Hindawi Publishing Corporation Advances in Mathematical Physics Volume 2016, Article ID 7307614, 9 pages http://dx.doi.org/10.1155/2016/7307614
Research Article Existence of Positive Solutions for Two-Point Boundary Value Problems of Nonlinear Finite Discrete Fractional Differential Equations and Its Application Caixia Guo, Jianmin Guo, Ying Gao, and Shugui Kang School of Mathematics and Computer Sciences, Shanxi Datong University, Shanxi, Datong 037009, China Correspondence should be addressed to Shugui Kang;
[email protected] Received 13 February 2016; Accepted 5 April 2016 Academic Editor: Ivan Avramidi Copyright © 2016 Caixia Guo 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. This paper is concerned with the two-point boundary value problems of nonlinear finite discrete fractional differential equations. On one hand, we discuss some new properties of the Green function. On the other hand, by using the main properties of Green function and the Krasnoselskii fixed point theorem on cones, some sufficient conditions for the existence of at least one or two positive solutions for the boundary value problem are established.
1. Introduction Since 1974, the development of fractional differential equations is driven by an extremely wide application background. In those thirty years, a lot of papers and monographs have been produced and several international conferences were held about fractional calculus and fractional differential equation theory and application. At the same time, due to its application in physics, it also caused the fierce debate to classical laws of physics (see [1–11] and the references therein). However, it was basically confined to the fractional order differential equations and there is less literature available on paper concerned with discrete fractional differential equations. In this paper, we will be interested in the nonlinear finite discrete FBVP given by −Δ] 𝑦 (𝑡) = 𝑓 (𝑡 + ] − 1, 𝑦 (𝑡 + ] − 1)) , 𝑦 (] − 2) = 0 = 𝑦 (] + 𝑏 + 1) ,
(1)
where 1 < ] ⩽ 2, 𝑡 ∈ [0, 𝑏 + 1]N0 , 𝑓 : [] − 1, ] + 𝑏]N]−1 × R → R is continuous, 𝑏 ∈ N0 . In Section 2, we will deduce some new properties of the Green function. In Section 3, by means of using the properties of the Green function and Krasnoselskii fixed point theorem on cones, the existence of two positive solutions and one positive solution of problem
(1) is established. At last, an example is given to illustrate the main results of this paper. The difference with other literatures is that we get some new results about Green’s function firstly, and then with the help of the nature of Green’s function we introduce a new function, thus establishing the new operator equation. By discussing the fixed point of the new operator equation, we gain some existence results of solution in (1).
2. Preliminaries For the convenience of the reader, we present here the necessary definitions from fractional calculus theory. Definition 1. The ]th fractional sum of a function 𝑓 defined on N𝑎 fl {𝑎, 𝑎 + 1, . . .}, for ] > 0, is defined to be Δ−] 𝑓 (𝑡) fl
𝑡−] 1 = ∑ (𝑡 − 𝑠 − 1)]−1 𝑓 (𝑠) , Γ (]) 𝑠=𝑎
(2)
where 𝑡 ∈ {𝑎 + ], 𝑎 + ] + 1, . . .} fl N𝑎+] . We also define the ]th fractional difference, where ] > 0 and 0 ⩽ 𝑁 − 1 < ] ⩽ 𝑁 with 𝑁 ∈ N to be Δ] 𝑓 (𝑡) = Δ𝑁Δ−(𝑁−]) 𝑓 (𝑡) , where 𝑡 ∈ N𝑎+𝑁−] .
(3)
2
Advances in Mathematical Physics
Definition 2. We define 𝑡] =
Γ (𝑡 + 1) Γ (𝑡 + 1 − ])
(4)
Lemma 7. Let 𝐸 be a Banach space and let 𝑃 ⊂ 𝐸 be a cone. Assume that Ω1 and Ω2 are open sets contained in 𝐸 such that 0 ∈ Ω1 and Ω1 ⊂ Ω2 . Assume, further, that 𝑇 : 𝑃∩(Ω2 \Ω1 ) → 𝐾 is a completely continuous operator. If either
for any 𝑡 and ] for which the right-hand side is defined. One appeals to the convention that if 𝑡+1−] is a pole of the Gamma function and 𝑡 + 1 is not a pole, then 𝑡] = 0.
(𝐻1 ) ‖𝑇𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ∩ 𝜕Ω1 and ‖𝑇𝑢‖ ⩾ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ∩ 𝜕Ω2 or
Lemma 3. Let 𝑡 and ] be any numbers for which 𝑡] and 𝑡]−1 are defined. Then,
(𝐻2 ) ‖𝑇𝑢‖ ≤ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ∩ 𝜕Ω2 and ‖𝑇𝑢‖ ⩾ ‖𝑢‖, ∀𝑢 ∈ 𝑃 ∩ 𝜕Ω1
Δ𝑡] = ]𝑡]−1 .
(5)
then 𝑇 has at least one fixed point in 𝑃 ∩ (Ω2 \ Ω1 ).
Lemma 4. Let 𝑓 be a real-valued function, and let 𝜇, ] > 0. Then, for all 𝑡 such that 𝑡 = 𝑎 + 𝜇 + ], we have
Moreover, there is one paper [12] in which the following two statements have been shown.
Δ−] [Δ−𝜇 𝑓 (𝑡)] = Δ−(]+𝜇) 𝑓 (𝑡) = Δ−𝜇 [Δ−] 𝑓 (𝑡)] ,
Lemma 8 (see [13]). The unique solution of the FBVP (1) is given by
(6)
where 𝑡 ∈ N𝑎+]+𝜇 .
𝑏+1
Lemma 5. Let 0 ⩽ 𝑁 − 1 < ] ⩽ 𝑁. Then,
𝑦 (𝑡) fl ∑ 𝐺 (𝑡, 𝑠) ℎ (𝑠 + ] − 1) ,
Δ−] [Δ] 𝑓 (𝑡)] = 𝑓 (𝑡) + 𝐶1 𝑡]−1 + 𝐶2 𝑡]−2 + ⋅ ⋅ ⋅ + 𝐶𝑁𝑡]−𝑁,
(7)
where 𝐺(𝑡, 𝑠) is Green’s function for the problem
for some 𝐶𝑖 ∈ R with 1 ⩽ 𝑖 ⩽ 𝑁.
−Δ] 𝑦 (𝑡) = ℎ (𝑡 + ] − 1) ,
Definition 6. Let 𝐸 be a real Banach space. A nonempty convex closed set 𝑃 is called a cone provided that (1) 𝑎𝑢 ∈ 𝑃, for all 𝑢 ∈ 𝑃; 𝑎 ⩾ 0 and (2) 𝑢, −𝑢 ∈ 𝑃 implies 𝑢 = 0.
𝑦 (] − 2) = 0 = 𝑦 (] + 𝑏 + 1) ,
Lemma 9 (see [13]). Let 𝐺(𝑡, 𝑠) be given as in the statement of Lemma 8. Then, we find that (1) 𝐺(𝑡, 𝑠) > 0 for each (𝑡, 𝑠) ∈ []−1, ]+𝑏]N]−1 ×[0, 𝑏+1]N0 ;
Δ 𝑡 𝐺 (𝑡, 𝑠) =
(3) there exists a number 𝛾 ∈ (0, 1) such that max
𝑡∈[]−1,]+𝑏]N]−1
(11)
= 𝛾𝐺 (𝑠 + ] − 1, 𝑠)
Lemma 10. For given 𝑠 ∈ [0, 𝑏 + 1]N0 , 𝐺(𝑡, 𝑠) is decreasing and Δ 𝑡 𝐺(𝑡, 𝑠) is increasing with respect to 𝑡 for 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1, where Δ 𝑡 𝐺(𝑡, 𝑠) = 𝑑𝐺(𝑡, 𝑠)/𝑑𝑡.
(] − 1) Γ (𝑏 + 3) Γ (]) Γ (] + 𝑏 + 2)
(12)
Γ (𝑡 + 1) Γ (] + 𝑏 − 𝑠 + 1) ⋅[ Γ (𝑡 − ] + 3) Γ (𝑏 − 𝑠 + 2)
𝐺 (𝑡, 𝑠)
for 𝑠 ∈ [0, 𝑏 + 1]N0 .
] − 1 𝑡]−2 (] + 𝑏 − 𝑠)]−1 [ Γ (]) (] + 𝑏 + 1)]−1
− (𝑡 − 𝑠 − 1)]−2 ] =
]−1
𝐺 (𝑡, 𝑠) ⩾ 𝛾 ⋅
(10)
Proof. For 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1, we find that
(2) max𝑡∈[]−1,]+𝑏]N 𝐺(𝑡, 𝑠) = 𝐺(𝑠+]−1, 𝑠), 𝑠 ∈ [0, 𝑏+1]N0 ;
min
(9)
where 1 < ] ⩽ 2, which is given by
𝑡]−1 (] + 𝑏 − 𝑠)]−1 { { − (𝑡 − 𝑠 − 1)]−1 , 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1, { ]−1 { { + 𝑏 + 1) (] 1 𝐺 (𝑡, 𝑠) = { Γ (]) { { 𝑡]−1 (] + 𝑏 − 𝑠)]−1 { { , 0 ⩽ 𝑡 − ] + 1 < 𝑠 ⩽ 𝑏 + 1. ]−1 { (] + 𝑏 + 1)
𝑡∈[(𝑏+])/4,3(𝑏+])/4]
(8)
𝑠=0
−
Γ (𝑡 − 𝑠) Γ (] + 𝑏 + 2) ]. Γ (𝑡 − 𝑠 − ] + 2) Γ (𝑏 + 3)
Similar to the proof of [14], we have Γ (𝑡 − 𝑠) Γ (] + 𝑏 + 2) Γ (𝑡 − ] + 3) Γ (𝑏 − 𝑠 + 2) Γ (𝑡 + 1) Γ (] + 𝑏 − 𝑠 + 1) Γ (𝑡 − 𝑠 − ] + 2) Γ (𝑏 + 3) > 1.
(13)
Advances in Mathematical Physics
3
So, it shows that
we get
Δ 𝑡 𝐺 (𝑡, 𝑠) < 0, 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1.
(14) Δ𝑡 [
Then, 𝐺(𝑡, 𝑠) is decreasing with respect to 𝑡 for 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1. This implies that Δ 𝑡𝑡 𝐺 (𝑡, 𝑠) fl Δ 𝑡 (Δ 𝑡 𝐺 (𝑡, 𝑠)) =
(] − 1) (] − 2) 𝑡]−3 (] + 𝑏 − 𝑠)]−1 [ Γ (]) (] + 𝑏 + 1)]−1
(15)
− (𝑡 − 𝑠 − 1)]−3 ] . Similar to the proof of [14], we have
𝑡−]+3 𝑡−]−𝑠+2
Γ (𝑡 − 𝑠) Γ (] + 𝑏 + 2) Γ (𝑡 − ] + 3) Γ (𝑏 − 𝑠 + 2) Γ (𝑡 + 1) Γ (] + 𝑏 − 𝑠 + 1) Γ (𝑡 − 𝑠 − ] + 2) Γ (𝑏 + 3) (16) 𝑡−]+3 > 𝑡−]−𝑠+2 Γ (𝑡 − 𝑠) Γ (] + 𝑏 + 2) Γ (𝑡 − ] + 3) Γ (𝑏 − 𝑠 + 2) Γ (𝑡 + 1) Γ (] + 𝑏 − 𝑠 + 1) Γ (𝑡 − 𝑠 − ] + 2) Γ (𝑏 + 3)
> 1.
0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1.
(17)
Then, Δ 𝑡 𝐺(𝑡, 𝑠) is increasing with respect to 𝑡 for 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1. Lemma 11. For given 𝑠 ∈ [0, 𝑏 + 1]N0 , 𝐺(𝑡, 𝑠)/(𝑏 + ] + 1 − 𝑡) is decreasing with respect to 𝑡 for 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1. Proof. By Lemma 10, for any given 𝑠 ∈ [0, 𝑏 + 1]N0 , 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1, we have 𝐺 (𝑡, 𝑠) ⩾ 𝐺 (𝑏 + ] + 1, 𝑠) = 0, 𝑠 < 𝑡 − ] + 1, Δ 𝑡 𝐺 (𝑡, 𝑠) ⩽ Δ 𝑡 𝐺 (𝜃, 𝑠) = 0,
𝑡 − ] + 1 ⩽ 𝜃 ⩽ 𝑏 + 1.
(18)
Δ 𝑡 𝐺 (𝑡, 𝑠) + 𝐺 (𝑡, 𝑠) / (𝑏 + ] + 1 − 𝑡) (𝑏 + ] + 1 − 𝑡)
=
Δ 𝑡 𝐺 (𝑡, 𝑠) − (𝐺 (𝑏 + ] + 1, 𝑠) − 𝐺 (𝑡, 𝑠)) / (𝑏 + ] + 1 − 𝑡) (𝑏 + ] + 1 − 𝑡)
=
Δ 𝑡 𝐺 (𝑡, 𝑠) − Δ 𝑡 𝐺 (𝜃, 𝑠) ⩽ 0. (𝑏 + ] + 1 − 𝑡)
𝑡]−1 (] + 𝑏 − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) (] − 1) ⩽ 𝐺 (𝑡, 𝑠) (𝑏 + 2) (𝑏 + 3) Γ (]) (𝑏 + ] + 1)]−1 𝑡]−1 (𝑏 + ] − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) ⩽ . Γ (]) (𝑏 + ] + 1)]−1
(20)
(21)
𝐺 (𝑡, 𝑠) 𝐺 (𝑡, 𝑠) − 𝐺 (𝑏 + ] + 1, 𝑠) = 𝑡−𝑏−]−1 𝑡−𝑏−]−1 𝑡 − ] + 1 < 𝜃 < 𝑏 + 1,
𝐺 (𝑡, 𝑠) 𝐺 (𝑠 + ] − 1, 𝑠) ⩽ 𝑏+]+1−𝑡 𝑏+2−𝑠 ⩽
(𝑠 + ] − 1)]−1 (𝑏 + ] − 𝑠)]−1 Γ (]) (𝑏 + ] + 1)]−1
⩽
𝑡]−1 (𝑏 + ] − 𝑠)]−1 , Γ (]) (𝑏 + ] + 1)]−1
𝐺 (𝑡, 𝑠) 𝐺 (𝑡, 𝑠) ⩾ lim − 𝑏 + ] + 1 − 𝑡 𝑡→(𝑏+]+1) 𝑏 + ] + 1 − 𝑡 =
Because
= Δ 𝑡 𝐺 (𝜃, 𝑠) ,
=
Proof. By the definition of 𝐺(𝑡, 𝑠), it was obvious that 𝐺(𝑡, 𝑠) = 𝐺(] + 𝑏 − 𝑠, ] + 𝑏 − 𝑡) for 𝑠 ∈ [0, 𝑏 + 1]N0 . By Lemma 11, for 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1, we get
So, since ] − 2 ⩽ 0, we have Δ 𝑡𝑡 𝐺 (𝑡, 𝑠) > 0,
(𝑏 + ] + 1 − 𝑡) Δ 𝑡 𝐺 (𝑡, 𝑠) − 𝐺 (𝑡, 𝑠) Δ 𝑡 (𝑏 + ] + 1 − 𝑡) (𝑏 + ] + 1 − 𝑡) (𝑏 + ] + 1 − 𝑡)
Theorem 12. The Green function 𝐺(𝑡, 𝑠) defined by Lemma 8 has the following: 𝐺(𝑡, 𝑠) = 𝐺(] + 𝑏 − 𝑠, ] + 𝑏 − 𝑡) and for each (𝑡, 𝑠) ∈ [] − 1, ] + 𝑏]N]−1 × [0, 𝑏 + 1]N0 ,
⋅
⋅
=
Namely, for given 𝑠 ∈ [0, 𝑏 + 1]N0 , 𝐺(𝑡, 𝑠)/(𝑏 + ] + 1 − 𝑡) is decreasing with respect to 𝑡 for 0 ⩽ 𝑠 < 𝑡 − ] + 1 ⩽ 𝑏 + 1.
Γ (𝑡 − 𝑠) Γ (] + 𝑏 + 2) Γ (𝑡 − ] + 4) Γ (𝑏 − 𝑠 + 2) Γ (𝑡 + 1) Γ (] + 𝑏 − 𝑠 + 1) Γ (𝑡 − 𝑠 − ] + 3) Γ (𝑏 + 3) =
𝐺 (𝑡, 𝑠) ] 𝑏+]+1−𝑡
(19)
lim
𝑡→(𝑏+]+1)−
(−Δ 𝑡 𝐺 (𝑡, 𝑠))
=
𝑏−𝑠+1 ]−1 (] + 𝑏 − 𝑠)]−1 Γ (]) 𝑏−𝑠+2
⩾
(] − 1) (] + 𝑏 − 𝑠)]−1 𝑡]−1 . Γ (]) (𝑏 + 2) (𝑏 + 3) (] + 𝑏 + 1)]−1
(22)
4
Advances in Mathematical Physics For 0 ⩽ 𝑡 − ] + 1 < 𝑠 ⩽ 𝑏 + 1, we get 𝐺 (𝑡, 𝑠) = ⩽
𝑡]−1 (] + 𝑏 − 𝑠)]−1 Γ (]) (] + 𝑏 + 1)]−1 𝑡]−1 (𝑏 + ] − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) , Γ (]) (𝑏 + ] + 1)]−1
𝑡]−1 (] + 𝑏 − 𝑠)]−1 𝐺 (𝑡, 𝑠) = Γ (]) (] + 𝑏 + 1)]−1 ⩾
boundary value problems of discrete fractional differential equation (see the proofs of Theorems 15 and 17). Theorems 12 and 13 extend the results of integer-order Dirichlet boundary value problems.
3. Main Results and Proofs (23)
𝑡]−1 (] + 𝑏 − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) (] − 1) , (𝑏 + 2) (𝑏 + 3) Γ (]) (] + 𝑏 + 1)]−1
In this section, we will give the existence results of positive solution to the boundary value problem (1) on the basis of Theorem 13 and Lemma 7 and make the following assumptions: (𝐻0 ) There exist 𝑔 ∈ 𝐶([0, +∞) × [0, +∞)), 𝑞1 , 𝑞2 ∈ 𝐶([] − 1, ] + 𝑏]N]−1 × (0, +∞)) such that
so 𝑡]−1 (] + 𝑏 − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) (] − 1) ⩽ 𝐺 (𝑡, 𝑠) (𝑏 + 2) (𝑏 + 3) Γ (]) (] + 𝑏 + 1)]−1 𝑡]−1 (𝑏 + ] − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) ⩽ . Γ (]) (𝑏 + ] + 1)]−1
𝑞1 (𝑡) 𝑔 (𝑦) ⩽ 𝑓 (𝑡,
𝑡]−1 (𝑏 + ] − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) ⩽ . Γ (]) (𝑏 + ] + 1)]−1
𝑏+1
∑ 𝑞𝑖 (𝑠 + ] − 1) < +∞, 𝑖 = 1, 2.
For convenience, we introduce the following notations: (25)
(26)
𝑡]−1 (] + 𝑏 − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) (] − 1) ⩽ 𝐺 (𝑡, 𝑠) (𝑏 + 2) (𝑏 + 3) Γ (]) (] + 𝑏 + 1)]−1
⩽
(27)
⩽
(𝑏 + ] − 𝑠) (𝑠 + 1) . Γ (]) (𝑏 + ] + 1)]−1
𝑔 (𝑦) . 𝑔0 = lim inf 𝑦→0 𝑦
(31)
(𝐻1 ) 𝑔0 = ∞, 𝑔∞ = ∞. (𝐻2 ) 𝑔0 = 0, 𝑔∞ = 0. (𝐻3 ) There exists 𝑝 > 0 such that 𝑔(𝑦) < 𝑀1 𝑝 for 0 ⩽ 𝑦 ⩽ 𝑝, where 𝑀1 = [(1/Γ(])(𝑏+]+1)]−1 ) ∑𝑏+1 𝑠=0 (𝑏+ ] − 𝑠)]−1 (𝑠 + 1)𝑞2 (𝑠 + ] − 1)]−1 . > 0 such that (𝐻4 ) There exists 𝑝 𝑔(𝑦) > 𝑀2 𝑝 for 𝑝(𝛼 − 1)/16 ⩽ 𝑦 ⩽ 𝑝, ∗ where 𝑀2 = [∑3(𝑏+])/4 𝑠=(𝑏+])/4 𝐺 ((𝑏 + ])/2, −1 𝑠)𝑞1 (𝑠 + ] − 1)] . We note that 𝑦 is a solution of the boundary value problem (1) if and only if 𝑦 is a fixed point of the operator 𝑇𝑦 fl ∑ 𝐺∗ (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑠=0
𝑦 (𝑠 + ] − 1) ), 𝑠]−1
(32)
where 𝐺∗ is Green’s function derived in this paper and 𝑇 : 𝐵 → 𝐵 and 𝐵 is the Banach space
Let 𝐺∗ (𝑡, 𝑠) = (1/𝑡]−1 )𝐺(𝑡, 𝑠). It is clear that
]−1
𝑔 (𝑦) , 𝑦
𝑏+1
(𝑏 + ] − 𝑠)]−1 𝑡]−1 (𝑠 + 1) . Γ (]) (𝑏 + ] + 1)]−1
(] + 𝑏 − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) (] − 1) ⩽ 𝐺∗ (𝑡, 𝑠) ]−1 + 2) + 3) (𝑏 (𝑏 Γ (]) (] + 𝑏 + 1)
𝑔∞ = lim sup 𝑦→∞
Proof. By Theorem 12, we get
= 𝐺 (] + 𝑏 − 𝑠, ] + 𝑏 − 𝑡)
(30)
𝑠=0
]−1
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) . ⩽ Γ (]) (𝑏 + ] + 1)]−1
(29)
here
Theorem 13. The function 𝐺∗ (𝑡, 𝑠) = (1/𝑡]−1 )𝐺(𝑡, 𝑠) has the following properties: (𝑏 + ] + 1 − 𝑡) (] − 1) (] + 𝑏 − 𝑠) ⩽ 𝐺∗ (𝑡, 𝑠) (𝑏 + 2) (𝑏 + 3) Γ (]) (] + 𝑏 + 1)]−1
) ⩽ 𝑞2 (𝑡) 𝑔 (𝑦) ,
𝑡 ∈ [] − 1, ] + 𝑏]N]−1 , 𝑦 ∈ [0, +∞) ;
(24)
Hence, for each (𝑡, 𝑠) ∈ [] − 1, ] + 𝑏]N]−1 × [0, 𝑏 + 1]N0 , 𝑡]−1 (] + 𝑏 − 𝑠)]−1 (𝑏 + ] + 1 − 𝑡) (] − 1) ⩽ 𝐺 (𝑡, 𝑠) (𝑏 + 2) (𝑏 + 3) Γ (]) (] + 𝑏 + 1)]−1
𝑦 𝑡]−1
𝐵 fl {𝑦 : [] − 2, ] + 𝑏 + 1]N]−2 → R | 𝑦 (] − 2) = 0 (28)
It is worth noting that two inequalities of Theorem 13 are very useful in the research of positive solutions to Dirichlet-type
= 𝑦 (] + 𝑏 + 1)} equipped with the norm 𝑦 (𝑡) š max 𝑡∈[]−1,]+𝑏]
N]−1
𝑦 (𝑡) .
(33)
(34)
Advances in Mathematical Physics
5
Define the cone 𝐾 ⊂ 𝐵 by
Let 𝑀 = max0⩽𝑦⩽𝐿 |𝑔(𝑦)| + 1; then for 𝑦 ∈ 𝐷, by Theorem 13 and (𝐻0 ), we have
𝐾 = {𝑦 ∈ 𝐵 : 𝑦 (𝑡) ⩾ 0, 𝑦 (𝑡) (35)
(] − 1) (𝑏 + ] + 1 − 𝑡) ⩾ 𝑦} . (𝑏 + 2)2 (𝑏 + 3)
𝑏+1 𝑦 (𝑠 + ] − 1) ∗ ) 𝑇𝑦 (𝑡) ⩽ ∑ 𝐺 (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑠]−1 𝑠=0 𝑏+1
⩽ ∑ 𝐺∗ (𝑡, 𝑠) 𝑞2 (𝑠 + ] − 1) 𝑔 (𝑦 (𝑠 + ] − 1)) (40) 𝑠=0
Lemma 14. Assume 𝑇 : 𝐾 → 𝐵 is the operator defined by 𝑏+1
𝑇𝑦 fl ∑ 𝐺∗ (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑠=0
𝑦 (𝑠 + ] − 1) ). 𝑠]−1
𝑏+1
(36)
Thus, 𝑇(𝐷) is bounded. Finally, we prove that 𝑇 is equicontinuous. Note that 𝐺∗ (𝑡, 𝑠) is uniformly continuous on (𝑡, 𝑠) ∈ [] − 1, ] + 𝑏]N]−1 × [0, 𝑏 + 1]N0 ; then, for 𝜀 > 0, each 𝑢 ∈ 𝐷, 𝑡1 , 𝑡2 ∈ []−1, ]+𝑏]N]−1 , 𝑡1 < 𝑡2 , there exists 𝜂 > 0 such that |𝐺∗ (𝑡2 , 𝑠)− 𝐺∗ (𝑡1 , 𝑠)| < 𝜀/𝑀 ∑𝑏+1 𝑠=0 𝑞2 (𝑠 + ] − 1) for 𝑡2 − 𝑡1 < 𝜂. Then, we get
Then, the mapping 𝑇 : 𝐾 → 𝐾 is completely continuous. Proof (𝐾 ⊂ 𝐵). Hence, 𝑇 : 𝐾 → 𝐾. On one hand, 𝑏+1
(𝑇𝑦) (𝑡) = ∑ 𝐺∗ (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑠=0
⩾
𝑦 (𝑠 + ] − 1) ) 𝑠]−1
(𝑏 + ] + 1 − 𝑡) (] − 1) (𝑏 + 2) (𝑏 + 3)
𝑏+1
(] + 𝑏 − 𝑠)]−1 ⋅∑ 𝑓 (𝑠 + ] ]−1 𝑠=0 Γ (]) (] + 𝑏 + 1) − 1,
(37)
𝑦 (𝑠 + ] − 1) ). 𝑠]−1
𝑇𝑦 (𝑡2 ) − 𝑇𝑦 (𝑡1 ) 𝑏+1 𝑦 (𝑠 + ] − 1) = ∑ 𝐺∗ (𝑡2 , 𝑠) 𝑓 (𝑠 + ] − 1, ) 𝑠=0 𝑠]−1 𝑏+1 𝑦 (𝑠 + ] − 1) ) − ∑ 𝐺∗ (𝑡1 , 𝑠) 𝑓 (𝑠 + ] − 1, 𝑠]−1 𝑠=0 𝑏+1
⩽ ∑ 𝐺∗ (𝑡2 , 𝑠) − 𝐺∗ (𝑡1 , 𝑠)
(41)
𝑠=0
On the other hand, max 𝑇𝑦 = 𝑡∈[]−1,]+𝑏+1]
N]−1
𝑇𝑦 (𝑡)
⋅ 𝑓 (𝑠 + ] − 1, 𝑏+1
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) ⩽∑ 𝑓 (𝑠 + ] ]−1 𝑠=0 Γ (]) (𝑏 + ] + 1) 𝑦 (𝑠 + ] − 1) ) ⩽ (𝑏 + 2) 𝑠]−1
𝑠=0
(38)
⋅ 𝑞2 (𝑠 + ] − 1) 𝑔 (𝑦 (𝑠 + ] − 1)) < 𝜀. Hence, we get 𝑇𝑦 (𝑡2 ) − 𝑇𝑦 (𝑡1 ) < 𝜀.
𝑏+1
(] + 𝑏 − 𝑠)]−1 𝑓 (𝑠 + ] ]−1 𝑠=0 Γ (]) (] + 𝑏 + 1)
⋅∑
Theorem 15. Suppose that (𝐻0 ), (𝐻1 ), and (𝐻3 ) are satisfied. Then, problem (1) has two positive solutions.
Then, we get (] − 1) (𝑏 + ] + 1 − 𝑡) 𝑇𝑦 . (𝑏 + 2)2 (𝑏 + 3)
(42)
In view of the Arzela-Ascoli theorem; it is easy to see that 𝑇 : 𝐾 → 𝐾 is completely continuous.
𝑦 (𝑠 + ] − 1) − 1, ). 𝑠]−1
(𝑇𝑦) (𝑡) ⩾
𝑦 (𝑠 + ] − 1) ) 𝑠]−1
⩽ ∑ 𝐺∗ (𝑡2 , 𝑠) − 𝐺∗ (𝑡1 , 𝑠)
𝑏+1
− 1,
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞 (𝑠 + ] − 1) . ]−1 2 𝑠=0 Γ (]) (𝑏 + ] + 1)
⩽ 𝑀∑
(39)
And by means of the expression of 𝐺∗ (𝑡, 𝑠), there are clearly (𝑇𝑦)(𝑡) ⩾ 0 whenever 𝑦 ∈ 𝐾, whence 𝑇𝑦 ∈ 𝐾. Next, we show that 𝑇 is uniformly bounded. Let 𝐷 ⊂ 𝐾 be bounded; that is, there exists a positive real number 𝐿 > 0 such that ‖𝑦‖ ⩽ 𝐿 for all 𝑦 ∈ 𝐷.
Proof. Assume that (𝐻1 ) holds. Since 𝑔0 = ∞, there exists 𝑟 ∈ (0, 𝑝) such that 𝑔(𝑦) ⩾ 𝑁𝑦 for 0 < 𝑦 < 𝑟, where 𝑁 > 0 satisfies 3(𝑏+])/4−]+1
∑
𝑠=(𝑏+])/4−]+1
𝐺∗ ({⌊
𝑏+1 ⌋ + ]} , 𝑠) 𝑞1 (𝑠 + ] − 1) 𝑁 2
(] − 1) ] ⋅ ⩾ 1. (𝑏 + 2)2 (𝑏 + 3)
(43)
6
Advances in Mathematical Physics
Let Ω𝑟 = {𝑦 ∈ 𝐵 | ‖𝑦‖ < 𝑟} for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑟 ; we note that {⌊(𝑏 + 1)/2⌋ + ]} ⊂ [(𝑏 + ])/4, 3(𝑏 + ])/4], where ⌊⋅⌋ is the usual so-called floor function. So we get
𝑦 (𝑡) ⩾
𝑏+1 ⌋ + ]}) 𝑇𝑦 ⩾ 𝑇𝑦 ({⌊ 2 𝑏+1
= ∑ 𝐺∗ ({⌊ 𝑠=0
𝑏 + ] 3 (𝑏 + ]) 𝑡∈[ , ], 4 4
𝑦 (𝑠 + ] − 1) ) 𝑠]−1
𝑏+1
𝑏+] 𝑏+] ) = ∑ 𝐺∗ ( , 𝑠) 𝑇𝑦 ⩾ 𝑇𝑦 ( 2 2
∗
𝑏+1
𝑏+1 ⋅ 𝑔 (𝑦 (𝑠 + ] − 1)) ⩾ ∑ 𝐺 ({⌊ ⌋ + ]} , 𝑠) 2 𝑠=0
𝑠=0
⋅ 𝑓 (𝑠 + ] − 1,
∗
(44)
∑
𝐺∗ ({⌊
𝑠=(𝑏+])/4−]+1
⋅ 𝑞1 (𝑠 + ] − 1) 𝑁 3(𝑏+])/4−]+1
⩾
∑
𝑠=(𝑏+])/4−]+1
⋅ 𝑞1 (𝑠 + ] − 1) 𝑁
𝑏+1 ⌋ + ]} , 𝑠) 2
⩾
𝐺∗ (
𝑏+] , 𝑠) 𝑞1 (𝑠 + ] − 1) 2 3(𝑏+])/4−]+1
∑
𝐺∗ (
𝑠=(𝑏+])/4−]+1
𝑏+1 ⌋ + ]} , 𝑠) 2
(47)
𝑏+] , 𝑠) 2
⋅ 𝑞1 (𝑠 + ] − 1) 𝜌𝑦 (𝑠 + ] − 1) ⩾
(] − 1) ] 𝑦 ⩾ 𝑦 . (𝑏 + 2)2 (𝑏 + 3)
𝑏+1 (𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑓 (𝑠 + ] 𝑇𝑦 ⩽ ∑ ]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
𝜌 (] − 1) ((𝑏 + ]) /4 + 1) (𝑏 + 2)2 (𝑏 + 3)
𝐺∗ (
𝑏+] , 𝑠) 𝑞1 (𝑠 + ] 2
− 1) 𝑦 ⩾ 𝑦 . Hence, for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑅 , ‖𝑇𝑦‖ ⩾ ‖𝑦‖. Thus, by Lemma 7, it follows that 𝑇 has a fixed point 𝑦2 in 𝐾 ∩ (𝜕Ω𝑅 \ Ω𝑝 ) such that 𝑝 ⩽ ‖𝑦2 ‖ ⩽ 𝑅. Then, 𝑇 has a fixed point 𝑦1 in 𝐾 ∩ (𝜕Ω𝑝 \ Ω𝑟 ) and a fixed point 𝑦2 in 𝐾∩(𝜕Ω𝑅 \Ω𝑝 ) such that 𝑟 ⩽ ‖𝑦1 ‖ ⩽ 𝑝 ⩽ ‖𝑦2 ‖ ⩽ 𝑅, and satisfy
𝑏+1 𝑦 (𝑠 + ] − 1) (𝑏 + ] − 𝑠)]−1 (𝑠 + 1) ) ⩽ 𝑞 (𝑠 ∑ ]−1 2 𝑠]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
+ ] − 1) 𝑔 (𝑦 (𝑠 + ] − 1))
∑
⋅ 𝑔 (𝑦 (𝑠 + ] − 1)) ⩾
Then, this implies ‖𝑇𝑦‖ ⩾ ‖𝑦‖ for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑟 . Now, let Ω𝑝 = {𝑦 ∈ 𝐵 | ‖𝑦‖ < 𝑝} for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑝 . It follows from (𝐻3 ) that we get
− 1,
3(𝑏+])/4−]+1 𝑠=(𝑏+])/4−]+1
(] − 1) (𝑏 + ] + 1 − 𝑡) 𝑦 (𝑏 + 2)2 (𝑏 + 3)
𝐺∗ ({⌊
𝑏+1 𝑦 (𝑠 + ] − 1) 𝑏+] ) ⩾ ∑ 𝐺∗ ( , 𝑠) ]−1 𝑠 2 𝑠=0
⋅ 𝑞1 (𝑠 + ] − 1) 𝑔 (𝑦 (𝑠 + ] − 1))
⋅ 𝑞1 (𝑠 + ] − 1) 𝑁𝑦 (𝑠 + ] − 1) 3(𝑏+])/4−]+1
(46)
and so
𝑏+1 ⩾ ∑ 𝐺 ({⌊ ⌋ + ]} , 𝑠) 𝑞1 (𝑠 + ] − 1) 2 𝑠=0
⩾
(] − 1) ((𝑏 + ]) /4 + 1) ∗ 𝑦 > 𝑀 , (𝑏 + 2)2 (𝑏 + 3)
𝑏+1 ⌋ + ]} , 𝑠) 2
⋅ 𝑓 (𝑠 + ] − 1, 𝑏+1
Let 𝑅 = 𝑝 + (4(𝑏 + 2)2 /(] − 1)(𝑏 + ]))𝑀∗ and Ω𝑅 = {𝑦 ∈ 𝐵 | ‖𝑦‖ < 𝑅}, for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑅 ; we get
(45)
𝑏+1
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞 (𝑠 + ] − 1) 𝑀1 𝑝 = 𝑝 0 such that 𝑔(𝑦) ⩾ 𝜌𝑦 for 𝑦 > 𝑀∗ , where 𝜌 > 0 is chosen so that (𝜌(] − 1)((𝑏+])/4+1)/(𝑏+2)2 (𝑏+3))𝐺∗ ((𝑏+])/2, 𝑠)𝑞1 (𝑠+]−1) ⩾ 1.
𝑏+1
𝑦𝑖 (𝑡) = ∑ 𝐺∗ (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑠=0
𝑦𝑖 (𝑠 + ] − 1) ), 𝑠]−1
(48)
𝑡 ∈ [] − 1, ] + 𝑏]N]−1 , 𝑖 = 1, 2, and ‖𝑦𝑖 ‖ ⩽ 𝑅. It is clear that 𝑢𝑖 (𝑡) = (1/𝑡]−1 )𝑦𝑖 (𝑡) are two solutions of (1); that is, 𝑏+1
𝑢𝑖 (𝑡) = ∑ 𝐺 (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑢𝑖 (𝑠 + ] − 1)) , 𝑠=0
𝑡 ∈ [] − 1, ] + 𝑏]N]−1 , 𝑖 = 1, 2.
(49)
Advances in Mathematical Physics
7
Next, we will prove 𝑢𝑖 (0) = 0 (𝑖 = 1, 2). By means of (𝐻0 ), we get
⩾
lim 𝑢𝑖 (𝑡) 𝑏+1
𝑏+1 𝑠=0
(50)
𝑠=0
𝑏+1
⩽ lim inf ∑ 𝐺 (𝑡, 𝑠) 𝑞2 (𝑠 + ] − 1) 𝑠=0
⋅ max 𝑔 (𝑦𝑖 (𝑠 + ] − 1)) = 0, 𝑖 = 1, 2. ‖𝑦𝑖 ‖⩽𝑅
Therefore 𝑢𝑖 (0) = 0, 𝑖 = 1, 2, and 𝑢𝑖 (𝑡) = (1/𝑡]−1 )𝑦𝑖 (𝑡) is a solution of problem (1). Theorem 16. Suppose that (H0 ), (H2 ), and (H4 ) are satisfied. Then, problem (1) has two positive solutions. Proof. Assume that (𝐻2 ) holds. Since 𝑔0 = 0, there exists 𝑟 ∈ (0, 𝑝) such that 𝑔(𝑦) ⩽ 𝜀𝑦 for 0 < 𝑦 < 𝑟, where 𝜀 satisfy
Then, we see that ‖𝑇𝑦‖ > ‖𝑦‖ for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑝 . As a result, by Lemma 7, we obtain that 𝑇 has a fixed point in 𝐾 ∩ (𝜕Ω𝑝 \ Ω𝑟 ), such that 𝑟 ⩽ ‖𝑦1 ‖ < 𝑝. Furthermore, because 𝑔∞ = 0, there exists 𝑀 > 0 such that 𝑔(𝑦) ⩽ 𝜀𝑦 for 𝑦 > 𝑀, where 𝜀 > 0 is chosen so that ]−1 (𝑠 + 1)/Γ(])(𝑏 + ] + 1)]−1 )𝑞2 (𝑠 + ] − 1) < 1. 𝜀 ∑𝑏+1 0 ((𝑏 + ] − 𝑠) ]−1 (𝑠 + Let 𝑅 = max{max0⩽𝑦(𝑠)⩽𝑀{𝑔(𝑦)} ∑𝑏+1 0 ((𝑏 + ] − 𝑠) ]−1 1)/Γ(])(𝑏 + ] + 1)]−1 )𝑞2 (𝑠 + ] − 1)/(1 − 𝜀 ∑𝑏+1 (𝑠 + 0 ((𝑏 + ] − 𝑠) ]−1 1)/Γ(])(𝑏 + ] + 1) )𝑞2 (𝑠 + ] − 1)), 𝑝 + 𝑀} and Ω𝑅 = {𝑦 ∈ 𝐵 | ‖𝑦‖ < 𝑅}, for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑅 ; we get 𝑏+1
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑓 (𝑠 + ] ]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
𝑇𝑦 (𝑡) ⩽ ∑
𝜀 (𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞2 (𝑠 + ] − 1) ⩽ 1. ]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
(51)
Let Ω𝑟 = {𝑦 ∈ 𝐵 | ‖𝑦‖ < 𝑟} for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑟 ; we get
𝑏+1
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞 (𝑠 + ] − 1) ]−1 2 𝑠=0 Γ (]) (𝑏 + ] + 1)
⩽∑
⋅ 𝑔 (𝑦 (𝑠 + ] − 1)) ⩽(
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑓 (𝑠 + ] 𝑇𝑦 (𝑡) ⩽ ∑ ]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
𝑏+1
⩽∑ 0
(52)
+ ] − 1) 𝜀𝑦 (𝑠 + ] − 1)
∑
)
𝑀⩽𝑦(𝑠)⩽𝑅
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) Γ (]) (𝑏 + ] + 1)]−1 (54)
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞2 (𝑠 + ] − 1) Γ (]) (𝑏 + ] + 1)]−1
⋅ max 𝑔 (𝑦 (𝑠 + ] − 1)) 0⩽𝑦(𝑠)⩽𝑀
𝑏+1
𝜀 (𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞2 (𝑠 + ] − 1) 𝑦 ]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
⩽∑
⩽ 𝑦 . Then, this implies that ‖𝑇𝑦‖ ⩽ ‖𝑦‖ for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑟 . Now, let Ω𝑝 = {𝑦 ∈ 𝐵 | ‖𝑦‖ < 𝑝} for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑝 ; then, 𝑦(𝑡) ⩾ ((𝛼 − 1)/16)𝑝, 𝑡 ∈ [(𝑏 + ])/4, 3(𝑏 + ])/4]; by (𝐻4 ), we get
⋅ 𝑓 (𝑠 + ] − 1,
+
⋅ 𝑞2 (𝑠 + ] − 1) 𝑔 (𝑦 (𝑠 + ] − 1))
𝑏+1 𝑦𝑖 (𝑠 + ] − 1) (𝑏 + ] − 𝑠)]−1 (𝑠 + 1) ) ⩽ 𝑞 (𝑠 ∑ ]−1 2 𝑠]−1 𝑠=0 Γ (]) (𝑏 + ] + 1)
𝑏+] )⩾ 𝑇𝑦 ⩾ 𝑇𝑦 ( 2
∑
0⩽𝑦(𝑠)⩽𝑀
𝑏+1
− 1,
𝑦𝑖 (𝑠 + ] − 1) ) 𝑠]−1
− 1,
𝑏+1
∑
𝑏+] , 𝑠) 2
(53)
⩽ lim inf ∑ 𝐺 (𝑡, 𝑠) 𝑞2 (𝑠 + ] − 1) 𝑔 (𝑦𝑖 (𝑠 + ] − 1))
𝑡→0
∑ 𝐺∗ (
⋅ 𝑞1 (𝑠 + ] − 1) 𝑀2 𝑝 = 𝑝 = 𝑦 .
𝑦𝑖 (𝑠 + ] − 1) ) 𝑠]−1
𝑏+1
3(𝑏+])/4
𝑠=(𝑏+])/4
𝑠=0
= lim inf ∑ 𝐺 (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1,
𝑡→0
𝑏+] , 𝑠) 𝑞1 (𝑠 + ] − 1) 2
⋅ 𝑔 (𝑦 (𝑠 + ] − 1)) >
= lim inf ∑ 𝐺 (𝑡, 𝑠) 𝑓 (𝑠 + ] − 1, 𝑢𝑖 (𝑠 + ] − 1))
𝑡→0
∑ 𝐺∗ (
𝑠=(𝑏+])/4
𝑡→0−
𝑡→0
3(𝑏+])/4
3(𝑏+])/4
∑ 𝐺∗ (
𝑠=(𝑏+])/4
𝑦 (𝑠 + ] − 1) ) 𝑠]−1
𝑏+] , 𝑠) 2
+
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞 (𝑠 + ] − 1) 𝜀𝑦 (𝑠 ]−1 2 𝑀⩽𝑦(𝑠)⩽𝑅 Γ (]) (𝑏 + ] + 1) ∑
+ ] − 1) ⩽ ( max {𝑔 (𝑦)} + 𝜀 𝑦) 0⩽𝑦(𝑠)⩽𝑀
𝑏+1
⋅∑ 0
(𝑏 + ] − 𝑠)]−1 (𝑠 + 1) 𝑞2 (𝑠 + ] − 1) ⩽ 𝑅 = 𝑦 . Γ (]) (𝑏 + ] + 1)]−1
Therefore, for 𝑦 ∈ 𝐾 ∩ 𝜕Ω𝑅 we get ‖𝑇𝑦‖ ⩽ ‖𝑦‖. So it suffices to obtain that 𝑇 has a fixed point in 𝐾∩(𝜕Ω𝑅 \ Ω𝑝 ) such that 𝑝 ⩽ ‖𝑦1 ‖ < 𝑅 in view of Lemma 7. Then, 𝑇 has fixed points 𝑦1 in 𝐾 ∩ (𝜕Ω𝑝 \ Ω𝑟 ) and 𝑦2 in 𝐾 ∩ (𝜕Ω𝑅 \ Ω𝑝 ) such that 𝑟 ⩽ ‖𝑦1 ‖ < 𝑝 ⩽ ‖𝑦2 ‖ < 𝑅.
8
Advances in Mathematical Physics
Therefore, problem (1) gets two positive solutions 𝑢1 = (1/𝑡]−1 )𝑦1 (𝑡) and 𝑢2 = (1/𝑡]−1 )𝑦2 (𝑡). Theorem 17. Suppose that (H0 ) is satisfied. Furthermore, either 𝑔0 = 0,
So, for 𝜇 < 𝜇∗ , we get 𝑔(𝑦) = 𝜇(𝑦𝑎 + 𝑦𝑏 ) ⩽ 𝜇(𝑝𝑎 + 𝑝𝑏 ) < 𝑀1 𝑝, 0 ⩽ 𝑦 ⩽ 𝑝. Therefore, (𝐻3 ) holds. According to Theorem 15, problem (57) has two positive solutions.
Competing Interests The authors declare that they have no competing interests.
𝑔∞ = ∞
(55)
Authors’ Contributions
(𝑠𝑢𝑝𝑙𝑖𝑛𝑒𝑎𝑟)
All four authors read and approved the final paper.
or 𝑔0 = ∞, 𝑔∞ = 0
(56)
Acknowledgments
Then, problem (1) has one positive solution.
The research was supported by a grant from of the National Natural Science Foundation of China (no. 11271235) and the Foundation of Datong University (2010-B-01, 2014Q10, and XJY-2012211).
The proof of Theorem 17 is similar to the proof of Theorems 15 and 16.
References
(𝑠𝑢𝑏𝑙𝑖𝑛𝑒𝑎𝑟) .
[1] A. A. Kilbas, H. H. Srivastava, and J. J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier, Amsterdam, The Netherlands, 2006.
4. Example Consider the following boundary value problem:
[2] K. S. Miller and B. Ross, An Introduction to the Fractional Calculus and Differential Equations, John Wiley & Sons, New York, NY, USA, 1993.
]
− Δ 𝑦 (𝑡) 𝑎
𝑏
= 𝜇 (𝑎 (𝑡]−1 𝑦 (𝑡 + ] − 1)) + 𝑏 (𝑡]−1 𝑦 (𝑡 + ] − 1)) ) , 1 0
