Hindawi Publishing Corporation
Discrete Dynamics in Nature and Society
Volume 2013, Article ID 684962, 10 pages
http://dx.doi.org/10.1155/2013/684962
Research Article
Positive Solutions for Fourth-Order Nonlinear Differential
Equation with Integral Boundary Conditions
Qi Wang, Yanping Guo, and Yude Ji
Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, China
Correspondence should be addressed to Yanping Guo; guoyanping65@sohu.com
Received 5 November 2012; Accepted 8 January 2013
Academic Editor: Cengiz Çinar
Copyright © 2013 Qi Wang 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 investigates the existence and nonexistence of positive solutions for a class of fourth-order nonlinear differential equation
with integral boundary conditions. The associated Green’s function for the fourth-order boundary value problems is first given, and
the arguments are based on Krasnoselskii’s fixed point theorem for operators on a cone.
1. Introduction
In this paper, we consider the following fourth-order boundary value problems (BVPs) with integral boundary conditions
𝑥(4) (𝑡) = 𝑤 (𝑡) 𝑓 (𝑡, 𝑥 (𝑡) , 𝑥󸀠󸀠 (𝑡)) ,
𝑡 ∈ (0, 1) ,
1
𝑥 (0) = ∫ ℎ1 (𝑠) 𝑥 (𝑠)d𝑠,
0
1
𝑥 (1) = ∫ 𝑘1 (𝑠) 𝑥 (𝑠)d𝑠,
0
(1)
ℎ1 , ℎ2 , 𝑘1 , 𝑘2 = 0, BVP (1) reduces to the following two-point
BVP
𝑥(4) (𝑡) = 𝑤 (𝑡) 𝑓 (𝑡, 𝑥 (𝑡) , 𝑥󸀠󸀠 (𝑡)) , 𝑡 ∈ (0, 1) ,
𝑥 (0) = 𝑥 (1) = 𝑥󸀠󸀠 (0) = 𝑥󸀠󸀠 (1) = 0.
A great deal of research has been devoted to the existence
of solutions for problem (2) by using the Leray-Schauder
continuation method, topological degree, and the method of
lower and upper solutions.
For the case of ℎ1 = 𝑔, ℎ2 = ℎ, 𝑘1 = 0, 𝑘2 = 0 or ℎ1 =
0, ℎ2 = 0, 𝑘1 = 𝑔, and 𝑘2 = ℎ, BVP (1) reduces to the following BVP with integral boundary conditions
1
𝑥(4) (𝑡) = 𝑤 (𝑡) 𝑓 (𝑡, 𝑥 (𝑡) , 𝑥󸀠󸀠 (𝑡)) ,
𝑥󸀠󸀠 (0) = ∫ ℎ2 (𝑠) 𝑥󸀠󸀠 (𝑠)d𝑠,
0
󸀠󸀠
1
1
𝑥 (0) = ∫ 𝑔 (𝑠) 𝑥 (𝑠)d𝑠,
󸀠󸀠
𝑥 (1) = ∫ 𝑘2 (𝑠) 𝑥 (𝑠)d𝑠,
0
0
where 𝑤 may be singular at 𝑡 = 0 and(or) 𝑡 = 1, 𝑓 :
[0, 1] × [0, +∞) × (−∞, 0] → [0, +∞) is continuous, and
ℎ1 , ℎ2 , 𝑘1 , 𝑘2 ∈ 𝐿1 [0, 1] are nonnegative.
The existence of solutions for nonlinear higher-order
nonlocal BVPs has been studied by several authors, for example, see [1–9] and the references therein. However, there are
a few papers dealing with the existence of positive solutions for the fourth-order boundary value problems with
integral boundary conditions, see [7–15]. For the case of
(2)
1
𝑥󸀠󸀠 (0) = ∫ ℎ (𝑠) 𝑥󸀠󸀠 (𝑠)d𝑠,
0
𝑡 ∈ (0, 1) ,
𝑥 (1) = 0,
(3)
𝑥󸀠󸀠 (1) = 0,
or
𝑥(4) (𝑡) = 𝑤 (𝑡) 𝑓 (𝑡, 𝑥 (𝑡) , 𝑥󸀠󸀠 (𝑡)) ,
𝑥 (0) = 0,
𝑥󸀠󸀠 (0) = 0,
𝑡 ∈ (0, 1) ,
1
𝑥 (1) = ∫ 𝑔 (𝑠) 𝑥 (𝑠)d𝑠,
0
1
𝑥󸀠󸀠 (1) = ∫ ℎ (𝑠) 𝑥󸀠󸀠 (𝑠)d𝑠.
0
(4)
2
Discrete Dynamics in Nature and Society
By using the fixed point theorem of cone expansion and
compression of norm type, Zhang and Ge [11] have obtained
the existence and multiplicity of positive solutions for BVP
(3) and (4).
The following assumptions will stand throughout this
paper:
(𝐴1) 𝜔 is nonnegative, and 𝜔 ∈ 𝐿1 [0, 1] may have singularities at 𝑡 = 0 and(or) 𝑡 = 1;
(𝐴2) 𝑓 ∈ 𝐶([0, 1] × [0, +∞) × (−∞, 0], [0, +∞));
has a unique solution 𝑥 which is given by
1
̃ (𝑡, 𝑠) 𝑦 (𝑠)d𝑠,
𝑥 (𝑡) = ∫ 𝐻
0
where
1
̃ (𝑡, 𝑠) = 𝐺 (𝑡, 𝑠) + 𝑚 + 𝜇𝑡 ∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝐻
𝑚𝜈 + 𝑛𝜇 0
(𝐴3) ℎ1 , ℎ2 , 𝑘1 , 𝑘2 ∈ 𝐿1 [0, 1] are nonnegative and
𝑛 − 𝜈𝑡 1
+
∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏)d𝜏,
𝑚𝜈 + 𝑛𝜇 0
1
𝜇1 = 1 − ∫ ℎ1 (𝑠)d𝑠 > 0,
0
𝐺 (𝑡, 𝑠) = {
1
𝜈1 = 1 − ∫ 𝑘1 (𝑠)d𝑠 > 0,
0
0
The paper is organized as follows. In Section 2, we provide
some necessary background material such as the Krasnoselskii’s fixed point theorem in cones. The associated Green’s
function for the fourth-order boundary value problem is first
given, and we also look at some properties of the Green’s
function associated with the problem (1). In Section 3, the
main results of problem (1) will be stated and proved. In
Section 4, we give an example to illustrate how the main
results can be used in practice.
0
1
∫0 𝑘(𝑠)𝑥(𝑠)d𝑠, it follows that
1
𝜏
𝐵 = − ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠) 𝑦 (𝑠)d𝑠 d𝜏
1
Lemma 2. If ℎ, 𝑘 ∈ 𝐿 [0, 1] are nonnegative, and 𝜇 = 1 −
1
1
∫0 ℎ(𝑠)𝑑𝑠 > 0, 𝜈 = 1 − ∫0 𝑘(𝑠)d𝑠 > 0, then for any 𝑦 ∈ 𝐶(0, 1),
the BVP
𝑡 ∈ (0, 1) ,
0
1
1
0
0
+ 𝐴 ∫ 𝜏ℎ (𝜏)d𝜏 + 𝐵 ∫ ℎ (𝜏)d𝜏,
1
Lemma 1 (in [16]). Let Ω1 and Ω2 be two bounded open sets
in a real Banach space 𝐸, such that 𝜃 ∈ Ω1 and Ω1 ⊂ Ω2 . Let
operator 𝐴 : 𝑃∩(Ω2 \Ω1 ) → 𝑃 be continuous, where 𝜃 denotes
the zero element of 𝐸 and 𝑃 is a cone in 𝐸. Suppose that one of
the two conditions
(1) ‖𝐴𝑥‖ ≤ ‖𝑥‖, ∀𝑥 ∈ 𝑃 ∩ 𝜕Ω1 and ‖𝐴𝑥‖ ≥ ‖𝑥‖, ∀𝑥 ∈
𝑃 ∩ 𝜕Ω2 or
(2) ‖𝐴𝑥‖ ≥ ‖𝑥‖, ∀𝑥 ∈ 𝑃 ∩ 𝜕Ω1 and ‖𝐴𝑥‖ ≤ ‖𝑥‖, ∀𝑥 ∈
𝑃 ∩ 𝜕Ω2 is satisfied. Then, 𝐴 has at least one fixed point
in 𝑃 ∩ (Ω2 \ Ω1 ).
(11)
1
In our main results, we will make use of the following lemmas.
− ∫ (1 − 𝑠) 𝑦 (𝑠)d𝑠 + 𝐴 + 𝐵
(12)
0
1
𝜏
= − ∫ 𝑘 (𝜏) ∫ (𝜏 − 𝑠) 𝑦 (𝑠)d𝑠 d𝜏
0
0
1
1
0
0
+ 𝐴 ∫ 𝜏𝑘 (𝜏)d𝜏 + 𝐵 ∫ 𝑘 (𝜏)d𝜏,
that is,
1
1
0
0
𝐴 ∫ 𝜏ℎ (𝜏) d𝜏 − 𝐵 (1 − ∫ ℎ (𝜏) d𝜏)
1
𝜏
0
0
= ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠) 𝑦 (𝑠)d𝑠 d𝜏,
1
1
0
0
𝐴 (1 − ∫ 𝜏𝑘 (𝜏)d𝜏) + 𝐵 (1 − ∫ 𝑘 (𝜏)d𝜏)
1
0
(10)
Now, we solve for 𝐴, 𝐵 by 𝑥(0) = ∫0 ℎ(𝑠)𝑥(𝑠)d𝑠 and 𝑥(1) =
0
𝑥 (1) = ∫ 𝑘 (𝑠) 𝑥 (𝑠)d𝑠,
(9)
𝑡
𝑥 (𝑡) = − ∫ (𝑡 − 𝑠) 𝑦 (𝑠)d𝑠 + 𝐴𝑡 + 𝐵.
2. Preliminary Results
1
0
(8)
Proof. The general solution of −𝑥󸀠󸀠 (𝑡) = 𝑦(𝑡) can be written
as
1
𝜈2 = 1 − ∫ 𝑘2 (𝑠)d𝑠 > 0.
0
𝑛 = 1 − ∫ 𝑠𝑘 (𝑠)d𝑠.
0
0
𝑥 (0) = ∫ ℎ (𝑠) 𝑥 (𝑠)d𝑠,
1
𝑚 = ∫ 𝑠ℎ (𝑠)𝑑𝑠,
𝜇2 = 1 − ∫ ℎ2 (𝑠)d𝑠 > 0,
−𝑥󸀠󸀠 (𝑡) = 𝑦 (𝑡) ,
𝑠 (1 − 𝑡) , 0 ≤ 𝑠 ≤ 𝑡 ≤ 1,
𝑡 (1 − 𝑠) , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1,
1
(5)
1
(7)
(6)
1
1
𝜏
0
0
0
= ∫ (1 − 𝑠) 𝑦 (𝑠)d𝑠 − ∫ 𝑘 (𝜏) ∫ (𝜏 − 𝑠) 𝑦 (𝑠)d𝑠 d𝜏.
(13)
Discrete Dynamics in Nature and Society
3
1
Solving the above equations, we get
× ∫ (1 − 𝑠)𝑦(𝑠)d𝑠 − 𝜇𝑡
0
1
𝜏
1
𝐴=
[𝜈 ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏
𝑚𝜈 + 𝑛𝜇
0
0
1
𝜏
0
0
1
+ 𝜇 (∫ (1 − 𝑠)𝑦(𝑠)d𝑠
1
0
1
𝜏
0
0
− ∫ 𝑘 (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)] ,
1
[𝑚 (∫ (1 − 𝑠)𝑦(𝑠) d𝑠
𝑚𝜈 + 𝑛𝜇
0
0
0
(14)
0
0
1
1
1
0
0
1
1
1
0
0
0
𝑡
1
0
𝑡
0
0
1
𝜏
0
0
𝜏
× ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
0
1
1
0
0
(15)
𝜏
The unique solution of (6) be expressed as
1
𝑥 (𝑡) = ∫ 𝑠 (1 − 𝑡) 𝑦 (𝑠)d𝑠 + ∫ 𝑡 (1 − 𝑠) 𝑦 (𝑠)d𝑠 +
𝑚𝜈 + 𝑛𝜇
0
𝑡
1
1
0
0
𝜏
𝜏
0
1
1
+ ∫ ℎ (𝜏) ∫ 𝜏 (1 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
0
1
𝜏
𝜏
+ (∫ ℎ (𝜏) ∫ 𝑠 (1 − 𝜏)𝑦(𝑠)d𝑠 d𝜏
0
0
1
1
0
𝜏
+ ∫ ℎ (𝜏) ∫ 𝜏 (1 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
−𝑛 ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏] .
1
0
0
0
𝑡
1
1
× ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
0
1
− 𝜈𝑡 (∫ ℎ (𝜏) ∫ 𝑠 (1 − 𝜏)𝑦(𝑠)d𝑠 d𝜏
+ 𝑚 (∫ (1 − 𝑠)𝑦(𝑠)d𝑠 − ∫ 𝑘 (𝜏)
0
1
𝑚𝜈 + 𝑛𝜇
+ ∫ 𝑘 (𝜏) ∫ 𝜏 (1 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
+ 𝜇𝑡 (∫ (1 − 𝑠) 𝑦 (𝑠)d𝑠 − ∫ 𝑘 (𝜏)
𝜏
0
1
× [𝜇𝑡 (∫ 𝑘 (𝜏) ∫ 𝑠 (1 − 𝜏)𝑦(𝑠)d𝑠 d𝜏
× [𝜈𝑡 ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏
1
1
= ∫ 𝑠 (1 − 𝑡)𝑦(𝑠)d𝑠 + ∫ 𝑡 (1 − 𝑠) 𝑦 (𝑠)d𝑠 +
1
𝑥 (𝑡) = − ∫ (𝑡 − 𝑠) 𝑦 (𝑠)d𝑠 +
𝑚𝜈 + 𝑛𝜇
0
0
𝜏
+ ∫ 𝜏ℎ (𝜏)d𝜏 ∫ 𝜏𝑘 (𝜏)d𝜏 ∫ (1 − 𝑠)𝑦(𝑠)d𝑠]
𝑡
0
0
1
× ∫ 𝜏ℎ (𝜏)d𝜏 ∫ (1 − 𝑠)𝑦(𝑠)d𝑠
Therefore, (6) has a unique solution
𝜏
0
1
× ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏 − ∫ 𝜏𝑘 (𝜏)d𝜏
−𝑛 ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏] .
1
0
0
0
𝜏
0
0
1
× ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏 + ∫ 𝜏𝑘 (𝜏)d𝜏
− ∫ 𝑘 (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
0
𝜏
𝜏
𝜏
1
0
1
× ∫ 𝑘 (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏 − ∫ ℎ (𝜏)
1
1
1
+ ∫ 𝜏ℎ (𝜏)d𝜏 ∫ (1 − 𝑠)𝑦(𝑠)d𝑠 − ∫ 𝜏ℎ (𝜏)d𝜏
0
𝐵=
1
× ∫ 𝑘 (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏
1
− ∫ 𝜏𝑘 (𝜏)d𝜏
0
1
𝜏
× (∫ ℎ (𝜏) ∫ 𝑠 (1 − 𝜏)𝑦(𝑠)d𝑠 d𝜏
0
0
1
1
0
𝜏
+ ∫ ℎ (𝜏) ∫ 𝜏 (1 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)
× [𝜇𝑡 ∫ 𝜏𝑘 (𝜏)d𝜏 ∫ (1 − 𝑠) 𝑦 (𝑠)d𝑠 − 𝜇𝑡
1
1
× ∫ (1 − 𝑠) 𝑦 (𝑠)d𝑠 − 𝜈𝑡
0
1
1
0
0
× ∫ 𝜏ℎ (𝜏)d𝜏 ∫ (1 − 𝑠)𝑦(𝑠)d𝑠
1
𝜏
0
0
+ 𝜈𝑡 ∫ ℎ (𝜏) ∫ (𝜏 − 𝑠)𝑦(𝑠)d𝑠 d𝜏 + 𝜇𝑡
+ ∫ 𝜏ℎ (𝜏)d𝜏
0
1
𝜏
0
0
× (∫ 𝑘 (𝜏) ∫ 𝑠 (1 − 𝜏)𝑦(𝑠)d𝑠 d𝜏
1
1
0
𝜏
+ ∫ 𝑘 (𝜏) ∫ 𝜏 (1 − 𝑠)𝑦(𝑠)d𝑠 d𝜏)]
4
Discrete Dynamics in Nature and Society
1
Proof.
(1) From (8) and Lemma 3, we get (1).
(2) For all 𝑡, 𝑠 ∈ [0, 1], by (8) and Lemma 3, we have
= ∫ 𝐺 (𝑡, 𝑠) 𝑦 (𝑠) d𝑠
0
1
1
1
× [𝜇𝑡 ∫ 𝑘 (𝜏) ∫ 𝐺 (𝑠, 𝜏)𝑦(𝑠)d𝑠 d𝜏
𝑚𝜈 + 𝑛𝜇
0
0
+
1
1
0
0
1
̃ (𝑡, 𝑠) = 𝐺 (𝑡, 𝑠) + 𝑚 + 𝜇𝑡 ∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝐻
𝑚𝜈 + 𝑛𝜇 0
− 𝜈𝑡 ∫ ℎ (𝜏) ∫ 𝐺 (𝑠, 𝜏)𝑦(𝑠)d𝑠 d𝜏
1
1
0
0
+
+ ∫ ℎ (𝜏) ∫ 𝐺 (𝑠, 𝜏)𝑦(𝑠)d𝑠 d𝜏
1
1
0
0
≥
− ∫ 𝜏𝑘 (𝜏)d𝜏 ∫ ℎ (𝜏)
1
0
1
1
0
0
𝑚 + 𝜇𝑡 1
∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
+
× ∫ 𝐺 (𝑠, 𝜏)𝑦(𝑠)d𝑠 d𝜏
≥
+ ∫ 𝜏ℎ (𝜏)d𝜏 ∫ 𝑘 (𝜏)
𝑛 − 𝜈𝑡 1
∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
𝑛 − 𝜈𝑡 1
∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
1
𝑒 (𝑠)
(𝜇𝑡 ∫ 𝑒 (𝜏) 𝑘 (𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇
0
1
+ (1 − 𝑡) 𝜈 ∫ 𝑒 (𝜏) ℎ (𝜏)d𝜏)
1
0
× ∫ 𝐺 (𝑠, 𝜏)𝑦(𝑠)d𝑠 d𝜏]
0
≥
1
= ∫ 𝐺 (𝑡, 𝑠) 𝑦 (𝑠)d𝑠
0
= 𝜌𝑒 (𝑡) 𝑒 (𝑠) ,
1
𝑚 + 𝜇𝑡 1
+
∫ 𝑦 (𝑠) ∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏 d𝑠
𝑚𝜈 + 𝑛𝜇 0
0
+
1
1
𝑒 (𝑠) 𝑒 (𝑡)
(𝜇 ∫ 𝑒 (𝜏) 𝑘 (𝜏)d𝜏 + 𝜈 ∫ 𝑒 (𝜏) ℎ (𝜏)d𝜏)
𝑚𝜈 + 𝑛𝜇
0
0
1
̃ (𝑡, 𝑠) = 𝐺 (𝑡, 𝑠) + 𝑚 + 𝜇𝑡 ∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝐻
𝑚𝜈 + 𝑛𝜇 0
1
𝑛 − 𝜈𝑡 1
∫ 𝑦 (𝑠) ∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏) d𝜏 d𝑠.
𝑚𝜈 + 𝑛𝜇 0
0
+
(16)
Therefore, the unique solution of (6) is 𝑥(𝑡)
1
̃ 𝑠)𝑦(𝑠)d𝑠.
∫0 𝐻(𝑡,
=
≤ 𝐺 (𝑠, 𝑠) +
+
Lemma 3 (in [11]). If ℎ, 𝑘 ∈ 𝐿1 [0, 1] are nonnegative, then,
(1) 𝑛 > 𝜈,
(2) 𝐺(𝑡, 𝑠) > 0, for all 𝑡, 𝑠 ∈ (0, 1), 𝐺(𝑡, 𝑠) ≥ 0, for all
𝑡, 𝑠 ∈ [0, 1],
(3) Letting 𝑒(𝑡) = 𝑡(1 − 𝑡), for all 𝑡, 𝑠 ∈ [0, 1], one has
𝑒(𝑡)𝑒(𝑠) ≤ 𝐺(𝑡, 𝑠) ≤ 𝐺(𝑡, 𝑡) = 𝑒(𝑡) ≤ 1/4,
(4) Letting 𝛿 = (0, 1/2), 𝐽𝛿 = [𝛿, 1 − 𝛿], for all 𝑡 ∈
𝐽𝛿 , 𝑠 ∈ [0, 1], we have 𝐺(𝑡, 𝑠) ≥ 𝛿𝐺(𝑠, 𝑠), where 𝐺(𝑡, 𝑠)
is defined by (9).
Lemma 4. If ℎ, 𝑘 ∈ 𝐿1 [0, 1] are nonnegative, and 𝜇 = 1 −
1
1
̃ 𝑠), we
∫0 ℎ(𝑠)d𝑠 > 0, 𝜈 = 1 − ∫0 𝑘(𝑠)d𝑠 > 0, for function 𝐻(𝑡,
have the following properties:
̃ 𝑠) > 0, for all 𝑡, 𝑠 ∈ (0, 1), 𝐻(𝑡,
̃ 𝑠)
(1) 𝐻(𝑡,
0, for all 𝑡, 𝑠 ∈ [0, 1],
̃ 𝑠) ≤ 𝜂𝑒(𝑠), for all 𝑡, 𝑠 ∈ [0, 1],
(2) 𝜌𝑒(𝑡)𝑒(𝑠) ≤ 𝐻(𝑡,
̃ 𝑠) ≥ 𝛿𝐻(𝑠,
̃ 𝑠), 𝑡 ∈ 𝐽𝛿 , 𝑠 ∈ [0, 1],
(3) 𝐻(𝑡,
1
𝑛 − 𝜈𝑡 1
∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
𝑛 − 𝜈𝑡 1
∫ ℎ (𝜏) 𝐺 (𝑠, 𝑠)d𝜏
𝑚𝜈 + 𝑛𝜇 0
= 𝑒 (𝑠) (1 +
+
≤
𝑚 + 𝜇𝑡 1
∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝑠)d𝜏
𝑚𝜈 + 𝑛𝜇 0
𝑚 + 𝜇𝑡 1
∫ 𝑘 (𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
𝑛 − 𝜈𝑡 1
∫ ℎ (𝜏)d𝜏)
𝑚𝜈 + 𝑛𝜇 0
𝑒 (𝑠)
(𝑚𝜈 + 𝑛𝜇 + (𝑚 + 𝜇)
𝑚𝜈 + 𝑛𝜇
1
1
0
0
× ∫ 𝑘 (𝜏)d𝜏 + 𝑛 ∫ ℎ (𝜏)d𝜏)
= 𝜂𝑒 (𝑠) .
(17)
≥
(3) For all 𝑡 ∈ 𝐽𝛿 , 𝑠 ∈ [0, 1], by (8) and Lemma 3, we have
1
̃ (𝑡, 𝑠) = 𝐺 (𝑡, 𝑠) + 𝑚 + 𝜇𝑡 ∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝐻
𝑚𝜈 + 𝑛𝜇 0
1
where 𝜌 = (1/(𝑚𝜈 + 𝑛𝜇))(𝜇 ∫0 𝑒(𝜏)𝑘(𝜏)d𝜏 + 𝜈 ∫0 𝑒(𝜏)ℎ(𝜏)d𝜏),
𝜂 = (𝑚 + 𝑛 + 𝜇(1 − 𝜈))/(𝑚𝜈 + 𝑛𝜇).
+
𝑛 − 𝜈𝑡 1
∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
Discrete Dynamics in Nature and Society
𝛿 (𝑚 + 𝜇𝑠) 1
∫ 𝑘 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
≥ 𝛿𝐺 (𝑠, 𝑠) +
+
5
𝜌1 =
𝛿 (𝑛 − 𝜈𝑠) 1
∫ ℎ (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
𝑚𝜈 + 𝑛𝜇 0
1
𝑚1 𝜈1 + 𝑛1 𝜇1
1
1
0
0
× (𝜇1 ∫ 𝑒 (𝜏) 𝑘1 (𝜏) d𝜏 + 𝜈1 ∫ 𝑒 (𝜏) ℎ1 (𝜏) d𝜏) ,
̃ (𝑠, 𝑠) .
= 𝛿𝐻
𝜌2 =
(18)
1
𝑚2 𝜈2 + 𝑛2 𝜇2
1
1
0
0
× (𝜇2 ∫ 𝑒 (𝜏) 𝑘2 (𝜏) d𝜏 + 𝜈2 ∫ 𝑒 (𝜏) ℎ2 (𝜏) d𝜏) .
Lemma 5. Assume that (A1)–(A3) holds. If 𝑥(𝑡) ∈ 𝐶2 [0, 1] is
a solution of the following integral equation
1
𝑥 (𝑡) = (𝑇𝑥) (𝑡) ≜ ∫ 𝐻 (𝑡, 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝑠,
0
(19)
(24)
Proof. By using Lemma 3, the conclusions for (1) and the first
part of (2) are obvious.
By using by (21) and Lemma 3, we have
then 𝑥(𝑡) ∈ 𝐶2 [0, 1] ∩ 𝐶4 (0, 1) is a solution of BVP (1), where
+
1
𝐻 (𝑡, 𝑠) = ∫ 𝐻1 (𝑡, 𝜏) 𝐻2 (𝜏, 𝑠)d𝜏,
0
(20)
≤
1
𝑚1 + 𝜇1 𝑡
𝐻1 (𝑡, 𝜏) = 𝐺 (𝑡, 𝜏) +
∫ 𝑘 (𝜈) 𝐺 (𝜏, 𝜈)d𝜈
𝑚1 𝜈1 + 𝑛1 𝜇1 0 1
𝑛1 − 𝜈1 𝑡
∫ ℎ (𝜈) 𝐺 (𝜏, 𝜈)d𝜈,
𝑚1 𝜈1 + 𝑛1 𝜇1 0 1
𝐻2 (𝜏, 𝑠) = 𝐺 (𝜏, 𝑠) +
(21)
1
𝑚2 + 𝜇2 𝜏
∫ 𝑘2 (𝜈) 𝐺 (𝑠, 𝜈)d𝜈
𝑚2 𝜈2 + 𝑛2 𝜇2 0
1
𝑛2 − 𝜈2 𝜏
+
∫ ℎ2 (𝜈) 𝐺 (𝑠, 𝜈)d𝜈,
𝑚2 𝜈2 + 𝑛2 𝜇2 0
1
𝑚1 = ∫ 𝑠ℎ1 (𝑠)d𝑠,
0
1
𝑚2 = ∫ 𝑠ℎ2 (𝑠)d𝑠,
0
(22)
𝑛1 = 1 − ∫ 𝑠𝑘1 (𝑠)d𝑠,
1
(25)
1
𝑛1 − 𝜈1 𝑡
1
∫ ℎ1 (𝑠)d𝑠
𝑚1 𝜈1 + 𝑛1 𝜇1 0 4
𝜂1
.
4
Similarly, by (22), we get
≤
𝜂2
.
4
So, by (20), (25), and (26), we concluded
𝐻2 (𝜏, 𝑠) ≤
1
𝐻 (𝑡, 𝑠) = ∫ 𝐻1 (𝑡, 𝜏) 𝐻2 (𝜏, 𝑠)d𝜏 ≤
0
1
0
1
𝑛1 − 𝜈1 𝑡
∫ ℎ1 (𝜈) 𝐺 (𝜏, 𝜈) d𝜈
𝑚1 𝜈1 + 𝑛1 𝜇1 0
1
𝑚1 + 𝜇1 𝑡
1
1
+
∫ 𝑘1 (𝑠)d𝑠
4 𝑚1 𝜈1 + 𝑛1 𝜇1 0 4
+
1
+
1
𝑚1 + 𝜇1 𝑡
∫ 𝑘1 (𝜈) 𝐺 (𝜏, 𝜈)d𝜈
𝑚1 𝜈1 + 𝑛1 𝜇1 0
𝐻1 (𝑡, 𝜏) = 𝐺 (𝑡, 𝜏) +
(26)
𝜂1 𝜂2
.
16
(27)
By using Lemmas 3 and 4, we have
(23)
𝑛2 = 1 − ∫ 𝑠𝑘2 (𝑠)d𝑠.
0
1
1
1
𝐻 ( , 𝑠) = ∫ 𝐻1 ( , 𝜏) 𝐻2 (𝜏, 𝑠)d𝜏
2
2
0
Proof. By using Lemma 2, the conclusion is obvious.
≥
𝜌1 1
∫ 𝑒 (𝜏) 𝐻2 (𝜏, 𝑠)d𝜏
4 0
Lemma 6. If (A3) holds, then one has the following three
properties:
≥
1
𝜌𝜌
𝜌1 𝜌2
𝑒 (𝑠) ∫ 𝑒2 (𝜏)d𝜏 = 1 2 𝑒 (𝑠) .
4
120
0
(28)
(1) 𝐻(𝑡, 𝑠) > 0, for all 𝑡, 𝑠 ∈ (0, 1),
(2) 0 ≤ 𝐻(𝑡, 𝑠) ≤ 𝜂1 𝜂2 /16, 𝑡, 𝑠 ∈ [0, 1],
(3) 𝐻(1/2, 𝑠) ≥ (𝜌1 𝜌2 /120)𝑒(𝑠), for all 𝑠 ∈ [0, 1],
where
𝜂1 =
𝑚1 + 𝑛1 + 𝜇1 (1 − 𝜈1 )
,
𝑚1 𝜈1 + 𝑛1 𝜇1
𝑚 + 𝑛2 + 𝜇2 (1 − 𝜈2 )
𝜂2 = 2
,
𝑚2 𝜈2 + 𝑛2 𝜇2
Let 𝐸 = 𝐶2 [0, 1]. Then 𝐸 is a Banach space with the norm
‖𝑥‖2 = ‖𝑥‖ + ‖𝑥󸀠󸀠 ‖, where ‖𝑥‖ = max𝑡∈[0,1] |𝑥(𝑡)|, ‖𝑥󸀠󸀠 ‖ =
max𝑡∈[0,1] |𝑥󸀠󸀠 (𝑡)|.
For a fixed 𝛿 = (0, 1/2), 𝐽𝛿 = [𝛿, 1 − 𝛿], define the cone
𝐾 ⊂ 𝐸 by
𝐾 ≜ {𝑥 ∈ 𝐸 : 𝑥 ≥ 0, 𝑥󸀠󸀠 ≤ 0, min 𝑥 (𝑡) ≥ 𝛿2 ‖𝑥‖ ,
𝑡∈𝐽𝛿
󵄩 󵄩
max 𝑥 (𝑡) ≤ −𝛿 󵄩󵄩󵄩󵄩𝑥󸀠󸀠 󵄩󵄩󵄩󵄩} .
𝑡∈𝐽𝛿
󸀠󸀠
2
(29)
6
Discrete Dynamics in Nature and Society
It’s easy to prove that 𝐾 is a closed convex cone, and
󵄨
󵄨
|𝑥 (𝑡)| + 󵄨󵄨󵄨󵄨𝑥󸀠󸀠 (𝑡)󵄨󵄨󵄨󵄨 ≥ 𝛿2 ‖𝑥‖2 ,
1
1
0
0
≥ 𝛿 ∫ ∫ [𝐺 (𝜏, 𝜏) +
𝑡 ∈ 𝐽𝛿 , 𝑥 ∈ 𝐾.
𝑚1 + 𝜇1 𝛿
𝑚1 𝜈1 + 𝑛1 𝜇1
1
(30)
× ∫ 𝑘1 (𝜈) 𝐺 (𝜈, 𝜈)d𝜈 +
0
1
Lemma 7. If (A1)–(A3) hold, then 𝑇(𝐾) ⊂ 𝐾, and 𝑇 : 𝐾 →
𝐾 is completely continuous.
× ∫ ℎ1 (𝜈) 𝐺 (𝜈, 𝜈) d𝜈]
0
× 𝐻2 (𝜏, 𝑠) 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝜏 d𝑠
2
Proof. By the properties of Green’s function, if 𝑥 ∈ 𝐶 [0, 1]
then 𝑇𝑥 ∈ 𝐶2 [0, 1] and
1
1
0
0
≥ 𝛿2 ∫ ∫ [𝐺 (𝜏, 𝜏) +
1
(𝑇𝑥)󸀠󸀠 (𝑡) = − ∫ 𝐻2 (𝑡, 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝑠.
0
𝑛1 − 𝜈1 (1 − 𝛿)
𝑚1 𝜈1 + 𝑛1 𝜇1
(31)
𝑚1 + 𝜇1
𝑚1 𝜈1 + 𝑛1 𝜇1
1
× ∫ 𝑘1 (𝜈) 𝐺 (𝜈, 𝜈)d𝜈
0
For all 𝑥 ∈ 𝐾, we have by (19) and (31)
(𝑇𝑥) (𝑡) ≥ 0,
1
1
0
0
‖(𝑇𝑥)‖ ≤ ∫ ∫ [𝐺 (𝜏, 𝜏) +
+
(𝑇𝑥)󸀠󸀠 (𝑡) ≤ 0,
(32)
× 𝐻2 (𝜏, 𝑠) 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝜏 d𝑠
𝑚1 + 𝜇1
𝑚1 𝜈1 + 𝑛1 𝜇1
≥ 𝛿2 ‖𝑇𝑥‖ .
(35)
1
× ∫ 𝑘1 (𝜈) 𝐺 (𝜈, 𝜈)d𝜈
Similarly, by (31), (34), and Lemma 3, we obtain
0
1
𝑛1
+
∫ ℎ1 (𝜈) 𝐺 (𝜈, 𝜈)d𝜈]
𝑚1 𝜈1 + 𝑛1 𝜇1 0
× 𝐻2 (𝜏, 𝑠) 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝜏 d𝑠,
max(𝑇𝑥)󸀠󸀠 (𝑡)
𝑡∈𝐽𝛿
1
= −min ∫ 𝐻2 (𝑡, 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝑠
𝑡∈𝐽𝛿
(33)
0
1
= −min ∫ [𝐺 (𝑡, 𝑠) +
𝑡∈𝐽𝛿
1
𝑚2 + 𝜇2
󵄩󵄩
󵄩
󵄩󵄩(𝑇𝑥)󸀠󸀠 󵄩󵄩󵄩 ≤ ∫ [𝐺 (𝑠, 𝑠) +
󵄩
󵄩
𝑚2 𝜈2 + 𝑛2 𝜇2
0
0
𝑚2 + 𝜇2 𝑡
𝑚2 𝜈2 + 𝑛2 𝜇2
1
× ∫ 𝑘2 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏
0
1
× ∫ 𝑘2 (𝜏) 𝐺 (𝜏, 𝜏)d𝜏
0
+
1
𝑛1
∫ ℎ1 (𝜈) 𝐺 (𝜈, 𝜈) d𝜈]
𝑚1 𝜈1 + 𝑛1 𝜇1 0
(34)
1
𝑛2
∫ ℎ2 (𝜏) 𝐺 (𝜏, 𝜏)d𝜏]
𝑚2 𝜈2 + 𝑛2 𝜇2 0
× 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝑠.
+
1
𝑛2 − 𝜈2 𝑡
∫ ℎ2 (𝜏) 𝐺 (𝑠, 𝜏)d𝜏]
𝑚2 𝜈2 + 𝑛2 𝜇2 0
× 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝑠
1
≤ −𝛿2 ∫ [𝐺 (𝑠, 𝑠) +
0
On the other hand, by (19), (33), and Lemma 3, we obtain
𝑚2 + 𝜇2
𝑚2 𝜈2 + 𝑛2 𝜇2
1
× ∫ 𝑘2 (𝜏) 𝐺 (𝜏, 𝜏)d𝜏
0
min (𝑇𝑥) (𝑡)
𝑡∈𝐽𝛿
+
1
1
0
0
= min ∫ ∫ [𝐺 (𝑡, 𝜏) +
𝑡∈𝐽𝛿
𝑚1 + 𝜇1 𝑡
𝑚1 𝜈1 + 𝑛1 𝜇1
1
× ∫ 𝑘1 (𝜈) 𝐺 (𝜏, 𝜈)d𝜈 +
0
1
𝑛2
∫ ℎ2 (𝜏) 𝐺 (𝜏, 𝜏)d𝜏]
𝑚2 𝜈2 + 𝑛2 𝜇2 0
× 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝑠
𝑛1 − 𝜈1 𝑡
𝑚1 𝜈1 + 𝑛1 𝜇1
1
× ∫ ℎ1 (𝜈) 𝐺 (𝜏, 𝜈)d𝜈]
0
× 𝐻2 (𝜏, 𝑠) 𝜔 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑥󸀠󸀠 (𝑠))d𝜏 d𝑠
󵄩
󵄩
≤ −𝛿2 󵄩󵄩󵄩󵄩(𝑇𝑥)󸀠󸀠 󵄩󵄩󵄩󵄩 .
(36)
So 𝑇𝑥 ⊂ 𝐾, and then 𝑇(𝐾) ⊂ 𝐾.
Clearly, the operator 𝑇 is continuous; thus by the ArzelaAscoli theorem, it follows that 𝑇 is completely continuous.
Discrete Dynamics in Nature and Society
7
3. Main Results
Case 2. The condition (𝐻2 ) holds. Assume towards a contraction that 𝑥∗∗ (𝑡) is a positive solution of BVP (1), from the
proof of Lemma 7 and (29), then 𝑥∗∗ ∈ 𝐾, 𝑥∗∗ (𝑡) > 0, for
0 < 𝑡 < 1. By Lemmas 4 and 6 and (30), we have
For convenience, we introduce the following notations:
𝐾𝑟 = {𝑥 ∈ 𝐾 : ‖𝑥‖2 ≤ 𝑟} , 𝜕𝐾𝑟 = {𝑥 ∈ 𝐾 : ‖𝑥‖2 = 𝑟} ,
󵄨 ∗∗ 󵄨
󵄨 ∗∗ 󵄨
󵄩󵄩 ∗∗ 󵄩󵄩
󵄩󵄩𝑥 󵄩󵄩 = max 󵄨󵄨󵄨𝑥 (𝑡)󵄨󵄨󵄨 = max 󵄨󵄨󵄨𝑇𝑥 (𝑡)󵄨󵄨󵄨
𝑡∈[0,1]
𝑡∈[0,1]
𝐾𝑟,𝑅 = {𝑥 ∈ 𝐾 : 𝑟 ≤ ‖𝑥‖2 ≤ 𝑅} ,
𝑓 (𝑡, 𝑥, 𝑦)
󵄨󵄨 󵄨󵄨 ,
|𝑥|+|𝑦| → 𝛽 𝑡∈[0,1] |𝑥| + 󵄨󵄨𝑦󵄨󵄨
1
󸀠󸀠
1
≥ ∫ 𝐻 ( , 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥∗∗ (𝑠) , 𝑥∗∗ (𝑠))d𝑠
2
0
𝑓𝛽 = lim sup max
𝑓 (𝑡, 𝑥, 𝑦)
󵄨 󵄨 ,
|𝑥|+|𝑦| → 𝛽 𝑡∈[0,1] |𝑥| + 󵄨󵄨𝑦󵄨󵄨
󵄨 󵄨
𝑓𝛽 = lim inf min
𝑀𝛽 =
𝑡∈[0,1], 0<|𝑥|+|𝑦|≤𝛽
𝐿=(
𝑁=(
max
>
(37)
𝑓 (𝑡, 𝑥, 𝑦) ,
≥
1
𝜂1 𝜂2 𝜂2
+ ) ∫ 𝑤 (𝑠)d𝑠,
16
4
0
𝜌1 𝜌2 1
󵄨 󵄨󵄨 󸀠󸀠 󵄨󵄨
󵄨
∫ 𝑒 (𝑠) 𝑤 (𝑠) (󵄨󵄨󵄨𝑥∗∗ (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑥∗∗ (𝑠)󵄨󵄨󵄨)d𝑠
󵄨
󵄨
120𝑁 0
𝜌1 𝜌2 1−𝛿
󵄨 󵄨󵄨 󸀠󸀠 󵄨󵄨
󵄨
∫ 𝑒 (𝑠) 𝑤 (𝑠) (󵄨󵄨󵄨𝑥∗∗ (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑥∗∗ (𝑠)󵄨󵄨󵄨)d𝑠
󵄨
󵄨
120𝑁 𝛿
𝜌1 𝜌2 2 󵄩󵄩 ∗∗ 󵄩󵄩 1−𝛿
𝛿 󵄩𝑥 󵄩 ∫ 𝑒 (𝑠) 𝑤 (𝑠)d𝑠,
120𝑁 󵄩 󵄩2 𝛿
󵄩󵄩 ∗∗󸀠󸀠 󵄩󵄩
󵄨 󸀠󸀠 󵄨
󵄩󵄩𝑥 󵄩󵄩 = max 󵄨󵄨󵄨𝑥∗∗ (𝑡)󵄨󵄨󵄨 = max 󵄨󵄨󵄨󵄨(𝑇𝑥∗∗ )󸀠󸀠 (𝑡)󵄨󵄨󵄨󵄨
󵄩󵄩
󵄨󵄨 𝑡∈[0,1] 󵄨
󵄩󵄩 𝑡∈[0,1] 󵄨󵄨
󵄨
≥
𝜌1 𝜌2 𝜌2 2 1−𝛿
+ ) 𝛿 ∫ 𝑒 (𝑠) 𝑤 (𝑠)d𝑠.
120
4
𝛿
Theorem 8. Assume that (A1)–(A3) hold. In addition, one
supposes that one of the following conditions is satisfied
(𝐻1 ) 𝐿𝑓(𝑡, 𝑥, 𝑦) < |𝑥| + |𝑦|, for all |𝑥| + |𝑦| ≠ 0, 𝑡 ∈ [0, 1],
(𝐻2 ) 𝑁𝑓(𝑡, 𝑥, 𝑦) > |𝑥| + |𝑦|, for all |𝑥| + |𝑦| ≠ 0, 𝑡 ∈ [0, 1].
󵄨󵄨 1
󵄨󵄨
󸀠󸀠
󵄨
󵄨
= max 󵄨󵄨󵄨󵄨∫ 𝐻2 (𝑡, 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥∗∗ (𝑠) , 𝑥∗∗ (𝑠))d𝑠󵄨󵄨󵄨󵄨
𝑡∈[0,1] 󵄨󵄨 0
󵄨󵄨
1
󸀠󸀠
1
≥ ∫ 𝐻2 ( , 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥∗∗ (𝑠) , 𝑥∗∗ (𝑠))d𝑠
2
0
Then, the BVP (1) has no positive solution.
>
Proof. Case 1. The condition (𝐻1 ) holds. Assume towards a
contraction that 𝑥∗ (𝑡) is a positive solution of BVP (1), from
the proof of Lemma 7 and (29), then 𝑥∗ ∈ 𝐾, 𝑥∗ (𝑡) > 0 for
0 < 𝑡 < 1. By Lemma 6 and (29), we have
󵄨 ∗ 󵄨
󵄨 ∗ 󵄨
󵄩󵄩 ∗ 󵄩󵄩
󵄩󵄩𝑥 󵄩󵄩 = max 󵄨󵄨󵄨𝑥 (𝑡)󵄨󵄨󵄨 = max 󵄨󵄨󵄨𝑇𝑥 (𝑡)󵄨󵄨󵄨
𝑡∈[0,1]
𝑡∈[0,1]
≥
𝜌2 1
󵄨 󵄨󵄨 󸀠󸀠 󵄨󵄨
󵄨
∫ 𝑒 (𝑠) 𝑤 (𝑠) (󵄨󵄨󵄨𝑥∗∗ (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑥∗∗ (𝑠)󵄨󵄨󵄨)d𝑠
󵄨
󵄨
4𝑁 0
𝜌2 2 󵄩󵄩 ∗∗ 󵄩󵄩 1−𝛿
𝛿 󵄩𝑥 󵄩 ∫ 𝑒 (𝑠) 𝑤 (𝑠)d𝑠.
4𝑁 󵄩 󵄩2 𝛿
(39)
So, ‖𝑥∗∗ ‖2 = ‖𝑥∗∗ ‖ + ‖𝑥∗∗󸀠󸀠 ‖ > (1/𝑁)(𝜌1 𝜌2 /120 + 𝜌2 /4)𝛿2
󵄨󵄨 1
󵄨󵄨
󸀠󸀠
󵄨
󵄨
= max 󵄨󵄨󵄨󵄨∫ 𝐻 (𝑡, 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥∗ (𝑠) , 𝑥∗ (𝑠))d𝑠󵄨󵄨󵄨󵄨
𝑡∈[0,1] 󵄨󵄨 0
󵄨󵄨
‖𝑥∗∗ ‖2 ∫𝛿 𝑒(𝑠)𝑤(𝑠)d𝑠 = ‖𝑥∗∗ ‖2 , which is a contradiction,
and this completes the proof.
𝜂1 𝜂2 1
󵄨 󵄨󵄨 󸀠󸀠 󵄨󵄨
󵄨
∫ 𝑤 (𝑠) (󵄨󵄨󵄨𝑥∗ (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑥∗ (𝑠)󵄨󵄨󵄨)d𝑠
󵄨
󵄨
16𝐿 0
Theorem 9. Assume that (A1)–(A3) hold. In addition, one
supposes that one of the following conditions is satisfied
<
1−𝛿
𝜂1 𝜂2 󵄩󵄩 ∗ 󵄩󵄩 1
󵄩𝑥 󵄩 ∫ 𝑤 (𝑠)d𝑠,
16𝐿 󵄩 󵄩2 0
󵄩󵄩 ∗󸀠󸀠 󵄩󵄩
󵄩 󸀠󸀠 󵄩
󵄩󵄩𝑥 󵄩󵄩 = max 󵄩󵄩󵄩𝑥∗ (𝑡)󵄩󵄩󵄩 = max 󵄨󵄨󵄨󵄨(𝑇𝑥∗ )󸀠󸀠 (𝑡)󵄨󵄨󵄨󵄨
󵄩󵄩 󵄩󵄩 𝑡∈[0,1] 󵄩󵄩
󵄩󵄩 𝑡∈[0,1] 󵄨
󵄨
(𝐻3 ) 𝐿𝑓0 < 1 < 𝑁𝑓∞ ,
≤
(𝐻4 ) 𝐿𝑓∞ < 1 < 𝑁𝑓0 .
󵄨󵄨
󵄨󵄨 1
󸀠󸀠
󵄨
󵄨
= max 󵄨󵄨󵄨󵄨∫ 𝐻2 (𝑡, 𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥∗ (𝑠) , 𝑥∗ (𝑠))d𝑠󵄨󵄨󵄨󵄨
𝑡∈[0,1] 󵄨󵄨 0
󵄨󵄨
<
≤
𝜂2 1
󵄨 󵄨󵄨 󸀠󸀠 󵄨󵄨
󵄨
∫ 𝑤 (𝑠) (󵄨󵄨󵄨𝑥∗ (𝑠)󵄨󵄨󵄨 + 󵄨󵄨󵄨𝑥∗ (𝑠)󵄨󵄨󵄨)d𝑠
󵄨
󵄨
4𝐿 0
𝜂2 󵄩󵄩 ∗ 󵄩󵄩 1
󵄩𝑥 󵄩 ∫ 𝑤 (𝑠)d𝑠.
4𝐿 󵄩 󵄩2 0
(38)
So, ‖𝑥∗ ‖2 = ‖𝑥∗ ‖ + ‖𝑥∗󸀠󸀠 ‖ < (1/𝐿)(𝜂1 𝜂2 /16 + 𝜂2 /4)‖𝑥∗ ‖2
1
∫0 𝑤(𝑠)d𝑠 = ‖𝑥∗ ‖2 , which is a contradiction.
Then, the BVP (1) has at least one positive solution.
Proof. Case 1. The condition (𝐻3 ) holds.
Considering 𝐿𝑓0 < 1, then there exists 𝑟1 > 0, such that
𝑓(𝑡, 𝑥, 𝑦) ≤ (𝑓0 + 𝜀1 )(|𝑥| + |𝑦|), for 𝑡 ∈ [0, 1], (𝑥, 𝑦) ∈ {(𝑥, 𝑦) :
0 < |𝑥| + |𝑦| ≤ 𝑟1 }, where 𝜀1 > 0, satisfies 𝐿(𝑓0 + 𝜀1 ) ≤ 1.
Then, for 𝑡 ∈ [0, 1], 𝑥 ∈ 𝜕𝐾𝑟1 , by (19), (26), and Lemma 6,
we have
󵄩
󵄩
‖𝑇𝑥‖2 = ‖𝑇𝑥‖ + 󵄩󵄩󵄩󵄩(𝑇𝑥)󸀠󸀠 󵄩󵄩󵄩󵄩
≤ (
1
𝜂1 𝜂2 𝜂2
+ ) ∫ 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑠󸀠󸀠 (𝑠))d𝑠
16
4
0
8
Discrete Dynamics in Nature and Society
≤ (
𝜂1 𝜂2 𝜂2
+ ) (𝑓0 + 𝜀1 )
16
4
which means that
‖𝑇𝑥‖2 ≥ ‖𝑥‖2 ,
1
󵄨
󵄨
× ∫ 𝑤 (𝑠) (|𝑥 (𝑠)| + 󵄨󵄨󵄨󵄨𝑥󸀠󸀠 (𝑠)󵄨󵄨󵄨󵄨)d𝑠
0
≤ (
1
𝜂1 𝜂2 𝜂2
+ ) (𝑓0 + 𝜀1 ) ‖𝑥‖2 ∫ 𝑤 (𝑠)d𝑠 ≤ ‖𝑥‖2 ,
16
4
0
(40)
for 𝑥 ∈ 𝜕𝐾𝑟1 .
(41)
Next, turning to 𝑁𝑓∞ > 1, then there exists 𝑅1 > 0, such
that 𝑓(𝑡, 𝑥, 𝑦) ≥ (𝑓∞ − 𝜀2 )(|𝑥| + |𝑦|), for 𝑡 ∈ [0, 1], (𝑥, 𝑦) ∈
{(𝑥, 𝑦) : 0 < |𝑥| + |𝑦| > 𝑅1 }, where 𝜀2 > 0, satisfies 𝑁(𝑓∞ −
𝜀2 ) ≥ 1.
Let 𝑅1 = max{2𝑟1 , 𝑅1 /𝛿2 }, for all 𝑡 ∈ 𝐽𝛿 , 𝑥 ∈ 𝜕𝐾𝑅1 . By
(30) and Lemma 6, we have
1 󵄨󵄨 󵄨󵄨
1 󵄨󵄨
󵄩 󵄨󵄨
󵄩
‖𝑇𝑥‖2 = ‖𝑇𝑥‖ + 󵄩󵄩󵄩󵄩(𝑇𝑥)󸀠󸀠 󵄩󵄩󵄩󵄩 ≥ 󵄨󵄨󵄨󵄨𝑇𝑥 ( )󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨(𝑇𝑥)󸀠󸀠 ( )󵄨󵄨󵄨󵄨
2 󵄨 󵄨
2 󵄨
󵄨
≥ (
1
𝜌1 𝜌2 𝜌2
+ ) ∫ 𝑒 (𝑠) 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑠󸀠󸀠 (𝑠))d𝑠
120
4
0
≥ (
𝜌1 𝜌2 𝜌2
+ ) (𝑓∞ − 𝜀2 )
120
4
𝑀𝑅2 =
(43)
Case 2. The condition (𝐻4 ) holds.
Considering 𝑁𝑓0 > 1, then there exists 𝑟2 > 0, such that
𝑓(𝑡, 𝑥, 𝑦) ≥ (𝑓0 − 𝜀3 )(|𝑥| + |𝑦|), for 𝑡 ∈ [0, 1], (𝑥, 𝑦) ∈ {(𝑥, 𝑦) :
0 < |𝑥| + |𝑦| ≤ 𝑟2 }, where 𝜀3 > 0, satisfies 𝑁(𝑓0 − 𝜀3 ) ≥ 1.
Then, for 𝑡 ∈ 𝐽𝛿 , 𝑥 ∈ 𝜕𝐾𝑟2 , by (30) and Lemma 6, we have
󵄨󵄨
1 󵄨󵄨 󵄨󵄨
1 󵄨󵄨
‖𝑇𝑥‖2 ≥ 󵄨󵄨󵄨󵄨𝑇𝑥 ( )󵄨󵄨󵄨󵄨 + 󵄨󵄨󵄨󵄨(𝑇𝑥)󸀠󸀠 ( )󵄨󵄨󵄨󵄨
2 󵄨 󵄨
2 󵄨
󵄨
𝜌𝜌
𝜌
≥ ( 1 2 + 2 ) (𝑓0 − 𝜀3 )
120
4
0
≥ (
𝜌1 𝜌2 𝜌2
+ ) (𝑓0 − 𝜀3 ) 𝛿2 ‖𝑥‖2
120
4
×∫
1−𝛿
𝛿
𝑒 (𝑠) 𝑤 (𝑠)d𝑠 ≥ ‖𝑥‖2 ,
≤ (
1
𝜂1 𝜂2 𝜂2
+ ) ∫ 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑠󸀠󸀠 (𝑠))d𝑠
16
4
0
𝜂1 𝜂2 𝜂2
+ )
16
4
1
× (𝑀𝑅2 ∫ 𝑤 (𝑠)d𝑠 + (𝑓∞ + 𝜀4 )
0
1
󵄨
󵄨
× ∫ 𝑤 (𝑠) (|𝑥 (𝑠)| + 󵄨󵄨󵄨󵄨𝑥󸀠󸀠 (𝑠)󵄨󵄨󵄨󵄨)d𝑠)
‖𝑇𝑥‖2 ≤ ‖𝑥‖2 ,
which means that
1
󵄨
󵄨
× ∫ 𝑒 (𝑠) 𝑤 (𝑠) (|𝑥 (𝑠)| + 󵄨󵄨󵄨󵄨𝑥󸀠󸀠 (𝑠)󵄨󵄨󵄨󵄨)d𝑠
‖𝑇𝑥‖2 ≤ (
which means that
(42)
for 𝑥 ∈ 𝜕𝐾𝑅1 .
for 𝑡 ∈ [0, 1], 𝑓(𝑡, 𝑥, 𝑦) ≤ 𝑀𝑅2 + (𝑓∞ + 𝜀4 )(|𝑥| + |𝑦|).
Let 𝑅2 = max{2𝑟2 , 𝑅2 , 𝐿𝑀𝑅2 /(1 − 𝐿(𝑓∞ + 𝜀4 ))}, for 𝑡 ∈
[0, 1], 𝑥 ∈ 𝜕𝐾𝑅2 . By (26) and Lemma 6, we have
(47)
𝑒 (𝑠) 𝑤 (𝑠) d𝑠 ≥ ‖𝑥‖2
‖𝑇𝑥‖2 ≥ ‖𝑥‖2 ,
(46)
+ ‖𝑥‖2 𝐿 (𝑓∞ + 𝜀4 ) = ‖𝑥‖2 ,
𝜌𝜌
𝜌
≥ ( 1 2 + 2 ) (𝑓∞ − 𝜀2 ) 𝛿2 ‖𝑥‖2
120
4
𝛿
𝑓 (𝑡, 𝑥, 𝑦) ,
≤ 𝐿𝑀𝑅2 + ‖𝑥‖2 Ł (𝑓∞ + 𝜀4 ) ≤ 𝑅2 − 𝐿 (𝑓∞ + 𝜀4 ) 𝑅2
0
1−𝛿
max
𝑡∈[0,1], 0<|𝑥|+|𝑦|≤𝑅2
0
1
󵄨
󵄨
× ∫ 𝑒 (𝑠) 𝑤 (𝑠) (|𝑥 (𝑠)| + 󵄨󵄨󵄨󵄨𝑥󸀠󸀠 (𝑠)󵄨󵄨󵄨󵄨)d𝑠
×∫
(45)
Next, turning to 𝐿𝑓∞ < 1, then there exists 𝑅2 > 0, such
that 𝑓(𝑡, 𝑥, 𝑦) ≤ (𝑓∞ + 𝜀4 )(|𝑥| + |𝑦|), for 𝑡 ∈ [0, 1], (𝑥, 𝑦) ∈
{(𝑥, 𝑦) : |𝑥| + |𝑦| > 𝑅2 }, where 𝜀4 > 0, satisfies 𝐿(𝑓∞ + 𝜀4 ) ≤ 1.
Let
which means that
‖𝑇𝑥‖2 ≤ ‖𝑥‖2 ,
for 𝑥 ∈ 𝜕𝐾𝑟2 .
for 𝑥 ∈ 𝜕𝐾𝑅2 .
(48)
Applying Lemma 1 to (41) and (43), or (45), and (48)
yields that 𝑇 has a fixed point 𝑥∗ ∈ 𝐾𝑟𝑖 ,𝑅𝑖 (𝑖 = 1, 2) with
𝑥∗ (𝑡) ≥ 𝛿‖𝑥∗ ‖2 , 𝑡 ∈ (0, 1). Thus, it follows that BVP (1) has
at least one positive solution, and the theorem is proved.
Theorem 10. Assume that (A1)–(A3) hold, as do the following
two conditions:
(𝐻5 ) 𝑁𝑓0 > 1 and 𝑁𝑓∞ > 1,
(𝐻6 ) there exists 𝑏 > 0, such that max𝑡∈[0,1], 0<|𝑥|+|𝑦|≤𝑏 𝑓(𝑡,
𝑥, 𝑦) < 𝑏/𝐿.
Then, the BVP (1) has at least two positive solutions
𝑥∗ (𝑡), 𝑥∗∗ (𝑡), which satisfy
󵄩 󵄩
󵄩 󵄩
0 < 󵄩󵄩󵄩𝑥∗ 󵄩󵄩󵄩2 < 𝑏 < 󵄩󵄩󵄩𝑥∗∗ 󵄩󵄩󵄩2 .
(44)
(49)
Proof. We choose 𝑟, 𝑅 with 0 < 𝑟 < 𝑏 < 𝑅.
If 𝑁𝑓0 > 1, then by the proof of (45), we have
‖𝑇𝑥‖2 ≥ ‖𝑥‖2 ,
for ‖𝑥‖2 = 𝑟.
(50)
If 𝑁𝑓∞ > 1, then by the proof of (43), we have
‖𝑇𝑥‖2 ≥ ‖𝑥‖2 ,
for ‖𝑥‖2 = 𝑅.
(51)
Discrete Dynamics in Nature and Society
9
On the other hand, by (𝐻6 ), for 𝑥 ∈ 𝜕𝐾𝑏 , we have
‖𝑇𝑥‖2 ≤ (
By calculation, we obtain
1
𝜂1 𝜂2 𝜂2
+ ) ∫ 𝑤 (𝑠) 𝑓 (𝑠, 𝑥 (𝑠) , 𝑠󸀠󸀠 (𝑠))d𝑠
16
4
0
1
𝜂
𝜂𝜂
≤ ( 1 2 + 2 ) 𝑀𝑏 ∫ 𝑤 (𝑠)d𝑠 = 𝐿𝑀𝑏 ,
16
4
0
(52)
where 𝑀𝑏 = max𝑡∈[0,1], 0<|𝑥|+|𝑦|≤𝑏 𝑓(𝑡, 𝑥, 𝑦) < 𝑏/𝐿.
By (52), we have
‖𝑇𝑥‖2 < 𝑏 = ‖𝑥‖2 .
(53)
Theorem 11. Assume that (A1)–(A3) hold, as do the following
two conditions:
(𝐻7 ) 𝐿𝑓0 < 1 and 𝐿𝑓∞ < 1,
(𝐻8 ) there exist 𝛿 ∈ (0, 1/2) and 𝐵 > 0 such that 𝑓(𝑡, 𝑥, 𝑦) >
𝛿2 𝐵/𝑁 for all 𝑡 ∈ 𝐽𝛿 , 𝑥 ∈ [𝛿2 𝐵, 𝐵] and 𝑦 ∈
[−𝐵, −𝛿2 𝐵].
Then, the BVP (1) has at least two positive solutions
𝑥∗ (𝑡), 𝑥∗∗ (𝑡), which satisfy
󵄩 󵄩
󵄩 󵄩
0 < 󵄩󵄩󵄩𝑥∗ 󵄩󵄩󵄩2 < 𝐵 < 󵄩󵄩󵄩𝑥∗∗ 󵄩󵄩󵄩2 .
(54)
Proof. We choose 𝑟, 𝑅 with 0 < 𝑟 < 𝐵 < 𝑅.
If 𝐿𝑓0 < 1, then by the proof of (41), we have
for ‖𝑥‖2 = 𝑟.
(55)
If 𝐿𝑓∞ < 1, then by the proof of (48), we have
‖𝑇𝑥‖2 ≤ ‖𝑥‖2 ,
for ‖𝑥‖2 = 𝑅.
(56)
On the other hand, by (𝐻8 ), for 𝑥 ∈ 𝜕𝐾𝐵 , we have
‖𝑇𝑥‖2 ≥ (
1
𝜌1 𝜌2 𝜌2
+ ) ∫ 𝑒 (𝑠) 𝑤 (𝑠)
120
4
0
Applying Lemma 1 to (55), (56), and (57) yields that 𝑇
has a fixed point 𝑥∗ ∈ 𝐾𝑟,𝐵 , and a fixed point 𝑥∗∗ ∈ 𝐾𝐵,𝑅 .
Thus it follows that BVP (1) has at least two positive solutions
𝑥∗ and 𝑥∗∗ . Noticing (57), we have ‖𝑥∗ ‖2 ≠ 𝐵 and ‖𝑥∗∗ ‖2 ≠ 𝐵.
Therefore, (54) holds, and the proof is complete.
4. Example
Example 12. Consider the following fourth-order BVP
𝑡 ∈ (0, 1) ,
1
1
𝑥 (0) = ∫ 𝑠𝑥 (𝑠)d𝑠,
0
1
𝑥󸀠󸀠 (0) = ∫ 𝑠2 𝑥󸀠󸀠 (𝑠)d𝑠,
0
𝑥 (1) = ∫ 𝑠𝑥 (𝑠)d𝑠,
0
1
𝑥󸀠󸀠 (1) = ∫ 𝑠2 𝑥󸀠󸀠 (𝑠)d𝑠.
0
1
2
𝑛1 = 1 − ∫ 𝑠2 d𝑠 = ,
3
0
5
𝜂1 = ,
2
𝑁=
𝜂2 =
11
,
6
𝐿=
143
,
96
1
2
𝜈2 = 1 − ∫ 𝑠2 d𝑠 = ,
3
0
1
3
𝑛2 = 1 − ∫ 𝑠3 d𝑠 = ,
4
0
1
𝜌1 = ,
6
𝜌2 =
(59)
1
,
10
1
𝛿= ,
3
181√3
(3√2 − 2) ≈ 0.00032.
2187000
(1) About the nonexistence of positive solution, we consider BVP (58) with
󵄨 󵄨󵄨
󵄨
󵄨 󵄨
(60)
𝑓 (𝑡, 𝑥, 𝑦) = 𝑘1 𝑒𝑡 (|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨) + 𝑘2 󵄨󵄨󵄨sin (|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨)󵄨󵄨󵄨 ,
where 𝑘1 and 𝑘2 are two positive real numbers, then
𝑓 (𝑡, 𝑥, 𝑦)
󵄨 󵄨 < 𝑒𝑘1 + 𝑘2 ,
|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨
𝑓 (𝑡, 𝑥, 𝑦)
󵄨 󵄨 > 𝑘1 .
|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨
(61)
If 𝑒𝑘1 + 𝑘2 < 1/𝐿 or 𝑘1 > 1/𝑁, by Theorem 8, BVP
(58) has no positive solution.
(2) About the existence of positive solution, we consider
BVP (58) with
󵄨 󵄨󵄨
󵄨
󵄨 󵄨
𝑓 (𝑡, 𝑥, 𝑦) = 𝑘1 𝑒𝑡−1 (|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨) + 𝑘2 󵄨󵄨󵄨sin (|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨)󵄨󵄨󵄨 , (62)
where 𝑘1 and 𝑘2 are two positive real numbers, then
𝑓 (𝑡, 𝑥, 𝑦) 𝑘1
󵄨 󵄨 = 𝑒 + 𝑘2 ,
|𝑥|+|𝑦| → 0 𝑡∈[0,1] |𝑥| + 󵄨󵄨𝑦󵄨󵄨
󵄨 󵄨
𝑓0 = lim inf min
× 𝑓 (𝑠, 𝑥 (𝑠) , 𝑠 (𝑠))d𝑠 > 𝐵 = ‖𝑥‖2 .
1
𝑓 (𝑡, 𝑥 (𝑡) , 𝑥󸀠󸀠 (𝑡)) ,
√𝑡
1
1
𝑚2 = ∫ 𝑠3 d𝑠 = ,
4
0
(57)
󸀠󸀠
𝑥(4) (𝑡) =
1
1
𝑚1 = ∫ 𝑠2 d𝑠 = ,
3
0
1
1
𝜈1 = 1 − ∫ 𝑠 d𝑠 = ,
2
0
1
2
𝜇2 = 1 − ∫ 𝑠2 d𝑠 = ,
3
0
Applying Lemma 1 to (50), (51), and (53) yields that 𝑇
has a fixed point 𝑥∗ ∈ 𝐾𝑟,𝑏 , and a fixed point 𝑥∗∗ ∈ 𝐾𝑏,𝑅 .
Thus it follows that BVP (1) has at least two positive solutions
𝑥∗ and 𝑥∗∗ . Noticing (53), we have ‖𝑥∗ ‖2 ≠ 𝑏 and ‖𝑥∗∗ ‖2 ≠ 𝑏.
Therefore, (49) holds, and the proof is complete.
‖𝑇𝑥‖2 ≤ ‖𝑥‖2 ,
1
1
𝜇1 = 1 − ∫ 𝑠 d𝑠 = ,
2
0
𝑓
∞
𝑓 (𝑡, 𝑥, 𝑦)
= lim sup max
󵄨 󵄨 = 𝑘1 .
𝑡∈[0,1]
|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨
|𝑥|+|𝑦| → ∞
(63)
If 𝑘1 < 1/𝐿 and 𝑘2 > 1/𝑁, then we have 𝐿𝑓∞ < 1 <
𝑁𝑓0 . By Theorem 9, BVP (58) has at least one positive
solution.
(3) About the multiplicity of positive solution, we consider BVP (58) with
𝑓 (𝑡, 𝑥, 𝑦)
(58)
󵄨 󵄨
󵄨
󵄨 󵄨󵄨
𝑒𝑡 󵄨󵄨󵄨sin 𝑘1 (|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨)󵄨󵄨󵄨 ,
|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨 ≤ 𝑏,
{
{
{
{
{ 𝑡 󵄨󵄨
󵄨󵄨 󵄨󵄨 󵄨󵄨
= { 𝑒 󵄨󵄨sin 𝑘1 (|𝑥| + 󵄨󵄨𝑦󵄨󵄨)󵄨󵄨
2
󵄨󵄨 󵄨󵄨
{
{
{
{ +𝑘 (1 + 𝑡) (|𝑥| + 󵄨󵄨𝑦󵄨󵄨 − 𝑏) , |𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨 > 𝑏,
󵄨 󵄨
2
󵄨 󵄨
|𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨 + 1
{
(64)
10
Discrete Dynamics in Nature and Society
where 𝑘1 and 𝑘2 are two positive real numbers, then
𝑓 (𝑡, 𝑥, 𝑦)
𝑓0 = lim inf min
󵄨 󵄨 = 𝑘1 ,
|𝑥|+|𝑦| → 0 𝑡∈[0,1] |𝑥| + 󵄨󵄨󵄨𝑦󵄨󵄨󵄨
𝑓 (𝑡, 𝑥, 𝑦)
𝑓∞ = lim sup max
󵄨 󵄨 = 2𝑘2 .
𝑡∈[0,1] |𝑥| + 󵄨󵄨𝑦󵄨󵄨
|𝑥|+|𝑦| → ∞
󵄨 󵄨
(65)
If 𝑘1 > 1/𝑁 and 𝑘2 > 1/2𝑁, then we have 𝑁𝑓0 > 1 and
𝑁𝑓∞ > 1.
Furthermore, letting 𝑏 > 𝐿𝑀𝑏 > 143𝑒/96, then we have
𝑀𝑏 < 𝑏/𝐿. By Theorem 10, BVP (58) has at least two positive
solutions.
Remark 13. In [4, 7–10], the existence of solutions for fourthorder ordinary differential equations BVP has been treated
but did not discuss problems with singularities. Although,
[11] seems to have considered the existence of positive
solutions for the fourth-order BVP with integral boundary
conditions for the singularity allowed to appear at 𝑡 = 0
and/or 𝑡 = 1. However, in [11], only the boundary conditions
𝑥(0) = 0, 𝑥󸀠󸀠 (0) = 0, or 𝑥(1) = 0, 𝑥󸀠󸀠 (1) = 0 have
been considered. It is clear that the boundary conditions of
Example 12 are 𝑥(0) ≠ 0, 𝑥󸀠󸀠 (0) ≠ 0, and 𝑥(1) ≠ 0, 𝑥󸀠󸀠 (1) ≠ 0.
Hence, we generalize the fourth-order ordinary differential
equations BVP.
Acknowledgments
This work is partially supported by the National Natural
Science Foundation of PR China (10971045), the Natural
Science Foundation of Hebei Province (A2013208147) and
(A2011208012), and the Foundation of Hebei University of
Science and Technology (XL201246). The authors thank the
referee for his/her careful reading of the paper and useful
suggestions that have greatly improved this paper.
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