Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2013, Article ID 679316, 12 pages
http://dx.doi.org/10.1155/2013/679316
Research Article
Multiple Positive Solutions to Multipoint Boundary Value
Problem for a System of Second-Order Nonlinear Semipositone
Differential Equations on Time Scales
Gang Wu,1,2 Longsuo Li,1 Xinrong Cong,1 and Xiufeng Miao1
1
2
Department of Mathematics, Harbin Institute of Technology, Harbin, Heilongjiang 150001, China
Basic Science, Harbin University of Commerce, Harbin, Heilongjiang 150076, China
Correspondence should be addressed to Longsuo Li; lilongsuo6982@126.com
Received 13 November 2012; Revised 31 January 2013; Accepted 31 January 2013
Academic Editor: Naseer Shahzad
Copyright © 2013 Gang Wu 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 study a system of second-order dynamic equations on time scales (𝑝1 𝑢1∇ )Δ (𝑡) − 𝑞1 (𝑡)𝑢1 (𝑡) + 𝜆𝑓1 (𝑡, 𝑢1 (𝑡), 𝑢2 (𝑡)) = 0, 𝑡 ∈
(𝑡1 , 𝑡𝑛 ), (𝑝2 𝑢2∇ )Δ (𝑡) − 𝑞2 (𝑡)𝑢2 (𝑡) + 𝜆𝑓2 (𝑡, 𝑢1 (𝑡), 𝑢2 (𝑡)) = 0, satisfying four kinds of different multipoint boundary value conditions, 𝑓𝑖 is
continuous and semipositone. We derive an interval of 𝜆 such that any 𝜆 lying in this interval, the semipositone coupled boundary
value problem has multiple positive solutions. The arguments are based upon fixed-point theorems in a cone.
𝑛−2
1. Introduction
𝛼1 𝑢1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑢1∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑢1 (𝑡𝑖 ) ,
In this paper, we consider the following dynamic equations
on time scales:
𝑖=2
𝛾1 𝑢1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑢1∇ (𝑡𝑛 ) = 0,
Δ
(𝑝1 𝑢1∇ ) (𝑡) − 𝑞1 (𝑡) 𝑢1 (𝑡) + 𝜆𝑓1 (𝑡, 𝑢1 (𝑡) , 𝑢2 (𝑡)) = 0,
𝑡 ∈ (𝑡1 , 𝑡𝑛 ) , 𝜆 > 0,
𝑛−2
𝛼2 𝑢2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑢2∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑢2 (𝑡𝑖 ) ,
𝛾2 𝑢2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑢2∇ (𝑡𝑛 ) = 0,
Δ
𝛼1 𝑢1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑢1∇ (𝑡1 ) = 0,
satisfying one of the boundary value conditions
𝑛−2
𝛾1 𝑢1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑢1∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑢1 (𝑡𝑖 ) ,
𝛼1 𝑢1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑢1∇ (𝑡1 ) = 0,
𝛾1 𝑢1 (𝑡𝑛 ) +
𝑖=2
(1)
(𝑝2 𝑢2∇ ) (𝑡) − 𝑞2 (𝑡) 𝑢2 (𝑡) + 𝜆𝑓2 (𝑡, 𝑢1 (𝑡) , 𝑢2 (𝑡)) = 0,
𝛿1 𝑝1 (𝑡𝑛 ) 𝑢1∇
𝑖=2
𝑛−2
(𝑡𝑛 ) = ∑ 𝑏𝑖 𝑢1 (𝑡𝑖 ) ,
𝑖=2
𝛼2 𝑢2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑢2∇ (𝑡1 ) = 0,
𝑛−2
𝛾2 𝑢2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑢2∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑢2 (𝑡𝑖 ) ,
𝑖=2
(3)
𝛼2 𝑢2 (𝑡1 ) −
(2)
𝛽2 𝑝2 (𝑡1 ) 𝑢2∇
𝑛−2
(𝑡1 ) = ∑ 𝑎𝑖 𝑢2 (𝑡𝑖 ) ,
𝑖=2
𝛾2 𝑢2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑢2∇ (𝑡𝑛 ) = 0,
𝑛−2
𝛼1 𝑢1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑢1∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑢1 (𝑡𝑖 ) ,
𝑖=2
(4)
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Journal of Applied Mathematics
𝛾1 𝑢1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑢1∇ (𝑡𝑛 ) = 0,
existence of multiple positive solutions for the boundary
value problem on time scales.
In 2009, Topal and Yantir [3] studied the second-order
nonlinear 𝑚-point boundary value problems
𝛼2 𝑢2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑢2∇ (𝑡1 ) = 0,
𝑛−2
𝛾2 𝑢2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑢2∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑢2 (𝑡𝑖 ) ,
𝑖=2
(5)
𝑢Δ∇ (𝑡) + 𝑎 (𝑡) 𝑢Δ (𝑡) + 𝑏 (𝑡) 𝑢 (𝑡)
+ 𝜆𝑞 (𝑡) 𝑓 (𝑡, 𝑢 (𝑡)) = 0,
where
𝑝𝑖 , 𝑞𝑖 : [𝑡1 , 𝑡𝑛 ] 󳨀→ (0, +∞)
Δ
with 𝑝𝑖 ∈ 𝐶 [𝑡1 , 𝑡𝑛 ) ,
𝑢 (𝜌 (0)) = 0,
𝑞𝑖 ∈ 𝐶 [𝑡1 , 𝑡𝑛 ]
for 𝑖 = 1, 2;
𝛼𝑖 , 𝛽𝑖 , 𝛾𝑖 , 𝛿𝑖 ∈ [0, +∞)
with 𝛼𝑖 𝛾𝑖 +𝛼𝑖 𝛿𝑖 +𝛽𝑖 𝛾𝑖 > 0
𝑢 (𝜎 (1)) = ∑ 𝛼𝑖 𝑢 (𝜂𝑖 ) ,
(6)
and 𝑓𝑖 is continuously and nonegative functionsquad, 𝑎𝑖 , 𝑏𝑖 ∈
[0, +∞) for 𝑖 ∈ {1, 2, . . . , 𝑛}; the points 𝑡𝑖 ∈ T𝜅𝜅 for 𝑖 ∈
{1, 2, . . . , 𝑛} with 𝑡1 < 𝑡2 < ⋅ ⋅ ⋅ < 𝑡𝑛 .
In the past few years, the boundary value problems of
dynamic equations on time scales have been studied by many
authors (see [1–19] and references). Recently, multipoint
boundary value problems on time scale have been studied,
for instance, see [1–12].
In 2006, Anderson and Ma [1] studied the second-order
multiple time-scale eigenvalue problem:
Δ
(𝑝𝑦∇ ) (𝑡) − 𝑞 (𝑡) 𝑦 (𝑡) + 𝜆ℎ (𝑡) 𝑓 (𝑦) = 0,
𝑡 ∈ (𝑡1 , 𝑡𝑛 ) , 𝜆 > 0,
𝑛−2
(9)
𝑚−2
𝑖=1
for 𝑖 = 1, 2,
𝛼𝑦 (𝑡1 ) − 𝛽𝑝 (𝑡1 ) 𝑦∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑦 (𝑡𝑖 ) ,
𝑡 ∈ (0, 1)T ,
(7)
𝑖=2
where 𝛼𝑖 ≥ 0, 0 < 𝜂𝑖 < 𝜂𝑖+1 < 1; for all 𝑖 = 1, 2, . . . , 𝑚 −
2; 𝑎 ∈ 𝐶([0, 1], [0, +∞)), 𝑏 ∈ 𝐶([0, 1], (−∞, 0]), 𝑓, 𝑞 are
continuously and nonegative functions. The authors deal with
determining the value of 𝜆, and the existences of multiple
positive solutions of the equation are obtained. In 2010, Yuan
and Liu [4] also study the second-order 𝑚-point boundary
value problems; Yuan and Liu shows the existence of multiple
positive solutions if 𝑓 is semipositone and superlinear.
Motivated by the above results mentioned, we study the
second-order nonlinear 𝑚-point boundary value problem (1)
with boundary condition (𝑘), and nonlinear term may be
singularity and semipositone.
In this paper, the nonlinear term 𝑓𝑖 of (1) is suit to and
semipositone and the superlinear case, we shall prove our
two existence results for the problem (1) with (𝑘) by using
a nonlinear alternative of Leray-Schauder type and Krasnosel’skii fixed-point theorem. This paper is organized as
follows. In Section 2, we start with some preliminary lemmas.
In Section 3, we give the main result which state the sufficient
conditions for (1) with 𝑚-point boundary value (𝑘) to have
existence of positive solutions (𝑘 = 2, . . . , 5).
𝑛−2
𝛾𝑦 (𝑡𝑛 ) + 𝛿𝑝 (𝑡𝑛 ) 𝑦∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑦 (𝑡𝑖 ) ,
2. Preliminaries
𝑖=2
where the functions 𝑓 : [0, +∞) → [0, +∞) and ℎ :
[𝑡1 , 𝑡𝑛 ] → [0, +∞) are continuous. The authors discuss conditions for the existence of at least one positive solution to
the second-order Sturm-Liouville-type multiple eigenvalue
problem on time scales.
In 2009, Feng et al. [2] studied
Δ
(𝑝𝑦∇ ) (𝑡) − 𝑞 (𝑡) 𝑦 (𝑡) = 𝑓 (𝑡, 𝑦) ,
𝑡 ∈ (𝑡1 , 𝑡𝑛 ) ,
∇
𝑛−2
𝛼𝑦 (𝑡1 ) − 𝛽𝑝 (𝑡1 ) 𝑦 (𝑡1 ) = ∑ 𝑎𝑖 𝑦 (𝑡𝑖 ) ,
In this section, we state the preliminary information that we
need to prove the main results.
In this paper, for our constructions, we shall consider the
Banach space 𝐸 = 𝐶[𝜌(𝑡1 ), 𝑡𝑛 ] equipped with standard norm
‖𝑥‖ = max𝜌(𝑡1 )≤𝑡≤𝑡𝑛 |𝑥(𝑡)|, 𝑥 ∈ 𝐸; for each (𝑥, 𝑦) ∈ 𝐸 × 𝐸, we
write ||(𝑥, 𝑦)||1 = ||𝑥||+||𝑦||. Clearly, (𝐸×𝐸, ||⋅||1 ) is a Banach
space. Denote by 𝜙𝑖1 and 𝜙𝑖2 (𝑖 = 1, 2), the solutions of the
equation
Δ
(8)
(𝑝𝑖 𝑢𝑖∇ ) (𝑡) − 𝑞𝑖 (𝑡) 𝑢𝑖 (𝑡) = 0,
𝑡 ∈ [𝑡1 , 𝑡𝑛 ) ,
(10)
𝑖=2
∇
𝑛−2
under the initial conditions
𝛾𝑦 (𝑡𝑛 ) + 𝛿𝑝 (𝑡𝑛 ) 𝑦 (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑦 (𝑡𝑖 ) ,
𝑖=2
∑𝑛𝑗=1 𝑐𝑗 (𝑡)𝑦V𝑗 , 𝑐𝑗
where the functions 𝑓(𝑡, 𝑦) =
∈ 𝐶([𝑡1 , 𝑡𝑛 ],
[0, ∞)), V𝑖 ∈ [0, ∞), 𝑗 = 1, 2, . . . , 𝑛. This paper shows the
𝑢𝑖 (𝑡1 ) = 𝛽𝑖 ,
𝑝𝑖 (𝑡1 ) 𝑢𝑖∇ (𝑡1 ) = 𝛼𝑖 ,
𝑢𝑖 (𝑡𝑛 ) = 𝛿𝑖 ,
𝑝𝑖 (𝑡𝑛 ) 𝑢𝑖∇ (𝑡𝑛 ) = −𝛾𝑖 ,
(11)
Journal of Applied Mathematics
3
respectively. So that 𝜙𝑖1 and 𝜙𝑖2 (𝑖 = 1, 2) satisfy
∇ Δ
(𝑝𝑖 𝜙𝑖1
)
𝐻14 (𝑡, 𝑠)
= 𝐺1 (𝑡, 𝑠) +
(𝑡) − 𝑞𝑖 (𝑡) 𝜙𝑖1 (𝑡) = 0,
𝑡 ∈ [𝑡1 , 𝑡𝑛 ) ,
𝜙𝑖1 (𝑡1 ) = 𝛽𝑖 ,
∇
𝑝𝑖 (𝑡1 ) 𝜙𝑖1
(𝑡1 ) = 𝛼𝑖 ,
(12)
Δ
1
𝑛−1
1
𝑛−1
∑ 𝑏𝑖 𝐺1 (𝑡𝑖 , 𝑠) 𝜙11 (𝑡)
𝑑1 − ∑𝑛−1
𝑗=2 𝑏𝑖 𝜙11 (𝑡𝑖 ) 𝑗=2
𝐻24 (𝑡, 𝑠)
= 𝐺2 (𝑡, 𝑠) +
𝜙𝑖2 (𝑡𝑛 ) = 𝛿𝑖 ,
∑ 𝑎𝑖 𝐺2 (𝑡𝑖 , 𝑠) 𝜙22 (𝑡)
𝑑2 − ∑𝑛−1
𝑗=2 𝑎𝑖 𝜙22 (𝑡𝑖 ) 𝑗=2
= 𝐻23 (𝑡, 𝑠) ,
∇
𝑝𝑖 (𝑡𝑛 ) 𝜙𝑖2
(𝑡𝑛 ) = −𝛾𝑖 ,
𝐻15 (𝑡, 𝑠)
∇
respectively. For 𝑖 = 1, 2, set 𝑑𝑖 = 𝛼𝑖 𝜙𝑖2 (𝑡1 ) − 𝛽𝑖 𝑝𝑖 (𝑡1 )𝜙𝑖2
(𝑡1 ) =
∇
𝛾𝑖 𝜙𝑖1 (𝑡𝑛 ) + 𝛿𝑖 𝑝𝑖 (𝑡𝑛 )𝜙𝑖1 (𝑡𝑛 ), the Green’s function of the corresponding homogeneous boundary value problem is defined
by
= 𝐺1 (𝑡, 𝑠) +
𝑑1 −
∑𝑛−1
𝑗=2 𝑎𝑖 𝜙12
∑ 𝑎𝑖 𝐺1 (𝑡𝑖 , 𝑠) 𝜙12 (𝑡)
(𝑡𝑖 ) 𝑗=2
= 𝐻13 (𝑡, 𝑠) ,
𝐻25 (𝑡, 𝑠)
𝐺𝑖 (𝑡, 𝑠)
=
𝑛−1
= 𝐻12 (𝑡, 𝑠) ,
∇
(𝑝𝑖 𝜙𝑖2
) (𝑡) − 𝑞𝑖 (𝑡) 𝜙𝑖2 (𝑡) = 0,
𝑡 ∈ [𝑡1 , 𝑡𝑛 ) ,
1
1 𝜙𝑖2 (𝑡) 𝜙𝑖1 (𝑠) ,
{
𝑑𝑖 𝜙𝑖1 (𝑡) 𝜙𝑖2 (𝑠) ,
𝜌 (𝑡1 ) ≤ 𝑠 ≤ 𝑡 ≤ 𝑡𝑛 ,
𝜌 (𝑡1 ) ≤ 𝑡 ≤ 𝑠 ≤ 𝑡𝑛 ,
for 𝑖 = 1, 2.
(13)
= 𝐺2 (𝑡, 𝑠) +
1
𝑛−1
∑ 𝑏𝑖 𝐺2 (𝑡𝑖 , 𝑠) 𝜙21 (𝑡)
𝑑2 − ∑𝑛−1
𝑗=2 𝑏𝑖 𝜙21 (𝑡𝑖 ) 𝑗=2
= 𝐻22 (𝑡, 𝑠) .
(16)
From Lemmas 3.1 and 3.3 in [1], we have the following lemma.
Lemma 1. If 𝑑𝑖 ≠ 0, (𝑢1 , 𝑢2 ) is a solution of (1) with boundary
value condition (𝑘) if only and if
𝑡𝑛
𝑢1 (𝑡) = 𝜆 ∫ 𝐻1𝑘 (𝑡, 𝑠) 𝑓1 (𝑠, 𝑢1 (𝑠) , 𝑢2 (𝑠)) Δ𝑠,
𝑡1
(14)
For the rest of the paper, we need the following assumption:
𝑛−1
𝑛−1
𝑗=2
𝑗=2
0 < ∑ 𝑏𝑗 𝜙𝑖1 (𝑡𝑗 ) ,
∑ 𝑎𝑗 𝜙𝑖2 (𝑡𝑗 ) < 𝑑𝑖 ,
for 𝑖 = 1, 2.
(C)
𝑡 ∈ [𝜌 (𝑡1 ) , 𝑡𝑛 ] ,
𝑡𝑛
𝑢2 (𝑡) = 𝜆 ∫ 𝐻2𝑘 (𝑡, 𝑠) 𝑓2 (𝑠, 𝑢1 (𝑠) , 𝑢2 (𝑠)) Δ𝑠,
𝑡1
(15)
From 𝜙𝑖1 is nondecreasing on [𝜌(𝑡1 ), 𝑡𝑛 ], 𝜙𝑖2 is nonincreasing on [𝜌(𝑡1 ), 𝑡𝑛 ] (see [2, Proposition 2.3]), it is easy to verify
the following inequalities:
where 𝑘 = 2, . . . , 5, and
𝑑𝑖 𝐺𝑖 (𝑡, 𝑠) ≤ 𝜙𝑖1 (𝑡) 𝜙𝑖2 (𝑡) ,
𝐻𝑖2 (𝑡, 𝑠)
= 𝐺𝑖 (𝑡, 𝑠) +
𝑑𝑖 𝐺𝑖 (𝑡, 𝑠) ≤ 𝜙𝑖1 (𝑠) 𝜙𝑖2 (𝑠) ,
1
𝑑𝑖 −
∑𝑛−1
𝑗=2 𝑏𝑗 𝜙𝑖1
𝑛−1
∑ 𝑏𝑗 𝐺𝑖 (𝑡𝑗 , 𝑠) 𝜙𝑖1 (𝑡) ,
(𝑡𝑗 ) 𝑗=2
(𝑖 = 1, 2) ,
𝐻𝑖3 (𝑡, 𝑠)
= 𝐺𝑖 (𝑡, 𝑠) +
1
𝑛−1
∑ 𝑎𝑗 𝐺𝑖 (𝑡𝑗 , 𝑠) 𝜙𝑖2 (𝑡) ,
𝑑𝑖 − ∑𝑛−1
𝑗=2 𝑎𝑗 𝜙𝑖2 (𝑡𝑗 ) 𝑗=2
(𝑖 = 1, 2) ,
(17)
1
𝑑𝑖 𝐺𝑖 (𝑡, 𝑠) ≥ 󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝜙𝑖1 (𝑡) 𝜙𝑖2 (𝑡) 𝜙𝑖1 (𝑠) 𝜙𝑖2 (𝑠) .
󵄩󵄩𝜙𝑖1 󵄩󵄩 󵄩󵄩𝜙𝑖2 󵄩󵄩
Lemma 2. The Green’s function 𝐺𝑖 (𝑡, 𝑠) has properties
𝐺𝑖 (𝑡, 𝑠) ≤ 𝐺𝑖 (𝑡, 𝑡) ,
𝑑𝑖
󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩 𝐺𝑖 (𝑡, 𝑡) 𝐺𝑖 (𝑠, 𝑠) ≤ 𝐺𝑖 (𝑡, 𝑠) ≤ 𝐺𝑖 (𝑠, 𝑠) .
󵄩󵄩𝜙𝑖1 󵄩󵄩 󵄩󵄩𝜙𝑖2 󵄩󵄩
(18)
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Journal of Applied Mathematics
Lemma 3. For 𝐻𝑖𝑘 (𝑡, 𝑠), 𝑘 = 2, . . . , 5 and 𝑖 = 1, 2, one has the
conclusions 𝐻𝑖𝑘 (𝑡, 𝑠) ≤ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) and
𝑐∗ 𝜙𝑖1 (𝑡) 𝐺𝑖 (𝑠, 𝑠) ≤ 𝐻𝑖2 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙𝑖1 (𝑡) ,
(𝑖 = 1, 2) ,
𝑐∗ 𝜙𝑖2 (𝑡) 𝐺𝑖 (𝑠, 𝑠) ≤ 𝐻𝑖3 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙𝑖2 (𝑡) ,
(𝑖 = 1, 2) ,
∗
𝑐∗ 𝜙𝑖𝑖 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻𝑖4 (𝑡, 𝑠) ≤ 𝐶 𝜙𝑖𝑖 (𝑡) ,
𝑛−1
1
𝑑𝑖 −
≤
∑𝑛−1
𝑗=2 𝑎𝑗 𝜙𝑖2
∑ 𝑎𝑗 𝐺𝑖 (𝑡𝑗 , 𝑠) 𝜙𝑖2 (𝑡)
(𝑡𝑗 ) 𝑗=2
𝑛−1
1
󵄩 󵄩
∑ 𝑎𝑗 𝐺𝑖 (𝑠, 𝑠) 󵄩󵄩󵄩𝜙𝑖2 󵄩󵄩󵄩 ,
𝑑𝑖 − ∑𝑛−1
𝑎
𝜙
(𝑡
)
𝑗=2 𝑗 𝑖2 𝑗 𝑗=2
(21)
(𝑖 = 1, 2) , (19)
𝑐∗ 𝜙12 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻15 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙12 (𝑡) ,
we have
𝑐∗ 𝜙21 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻25 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙21 (𝑡) ,
𝐻𝑖𝑘 (𝑡, 𝑠) ≤ 𝐶1 𝐺𝑖 (𝑠, 𝑠) ≤ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) .
∗
where 𝐶 = 𝐶1 + 𝐶2 and
For 𝑘 = 2 or 3, we have
󵄩󵄩 󵄩󵄩
𝑛−1
{
󵄩󵄩𝜙𝑖1 󵄩󵄩
𝐶1 = max {1 +
∑ 𝑏𝑗
𝑛−1
𝑖=1,2
𝑑𝑖 − ∑𝑗=2 𝑏𝑗 𝜙𝑖1 (𝑡𝑗 ) 𝑖=1
{
+
󵄩
󵄩󵄩
󵄩𝜙 (𝑡)󵄩󵄩
1
𝐻𝑖2 (𝑡, 𝑠) ≤ 󵄩 𝑖2 󵄩 𝜙𝑖1 (𝑡) +
𝑛−1
𝑑𝑖
𝑑𝑖 − ∑𝑗=2 𝑏𝑖 𝜙𝑖1 (𝑡𝑖 )
󵄩󵄩 󵄩󵄩
𝑛−1 }
󵄩󵄩𝜙𝑖2 󵄩󵄩
∑ 𝑎𝑗 } ,
𝑑𝑖 − ∑𝑛−1
𝑗=2 𝑎𝑗 𝜙𝑖2 (𝑡𝑗 ) 𝑗=2 }
𝑛−1
× ∑ 𝑏𝑗 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) 𝜙𝑖1 (𝑡)
𝑗=2
{ 󵄩󵄩󵄩𝜙 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜙 󵄩󵄩󵄩
1
𝐶2 = max { 󵄩 𝑖1 󵄩 󵄩 𝑖2 󵄩 +
𝑛−1
𝑖=1,2
𝑑𝑖
𝑑𝑖 − ∑𝑗=2 𝑏𝑗 𝜙𝑖1 (𝑡𝑗 )
{
𝑛−1
× ∑ 𝑏𝑗 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) +
𝑗=2
≤ 𝐶∗ 𝜙𝑖1 (𝑡) ,
󵄩
󵄩󵄩
󵄩𝜙 (𝑡)󵄩󵄩
1
𝐻𝑖3 (𝑡, 𝑠) ≤ 󵄩 𝑖1 󵄩 𝜙𝑖2 (𝑡) +
𝑛−1
𝑑𝑖
𝑑𝑖 − ∑𝑗=2 𝑎𝑖 𝜙𝑖2 (𝑡𝑖 )
1
𝑑𝑖 −
∑𝑛−1
𝑗=2 𝑏𝑗 𝜙𝑖2
(𝑡𝑗 )
𝑛−1
}
× ∑ 𝑎𝑗 𝐺2 (𝑡𝑗 , 𝑡𝑗 )} ,
𝑗=2
}
{
𝑑
1
𝑐∗ = min { 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑖 󵄩󵄩
󵄩󵄩𝜙𝑖1 󵄩󵄩 󵄩󵄩𝜙𝑖2 󵄩󵄩 𝑑𝑖 − ∑𝑛−1
𝑗=2 𝑏𝑗 𝜙𝑖1 (𝑡𝑗 )
{
𝑛−1
× ∑ 𝑎𝑗 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) 𝜙𝑖2 (𝑡)
𝑗=2
(20)
≤ 𝐶∗ 𝜙𝑖2 (𝑡) ,
𝐻𝑖2 (𝑡, 𝑠) ≥
𝑗=2
𝑗=2
≥
1
𝑑𝑖 −
≤
∑𝑛−1
𝑗=2 𝑏𝑗 𝜙𝑖1
1
𝑑𝑖 −
∑𝑛−1
𝑗=2 𝑏𝑗 𝜙𝑖1
(𝑡𝑗 )
𝑛−1
𝑑
× ∑ 𝑏𝑗 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑖 󵄩󵄩 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) 𝐺𝑖 (𝑠, 𝑠) 𝜙𝑖1 (𝑡)
󵄩𝜙1 󵄩󵄩 󵄩󵄩𝜙2 󵄩󵄩
𝑗=2 󵄩
𝑛−1
Proof. From Lemma 2 and
(𝑡𝑗 )
× ∑ 𝑏𝑗 𝐺𝑖 (𝑡𝑗 , 𝑠) 𝜙𝑖1 (𝑡)
× ∑ 𝑏𝑗 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) ,
}
× ∑ 𝑎𝑗 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) ; 𝑖 = 1, 2} .
𝑗=2
}
𝑑𝑖 −
𝑛−1
𝑛−1
𝑑𝑖
1
󵄩󵄩 󵄩󵄩 󵄩󵄩 󵄩󵄩
󵄩󵄩𝜙𝑖1 󵄩󵄩 󵄩󵄩𝜙𝑖2 󵄩󵄩 𝑑𝑖 − ∑𝑛−1
𝑗=2 𝑎𝑗 𝜙𝑖2 (𝑡𝑗 )
1
∑𝑛−1
𝑗=2 𝑏𝑗 𝜙𝑖1
≥ 𝑐∗ 𝜙𝑖1 (𝑡) 𝐺𝑖 (𝑠, 𝑠) ,
𝐻𝑖3 (𝑡, 𝑠) ≥
𝑛−1
1
𝑑𝑖 −
∑𝑛−1
𝑗=2 𝑎𝑗 𝜙𝑖2
(𝑡𝑗 )
𝑛−1
∑ 𝑏𝑗 𝐺𝑖 (𝑡𝑗 , 𝑠) 𝜙𝑖1 (𝑡)
× ∑ 𝑎𝑗 𝐺𝑖 (𝑡𝑗 , 𝑠) 𝜙𝑖2 (𝑡)
(𝑡𝑗 ) 𝑗=2
𝑗=2
𝑛−1
1
󵄩 󵄩
∑ 𝑏𝑗 𝐺𝑖 (𝑠, 𝑠) 󵄩󵄩󵄩𝜙𝑖1 󵄩󵄩󵄩 ,
𝑑𝑖 − ∑𝑛−1
𝑏
𝜙
(𝑡
)
𝑗=2 𝑗 𝑖1 𝑗 𝑗=2
≥
1
𝑑𝑖 − ∑𝑛−1
𝑗=2 𝑎𝑗 𝜙𝑖1 (𝑡𝑗 )
(22)
Journal of Applied Mathematics
5
𝑛−1
𝑑
× ∑ 𝑎𝑗 󵄩󵄩 󵄩󵄩 󵄩󵄩𝑖 󵄩󵄩 𝐺𝑖 (𝑡𝑗 , 𝑡𝑗 ) 𝐺𝑖 (𝑠, 𝑠) 𝜙𝑖2 (𝑡)
𝜙
󵄩
󵄩
󵄩𝜙2 󵄩󵄩
𝑗=2 󵄩 1 󵄩 󵄩
⊂
(𝑡1 , 𝑡𝑛 ) such that
(H3 ) There exists [𝜃1 , 𝜃2 ]
lim𝑢1 +𝑢2 ↑+∞ min𝑡∈[𝜃1 ,𝜃2 ] (𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 )/(𝑢1 + 𝑢2 )) =
+∞ (𝑖 = 1, 2).
≥ 𝑐∗ 𝜙𝑖2 (𝑡) 𝐺𝑖 (𝑠, 𝑠) .
(23)
So, we have
𝑐∗ 𝜙𝑖1 (𝑡) 𝐺𝑖 (𝑠, 𝑠) ≤ 𝐻𝑖2 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙𝑖1 (𝑡) ,
(𝑖 = 1, 2) ,
𝑐∗ 𝜙𝑖2 (𝑡) 𝐺𝑖 (𝑠, 𝑠) ≤ 𝐻𝑖3 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙𝑖2 (𝑡) ,
(𝑖 = 1, 2) .
(24)
Since 𝐻14 (𝑡, 𝑠) = 𝐻12 (𝑡, 𝑠), 𝐻24 (𝑡, 𝑠) = 𝐻23 (𝑡, 𝑠), 𝐻15 (𝑡,
𝑠) = 𝐻13 (𝑡, 𝑠), 𝐻25 (𝑡, 𝑠) = 𝐻22 (𝑡, 𝑠), then we also have
𝑡
𝑡
(H4 ) ∫𝑡 𝑛 𝐺𝑖 (𝑠, 𝑠)𝑔(𝑠)∇𝑠 < +∞ and ∫𝑡 𝑛 𝐺𝑖 (𝑠, 𝑠)𝑓𝑖 (𝑠, 𝑧1 ,
1
1
𝑧2 )∇𝑠 < +∞ for any 𝑧𝑖 ∈ [0, 𝑚], 𝑚 > 0 is any constant
(𝑖 = 1, 2).
In fact, we only consider the system
Δ
∗
(𝑝1 𝑥1∇ ) (𝑡) − 𝑞1 (𝑡) 𝑥1 (𝑡) + 𝜆 (𝑓1 (𝑡, [𝑥1 (𝑡) − V1𝑘 (𝑡)] ,
𝑐∗ 𝜙11 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻14 (𝑡, 𝑠) = 𝐻12 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙11 (𝑡) ,
𝑐∗ 𝜙22 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻24 (𝑡, 𝑠) = 𝐻23 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙22 (𝑡) ,
𝑐∗ 𝜙12 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻15 (𝑡, 𝑠) = 𝐻13 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙12 (𝑡) ,
∗
[𝑥2 (𝑡) − V2𝑘 (𝑡)] )
(25)
+𝑔 (𝑡) ) = 0,
𝑐∗ 𝜙21 (𝑡) 𝐺1 (𝑠, 𝑠) ≤ 𝐻25 (𝑡, 𝑠) = 𝐻22 (𝑡, 𝑠) ≤ 𝐶∗ 𝜙21 (𝑡) .
The proof is complete.
𝜆 > 0,
Δ
The following theorems will play a major role in our next
analysis.
∗
(𝑝2 𝑥2∇ ) (𝑡) − 𝑞2 (𝑡) 𝑥2 (𝑡) + 𝜆 (𝑓2 (𝑡, [𝑥1 (𝑡) − V1𝑘 (𝑡)] ,
∗
[𝑥2 (𝑡) − V2𝑘 (𝑡)] )
Theorem 4 (see [20]). Let 𝑋 be a Banach space, and Ω ⊂ 𝑋
closed and convex. Assume 𝑈 is a relatively open subset of Ω
with 0 ∈ 𝑈, and let 𝑆 : 𝑈 → Ω be a compact, continuous
map. Then either
+𝑔 (𝑡) ) = 0,
𝜆 > 0,
(1) 𝑆 has a fixed point in 𝑈, or
(26)
(2) there exists 𝑢 ∈ 𝜕𝑈 and ] ∈ (0, 1), with 𝑢 = ]𝑆𝑢.
Theorem 5 (see [21]). Let 𝑋 be a Banach space, and let 𝑃 ⊂ 𝑋
be a cone in 𝑋. Let Ω1 , Ω2 be bounded open subsets of 𝑋 with
0 ∈ Ω1 ⊂ Ω1 ⊂ Ω2 , and let 𝑆 : 𝑃 → 𝑃 be a completely continuous operator such that, either
(1) ‖𝑆𝑤‖ ≤ ‖𝑤‖, 𝑤 ∈ 𝑃 ∩ 𝜕Ω1 , ‖𝑆𝑤‖ ≥ ‖𝑤‖, 𝑤 ∈ 𝑃 ∩ 𝜕Ω2 ,
or
(2) ‖𝑆𝑤‖ ≥ ‖𝑤‖, 𝑤 ∈ 𝑃 ∩ 𝜕Ω1 , ‖𝑆𝑤‖ ≤ ‖𝑤‖, 𝑤 ∈ 𝑃 ∩ 𝜕Ω2 .
Then 𝑆 has a fixed point in 𝑃 ∩ Ω2 \ Ω1 .
𝛼1 𝑥1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑥1∇ (𝑡1 ) = 0,
𝑛−2
𝛾1 𝑥1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑥1∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑥1 (𝜂𝑖 ) ,
𝑖=2
𝛼2 𝑥2 (𝑡1 ) −
𝛽2 𝑝2 (𝑡1 ) 𝑥2∇
(𝑡1 ) = 0,
𝑛−2
𝛾2 𝑥2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑥2∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑥2 (𝜂𝑖 ) ,
3. Main Results
𝑖=2
𝑛−2
We make the following assumptions:
(H1 ) 𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 ) ∈ 𝐶([𝑡1 , 𝑡𝑛 ] × [0, +∞)2 , (−∞, +∞)), moreover there exists a function 𝑔(𝑡) ∈ 𝐿1 ([𝑡1 , 𝑡𝑛 ], (0,
+∞)) such that 𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 ) ≥ −𝑔(𝑡), for any 𝑡 ∈
(𝑡1 , 𝑡𝑛 ), 𝑢𝑖 ∈ [0, +∞), 𝑖 = 1, 2.
(H∗1 )
with one of the boundary value conditions
2
𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 ) ∈ 𝐶((𝑡1 , 𝑡𝑛 ) × [0, +∞) , (−∞, +∞)) may
be singular at 𝑡 = 𝑡1 , 𝑡𝑛 ; moreover, there exists a function 𝑔(𝑡) ∈ 𝐿1 ((𝑡1 , 𝑡𝑛 ), (0, +∞)) such that 𝑓𝑖 (𝑡, 𝑢1 ,
𝑢2 ) ≥ −𝑔(𝑡), for any 𝑡 ∈ (𝑡1 , 𝑡𝑛 ), 𝑢𝑖 ∈ [0, +∞).
(H2 ) 𝑓𝑖 (𝑡, 0, 0) > 0, for 𝑡 ∈ [𝑡1 , 𝑡𝑛 ] (𝑖 = 1, 2).
𝛼1 𝑥1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑥1∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑥1 (𝜂𝑖 ) ,
𝑖=2
𝛾1 𝑥1 (𝑡𝑛 ) +
𝛿1 𝑝1 (𝑡𝑛 ) 𝑥1∇
(𝑡𝑛 ) = 0,
𝑛−2
𝛼2 𝑥2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑥2∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑥2 (𝜂𝑖 ) ,
𝑖=2
𝛾2 𝑥2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑥2∇ (𝑡𝑛 ) = 0,
𝛼1 𝑥1 (𝑡1 )−𝛽1 𝑝1 (𝑡1 ) 𝑥1∇ (𝑡1 ) = 0,
6
Journal of Applied Mathematics
𝑛−2
𝑛−2
𝛾1 𝑥1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑥1∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑥1 (𝜂𝑖 ) ,
𝛼1 V1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) V1∇ (𝑡1 ) = ∑ 𝑎𝑖 V1 (𝜂𝑖 ) ,
𝑖=2
𝑖=2
𝛾1 V1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) V1∇ (𝑡𝑛 ) = 0,
𝑛−2
𝛼2 𝑥2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑥2∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑥2 (𝜂𝑖 ) ,
𝑖=2
𝑛−2
𝛼2 V2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) V2∇ (𝑡1 ) = ∑ 𝑎𝑖 V2 (𝜂𝑖 ) ,
𝛾2 𝑥2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑥2∇ (𝑡𝑛 ) = 0,
𝑖=2
𝑛−2
𝛾2 V2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) V2∇ (𝑡𝑛 ) = 0,
𝑖=2
𝛼1 V1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) V1∇ (𝑡1 ) = 0,
𝛼1 𝑥1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑥1∇ (𝑡1 ) = ∑ 𝑎𝑖 𝑥1 (𝜂𝑖 ) ,
𝛾1 𝑥1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑥1∇ (𝑡𝑛 ) = 0,
𝛼2 𝑥2 (𝑡1 ) −
𝛽2 𝑝2 (𝑡1 ) 𝑥2∇
𝑛−2
𝛾1 V1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) V1∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 V1 (𝜂𝑖 ) ,
(𝑡1 ) = 0,
𝑖=2
𝑛−2
𝑛−2
𝛾2 𝑥2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑥2∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 𝑥2 (𝜂𝑖 ) ,
𝛼2 V2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) V2∇ (𝑡1 ) = ∑ 𝑎𝑖 V2 (𝜂𝑖 ) ,
𝑖=2
𝑖=2
(27)
𝛾2 V2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) V2∇ (𝑡𝑛 ) = 0,
𝑛−2
𝛼1 V1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑥1∇ (𝑡1 ) = ∑ 𝑎𝑖 V1 (𝜂𝑖 ) ,
where
𝑖=2
𝛾1 V1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑥1∇ (𝑡𝑛 ) = 0,
𝑦 (𝑡) , 𝑦 (𝑡) ≥ 0,
𝑦(𝑡)∗ = {
0,
𝑦 (𝑡) < 0,
𝛼2 V2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑥2∇ (𝑡1 ) = 0,
(28)
𝑛−2
𝛾2 V2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑥2∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 V2 (𝜂𝑖 ) .
𝑖=2
𝑡
and V𝑖𝑘 (𝑡) = 𝜆 ∫𝑡 𝑛 𝐻𝑖𝑘 (𝑡, 𝑠)𝑔(𝑠)Δ𝑠. For 𝑘 = 2, . . . , 5, from
1
Lemma 1, (V1𝑘 (𝑡), V2𝑘 (𝑡)) is the solution of the equation
Δ
(𝑝1 V1∇ ) (𝑡) − 𝑞1 (𝑡) V1 (𝑡) + 𝜆𝑔 (𝑡) = 0,
𝜆 > 0, 𝑡1 < 𝑡 < 𝑡𝑛 ,
Δ
(𝑝2 V2∇ ) (𝑡) − 𝑞2 (𝑡) V2 (𝑡) + 𝜆𝑔 (𝑡) = 0,
(29)
𝜆 > 0,
(30)
We will show that there exists a solution (𝑥1𝑘 , 𝑥2𝑘 ) to the
boundary value problem (𝑘) of the system (26) with 𝑥𝑖𝑘 (𝑡) ≥
V𝑖𝑘 (𝑡), 𝑡 ∈ [𝜌(𝑡1 ), 𝑡𝑛 ]. If this is true, then 𝑢𝑖𝑘 (𝑡) = 𝑥𝑖𝑘 (𝑡) − V𝑖𝑘 (𝑡)
is a nonnegative solution (positive on (𝜌(𝑡1 ), 𝑡𝑛 )) of the system
(1) with the boundary value problem (𝑘), (where 𝑖 = 1, 2; 𝑘 =
2, . . . , 5, 𝑘 = 𝑘 + 11). Since for any 𝑡 ∈ (𝑡1 , 𝑡𝑛 ), from
Δ
∇
∇
) (𝑡) − 𝑞𝑖 (𝑡) 𝑥𝑖𝑘
(𝑝𝑖 𝑥𝑖𝑘
(𝑡)
∇ Δ
∇
= (𝑝𝑖 (𝑢𝑖𝑘 + V𝑖𝑘 ) ) (𝑡) − 𝑞𝑖 (𝑡) (𝑢𝑖𝑘 + V𝑖𝑘 ) (𝑡)
respectively, satisfying the following boundary value conditions:
∗
∗
= −𝜆 (𝑓𝑖 (𝑡, [𝑥1𝑘 (𝑡)−V1𝑘 (𝑡)] , [𝑥2𝑘 (𝑡)−V2𝑘 (𝑡)] )+𝑔 (𝑡))
= −𝜆 (𝑓𝑖 (𝑡, 𝑢1𝑘 (𝑡) , 𝑢2𝑘 (𝑡)) + 𝑔 (𝑡)) ,
(31)
𝛼1 V1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) V1∇ (𝑡1 ) = 0,
𝑛−2
𝛾1 V1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) V1∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 V1 (𝜂𝑖 ) ,
𝑖=2
𝛼2 V2 (𝑡1 ) −
𝛽2 𝑝2 (𝑡1 ) V2∇
(𝑡1 ) = 0,
𝑛−2
𝛾2 V2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) V2∇ (𝑡𝑛 ) = ∑ 𝑏𝑖 V2 (𝜂𝑖 ) ,
𝑖=2
we have
Δ
∇
) (𝑡) − 𝑞𝑖 (𝑡) 𝑢𝑖𝑘 (𝑡) = −𝜆𝑓𝑖 (𝑡, 𝑢1𝑘 (𝑡) , 𝑢2𝑘 (𝑡)) . (32)
(𝑝𝑖 𝑢𝑖𝑘
As a result, we will concentrate our study on (26) with the
boundary value problem (𝑘).
Journal of Applied Mathematics
7
Employing Lemma 1, we note that (𝑥1𝑘 (𝑡), 𝑥2𝑘 (𝑡)) is a
solution of the system (26) with boundary value (𝑘) if and
only if
𝑡𝑛
For 𝑘 = 2, . . . , 5, from (35) and Lemma 3, we have
𝑇𝑘 (𝑥1𝑘 , 𝑥2𝑘 )(𝑡) ≥ 0 on [0, 1], for (𝑥1𝑘 , 𝑥2𝑘 ) ∈ 𝑃𝑖𝑗 × 𝑃𝑚𝑛 , we
have
𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑡)
∗
𝑥1𝑘 (𝑡) = 𝜆 ∫ 𝐻1𝑘 (𝑡, 𝑠) (𝑓1 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡𝑛
𝑡1
∗
= 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡1
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
+𝑔 (𝑠) ) Δ𝑠,
(37)
+𝑔 (𝑠)) Δ𝑠
𝑡 ∈ [𝜌 (𝑡1 ) , 𝑡𝑛 ] ,
(33)
𝑡𝑛
≤ 𝐶∗ 𝜆
∗
𝑥2𝑘 (𝑡) = 𝜆 ∫ 𝐻2𝑘 (𝑡, 𝑠) (𝑓2 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
×∫
𝑡1
𝜎(𝑡𝑛 )
𝜌(𝑡1 )
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥 (𝑠)−V (𝑠)]∗ ) + 𝑔 (𝑠)) Δ𝑠,
𝜎(𝑡 )
then ‖𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 )‖ ≤ 𝐶∗ 𝜆 ∫𝜌(𝑡 𝑛) 𝐺𝑖 (𝑠, 𝑠)(𝑓(𝑡, [𝑥(𝑡) − V(𝑡)]∗ ) +
1
𝑔(𝑡))Δ𝑠.
On the other hand, when 𝑘 = 2, we have
+𝑔 (𝑠) ) Δ𝑠.
We define a cone 𝑃𝑖𝑗 (𝑖, 𝑗 = 1, 2) by
𝑇𝑖2 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑡)
𝑃𝑖𝑗 = {𝑥 ∈ 𝐸 | 𝑥 (𝑡) ≥
𝑐∗
𝜙 (𝑡) ‖𝑥‖ , 𝑡 ∈ [𝜌 (𝑡1 ) , 𝑡𝑛 ]} .
𝐶∗ 𝑖𝑗
(34)
𝑡𝑛
∗
= 𝜆 ∫ 𝐻𝑖2 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡1
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) + 𝑔 (𝑠)) Δ𝑠
It is clearly that 𝑃𝑖𝑗 × 𝑃𝑚𝑛 is a cone of 𝐸 × 𝐸, (𝑖, 𝑗, 𝑚, 𝑛 = 1, 2).
Define the integral operator 𝑇2 : 𝑃11 × 𝑃21 → 𝐸 × 𝐸, 𝑇3 : 𝑃12 ×
𝑃22 → 𝐸 × 𝐸, 𝑇4 : 𝑃12 × 𝑃21 → 𝐸 × 𝐸, 𝑇5 : 𝑃21 × 𝑃12 → 𝐸 × 𝐸,
by
𝑇𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) = (𝑇1𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) , 𝑇2𝑘 (𝑥1𝑘 , 𝑥2𝑘 )) ,
(𝑘 = 2, . . . , 5) ,
𝑡𝑛
𝑡1
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
+𝑔 (𝑠)) Δ𝑠
(35)
≥
𝑐∗
𝜙 (𝑡) 𝜆
𝐶∗ 𝑖1
×∫
𝜎(𝑡𝑛 )
𝜌(𝑡1 )
where operators 𝑇𝑖𝑘 are defined by
≥
𝑡𝑛
∗
≥ 𝜆 ∫ 𝑐∗ 𝜙𝑖1 (𝑡) 𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠)−V1𝑘 (𝑠)] ,
∗
𝐶∗ 𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥 (𝑠)−V (𝑠)]∗ )+𝑔 (𝑠)) Δ𝑠
𝑐∗
󵄩
󵄩
𝜙 (𝑡) 󵄩󵄩𝑇 (𝑥 , 𝑥 )󵄩󵄩 .
𝐶∗ 𝑖1 󵄩 𝑖2 1𝑘 2𝑘 󵄩
(38)
𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑡) = 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡1
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
Thus, 𝑇𝑖2 (𝑃11 × 𝑃21 ) ⊂ 𝑃𝑖1 . Hence 𝑇2 (𝑃11 × 𝑃21 ) ⊂ 𝑃11 × 𝑃21 .
When 𝑘 = 3, we have
+𝑔 (𝑠) ) Δ𝑠,
𝑇𝑖3 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑡)
𝑡 ∈ [𝜌 (𝑡1 ) , 𝑡𝑛 ] ,
(36)
𝑡𝑛
∗
= 𝜆 ∫ 𝐻𝑖3 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡1
where 𝑖 = 1, 2. Clearly, if (𝑥1𝑘 , 𝑥2𝑘 ) is a fixed point of 𝑇𝑘 ,
then (𝑥1𝑘 , 𝑥2𝑘 ) is a solution of system (26) with (𝑘) (𝑘 =
2, . . . , 5, 𝑘 = 𝑘 + 11).
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
+𝑔 (𝑠) ) Δ𝑠
8
Journal of Applied Mathematics
𝑡𝑛
claim that ‖(𝑥1𝑘 , 𝑥2𝑘 )‖1 ≠ 𝑅0 . In fact for (𝑥1𝑘 , 𝑥2𝑘 ) ∈ 𝜕𝑈𝑘 and
‖(𝑥1𝑘 , 𝑥2𝑘 )‖1 = 𝑅0 , we have
∗
≥ 𝜆 ∫ 𝑐∗ 𝜙𝑖2 (𝑡) 𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠)−V1𝑘 (𝑠)] ,
𝑡1
∗
𝑥𝑖𝑘 = ]𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 )
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] )
𝑡𝑛
+𝑔 (𝑠)) Δ𝑠
≥
𝜎(𝑡𝑛 )
𝜌(𝑡1 )
≥
𝑡1
𝑐∗
𝜙 (𝑡) 𝜆
𝐶∗ 𝑖2
×∫
∗
≤ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) + 𝑔 (𝑠)) Δ𝑠
𝐶∗ 𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥 (𝑠)−V (𝑠)]∗ )+𝑔 (𝑠)) Δ𝑠
𝑡𝑛
𝑡1
𝑐∗
󵄩
󵄩
𝜙 (𝑡) 󵄩󵄩𝑇 (𝑥 , 𝑥 )󵄩󵄩 .
𝐶∗ 𝑖2 󵄩 𝑖3 1𝑘 2𝑘 󵄩
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) + 𝑔 (𝑠)) Δ𝑠
(39)
Thus, 𝑇𝑖3 (𝑃12 × 𝑃22 ) ⊂ 𝑃𝑖2 . Hence 𝑇3 (𝑃12 × 𝑃22 ) ⊂ 𝑃12 × 𝑃22 .
Similarly discussion, we also have 𝑇4 (𝑃12 ×𝑃21 ) ⊂ 𝑃12 ×𝑃21 ,
𝑇5 (𝑃21 × 𝑃12 ) ⊂ 𝑃21 × 𝑃12 . In addition, standard arguments
show that 𝑇𝑘 is a completely continuous operator.
For simplicity, we adopt the notation: 𝑃14 = 𝑃25 := 𝑃12
and 𝑃24 = 𝑃15 := 𝑃21 , then, we can write 𝑇𝑘 (𝑃1(𝑘−1) × 𝑃2(𝑘−1) ) ⊂
𝑃1(𝑘−1) × 𝑃2(𝑘−1) , that is, 𝑇𝑖𝑘 (𝑃1(𝑘−1) × 𝑃2(𝑘−1) ) ⊂ 𝑃𝑖(𝑘−1) , (𝑖 =
1, 2, 𝑘 = 2, . . . , 5).
Theorem 6. Suppose that (H1 )-(H2 ) hold. Then there exists
a constant 𝜆 > 0 such that, for any 0 < 𝜆 ≤ 𝜆, (1) with
boundary value condition (𝑘) has at least one positive solution
(𝑘 = 2, . . . , 5).
Proof. Fix 𝛿 ∈ (0, 1) and 𝑘 (𝑘 = 2, . . . , 5). From (H2 ) let 0 <
𝜀 < 1 be such that
𝑓𝑖 (𝑡, 𝑧1 , 𝑧2 ) ≥ 𝛿𝑓𝑖 (𝑡, 0, 0) ,
for 𝑡1 ≤ 𝑡 ≤ 𝑡𝑛 , 0 ≤ 𝑧𝑖 ≤ 𝜀, 𝑖 = 1, 2.
(40)
Let 𝑓(𝜀) = max𝑡1 ≤𝑡≤𝑡𝑛 ,0≤𝑧1 ,𝑧2 ≤𝜀 {max𝑖=1,2 {𝑓𝑖 (𝑡, 𝑧1 , 𝑧2 )} +
𝑡𝑛
≤ 𝜆 ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) 𝑓 (𝑅0 ) Δ𝑠
𝑡1
≤ 𝜆𝑐𝑓 (𝑅0 ) .
(44)
It follows that
1
(41)
𝑓 (𝑅0 )
𝑓 (𝑅0 )
1
1
≥
,
>
=
𝑅0
2𝑐𝜆 4𝑐𝜆
𝑅0
(46)
which implies that ‖(𝑥1𝑘 , 𝑥2𝑘 )‖1 ≠ 𝑅0 . By the nonlinear alternative of Leray-Schauder type, 𝑇𝑘 has a fixed point (𝑥𝑖1 , 𝑥𝑖2 ) ∈
𝑈𝑘 . Moreover combining (40) and the fact that 𝑅0 < 𝜀, we
obtain
𝑡𝑛
∗
𝑥𝑖𝑘 = 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝑓𝑖 (𝑡, [𝑥1𝑘 (𝑡) − V1𝑘 (𝑡)] ,
𝑡1
∗
[𝑥2𝑘 (𝑡) − V2𝑘 (𝑡)] ) + 𝑔 (𝑡)) Δ𝑠
≥ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝛿𝑓 (𝑠, 0, 0) + 𝑔 (𝑡)) Δ𝑠
𝑡1
𝑡𝑛
≥ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) 𝑔 (𝑡) Δ𝑠
𝑡1
= V𝑖𝑘 (𝑡)
for 𝑡 ∈ (𝜌 (𝑡1 ) , 𝑡𝑛 ) .
(47)
lim
(42)
𝑓 (𝜀)
1
<
,
𝜀
4𝑐𝜆
Then there exists a 𝑅0 ∈ (0, 𝜀] such that
𝑓 (𝑅0 )
1
=
.
𝑅0
4𝑐𝜆
(45)
𝑡𝑛
Set 𝜆 = 𝜀/4𝑐𝑓(𝜀), since for any 0 < 𝜆 ≤ 𝜆, fix the 𝜆 ∈
(0, 𝜆], we always have
𝑓 (𝑧)
= +∞,
𝑧↓0 𝑧
󵄩
󵄩
𝑅0 = 󵄩󵄩󵄩(𝑥1𝑘 , 𝑥2𝑘 )󵄩󵄩󵄩1 ≤ 2𝜆𝑐𝑓 (𝑅0 ) ,
that is,
𝑡
𝑔(𝑡)}, and 𝑐 = ∫𝑡 𝑛 𝐶∗ 𝐺𝑖 (𝑠, 𝑠)Δ𝑠. We have
𝑓 (𝑧)
lim
= +∞.
𝑧↓0 𝑧
∗
≤ 𝜆 ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
(43)
Let 𝑈𝑘 = {(𝑥1𝑘 , 𝑥2𝑘 ) ∈ 𝑃1(𝑘−1) × 𝑃2(𝑘−1) : ‖(𝑥1𝑘 , 𝑥2𝑘 )‖1 <
𝑅0 }, (𝑥1𝑘 , 𝑥2𝑘 ) ∈ 𝜕𝑈𝑘 and ] ∈ (0, 1) be such that (𝑥1𝑘 , 𝑥2𝑘 ) =
]𝑇𝑘 (𝑥1𝑘 , 𝑥2𝑘 ), that is, 𝑥𝑖𝑘 = ]𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑖 = 1, 2). We
Then 𝑇𝑘 has a positive fixed point (𝑥𝑖1 , 𝑥𝑖2 ) and
‖(𝑥𝑖1 , 𝑥𝑖2 )‖1 ≤ 𝑅0 < 1; that is, (𝑥𝑖1 , 𝑥𝑖2 ) is a positive solution
of the boundary value problem (26) with 𝑥𝑖𝑘 > V𝑖𝑘 (𝑡) for
𝑡 ∈ (𝑡1 , 𝑡𝑛 ).
Let 𝑢𝑖𝑘 (𝑡) = 𝑥𝑖𝑘 (𝑡)−V𝑖𝑘 (𝑡) ≥ 0 (𝑖 = 1, 2), then (𝑢1𝑘 , 𝑢2𝑘 ) is a
nonnegative solution (positive on (𝜌(𝑡1 ), 𝑡𝑛 )) of the boundary
value problem (1).
Theorem 7. Suppose that (H∗1 ) and (H3 )-(H4 ) hold. Then there
exists a constant 𝜆∗ > 0 such that, for any 0 < 𝜆 ≤ 𝜆∗ , (1) with
boundary value condition (𝑘) has at least one positive solution
(𝑘 = 2, . . . , 5).
Journal of Applied Mathematics
9
Proof. We fix 𝑘 (𝑘 = 2, . . . , 5). Let Ω1 = {(𝑥1𝑘 , 𝑥2𝑘 ) ∈ 𝐸 ×
𝐸 : ‖𝑥𝑖𝑘 ‖ < 𝑅1 , 𝑖 = 1, 2}, where 𝑅1 = max{1, 𝑟} and 𝑟 =
𝑡
2
(𝐶∗ /𝑐∗ ) ∫𝑡 𝑛 𝑔(𝑠)Δ𝑠. Choose
(𝑃1(𝑘−1) × 𝑃2(𝑘−1) ) ∩ 𝜕Ω2 , we have ‖𝑥1𝑘 ‖ = 𝑅2 or ‖𝑥2𝑘 ‖ = 𝑅2 .
Without loss of generality let ‖𝑥1𝑘 ‖ = 𝑅2 , so we have
1
𝑅
𝑅
𝜆 = min {1, 1 (𝑅 + 1)−1 , 1 } ,
2
2𝑟
∗
𝑡𝑛
(48)
𝑥1𝑘 (𝑡) − V1𝑘 (𝑡) = 𝑥1𝑘 (𝑡) − 𝜆 ∫ 𝐻1𝑘 (𝑡, 𝑠) 𝑔 (𝑠) Δ𝑠
𝑡1
≥ 𝑥1𝑘 (𝑡) − 𝜆 ∫
𝑡
where 𝑅 = ∫𝑡 𝑛 𝐶∗ 𝐺𝑖 (𝑠, 𝑠)(max0≤𝑧1 ,𝑧2 ≤𝑅1 𝑓𝑖 (𝑠, 𝑧1 , 𝑧2 ) + 𝑔(𝑠))Δ𝑠
1
and 𝑅 ≥ 0.
Then for any (𝑥1𝑘 , 𝑥2𝑘 ) ∈ (𝑃1(𝑘−1) × 𝑃2(𝑘−1) ) ∩ 𝜕Ω1 , 𝑥𝑖𝑘 (𝑠) −
V𝑖𝑘 (𝑠) ≤ 𝑥𝑖𝑘 (𝑠) ≤ ‖𝑥𝑖𝑘 ‖ ≤ 𝑅1 (𝑖 = 1, 2) and for 0 < 𝜆 ≤ 𝜆∗ , we
have
𝑡1
2
𝑥1𝑘 (𝑡)
󵄩󵄩 󵄩󵄩 𝑔 (𝑠) Δ𝑠
󵄩󵄩𝑥1𝑘 󵄩󵄩
𝑥 (𝑡)
≥ 𝑥1𝑘 (𝑡) − 󵄩󵄩1𝑘 󵄩󵄩 𝜆𝑟
󵄩󵄩𝑥1𝑘 󵄩󵄩
∗
≥ 𝑥1𝑘 (𝑡) −
∗
≥ (1 −
≤ 𝜆 ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡1
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) + 𝑔 (𝑠)) Δ𝑠
𝑡𝑛
≤ 𝜆 ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) ( max 𝑓𝑖 (𝑠, 𝑧1 , 𝑧2 ) + 𝑔 (𝑠)) Δ𝑠
≥
0≤𝑧1 ,𝑧2 ≤𝑅1
𝑡1
𝐶∗
𝑐∗
𝑥 (𝑡) 𝑡𝑛 𝐶∗ 2
= 𝑥1𝑘 (𝑡) − 𝜆 󵄩󵄩1𝑘 󵄩󵄩 ∫
𝑔 (𝑠) Δ𝑠
󵄩󵄩𝑥1𝑘 󵄩󵄩 𝑡1 𝑐∗
󵄩
󵄩󵄩
󵄩󵄩𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑡)󵄩󵄩󵄩
𝑡𝑛
𝑡𝑛
(53)
𝑥1𝑘 (𝑡)
𝜆𝑟
𝑅2
𝜆𝑟
) 𝑥1𝑘 (𝑡)
𝑅2
1
𝑥 (𝑡) ≥ 0,
2 1𝑘
𝑡 ∈ [𝜌 (𝑡1 ) , 𝑡𝑛 ] .
≤ 𝜆𝑅
≤
Thus
𝑅1
.
2
(49)
∗
∗
min {[𝑥1𝑘 (𝑡) − V1𝑘 (𝑡)] + [𝑥2𝑘 (𝑡) − V2𝑘 (𝑡)] }
𝜃1 ≤𝑡≤𝜃2
This implies
󵄩
󵄩
󵄩
󵄩󵄩
󵄩󵄩𝑇𝑘 (𝑥1𝑘 , 𝑥2𝑘 )󵄩󵄩󵄩1 ≤ 𝑅1 ≤ 󵄩󵄩󵄩(𝑥1𝑘 , 𝑥2𝑘 )󵄩󵄩󵄩1 ,
(𝑥1𝑘 , 𝑥2𝑘 ) ∈ (𝑃1(𝑘−1) × 𝑃2(𝑘−1) ) ∩ 𝜕Ω1 .
1
≥ min {𝑥1𝑘 (𝑡) − V1𝑘 (𝑡)} ≥ min { 𝑥1𝑘 (𝑡)}
𝜃1 ≤𝑡≤𝜃2
𝜃1 ≤𝑡≤𝜃2 2
(50)
𝜃1 ≤𝑡≤𝜃2
=
Choose a constant 𝑁 > 1 such that
𝜆𝑁𝛾
𝜃2
𝑐∗
𝐺 (𝑠, 𝑠) 𝜙𝑖1 (𝑠) 𝜙𝑖2 (𝑠) Δ𝑠 ≥ 1, (51)
∫
󵄩 󵄩 󵄩 󵄩
2 (󵄩󵄩󵄩𝜙𝑖1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜙𝑖2 󵄩󵄩󵄩) 𝜃1 𝑖
where 𝛾 = min𝑘 min𝜃1 ≤𝑡≤𝜃2 {𝜙1𝑘 (𝑡), 𝜙2𝑘 (𝑡)}.
By assumption (H3 ) and (H4 ), there exists a constant 𝐵 >
𝑅1 such that
𝑓𝑖 (𝑡, 𝑧1 , 𝑧2 )
> 𝑁,
𝑧1 + 𝑧2
≥ min {
𝑐∗
󵄩 󵄩 𝑐
󵄩 󵄩
𝜙 (𝑡) 󵄩󵄩𝑥 󵄩󵄩 , ∗ 𝜙 (𝑡) 󵄩󵄩𝑥 󵄩󵄩}
2𝐶∗ 1𝑘 󵄩 1𝑘 󵄩 2𝐶∗ 2𝑘 󵄩 1𝑘 󵄩
𝑐∗
𝑅 min {𝜙 (𝑡) , 𝜙2𝑘 (𝑡)} ≥ 𝐵 + 1 > 𝐵.
2𝐶∗ 2 𝜃1 ≤𝑡≤𝜃2 1𝑘
Now since 𝐵 > 𝑅1 , it follows that
𝑇𝑖𝑘 (𝑥1𝑘 , 𝑥2𝑘 ) (𝑡)
𝑡𝑛
∗
= 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝑡1
∗
that is, 𝑓𝑖 (𝑡, 𝑧1 , 𝑧2 ) > 𝑁 (𝑧1 + 𝑧2 ) ,
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) + 𝑔 (𝑠)) Δ𝑠
𝜃2
for 𝑡 ∈ [𝜃1 , 𝜃2 ] , 𝑧1 + 𝑧2 > 𝐵 (𝑖 = 1, 2) .
∗
≥ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) (𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
(52)
𝜃1
∗
Choose 𝑅2 = max{𝑅1 + 1, 2𝜆𝑟, 2𝐶∗ (𝐵 + 1)/𝑐∗ 𝛾} and let
Ω2 = {(𝑥1𝑘 , 𝑥2𝑘 ) ∈ 𝐸 × 𝐸 : ‖𝑥𝑖𝑘 ‖ < 𝑅2 , 𝑖 = 1, 2}. We note
that 𝑥(𝑡) ≥ (𝑐∗ /𝐶∗ )𝜙𝑖𝑗 (𝑡)‖𝑥‖ for all 𝑥 ∈ 𝑃𝑖𝑗 , by Lemma 3, we
2
have 𝐻𝑖𝑘 (𝑡, 𝑠) ≤ (𝐶∗ /𝑐∗ )(𝑥(𝑡)/‖𝑥‖). Then for any (𝑥1𝑘 , 𝑥2𝑘 ) ∈
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) + 𝑔 (𝑠)) Δ𝑠
𝜃2
∗
≥ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) 𝑓𝑖 (𝑠, [𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)] ,
𝜃1
∗
[𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) Δ𝑠
(54)
10
Journal of Applied Mathematics
𝜃2
≥ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) 𝑁 ([𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)]
Theorem 8. Suppose that (H1 )–(H3 ) hold. Then (1) with
boundary value condition (𝑘) has at least two positive solutions
for 𝜆 > 0 sufficiently small (𝑘 = 2, . . . , 5).
∗
𝜃1
∗
+ [𝑥2𝑘 (𝑠) − V2𝑘 (𝑠)] ) Δ𝑠
In fact with 0 < 𝜆 < min{𝜆, 𝜆∗ } then (1) with boundary
value condition (𝑘) has at least two positive solutions.
𝜃2
≥ 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, s) 𝑁 (𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)) Δ𝑠
𝜃1
𝜃2
≥ 𝜆 ∫ 𝑐∗ min {𝜙1𝑘 (𝑡) , 𝜙2𝑘 (𝑡)} 𝐺𝑖 (𝑠, 𝑠) 𝑁 (𝑥1𝑘 (𝑠) − V1𝑘 (𝑠)) Δ𝑠
𝜃1 ≤𝑡≤𝜃2
𝜃1
𝜃2
≥ 𝜆 ∫ 𝑐∗ min {𝜙1𝑘 (𝑡) , 𝜙2𝑘 (𝑡)} 𝐺𝑖 (𝑠, 𝑠)
𝜃1 ≤𝑡≤𝜃2
𝜃1
𝑁
𝑥 (𝑠) 𝑑𝑠
2 1𝑘
Remark 9. In Theorems 6–8, we use the assumption condition 16. If we have not the condition 16, that is, 𝑎𝑖 = 𝑏𝑖 = 0,
then the system (1) and boundary condition (𝑘) are
Δ
(𝑝1 𝑢1∇ ) (𝑡)−𝑞1 (𝑡) 𝑢1 (𝑡)+𝜆𝑓1 (𝑡, 𝑢1 (𝑡) , 𝑢2 (𝑡)) = 0,
≥ 𝜆 min {𝜙1𝑘 (𝑡) , 𝜙2𝑘 (𝑡)}
𝑡 ∈ (𝑡1 , 𝑡𝑛 ) , 𝜆 > 0,
𝜃1 ≤𝑡≤𝜃2
𝜃2
× ∫ 𝑐∗ 𝐺𝑖 (𝑠, 𝑠)
𝜃1
Δ
(𝑝2 𝑢2∇ ) (𝑡)−𝑞2 (𝑡) 𝑢2 (𝑡)+𝜆𝑓2 (𝑡, 𝑢1 (𝑡) , 𝑢2 (𝑡)) = 0,
𝑐∗ 𝑁
󵄩 󵄩
󵄩 󵄩 󵄩 󵄩 𝜙 (𝑠) 𝜙𝑖2 (𝑠) 󵄩󵄩󵄩𝑥1𝑘 󵄩󵄩󵄩 Δ𝑠
2 (󵄩󵄩󵄩𝜙𝑖1 󵄩󵄩󵄩 + 󵄩󵄩󵄩𝜙𝑖2 󵄩󵄩󵄩) 𝑖1
𝛼1 𝑢1 (𝑡1 ) − 𝛽1 𝑝1 (𝑡1 ) 𝑢1∇ (𝑡1 ) = 0,
𝜃2
𝑐
≥ 𝜆𝑁𝛾 󵄩󵄩 󵄩󵄩 ∗ 󵄩󵄩 󵄩󵄩 ∫ 𝐺𝑖 (𝑠, 𝑠) 𝜙𝑖1 (𝑠) 𝜙𝑖2 (𝑠) Δ𝑠𝑅2
2 (󵄩󵄩𝜙𝑖1 󵄩󵄩 + 󵄩󵄩𝜙𝑖2 󵄩󵄩) 𝜃1
≥ 𝑅2 ,
𝛾1 𝑢1 (𝑡𝑛 ) + 𝛿1 𝑝1 (𝑡𝑛 ) 𝑢1∇ (𝑡𝑛 ) = 0,
𝑡 ∈ [𝜃1 , 𝜃2 ] .
(55)
This implies
󵄩󵄩
󵄩
󵄩
󵄩
󵄩󵄩𝑇𝑘 (𝑥1𝑘 , 𝑥2𝑘 )󵄩󵄩󵄩1 ≥ 󵄩󵄩󵄩(𝑥1𝑘 , 𝑥2𝑘 )󵄩󵄩󵄩1 ,
(𝑥1𝑘 , 𝑥2𝑘 ) ∈ (𝑃1(𝑘−1) × 𝑃2(𝑘−1) ) ∩ 𝜕Ω2 .
(56)
For the Krasnosel’skii’s fixed point theorem, one deduces that
𝑇𝑘 has a fixed point (𝑥1𝑘 , 𝑥2𝑘 ) with 𝑅1 < ‖(𝑥1𝑘 , 𝑥2𝑘 )‖ < 𝑅2 ⇔
𝑅1 < ‖𝑥1𝑘 ‖ + ‖𝑥2𝑘 ‖ < 𝑅2 .
Since 𝑟 ≤ 𝑅1 < ‖𝑥𝑖𝑘 ‖ < 𝑅2 (𝑖 = 1, 2), then
𝑡𝑛
𝑥𝑖𝑘 (𝑡) − V𝑖𝑘 (𝑡) = 𝑥𝑖𝑘 (𝑡) − 𝜆 ∫ 𝐻𝑖𝑘 (𝑡, 𝑠) 𝑔 (𝑠) Δ𝑠
𝑡1
𝑡𝑛
≥ 𝑥𝑖𝑘 (𝑡) − 𝜆 ∫
𝑡1
∗2
𝐶
𝑐∗
𝛼2 𝑢2 (𝑡1 ) − 𝛽2 𝑝2 (𝑡1 ) 𝑢2∇ (𝑡1 ) = 0,
𝛾2 𝑢2 (𝑡𝑛 ) + 𝛿2 𝑝1 (𝑡𝑛 ) 𝑢2∇ (𝑡𝑛 ) = 0.
From Lemma 2, an argument similar to those in Theorems
6–8 yields the following theorems.
Theorem 10. Suppose that (H1 ) and (H2 ) hold. Then there
exists a constant 𝜆 > 0 such that, for any 0 < 𝜆 ≤ 𝜆, the
boundary value problem (58) has at least one positive solution.
Theorem 11. Suppose that (H∗1 ) and (H3 )-(H4 ) hold. Then
there exists a constant 𝜆∗ > 0 such that, for any 0 < 𝜆 ≤ 𝜆∗ , the
boundary value problem (58) has at least one positive solution.
Theorem 12. Suppose that (H1 )–(H4 ) hold. Then the boundary value problem (58) has at least two positive solutions for
𝜆 > 0 sufficiently small.
𝑥𝑖𝑘 (𝑡)
󵄩󵄩 󵄩󵄩 𝑔 (𝑠) Δ𝑠
󵄩󵄩𝑥𝑖𝑘 󵄩󵄩
𝑥 (𝑡)
= 𝑥𝑖𝑘 (𝑡) − 𝜆 󵄩󵄩𝑖𝑘 󵄩󵄩 𝑟
󵄩󵄩𝑥𝑖𝑘 󵄩󵄩
(58)
4. Example
(57)
≥ 𝑥𝑖𝑘 (𝑡) − 𝜆𝑥𝑖𝑘 (𝑡)
To illustrate the usefulness of the results, we give some
examples.
= (1 − 𝜆) 𝑥𝑖𝑘 (𝑡)
Example 13. Consider the boundary value problem
≥ (1 − 𝜆)
> 0,
𝑐∗ 𝜙𝑖1 (𝑡) 𝜙𝑖2 (𝑡) 󵄩󵄩 󵄩󵄩
󵄩𝑥 󵄩
𝐶∗ 󵄩󵄩󵄩󵄩𝜙𝑖1 󵄩󵄩󵄩󵄩 + 󵄩󵄩󵄩󵄩𝜙𝑖2 󵄩󵄩󵄩󵄩 󵄩 𝑖𝑘 󵄩
𝑢󸀠󸀠 − 𝑢 = − 𝜆 ((𝑢 + V)𝑎 +
1
𝑡 ∈ (𝜌 (𝑡1 ) , 𝑡𝑛 ) .
Thus (𝑥1𝑘 , 𝑥2𝑘 ) is a positive solution of the boundary value
problem (26) with 𝑥𝑖𝑘 (𝑡) > V𝑖𝑘 (𝑡) (𝑖 = 1, 2) for 𝑡 ∈ (𝜌(𝑡1 ), 𝑡𝑛 ).
Let 𝑢𝑖𝑘 (𝑡) = 𝑥𝑖𝑘 (𝑡)−V𝑖𝑘 (𝑡) ≥ 0 (𝑖 = 1, 2), then (𝑢1𝑘 , 𝑢2𝑘 ) is a
nonnegative solution (positive on (𝜌(𝑡1 ), 𝑡𝑛 )) of the boundary
value problem (1).
Since condition (H1 ) implies conditions (H∗1 ) and (H4 )
then from the proof of Theorems 6 and 7, we immediately
have the following theorem.
1/2
(𝑡 − 𝑡2 )
cos (2𝜋 (𝑢 + V))) ,
− 1 < 𝑡 < 1, 𝜆 > 0,
V󸀠󸀠 − V = −𝜆 ((𝑢 − 1)2 + V2 +
𝑢 (−1) = V (1) = 0,
1
1/2
(𝑡 − 𝑡2 )
𝑢 (1) = 𝑎𝑢 (0) ,
sin (2𝜋𝑢)) ,
V (−1) = 𝑏V (0) ,
(59)
where 𝑎 > 1. Then if 𝜆 > 0 is sufficiently small, (59) has a
positive solution 𝑢 with 𝑢(𝑡) > 0 for 𝑡 ∈ (0, 1).
Journal of Applied Mathematics
11
To see this, we will apply Theorem 7 with
𝑓1 (𝑡, 𝑢, V) = (𝑢 + V)𝑎 +
1
1/4
(𝑡2 − 𝑡4 )
𝑓2 (𝑡, 𝑢, V) = (𝑢 − 1)2 + V2 +
𝜆∗ = min {1,
cos (2𝜋 (𝑢 + V)) ,
1
sin (2𝜋𝑢) ,
1/4
(𝑡2 − 𝑡4 )
𝑔1 (𝑡) = 𝑔2 (𝑡) = 𝑔 (𝑡) =
Also let
1
(𝑡2 −
1/4
𝑡4 )
(60)
lim inf
𝑢+V↑+∞
Example 14. Consider the boundary value problem:
.
Δ
(𝑝1 𝑢1∇ ) (𝑡) − 𝑞1 (𝑡) 𝑢1 (𝑡)
= −𝜆 (𝑒𝑢1 + 𝑢22 + 7 cos (2𝜋𝑡𝑢1 )) ,
for 𝑡 ∈ (0, 1) 𝑖 = 1, 2,
𝑡1 < 𝑡 < 𝑡𝑛 ,
𝑓𝑖 (𝑡, 𝑢, V)
= +∞ for all 𝑡 ∈ [𝜃1 , 𝜃2 ] ⊂ (0, 1) .
𝑢+V
(61)
Now (H∗1 ), (H3 ), and (H4 ) hold. We note that the boundary
condition of (59) is in accord with (4), and from [1], we have
𝑒𝑡+1 − 𝑒−𝑡−1
𝜙11 = 𝜙21 =
,
2
(65)
Now, if 𝜆 < 𝜆∗ , Theorem 7 guarantees that (59) has a positive
solutions (𝑢, V) with ‖𝑢‖ ≥ 1 and ‖V‖ ≥ 1.
Clearly for 𝑡 ∈ (0, 1),
𝑓𝑖 (𝑡, 𝑢, V) + 𝑔 (𝑡) > 0,
−1 𝑅
𝑅1
(𝑅𝑎+2 + 𝜋) , 1 } .
2𝑎+2 𝐶∗ 𝑒4 1
2𝑟
𝑒−𝑡+1 − 𝑒𝑡−1
𝜙12 = 𝜙22 =
,
2
𝑑1 = 𝑑2 = sinh (2) .
(62)
𝜆 > 0,
(66)
Δ
(𝑝2 𝑢2∇ ) (𝑡) − 𝑞2 (𝑡) 𝑢2 (𝑡)
2
= −𝜆 ((𝑢1 − 1) + 𝑢22 + 5 sin (2𝜋𝑡𝑢2 ))
satisfying one of the boundary value conditions (𝑘), (𝑘 =
2, . . . , 5).
Then if 𝜆 > 0 is sufficiently small, (66) has two solutions
(𝑢11 , 𝑢12 ), (𝑢21 , 𝑢22 ) with 𝑢𝑖𝑗 (𝑡) > 0 for 𝑡 ∈ (0, 1), 𝑖, 𝑗 = 1, 2.
To see this, we will apply Theorem 8 with
𝑓1 (𝑡, 𝑢1 , 𝑢2 ) = 𝑒𝑢1 + 𝑢22 + 7 cos (2𝜋𝑡𝑢1 ) ,
Then
2
𝑓2 (𝑡, 𝑢1 , 𝑢2 ) = (𝑢1 − 1) + 𝑢22 + 5 sin (2𝜋𝑡𝑢2 ) ,
𝐺1 (𝑡, 𝑠) = 𝐺2 (𝑡, 𝑠)
=
1 𝜙12 (𝑡) 𝜙11 (𝑠) ,
{
𝑑1 𝜙11 (𝑡) 𝜙12 (𝑠) ,
𝐻24 (𝑡, 𝑠) = 𝐺2 (𝑡, 𝑠) +
𝑡𝑛
𝑔1 (𝑡) = 𝑔2 (𝑡) = 𝑔 (𝑡) = 8.
𝜌 (𝑡1 ) ≤ 𝑠 ≤ 𝑡 ≤ 𝑡𝑛 ,
𝜌 (𝑡1 ) ≤ 𝑡 ≤ 𝑠 ≤ 𝑡𝑛 ,
1
𝐻14 (𝑡, 𝑠) = 𝐺1 (𝑡, 𝑠) +
𝑎𝐺 (0, 𝑠) 𝜙11 (𝑡) ,
𝑑1 − 𝑎𝜙11 (0) 1
(63)
Clearly, for 𝑡 ∈ (0, 1),
𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 ) + 𝑔 (𝑡) ≥ 1 > 0,
1
𝑏𝐺 (0, 𝑠) 𝜙22 (𝑡) .
𝑑2 − 𝑏𝜙22 (0) 2
𝑓1 (𝑡, 0, 0) = 8 > 0,
Note 𝑟 = ∫𝑡 (𝐶 /𝑐∗ )𝑔(𝑠)Δ𝑠. Let 𝑅1 = 𝑟 + 1 and we have
1
lim inf
𝑢+V↑+∞
𝑡𝑛
𝑅 = ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) ( max 𝑓𝑖 (𝑠, 𝑧1 , 𝑧2 ) + 𝑔 (𝑠)) Δ𝑠
0≤𝑧1 ,𝑧2 ≤𝑅1
𝑡𝑛
≤ ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) [2𝑎+2 𝑅1𝑎+2 +
𝑡1
≤ ∫
−1
= ∫
0
1/4
(𝑠2 − 𝑠4 )
𝐶∗ 𝑒4 𝑎+2 𝑎+2
2
] Δ𝑠
[2 𝑅1 +
1/4
2
4
(𝑠 − 𝑠4 )
1
1
2
∗ 4
𝐶 𝑒
2
] Δ𝑠
[2𝑎+2 𝑅1𝑎+2 +
1/4
2
2
(𝑠 − 𝑠4 )
𝐶∗ 𝑒4 𝑎+2 𝑎+2
2
≤ ∫
] Δ𝑠
[2 𝑅1 +
1/2
2
0
(𝑠 − 𝑠2 )
1
𝑎+1
≤2
∗ 4
𝐶 𝑒
(𝑅1𝑎+2
+ 𝜋) .
(68)
𝑓2 (𝑡, 0, 0) = 3 > 0,
∗2
𝑡1
(67)
𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 )
= +∞,
𝑢1 + 𝑢2
𝑖 = 1, 2.
Now (H1 )–(H4 ) hold. Let 𝛿 = 1/100, 𝜀 = 1/8, and we have
𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 ) ≥ 𝛿𝑓𝑖 (𝑡, 0, 0) ,
] Δ𝑠
(64)
for 0 ≤ 𝑡 ≤ 1, 0 ≤ 𝑢𝑖 ≤ 𝜀, 𝑖 = 1, 2.
(69)
Furthermore
let
𝑓(𝜀)
=
max0≤𝑡≤1,0≤𝑢1 ,𝑢2 ≤𝜀 {max𝑖=1,2 𝑓𝑖 (𝑡, 𝑢1 , 𝑢2 ) + 𝑔(𝑡)}, and 𝑐 =
𝑡
∫𝑡 𝑛 𝐶∗ 𝐺𝑖 (𝑠, 𝑠)Δ𝑠. Note
1
𝜀
4𝑐𝑓 (𝜀)
≥
1
1
>
.
32𝑐 (𝑒 + 8) 352𝑐
(70)
Let 𝜆 = 1/352𝑐. Now, if 0 < 𝜆 < 𝜆 then 0 < 𝜆 < 𝜀/4𝑐𝑓(𝜀)
and Theorem 6 guarantees that (66) has positive solutions
(𝑢11 , 𝑢12 ) with ‖𝑢1𝑗 ‖ ≤ (1/8) (𝑗 = 1, 2).
12
Journal of Applied Mathematics
2
Next note 𝑟 = 8𝐶∗ (𝑡𝑛 − 𝑡1 )/𝑐∗ and let 𝑅1 = 𝑟 + 2 so we
have
1
𝑅 = ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) ( max 𝑓𝑖 (𝑠, 𝑧1 , 𝑧2 ) + 𝑔 (𝑠)) Δ𝑠
0≤𝑧1 ,𝑧2 ≤𝑅1
0
1
≤ ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) (𝑒𝑅1 + 2𝑅12 + 7 + 8) Δ𝑠
0
(71)
1
≤ ∫ 𝐶∗ 𝐺𝑖 (𝑠, 𝑠) Δ𝑠 (𝑒𝑅1 + 2𝑅12 + 15)
0
≤ (𝑒𝑅1 + 2𝑅12 + 15) 𝑐.
Also let
𝜆∗ = min {1,
𝑅1
𝑅
, 1}.
2 (𝑒𝑅1 + 2𝑅12 + 15) 𝑐 2𝑟
(72)
Now, if 𝜆 < 𝜆∗ , Theorem 7 guarantees that (59) has a positive
solutions (𝑢21 , 𝑢22 ) with ‖𝑢2𝑗 ‖ ≥ 2, 𝑗 = 1, 2.
Thus, if 𝜆 < min{𝜆, 𝜆∗ }, Theorem 8 guarantees that (66)
has two solutions (𝑢11 , 𝑢12 ) and (𝑢21 , 𝑢22 ) with 𝑢𝑖𝑗 > 0 for 𝑡 ∈
(0, 1), 𝑖, 𝑗 = 1, 2.
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