 𝑡𝛼−1 ‖̃ 𝑢𝑛 ‖ ≥ 𝑟̃1 𝑡𝛼−1 , we get 1

𝑢̃ (𝑡) = ∫ 𝐺 (𝑡, 𝑠) (𝑓 (𝑠, 𝑢̃ (𝑠) − 𝑤 (𝑠)) + 𝑘 (𝑠)) d𝑠, 0

(36)

0 < 𝑡 < 1. Let 𝑢(𝑡) = 𝑢̃(𝑡) − 𝑤(𝑡), and then from (36) one has that 𝑢(𝑡) is a positive solution for BVP (1)-(2). A multiplicity result follows if we replace (A2 ) and (A3 ) by the following: (A󸀠2 ) There exists 𝑟̃𝑖 > 𝜎‖𝑘‖1 (𝑖 = 1, 2, . . . , 𝑚) such that 1

∫ 𝐽 (𝑠) 𝜑 (𝑠, 𝑟̃𝑖 ) d𝑠 < 𝑟̃𝑖 ,

(37)

0

where 𝜑(𝑡, 𝑟̃𝑖 ) = max{𝑓(𝑡, 𝑢) : (̃𝑟𝑖 −𝜎‖𝑘‖1 )𝑡𝛼−1 ≤ 𝑢 ≤ 𝑟̃𝑖 }+𝑘(𝑡). ̃ 𝑖 (𝑖 = 1, 2, . . . , 𝑚) with (A󸀠3 ) There exists 𝑅 ̃ 1 < 𝑟̃2 < 𝑅 ̃ 2 < ⋅ ⋅ ⋅ < 𝑟̃𝑚 < 𝑅 ̃𝑚 0 < 𝑟̃1 < 𝑅

(38)

such that 1

̃ 𝑖 ) d𝑠 > 𝑅 ̃𝑖 , ∫ 𝐽 (𝑠) 𝜓 (𝑠, 𝑅

(39)

0

̃ 𝑖 − 𝜎‖𝑘‖1 )𝑡𝛼−1 ≤ 𝑢 ≤ 𝑅 ̃𝑖 } + ̃ 𝑖 ) = min{𝑓(𝑡, 𝑢) : (𝑅 where 𝜓(𝑡, 𝑅 𝑘(𝑡).

It is clear that (40)-(41) has the form of (1)-(2), where 𝛼 = 8/3, 𝑛 = 3, 𝑝 = 1/2, 𝑞 = 1/3, 𝑚 = 3, 𝜉1 = 1/2, 𝜉2 = 1/4, 𝜉3 = 4/5, 𝑎1 = 1/2, 𝑎2 = 2/3, and 𝑎3 = 1/4. Obviously, (A0 ) holds for 𝑘(𝑡) = 1/12√3 𝑡. After direct calculation, we have 𝛼 − 𝑝 − 1 = 7/6, 𝛼 − 𝑞 − 1 = 4/3, Γ(𝛼) = 1.5046, Γ(𝛼 − 𝑝) = 1.0823, Γ(𝛼 − 𝛼−𝑞−1 = 0.5290, Δ = 0.7217, 𝜎 = 1.2803, 𝑞) = 1.1906, ∑3𝑖=1 𝑎𝑖 𝜉𝑖 ‖𝑘‖1 = 0.1250, and 𝜎‖𝑘‖1 = 0.1600. It is clear that (A1 ) is valid for 𝛾𝑟1 ,𝑟2 = (1/16√3 𝑡2 (1 − 𝑡))[𝑟2 6 + (𝑟1 𝑡5/3 )−1/8 ] + 1/12√3 𝑡. Next, we check (A2 ) and (A3 ). Take 𝑟̃1 = 1 > 0.1600 = 𝜎‖𝑘‖1 , and then by Lemma 2 we have 1

1

∫ 𝐽 (𝑠) 𝜑 (𝑠, 1) d𝑠 ≤ ∫ [ 0

0

+

𝑚

1 𝛼−𝑞−1 ∑𝑎 𝜉 (1 − 𝑠)𝛼−𝑝−1 ] ΔΓ (𝛼 − 𝑞) 𝑖=1 𝑖 𝑖 1

⋅

16√𝑠2 3

max {(𝑢6 +

(1 − 𝑠)

1 ) : 0.8400𝑠5/3 ≤ 𝑢 √𝑢 8

1

(1 − 𝑠)7/6 (1 − (1 − 𝑠)1/2 )

0

Γ (𝛼)

≤ 1} d𝑠 = ∫ [ +

(1 − 𝑠)𝛼−𝑝−1 (1 − (1 − 𝑠)𝑝 ) Γ (𝛼)

∑3𝑖=1 𝑎𝑖 𝜉𝑖4/3 (1 − 𝑠)7/6 ] ΔΓ (𝛼 − 𝑞)

1 1 max {(𝑢6 + 8 ) : 0.8400𝑠5/3 ≤ 𝑢 √𝑢 16√3 𝑠2 (1 − 𝑠) (42) 1 1 7/6 ≤ 1} d𝑠 < ∫ [0.6646 (1 − 𝑠) − 0.6646 (1 16 0

⋅

− 𝑠)5/3 + 0.6156 × (1 − 𝑠)7/6 ] 𝑠−2/3 (1 − 𝑠)−1/3 (1 +

1 √0.8400𝑠5/3 8

) d𝑠
20. (43)

[10]

Thus, (A3 ) is valid. According to Theorem 6, BVP (40)-(41) admits at least one positive solution.

Conflicts of Interest The authors declare that they have no conflicts of interest.

[11]

[12]

Authors’ Contributions [13]

The authors contributed to each part of this work equally and read and approved the final version of the manuscript.

Acknowledgments The project is supported financially by the Innovative Training Program for College Students (Mathematical Modelling on Spread, Prevention and Control of HFMD 201610443081, cx2015072), a Project of Shandong Province Higher Educational Science and Technology Program (J15LI16), the Natural Science Foundation of Shandong Province of China (ZR2015AL002), the Foundation for Jining Medical College Natural Science (JY2015BS07), and the National Natural Science Foundation of China (11571197, 11571296, 11371221, and 11071141).

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Journal of Function Spaces [18] X. Zhang, “Positive solutions for a class of singular fractional differential equation with infinite-point boundary value conditions,” Applied Mathematics Letters. An International Journal of Rapid Publication, vol. 39, pp. 22–27, 2015. [19] Y. Cui, “Uniqueness of solution for boundary value problems for fractional differential equations,” Applied Mathematics Letters, vol. 51, pp. 48–54, 2016. [20] D.-x. Ma, “Positive solutions of multi-point boundary value problem of fractional differential equation,” Arab Journal of Mathematical Sciences, vol. 21, no. 2, pp. 225–236, 2015. [21] D. Guo and V. Lakshmikantham, Nonlinear Problems in Abstract Cones, Academic Press, San Diego, Calif, USA, 1988.

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