Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2013, Article ID 613672, 7 pages
http://dx.doi.org/10.1155/2013/613672
Research Article
Triple Positive Solutions of a Nonlocal Boundary Value Problem
for Singular Differential Equations with p-Laplacian
Jufang Wang, Changlong Yu, and Yanping Guo
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, China
Correspondence should be addressed to Jufang Wang; wangjufang1981@126.com and Changlong Yu; changlongyu@126.com
Received 22 October 2012; Revised 13 January 2013; Accepted 26 January 2013
Academic Editor: Bashir Ahmad
Copyright © 2013 Jufang 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.
We establish the existence of triple positive solutions of an m-point boundary value problem for the nonlinear singular secondorder differential equations of mixed type with a p-Laplacian operator by Leggett-William fixed point theorem. At last, we give an
example to demonstrate the use of the main result of this paper. The conclusions in this paper essentially extend and improve the
known results.
1. Introduction
The existence and multiplicity of positive solutions for differential equations boundary value problems (BVPs) with the pLaplacian operator subject to Dirichlet, Sturm-Liouville, or
nonlinear boundary value conditions have been extensively
investigated in recent years; see [1–10] and the references
therein. Particularly, the following differential equations with
one-dimensional p-Laplacian
󸀠
(𝜙𝑝 (𝑢󸀠 )) + 𝑞 (𝑡) 𝑓 (𝑡, 𝑢) = 0,
0≤𝑡≤1
󸀠
(𝜙𝑝 (𝑢󸀠 )) (𝑡) + 𝑞 (𝑡) 𝑓
× (𝑢 (𝑡) , 𝑢󸀠 (𝑡) , (𝑇𝑢) (𝑡) , (𝑆𝑢) (𝑡)) = 0,
𝑢󸀠 (0) = 𝛽𝑢󸀠 (𝜂) ,
0 ≤ 𝑡 ≤ 1, (3)
𝑢 (1) = 𝑔 (𝑢󸀠 (1)) .
(1)
have been studied subject to different kinds of boundary
conditions; see [1–4] and the references therein. The methods
mainly depend on Kransnosel’skii fixed point theorem, upper
and lower solution technique, Leggett-Williams fixed point
theorem, and some new fixed point theorems in cones, and
so forth.
Recently, in [9], Kong et al. have studied the existence of
triple positive solutions for the following BVP:
󸀠
Firstly, we confirm that the mistakes which have been
pointed out in [10] exist. At the same time, we think that the
value of 𝑀 designed in Theorem 3.1 in [10] is not suitable,
since the proof needs the condition 𝜙𝑞 (𝑀) ≤ 𝑀, but in fact
this condition does not always hold.
Motivated by the work above, in this paper, we will study
the following more extensive second-order m-point BVP:
󸀠
(𝜙𝑝 (𝑢󸀠 )) (𝑡) + 𝑞 (𝑡) 𝑓
(𝜙𝑝 (𝑢󸀠 )) (𝑡) + 𝑞 (𝑡) 𝑓
× (𝑢 (𝑡) , 𝑢󸀠 (𝑡) , (𝑇𝑢) (𝑡) , (𝑆𝑢) (𝑡)) = 0,
𝑢󸀠 (0) = 0,
and studied the existence of triple positive solutions for the
following BVP:
0 ≤ 𝑡 ≤ 1, (2)
𝑢 (1) = 𝑔 (𝑢󸀠 (1)) .
More recently, in [10], Hu and Ma have pointed out that
the equivalent integral equation of BVP (2) is wrong in [9]
× (𝑢 (𝑡) , 𝑢󸀠 (𝑡) , (𝑇𝑢) (𝑡) , (𝑆𝑢) (𝑡)) = 0,
𝑚−2
𝜙𝑝 (𝑢󸀠 (0)) = ∑ 𝛽𝑖 𝜙𝑝 (𝑢󸀠 (𝜂𝑖 )) ,
0 ≤ 𝑡 ≤ 1,
𝑢 (1) = 𝑔 (𝑢󸀠 (1)) ,
𝑖=1
(4)
2
Abstract and Applied Analysis
where 𝜙𝑝 (𝑠) = |𝑠|𝑝−2 𝑠(𝑝 > 1) is an increasing function,
(𝜙𝑝 )−1 (𝑠) = 𝜙𝑞 (𝑠), 1/𝑝 + 1/𝑞 = 1; 0 ≤ 𝛽𝑖 < 1, 𝑖 = 1, 2, . . . ,
𝑚 − 2, ∑𝑚−2
𝑖=1 𝛽𝑖 < 1, and 0 < 𝜂1 < 𝜂2 < ⋅ ⋅ ⋅ < 𝜂𝑚−2 ; 𝑇 and 𝑆 are
two linear operators defined by
𝑡
𝑇𝑢 (𝑡) = ∫ 𝑘 (𝑡, 𝑠) 𝑢 (𝑠) 𝑑𝑠,
0
1
𝑆𝑢 (𝑡) = ∫ ℎ (𝑡, 𝑠) 𝑢 (𝑠) 𝑑𝑠,
(5)
0
𝑢 ∈ 𝐶1 [0, 1] ,
in which 𝑘 ∈ 𝐶[𝐷, 𝑅+ ], ℎ ∈ 𝐶[𝐷0 , 𝑅+ ], 𝐷 = {(𝑡, 𝑠) ∈ 𝑅2 :
0 ≤ 𝑠 ≤ 𝑡 ≤ 1}, 𝐷0 = {(𝑡, 𝑠) ∈ 𝑅2 : 0 ≤ 𝑠, 𝑡 ≤ 1}, 𝑅+ =
[0, +∞), 𝑅 = (−∞, +∞), 𝑘0 = max{𝑘(𝑡, 𝑠) : (𝑡, 𝑠) ∈ 𝐷}, and
ℎ0 = max{ℎ(𝑡, 𝑠) : (𝑡, 𝑠) ∈ 𝐷0 }.
Obviously, when 𝑚 = 3, BVP (4) reduces to BVP (3), and
when 𝑚 = 3, 𝛽1 = 0, BVP (4) reduces to BVP (2), so BVP (2)
and BVP (3) are special cases of BVPs (4).
Throughout this paper, we always suppose the following
conditions hold:
(C1 ) 𝑓 ∈ 𝐶(𝑅+ × 𝑅 × 𝑅+ × 𝑅+ , (0, +∞));
(C2 ) 𝑞(𝑡) ∈ 𝐶([0, 1], 𝑅+ ) may be singular at 𝑡 = 0, 1 and 0 <
1
∫0 𝑞(𝑡)𝑑𝑡 < +∞, so it is easy to see that there exists a
1
constant 𝑀 > 0 such that 0 < ∫0 𝑞(𝑡)𝑑𝑡 < 𝜙𝑝 (𝑀);
(C3 ) 𝑔 : 𝑅 → 𝑅+ is nonincreasing and continuous, and
0 ≤ 𝑔(V) ≤ |V| for V ∈ 𝑅.
2. Preliminary Results
In this section, we firstly present some definitions, theorems,
and lemmas, which will be needed in the proof of the main
result.
Definition 1. Let 𝐸 be a real Banach space. A nonempty closed
convex set 𝑃 ⊂ 𝐸 is called a cone if it satisfies the following
two conditions:
(i) 𝑥 ∈ 𝑃, 𝜆 ≥ 0 implies 𝜆𝑥 ∈ 𝑃;
(ii) 𝑥 ∈ 𝑃, −𝑥 ∈ 𝑃 implies 𝑥 = 0.
Definition 2. Given a cone 𝑃 in a real Banach space 𝐸,
a continuous map 𝛼 is called a concave (resp., convex)
functional on 𝑃 if and only if, for all 𝑥, 𝑦 ∈ 𝑃 and 0 ≤ 𝑡 ≤ 1,
it holds:
𝛼(𝑡𝑥 + (1 − 𝑡)𝑦) ≥ 𝑡𝛼(𝑥) + (1 − 𝑡)𝛼(𝑦),
(resp., 𝛼(𝑡𝑥 + (1 − 𝑡)𝑦) ≤ 𝑡𝛼(𝑥) + (1 − 𝑡)𝛼(𝑦)).
We consider the Banach space 𝐸 = 𝐶1 [0, 1] equipped
with norm ‖𝑢‖ = max0≤𝑡≤1 {‖𝑢(𝑡)‖0 , ‖𝑢󸀠 (𝑡)‖0 }, where ‖𝑢‖0 =
max0≤𝑡≤1 |𝑢(𝑡)|.
We denote, for any fixed constants 𝑎, 𝑏, 𝑟,
𝐶+ [0, 1] = {𝑢 ∈ 𝐶[0, 1] : 𝑢(𝑡) ≥ 0, 𝑡 ∈ [0, 1]},
𝑃 = {𝑢 ∈ 𝐸 | 𝑢(𝑡) is concave and nonincreasing
on [0, 1]},
𝑃𝑟 = {𝑢 ∈ 𝑃 : ‖𝑢‖ < 𝑟},
𝑃(𝛼, 𝑎, 𝑏) = {𝑢 ∈ 𝑃 : 𝑎 ≤ 𝛼(𝑢), ‖𝑢‖ < 𝑏}.
It’s easy to see that 𝑃 is a cone in 𝐸.
Theorem 3 (Leggett-William). Let 𝐴 : 𝑃𝑐 → 𝑃𝑐 be
a completely continuous map and let 𝛼 be a nonnegative
continuous concave functional on 𝑃 with 𝛼(𝑢) ≤ ‖𝑢‖ for any
𝑢 ∈ 𝑃𝑐 . Suppose there exist constants 𝑎, 𝑏, and 𝑑 with 0 < 𝑎 <
𝑏 < 𝑑 ≤ 𝑐 such that
(i) {𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑑) : 𝛼(𝑢) > 𝑏} ≠ 𝜙 and 𝛼(𝐴𝑢) > 𝑏 for all
𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑑);
(ii) ‖𝐴𝑢‖ < 𝑎 for all 𝑢 ∈ 𝑃𝑎 ;
(iii) 𝛼(𝐴𝑢) > 𝑏 for all 𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑐) with ‖𝐴𝑢‖ > 𝑑.
Then 𝐴 has at least three fixed points 𝑢1 , 𝑢2 , and 𝑢3 satisfying
󵄩󵄩 󵄩󵄩
𝑏 < 𝛼 (𝑢2 ) ,
󵄩󵄩𝑢1 󵄩󵄩 < 𝑎,
(6)
󵄩󵄩 󵄩󵄩
𝛼 (𝑢3 ) < 𝑏.
󵄩󵄩𝑢3 󵄩󵄩 > 𝑎,
󸀠
Lemma 4. Suppose 𝑦 ∈ 𝐶1 [0, 1] with (𝜙𝑝 (𝑦󸀠 )) ∈ 𝐿1 [0, 1]
satisfies
󸀠
(𝜙𝑝 (𝑦󸀠 )) (𝑡) ≤ 0,
𝑚−2
𝜙𝑝 (𝑦󸀠 (0)) = ∑ 𝛽𝑖 𝜙𝑝 (𝑦󸀠 (𝜂𝑖 )) ,
0 ≤ 𝑡 ≤ 1,
𝑦 (1) = 𝑔 (𝑦󸀠 (1)) .
(7)
𝑖=1
Then, 𝑦(𝑡) ≥ 0 is concave and nonincreasing on [0, 1], that is,
𝑦 ∈ 𝑃.
󸀠
Proof. Since (𝜙𝑝 (𝑦󸀠 )) (𝑡) ≤ 0, we know that 𝜙𝑝 (𝑦󸀠 ) is
nonincreasing, that is, 𝑦󸀠 (𝑡) is nonincreasing, which means
𝑦(𝑡) is concave. At the same time, we have 𝑦󸀠 (𝑡) ≤ 𝑦󸀠 (0),
𝑚−2
󸀠
󸀠
so 𝑦󸀠 (0) = 𝜙𝑞 (∑𝑚−2
𝑖=1 𝛽𝑖 𝜙𝑝 (𝑦 (𝜂𝑖 ))) ≤ 𝜙𝑞 (∑𝑖=1 𝛽𝑖 𝜙𝑝 (𝑦 (0))) =
󸀠
󸀠
󸀠
𝜙𝑞 (∑𝑚−2
𝑖=1 𝛽𝑖 )𝑦 (0), namely 𝑦 (0) ≤ 0. Then, 𝑦 (𝑡) ≤ 0; that is
to say, 𝑦(𝑡) is nonincreasing. So 𝑦(𝑡) ≥ 𝑦(1) = 𝑔(𝑦󸀠 (1)) ≥ 0.
Above all, 𝑦 ∈ 𝑃. This completes the proof.
󸀠
Lemma 5. Let 𝑦 ∈ 𝐶[0, 1] and (𝜙𝑝 (𝑢󸀠 )) ∈ 𝐿1 [0, 1], then, BVP
󸀠
(𝜙𝑝 (𝑢󸀠 )) (𝑡) = −𝑦 (𝑡) ,
𝑚−2
𝜙𝑝 (𝑢󸀠 (0)) = ∑ 𝛽𝑖 𝜙𝑝 (𝑢󸀠 (𝜂𝑖 )) ,
0 ≤ 𝑡 ≤ 1,
𝑢 (1) = 𝑔 (𝑢󸀠 (1)) ,
(8)
𝑖=1
has a unique solution
1
𝑢 (𝑡) = ∫ 𝜙𝑞 (
𝑡
𝜂𝑖
𝑠
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑦 (𝑟) 𝑑𝑟 + ∫ 𝑦 (𝑟) 𝑑𝑟) 𝑑𝑠
1 − 𝛽 𝑖=1
0
0
+ 𝑔 (−𝜙𝑞 (
where 𝛽 = ∑𝑚−2
𝑖=1 𝛽𝑖 .
𝜂𝑖
1
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑦 (𝑟) 𝑑𝑟 + ∫ 𝑦 (𝑟) 𝑑𝑟)) ,
1 − 𝛽 𝑖=1
0
0
(9)
Abstract and Applied Analysis
3
Define the operator 𝐴 : 𝑃 → 𝐸 by
1
(𝐴𝑢) (𝑡) = ∫ 𝜙𝑞 (
𝑡
+ 𝜙𝑞 (
1
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
𝑠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
0
= 2𝜙𝑞 (
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
0
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
1
+𝑔 (−𝜙𝑞 (
𝑚−2
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
𝜂𝑖
1
∑ 𝛽 ∫ 𝑞 (𝑟)
1−𝛽 𝑖=1 𝑖 0
0
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
󵄨
󵄨
= 2 󵄨󵄨󵄨󵄨(𝐴𝑢)󸀠 (1)󵄨󵄨󵄨󵄨
󵄩
󵄩
= 2󵄩󵄩󵄩󵄩(𝐴𝑢)󸀠 󵄩󵄩󵄩󵄩0 .
(11)
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)) .
0
(10)
Obviously, 𝐴 is well defined and 𝑢 ∈ 𝐸 is a solution of BVP (4)
if and only if 𝑢 is a fixed point of 𝐴.
Lemma 6. 𝐴 : 𝑃 → 𝑃 is completely continuous.
Proof. It is similar to the proof of Lemma 2.2 in [9].
Lemma 7. For any 𝑢 ∈ 𝑃, one has ‖𝐴𝑢‖0 ≤ 2‖(𝐴𝑢)󸀠 ‖0 , ‖𝐴𝑢‖ ≤
2‖(𝐴𝑢)󸀠 ‖0 .
1
0
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
𝑠
󸀠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
0
+ 𝑔 (−𝜙𝑞 (
𝑚−2
𝜂𝑖
1
∑ 𝛽 ∫ 𝑞 (𝑟)
1 − 𝛽 𝑖=1 𝑖 0
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟))
0
𝜂𝑖
1 𝑚−2
≤ ∫ 𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
0
1
𝑠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
0
3. Main Results
For any 𝛿 ∈ (0, min{𝜂1 , 1/2}), we define a nonnegative
continuous concave function 𝛼 : 𝑃 → 𝑅+ by 𝛼(𝑢) =
min𝛿≤𝑡≤(1−𝛿) 𝑢(𝑡). Obviously, the following two conclusions
hold:
𝛼 (𝑢) = 𝑢 (1 − 𝛿) ≤ ‖𝑢‖0 ,
𝛼 (𝐴𝑢) = 𝐴𝑢 (1 − 𝛿) ,
∀𝑢 ∈ 𝑃.
(12)
The main result of this paper is following.
Proof. From (10), we obtain
‖(𝐴𝑢)‖0 = ∫ 𝜙𝑞 (
Since ‖𝐴𝑢‖ = max{‖𝐴𝑢‖0 , ‖(𝐴𝑢)󸀠 ‖0 }, so we have ‖𝐴𝑢‖ ≤
2‖(𝐴𝑢)󸀠 ‖0 , which completes the proof.
Theorem 8. Let 𝑚0 = min0≤𝑡≤1 𝑞(𝑡) and 𝛽 = ∑𝑚−2
𝑖=1 𝛽𝑖 < 1.
Suppose (C1 ), (C2 ), and (C3 ) hold. Suppose further that there
exist numbers 𝛿 ∈ (0, min{𝜂1 , 1/2}), 𝑎, 𝑏, 𝑐, and 𝑑 such that
0 < 𝑎 < 𝑏 ≤ 𝑚0 𝛿𝑑/𝑀 < 𝑑 ≤ 𝑐, and
(H1 ) 𝑓(𝑢, V, 𝑤, 𝑙) ≤ (1 − 𝛽)𝜙𝑝 (𝑎/2𝑀), for (𝑢, V, 𝑤, 𝑙) ∈
[0, 𝑎] × [−𝑎, 0] × [0, 𝑘0 𝑎] × [0, ℎ0 𝑎];
(H2 ) 𝑓(𝑢, V, 𝑤, 𝑙) ≤ (1 − 𝛽)𝜙𝑝 (𝑐/2𝑀), for (𝑢, V, 𝑤, 𝑙) ∈
[0, 𝑐] × [−𝑐, 0] × [0, 𝑘0 𝑐] × [0, ℎ0 𝑐];
(H3 ) 𝑓(𝑢, V, 𝑤, 𝑙) > 𝜙𝑝 (𝑏/𝛿𝐿), for (𝑢, V, 𝑤, 𝑙) ∈ [𝑏, 𝑑] ×
1−𝛿
[−𝑑, 0] × [0, 𝑘0 𝑑] × [0, ℎ0 𝑑], where 𝐿 = 𝜙𝑞 (∫0
1−𝛿
(H4 ) min(𝑢,V,𝑤,𝑙)∈𝐽 𝑓(𝑢, V, 𝑤, 𝑙)𝜙𝑝 (𝑀/(2𝑚0 )) ∫0
1
𝑞(𝑡)𝑑𝑡);
𝑞(𝑡)𝑑𝑡
≥
max(𝑢,V,𝑤,𝑙)∈𝐽 𝑓(𝑢, V, 𝑤, 𝑙) ∫0 𝑞(𝑡)𝑑𝑡, where 𝐽 = [0, 𝑐] ×
[−𝑐, 0] × [0, 𝑘0 𝑐] × [0, ℎ0 𝑐].
Then, BVP (4) has at least three positive solutions 𝑢1 , 𝑢2 , and
𝑢3 such that
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑢1 󵄩󵄩 < 𝑎,
󵄩󵄩 󵄩󵄩
󵄩󵄩𝑢3 󵄩󵄩 > 𝑎,
𝑏<
min 𝑢2 (𝑡) ,
𝛿≤𝑡<(1−𝛿)
min 𝑢3 (𝑡) < 𝑏.
𝛿≤𝑡<(1−𝛿)
(13)
4
Abstract and Applied Analysis
Proof. We divide the proof into three steps.
Step 1. We prove 𝐴𝑃𝑐 ⊂ 𝑃𝑐 , 𝐴𝑃𝑎 ⊂ 𝑃𝑎 ; that is, (ii) of
Theorem 3. By Lemma 6, we have 𝐴𝑃𝑐 ⊂ 𝑃, so ∀𝑢 ∈ 𝑃𝑐 ,
we get 0 ≤ 𝑢(𝑡) ≤ 𝑐, −𝑐 ≤ 𝑢󸀠 (𝑡) ≤ 0, 0 ≤ (𝑇𝑢)(𝑡) ≤ 𝑘0 𝑐,
0 ≤ (𝑆𝑢)(𝑡) ≤ ℎ0 𝑐. For 𝑡 ∈ [0, 1] and by (H2 )
‖(𝐴𝑢)‖0
≤ 𝜙𝑞 (
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
0
≤ 𝜙𝑞 (
1
= ∫ 𝜙𝑞 (
0
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟)
1 − 𝛽 𝑖=1
0
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
1
1
∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
1−𝛽 0
≤ 𝜙𝑞 (𝜙𝑝 (
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1
𝑐
)) 𝜙𝑞 (∫ 𝑞 (𝑟) 𝑑𝑟)
2𝑀
0
≤ 𝑐.
(14)
𝑠
󸀠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
0
+ 𝑔 (−𝜙𝑞 (
𝑚−2
𝜂𝑖
1
∑ 𝛽𝑖 ∫ 𝑞 (𝑟)
1 − 𝛽 𝑖=1
0
Hence, ‖𝐴𝑢‖ < 𝑐 and 𝐴𝑃𝑐 ⊂ 𝑃𝑐 . Similarly, we obtain 𝐴𝑃𝑎 ⊂
𝑃𝑎 .
Step 2. We show
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟))
0
1
≤ ∫ 𝜙𝑞 (
0
𝑚−2
𝜂𝑖
1
∑ 𝛽 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1−𝛽 𝑖=1 𝑖 0
𝑠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
0
{𝑢 ∈ 𝑃 (𝛼, 𝑏, 𝑑) : 𝛼 (𝑢) > 𝑏} ≠ 𝜙,
(15)
𝛼 (𝐴𝑢) > 𝑏,
(16)
that is, (i) of Theorem 3.
Let 𝑢 = (𝑏 + 𝑑)/2, then 𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑑), 𝛼(𝑢) = (𝑏 +
𝑑)/2 > 𝑏. Hence, (15) holds. For any 𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑑), we have
𝑏 ≤ 𝑢(𝑡) ≤ 𝑑, −𝑑 ≤ 𝑢󸀠 (𝑡) ≤ 0, 0 ≤ (𝑇𝑢)(𝑡) ≤ 𝑘0 𝑑, 0 ≤
(𝑆𝑢)(𝑡) ≤ ℎ0 𝑑, 𝑡 ∈ [0, 1 − 𝛿], so by (H3 ), we have
𝛼 (𝐴𝑢) = min (𝐴𝑢) (𝑡) = (𝐴𝑢) (1 − 𝛿)
𝑚−2
𝜂𝑖
1
+ 𝜙𝑞 (
∑ 𝛽 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1 𝑖 0
1
𝑡∈[𝛿,1−𝛿]
=∫
1
1−𝛿
𝜙𝑞 (
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
𝜂𝑖
1 𝑚−2
= 2𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1−𝛽 𝑖=1
0
1
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1−𝛽 𝑖=1
0
𝑠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
0
0
𝜂𝑖
1 𝑚−2
+𝑔 (−𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1−𝛽 𝑖=1
0
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟))
0
1
1
≤ 2𝜙𝑞 (
∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
1−𝛽 0
≤ 2𝜙𝑞 (𝜙𝑝 (
1
𝑐
)) 𝜙𝑞 (∫ 𝑞 (𝑟) 𝑑𝑟)
2𝑀
0
≤ 𝑐.
0
≥ 𝛿𝜙𝑞 (
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
+∫
1−𝛿
0
𝜂𝑖
1 𝑚−2
󵄩󵄩
󵄩
󵄩󵄩(𝐴𝑢)󸀠 󵄩󵄩󵄩 = 𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
󵄩
󵄩0
1−𝛽
0
𝑖=1
𝑠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
0
∀𝑢 ∈ 𝑃 (𝛼, 𝑏, 𝑑) ,
𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
𝜂𝑖
1 𝑚−2
+𝑔 (−𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1−𝛽 𝑖=1
0
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟))
0
Abstract and Applied Analysis
≥ 𝛿𝜙𝑞 (∫
1−𝛿
𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
0
≥ 𝛿𝜙𝑞 (𝜙𝑝 (
5
≥ 𝛿𝜙𝑞 (
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1 − 𝛽 𝑖=1
0
1−𝛿
𝑏
)) 𝜙𝑞 (∫ 𝑞 (𝑟) 𝑑𝑟)
𝛿𝐿
0
1−𝛿
+∫
0
= 𝑏.
(17)
Hence (16) holds.
Step 3. We show that 𝛼(𝐴𝑢) > 𝑏 for all 𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑐) with
‖𝐴𝑢‖ > 𝑑, that is, (iii) of Theorem 3.
If 𝑢 ∈ 𝑃(𝛼, 𝑏, 𝑐) with ‖𝐴𝑢‖ > 𝑑, we obtain 0 ≤ 𝑢(𝑡) ≤
𝑐, −𝑐 ≤ 𝑢󸀠 (𝑡) ≤ 0, 0 ≤ (𝑇𝑢)(𝑡) ≤ 𝑘0 𝑐, 0 ≤ (𝑆𝑢)(𝑡) ≤ ℎ0 𝑐, for
any 𝑡 ∈ [0, 1], and so by (H4 ), we have
1−𝛿
𝑀
𝜙𝑝 (
) ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
2𝑚0 0
1
≥ 𝛿𝜙𝑞 ((
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
0
× (𝜙𝑝 (
≥ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟.
=
Furthermore, we have
𝜙𝑝 (
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟)
1 − 𝛽 𝑖=1
0
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
(18)
0
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
𝑀
1
+ 𝜙𝑝 (
)
2𝑚0 1 − 𝛽
𝜂𝑖
𝑖=1
0
0
󸀠
× ∑ 𝛽𝑖 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
1
(19)
󸀠
=
2𝛿𝑚0
𝑀
=
2𝛿𝑚0 󵄩󵄩
󵄩
󵄩󵄩(𝐴𝑢)󸀠 󵄩󵄩󵄩
󵄩0
𝑀 󵄩
≥
𝛿𝑚0
𝑀
>
𝛿𝑚0
𝑑
𝑀
≥ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
0
𝑚−2
𝜂𝑖
1
∑ 𝛽 ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟.
1 − 𝛽 𝑖=1 𝑖 0
+
−1
𝑀
)) )
2𝑚0
𝜂𝑖
2𝛿𝑚0
1 𝑚−2
𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟)
𝑀
1 − 𝛽 𝑖=1
0
1−𝛿
𝑀
) ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
2𝑚0 0
𝑚−2
𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟)
Therefore, by Lemma 7, we have
󵄨󵄨
󵄨
󵄨󵄨(𝐴𝑢)󸀠 (1)󵄨󵄨󵄨
󵄨
󵄨
󵄩󵄩
󵄩
󵄩󵄩(𝐴𝑢)󸀠 󵄩󵄩󵄩
󵄩
󵄩
≥ 𝑏.
𝛼 (𝐴𝑢) = (𝐴𝑢) (1 − 𝛿)
=∫
1
1−𝛿
𝜙𝑞 (
(20)
𝜂𝑖
1 𝑚−2
∑ 𝛽𝑖 ∫ 𝑞 (𝑟)
1 − 𝛽 𝑖=1
0
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
𝑠
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟) 𝑑𝑠
Hence, by Theorem 3, the results of Theorem 8 hold. This
completes the proof of Theorem 8.
4. Example
0
𝜂𝑖
1 𝑚−2
+ 𝑔 (−𝜙𝑞 (
∑ 𝛽𝑖 ∫ 𝑞 (𝑟)
1 − 𝛽 𝑖=1
0
× 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟
Consider the following BVP:
󵄨 󵄨 󸀠
(󵄨󵄨󵄨󵄨𝑢󸀠 󵄨󵄨󵄨󵄨 𝑢󸀠 ) (𝑡) + 𝑞 (𝑡)
1
× 𝑓 (𝑢 (𝑡) , 𝑢󸀠 (𝑡) , (𝑇𝑢) (𝑡) , (𝑆𝑢) (𝑡)) = 0,
0
0 ≤ 𝑡 ≤ 1,
+ ∫ 𝑞 (𝑟) 𝑓 (𝑢, 𝑢󸀠 , 𝑇𝑢, 𝑆𝑢) 𝑑𝑟))
6
Abstract and Applied Analysis
2 󵄨󵄨
2
1 󵄨󵄨
󵄨󵄨 󸀠 󵄨󵄨 󸀠
󵄨󵄨𝑢 (0)󵄨󵄨 𝑢 (0) = 󵄨󵄨󵄨𝑢󸀠 ( )󵄨󵄨󵄨 𝑢󸀠 ( )
󵄨
󵄨
4 󵄨󵄨
5 󵄨󵄨
5
󵄨
󵄨
1 󵄨
1
1󵄨
+ 󵄨󵄨󵄨󵄨𝑢󸀠 ( )󵄨󵄨󵄨󵄨 𝑢󸀠 ( ) ,
4󵄨
2 󵄨
2
If 1 ≤ 𝑢 ≤ 1296, −1296 ≤ V ≤ 0, 0 ≤ 𝑤, 𝑙 ≤ 1296, then
𝑓 (𝑢, V, 𝑤, 𝑙) > 10 √1 +
10
𝑢 (1) = 𝑔 (𝑢󸀠 (1)) ,
(21)
max 𝑓 (𝑢, V, 𝑤, 𝑙) ≤ 10 √1600 +
10
√𝑙
2 + sin V √𝑤
{
{
+
+
,
10𝑢10 +
{
{
1000
1000 1000
{
{
{
0 ≤ 𝑢 ≤ 1, V ≤ 0, 𝑤, 𝑙 ≥ 0, (22)
𝑓 (𝑢, V, 𝑤, 𝑙) = {
3
√𝑙
{
2 + sin V √𝑤
{
10
{
√𝑢
+
+
+
,
10
{
{
1000
1000 1000
{
1 ≤ 𝑢, V ≤ 0, 𝑤, 𝑙 ≥ 0.
{
3
V1/2 ,
𝑔 (V) = { 2
V,
|V| ≥ 1,
|V| < 1.
Proof. Since 𝑀 = √2 and 𝑚0 = 1/300, 𝛽 = 1/2, 𝛿 =
9/25, 𝑎 = 1/2, 𝑏 = 1, 𝑑 = 1296, 𝑐 = 1600, 𝑘(𝑡, 𝑠) = 1 and
ℎ(𝑡, 𝑠) = 1, then we can obtain 0 < 𝑎 < 𝑏 ≤ (𝑚0 𝛿𝑑)/𝑀 < 𝑑 ≤
𝑐, and
1−9/25
𝐿 = √∫
0
𝜙𝑝 (
𝑞 (𝑡) 𝑑𝑡 = √ ∫
16/25
0
𝑐
1600
) = 320000,
) = 𝜙𝑝 (
𝜙𝑝 (
2𝑀
2√2
(23)
Next, we show that (H1 )–(H4 ) are satisfied.
If 0 ≤ 𝑢 ≤ 1/2, −1/2 ≤ V ≤ 0, 0 ≤ 𝑤, 𝑙 ≤ 1/2, then
3
1 10
2
+
+
2
1000 1000
< (1 − 𝛽) 𝜙𝑝 (
𝑎
1
)= .
2𝑀
64
(24)
So (H1 ) is satisfied.
If 0 ≤ 𝑢 ≤ 1600, −1600 ≤ V ≤ 0, 0 ≤ 𝑤, 𝑙 ≤ 1600, then
𝑓 (𝑢, V, 𝑤, 𝑙) < 10 √1600 +
3
40 × 2
+
1000
1000
< (1 − 𝛽) 𝜙𝑝 (
𝑐
) = 32000.
2𝑀
10
So (H2 ) is satisfied.
𝑀
) = 2 ⋅ 1502 ,
2𝑚0
1
∫ 𝑞 (𝑡) 𝑑𝑡 =
0
∫
1−𝛿
0
8
𝑞 (𝑡) 𝑑𝑡 = ,
5
(27)
1141
.
625
Hence, it’s easy to know that (H4 ) is satisfied.
So by Theorem 8, we conclude that the BVP (21) has three
positive solutions 𝑢1 , 𝑢2 , and 𝑢3 satisfying
󵄩󵄩 󵄩󵄩 1
󵄩󵄩𝑢1 󵄩󵄩 < ,
2
1<
min 𝑢2 (𝑡) ,
𝛿≤𝑡<(1−𝛿)
󵄩󵄩 󵄩󵄩 1
󵄩󵄩𝑢3 󵄩󵄩 >
2
(28)
min 𝑢3 (𝑡) < 1.
𝛿≤𝑡<(1−𝛿)
This paper is supported by the Natural Science Foundation of
China (10901045) and (11201112), the Natural Science Foundation of Hebei Province (A2009000664) and (A2011208012)
and the Foundation of Hebei University of Science and
Technology (XL200757).
References
252 5 3125
𝑏
1
𝜙𝑝 ( ) = 𝜙𝑝 (
)= 2 =
.
𝛿𝐿
9 8
648
(9/25) √8/5
𝑓 (𝑢, V, 𝑤, 𝑙) < 10
𝜙𝑝 (
3
40 × 2
+
,
1000
1000
Acknowledgments
8
𝑡−1/2 𝑑𝑡 = √ ,
5
1
𝑎
1/2
)= ,
) = 𝜙𝑝 (
√
2𝑀
32
2 2
1
,
1000
min 𝑓 (𝑢, V, 𝑤, 𝑙) ≥
16
0≤𝑡≤ ,
25
16
≤ 𝑡 ≤ 1.
25
(26)
So (H3 ) is satisfied.
For any (𝑢, V, 𝑤, 𝑙) ∈ [0, 1600] × [−1600, 0] × [0, 1600] ×
[0, 1600], we have
where
−1/2
{
{𝑡 ,
𝑞 (𝑡) = { 187
9359
{−
𝑡+
,
2700
{ 54
1
𝑏
3125
> 𝜙𝑝 ( ) =
.
1000
𝛿𝐿
648
(25)
[1] J. Wang, “The existence of positive solutions for the onedimensional 𝑝-Laplacian,” Proceedings of the American Mathematical Society, vol. 125, no. 8, pp. 2275–2283, 1997.
[2] L. Kong and J. Wang, “Multiple positive solutions for the onedimensional 𝑝-Laplacian,” Nonlinear Analysis, vol. 42, no. 8, pp.
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[3] X. He and W. Ge, “Twin positive solutions for the onedimensional 𝑝-Laplacian boundary value problems,” Nonlinear
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[4] D. Zhao, H. Wang, and W. Ge, “Existence of triple positive
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[5] A. Lakmeche and A. Hammoudi, “Multiple positive solutions
of the one-dimensional 𝑝-Laplacian,” Journal of Mathematical
Analysis and Applications, vol. 317, no. 1, pp. 43–49, 2006.
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a three-point boundary value problem with a 𝑝-Laplacian
operator,” Journal of Applied Mathematics & Computing, vol. 25,
no. 1-2, pp. 329–337, 2007.
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[7] H. Feng and W. Ge, “Existence of three positive solutions for
𝑚-point boundary-value problems with one-dimensional 𝑝Laplacian,” Nonlinear Analysis, vol. 68, no. 7, pp. 2017–2026,
2008.
[8] Y. Guo, C. Yu, and J. Wang, “Existence of three positive solutions
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