Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2009, Article ID 258090, 16 pages
doi:10.1155/2009/258090
Research Article
MultiPoint BVPs for Second-Order Functional
Differential Equations with Impulses
Xuxin Yang,1, 2 Zhimin He,3 and Jianhua Shen4
1
Department of Mathematics, Hunan Normal University, Changsha, Hunan 410081, China
Department of Mathematics, Hunan First Normal University, Changsha, Hunan 410205, China
3
College of Science, Zhejiang Forestry University, Hangzhou, Zhejiang 311300, China
4
Department of Mathematics, Hangzhou Normal University, Hangzhou, Zhejiang 310036, China
2
Correspondence should be addressed to Jianhua Shen, jhshen2ca@yahoo.com
Received 14 April 2009; Accepted 10 June 2009
Recommended by Fernando Lobo Pereira
This paper is concerned about the existence of extreme solutions of multipoint boundary value
problem for a class of second-order impulsive functional differential equations. We introduce
a new concept of lower and upper solutions. Then, by using the method of upper and lower
solutions introduced and monotone iterative technique, we obtain the existence results of extreme
solutions.
Copyright q 2009 Xuxin Yang 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.
1. Introduction
In this paper, we consider the multipoint boundary value problems for the impulsive
functional differential equation:
−u t ft, ut, uθt,
t ∈ J 0, 1, t /
tk ,
Δu tk Ik utk , k 1, . . . , m,
u1 bu 1 duξ,
u0 − au 0 cu η ,
1.1
where f ∈ CJ × R2 , R, 0 ≤ θt ≤ t, t ∈ J, θ ∈ CJ, a ≥ 0, b ≥ 0, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1,
0 < η, ξ < 1. 0 < t1 < t2 < · · · < tm < 1, f is continuous everywhere except at {tk } × R2 ;
ftk , ·, ·, and ft−k , ·, · exist with ft−k , ·, · ftk , ·, ·; Ik ∈ CR, R, Δu tk u tk − u t−k .
Denote J − J \{ti , i 1, 2, . . . , m}. Let P CJ, R {u : J → R; ut|J − is continuous, utk and
2
Mathematical Problems in Engineering
ut−k exist with ut−k utk , k 1, 2, . . . , m}; P C1 J, R {u : J → R; ut|J − is continuous
differentiable, u tk and u t−k exist with u t−k u tk , k 1, 2, . . . , m}. Let E P C1 J, R ∩
C2 J, R. A function u ∈ E is called a solution of BVP1.1 if it satisfies 1.1.
The method of upper and lower solutions combining monotone iterative technique
offers an approach for obtaining approximate solutions of nonlinear differential equations
1–3. There exist much literature devoted to the applications of this technique to general
boundary value problems and periodic boundary value problems, for example, see 1, 4–6
for ordinary differential equations, 7–11 for functional differential equations, and 12 for
differential equations with piecewise constant arguments. For the studies about some special
boundary value problems, for example, Lidston boundary value problems and antiperiodic
boundary value problems, one may see 13, 14 and the references cited therein.
Here, we hope to mention some papers where existence results of solutions of certain
boundary value problems of impulsive differential equations were studied 11, 15 and
certain multipoint boundary value problems also were studied 6, 16–21. These works
motivate that we study the multipoint boundary value problems for the impulsive functional
differential equation 1.1.
We also note that when Ik 0 and θt t, the boundary value problem 1.1 reduces
to multi-point boundary value problems for ordinary differential equations which have been
studied in many papers, see, for example, 6, 16–18 and the references cited therein. To our
knowledge, only a few papers paid attention to multi-point boundary value problems for
impulsive functional differential equations.
In this paper, we are concerned with the existence of extreme solutions for the
boundary value problem 1.1. The paper is organized as follows. In Section 2, we establish
two comparison principles. In Section 3, we consider a linear problem associated to 1.1 and
then give a proof for the existence theorem. In Section 4, we first introduce a new concept
of lower and upper solutions. By using the method of upper and lower solutions with a
monotone iterative technique, we obtain the existence of extreme solutions for the boundary
value problem 1.1.
2. Comparison Principles
In the following, we always assume that the following condition H is satisfied:
H
a ≥ 0, b ≥ 0, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1, 0 < η, ξ < 1, a c > 0, b d > 0.
For any given function g ∈ E, we denote
g0 − ag 0 − cg η g1 bg 1 − dgξ
Ag max
,
,
aπ c sin πη
bπ d sin πξ
Bg max Ag , 0 ,
cg t Bg sinπt,
We now present main results of this section.
r π 2.
2.1
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3
Theorem 2.1. Assume that u ∈ E satisfies
−u t Mut Nuθt ≤ 0,
t ∈ J, t /
tk ,
Δu tk ≥ Lk utk , k 1, . . . , m,
u1 bu 1 ≤ duξ,
u0 − au 0 ≤ cu η ,
2.2
where a ≥ 0, b ≥ 0, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1, 0 < η, ξ < 1, Lk ≥ 0 and constants M, N satisfy
M > 0, N ≥ 0,
m
MN Lk ≤ 1.
2
k1
2.3
Then ut ≤ 0 for t ∈ J.
Proof. Suppose, to the contrary, that ut > 0 for some t ∈ J.
If u1 maxt∈J ut > 0, then u 1 ≥ 0, u1 ≥ uξ, and
duξ ≤ u1 ≤ u1 bu 1 ≤ duξ.
2.4
So d 1 and uξ is a maximum value.
If u0 maxt∈J ut > 0, then u 0 ≤ 0, u0 ≥ uη, and
cu η ≤ u0 ≤ u0 − bu 0 ≤ cu η .
2.5
So c 1 and uη is a maximum value.
Therefore, there is a ρ ∈ 0, 1 such that
u ρ max ut > 0,
t∈J
u ρ ≤ 0.
2.6
Suppose that ut ≥ 0 for t ∈ J. From the first inequality of 2.2, we obtain that u t ≥
0 for t ∈ J. Hence
u0 max ut
t∈J
or u1 max ut.
t∈J
2.7
If u 0 ≥ 0, then u t ≥ 0, t ∈ ti , ti1 , it is easy to obtain that ut is nondecreasing.
Since u1 ≤ duξ ≤ u1, it follows that ut ≡ K K > 0 for t ∈ ξ, 1. From the first
inequality of 2.2, we have that when t ∈ ξ, 1,
0 < MK ≤ Mut Nuθt ≤ u t 0,
2.8
which is a contradiction.
If u 0 ≤ 0, then u0 maxt∈J ut > 0, or u1 maxt∈J ut > 0. If u0 maxt∈J ut >
0, then ut ≡ K K > 0 for t ∈ 0, η. If u1 maxt∈J ut > 0, then ut ≡ K for t ∈ ξ, 1.
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Mathematical Problems in Engineering
From the first inequality of 2.2, we have that when t ∈ ξ, 1,
0 < MK ≤ Mut Nuθt ≤ u t 0,
2.9
which is a contradiction.
Suppose that there exist t1 , t2 ∈ J such that ut1 > 0 and ut2 < 0. We consider two
possible cases.
Case 1 u0 > 0. Since ut2 < 0, there is κ > 0, ε > 0 such that uκ 0, ut ≥ 0 for t ∈ 0, κ
and ut < 0 for all t ∈ κ, κ ε. It is easy to obtain that u t ≥ 0 for t ∈ 0, κ. If t∗ < κ,
then 0 < Mut∗ ≤ u t∗ ≤ 0, a contradiction. Hence t∗ > κ ε. Let t∗ ∈ 0, t∗ such that
ut∗ mint∈0,t∗ ut, then ut∗ < 0. From the first inequality of 2.2, we have
u t ≥ M Nut∗ ,
Δu tk ≥ Lk utk ,
t ∈ 0, t∗ , t / tk ,
k 1, . . . , m.
2.10
Integrating the above inequality from st∗ ≤ s ≤ t∗ to t∗ , we obtain
u t∗ − u s ≥ t∗ − sM Nut∗ ≥ t∗ − sM Nut∗ Lk utk s<tk <t∗
m
Lk ut∗ .
2.11
k1
Hence
−u s ≥ t∗ − sM N m
Lk ut∗ ,
t∗ ≤ s ≤ t ∗ ,
2.12
k1
and then integrate from t∗ to t∗ to obtain
−ut∗ < ut∗ − ut∗ t∗
m
≤
s − t∗ M Nut∗ ds − Lk ut∗ t∗
m
MN ∗
t − t∗ 2 Lk
2
k1
m
MN Lk ut∗ .
≤−
2
k1
≤−
From 2.3, we have that ut∗ > 0. This is a contradiction.
k1
ut∗ 2.13
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5
Case 2 u0 ≤ 0. Let t∗ ∈ 0, t∗ such that ut∗ mint∈0,t∗ ut ≤ 0. From the first inequality
of 2.2, we have
u t ≥ M Nut∗ ,
Δu tk ≥ Lk utk ,
t ∈ 0, t∗ , t / tk ,
2.14
k 1, . . . , m.
The rest proof is similar to that of Case 1. The proof is complete.
Theorem 2.2. Assume that H holds and u ∈ E satisfies
−u t Mut Nuθt M rcu t Ncu θt ≤ 0,
Δu tk ≥ Lk utk Lk cu tk ,
t ∈ J, t /
tk ,
k 1, . . . , m,
2.15
where constants M, N satisfy 2.3, and Lk ≥ 0, then ut ≤ 0 for t ∈ J.
Proof. Assume that u0−au 0 ≤ cuη, u1bu 1 ≤ duξ, then cu t ≡ 0. By Theorem 2.1,
ut ≤ 0.
Assume that u0 − au 0 ≤ cuη, u1 bu 1 > duξ, then
cu t sinπt
u1 bu 1 − duξ .
bπ d sinπξ
2.16
Put yt ut cu t, t ∈ J, then yt ≥ ut for all t ∈ J, and
y t u t π cosπt u1 bu 1 − duξ ,
bπ d sinπξ
y t u t − rcu t,
t ∈ J,
2.17
t ∈ J.
Hence
y0 u0,
yξ uξ y1 u1,
sinπξ
u1 bu 1 − duξ ,
bπ d sinπξ
y 0 u 0 π
u1 bu 1 − duξ ,
bπ d sinπξ
y 1 u 1 −
π
u1 bu 1 − duξ ,
bπ d sinπξ
− y t Myt Nyθt −u t Mut Nuθt M rcu t Ncu θt ≤ 0,
y0 − ay 0 u0 − au 0 −
aπ
u1 bu 1 − duξ ≤ cu η ≤ cy η ,
bπ d sinπξ
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Mathematical Problems in Engineering
y1 by 1 − dyξ
bπ
u1 bu 1 − duξ
bπ d sinπξ
d sinπξ −
u1 bu 1 − duξ ≤ 0,
bπ d sin πξ
u1 bu 1 − duξ −
Δy tk Δu tk Δcu tk ≥ Lk utk Lk cu tk Lk ytk .
2.18
By Theorem 2.1, yt ≤ 0 for all t ∈ J, which implies that ut ≤ 0 for t ∈ J.
Assume that u0 − au 0 > cuη, u1 bu 1 ≤ duξ, then
cu t sin πt
u0 − au 0 − cu η .
aπ c sin πη
2.19
Put yt ut cu t, t ∈ J, then yt ≥ ut for all t ∈ J, and
y t u t π cosπt u0 − au 0 − cu η ,
aπ c sin πη
y t u t − rcu t,
t ∈ J,
2.20
t ∈ J.
Hence
y0 u0,
y1 u1,
sin πη
y η u η u0 − au 0 − cu η ,
aπ c sin πη
π
y 0 u 0 u0 − au 0 − cu η ,
aπ c sin πη
y 1 u 1 −
π
u0 − au 0 − cu η ,
aπ c sin πη
− y t Myt Nyθt −u t Mut Nuθt M rcu t Ncu θt ≤ 0,
y0 − ay 0 − cy η
u0 − au 0 − cu η −
aπ
u0 − au 0 − cu η
aπ c sin πη
c sin πη
−
u0 − au 0 − cu η ≤ 0,
aπ c sin πη
y1 by 1 u1 bu 1 −
bπ
u0 − au 0 − cu η ≤ duξ ≤ dyξ,
aπ c sin πη
Δy tk Δu tk Δcu tk ≥ Lk utk Lk cu tk Lk ytk .
2.21
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7
By Theorem 2.1, yt ≤ 0 for all t ∈ J, which implies that ut ≤ 0 for t ∈ J.
Assume that u0 − au 0 > cuη, u1 bu 1 > duξ, then cu t Au sinπt.
Put yt ut cu t, t ∈ J, then yt ≥ ut for all t ∈ J, and
y t u t Au π cosπt,
y t u t − rcu t,
t ∈ J,
t ∈ J.
2.22
Hence
y0 u0,
y1 u1,
y η u η Au sin πη ,
y 0 u 0 Au π,
yξ uξ Au sinπξ,
y 1 u 1 − Au π,
− y t Myt Nyθt
−u t Mut Nuθt M rcu t Ncu θt ≤ 0,
y0 − ay 0 − cy η u0 − au 0 − cu η − aAu π − cAu sin πη ≤ 0,
2.23
y1 by 1 − dyξ u1 bu 1 − duξ − bAu π − dAu sinπξ ≤ 0,
Δy tk Δu tk Δcu tk ≥ Lk utk Lk cu tk Lk ytk .
By Theorem 2.1, yt ≤ 0 for all t ∈ J, which implies that ut ≤ 0 for t ∈ J. The proof is
complete.
3. Linear Problem
In this section, we consider the linear boundary value problem
−u t Mut Nuθt σt,
t ∈ J, t / tk ,
Δu tk Lk utk ek , k 1, . . . , m,
u1 bu 1 duξ.
u0 − au 0 cu η ,
3.1
Theorem 3.1. Assume that H holds, σ ∈ CJ, ek ∈ R, and constants M, N satisfy 2.3 with
μ
m
1
1 2b 2
a1 2b
1 b2 Lk < 1.
M N 1 2a b 1 8 a b 1
a b 1 k1
3.2
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Mathematical Problems in Engineering
Further suppose that there exist α, β ∈ E such that
h1 α ≤ β on J,
h2 −α t Mαt Nαθt M rcα t Ncα θt ≤ σt,
Δα tk ≥ Lk αtk Lk cα tk ek ,
t ∈ J, t /
tk ,
k 1, . . . , m,
3.3
h3 −β t Mβt Nβθt − M rc−β t Nc−β θt ≥ σt,
Δβ tk ≤ Lk βtk − Lk c−β tk ek ,
t ∈ J, t /
tk ,
k 1, . . . , m.
3.4
Then the boundary value problem 3.1 has one unique solution ut and α ≤ u ≤ β for t ∈ J.
Proof. We first show that the solution of 3.1 is unique. Let u1 , u2 be the solution of 3.1 and
set v u1 − u2 . Thus,
−v t Mvt Nvθt 0,
t ∈ J, t /
tk ,
Δv tk Lk vtk , k 1, . . . , m,
v1 bv 1 dvξ.
v0 − av 0 cv η ,
3.5
By Theorem 2.1, we have that v ≤ 0 for t ∈ J, that is, u1 ≤ u2 on J. Similarly, one can obtain
that u2 ≤ u1 on J. Hence u1 u2 .
Next, we prove that if u is a solution of 3.1, then α ≤ u ≤ β. Let p α − u. From
boundary conditions, we have that cα t cp t for all t ∈ J. From h2 and 3.1, we have
−p t Mpt Npθt M rcp t Ncp θt ≤ 0,
Δp tk ≥ Lk ptk Lk cp tk ,
t ∈ J, t / tk ,
k 1, . . . , m.
3.6
By Theorem 2.1, we have that p α − u ≤ 0 on J. Analogously, u ≤ β on J.
Finally, we show that the boundary value problem 3.1 has a solution by five steps as
follows.
Step 1. Let αt αt cα t, βt βt − c−β t. We claim that
1
−α t Mαt Nαθt M rcα t Ncα θt ≤ σt for t ∈ J, t /
tk ,
Δα tk ≥ Lk αtk ek ,
k 1, . . . , m,
3.7
Mathematical Problems in Engineering
9
2
−β t Mβt Nβθt − M rc−β t Nc−β θt ≥ σt for t ∈ J, t / tk ,
Δβ tk ≤ Lk βtk ek , k 1, . . . , m,
3.8
3 αt ≤ αt ≤ βt ≤ βt for t ∈ J.
From h2 and h3 , we have
−α t Mαt Nαθt ≤ σt,
Δα tk ≥ Lk αtk ek ,
t ∈ J, t /
tk ,
k 1, . . . , m.
−β t Mβt Nβθt ≥ σt,
t ∈ J, t /
tk ,
Δβ tk ≤ Lk βtk ek , k 1, . . . , m,
α0 − aα 0 − cα η α0 − aα 0 − cα η − aπ c sin πη Bα ≤ 0,
α1 bα 1 − dαξ α1 bα 0 − dαξ − bπ d sinπξBα ≤ 0,
−β0 aβ 0 cβ η − aπ c sin πη B−β ≤ 0,
− β0 − aβ 0 − cβ η
− β1 bβ 1 − dβξ −β1 − bβ 0 dβξ − bπ d sinπξB−β ≤ 0.
3.9
3.10
3.11
3.12
3.13
3.14
From 3.9–3.14, we obtain that cα t c−β t ≡ 0, t ∈ J. Combining 3.9 and 3.10, we
obtain that 1 and 2 hold.
It is easy to see that α ≤ α, β ≤ β on J. We show that α ≤ β on J. Let p α − β, then
pt αt − βt cα t c−β t. From 3.9–3.14, we have
− p t Mpt Npθt ≤ 0, t ∈ J,
t
/ tk ,
Δp tk ≥ Lk ptk , k 1, . . . , m,
p0 − ap 0 − cp η α0 − aα 0 − cα η − aπ c sin I πη Bα
− β0 aβ 0 cβ η − aπ c sin πη B−β ≤ 0,
p1 bp 1 − dpξ α1 bα 1 − dαξ − bπ d sinπξBα
− β1 − bβ 1 dβ η − bπ d sinπξB−β ≤ 0,
Δp tk Δα tk − Δβ tk Δca tk Δc−β
tk ≥ Lk αtk − βtk Lk cα tk c−β tk Lk ptk .
By Theorem 2.1, we have that p ≤ 0 on J, that is, α ≤ β on J. So 3 holds.
3.15
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Step 2. Consider the boundary value problem
−u t Mut Nuθt σt,
Δu tk Lk utk ek ,
u0 − au 0 λ,
t ∈ J, t / tk ,
3.16
k 1, . . . , m,
u1 bu 1 δ,
where λ ∈ R, δ ∈ R. We show that the boundary value problem 3.16 has one unique
solution ut, λ, δ.
It is easy to check that the boundary value problem 3.16 is equivalent to the integral
equation:
δt 1 − tλ bλ aδ
ut ab1
1
Gt, sσs − Mus − Nuθsds
0
m
1
t − tk Lk utk ek −
t b 1 − tk bLk utk ek ,
ab1
0<tk <t
k1
3.17
where
⎧
⎨a t1 b − s, 0 ≤ t ≤ s ≤ 1,
1
Gt, s a b 1 ⎩a s1 b − t, 0 ≤ s ≤ t ≤ 1.
3.18
It is easy to see that P CJ, R with norm u
maxt∈J |ut| is a Banach space. Define a
mapping Φ : P CJ, R → P CJ, R by
δt 1 − tλ bλ aδ
Φut ab1
1
Gt, sσs − Mus − Nuθsds
0
t − tk Lk utk ek −
0<tk <t
m
1
t b 1 − tk bLk utk ek .
ab1
k1
3.19
For any x, y ∈ P CJ, R, we have
Φxt − Φy t
≤
1
Gt, s M ys − xs N yθs − xθs ds 0
≤
1
0
Gt, sds
x − y
M N 1 b2
1
ab1
1 b2
1
ab1
m
Lk x − y
.
k1
m
Lk x − y
k1
3.20
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11
Since
1
max
t∈J
Gt, sds 0
1
1 2b 2
a1 2b
,
2a b 1 8 a b 1
3.21
the inequality 3.2 implies that Φ : P CJ → P CJ is a contraction mapping. Thus there
exists a unique u ∈ P CJ such that Φu u. The boundary value problem 3.16 has a unique
solution.
Step 3. We show that for any t ∈ J, the unique solution ut, λ, δ of the boundary value
problem 3.16 is continuous in λ and δ. Let ut, λi , δi , i 1, 2, be the solution of
−u t Mut Nuθt σt,
Δu tk Lk utk ek ,
u0 − au 0 λi ,
t ∈ J, t /
tk ,
3.22
k 1, . . . , m,
u1 bu 1 δi ,
i 1, 2.
Then
δi t 1 − tλi bλi aδi
ut, λi , δi ab1
m
Gt, sσs − Mus, λi , δi − Nuθs, λi , δi ds
0
t − tk Lk utk ek −
0<tk <t
×
1
1
t b
ab1
1 − tk bLk utk ek ,
i 1, 2.
k1
3.23
From 3.23, we have that
ut, λ1 , δ1 − ut, λ2 , δ2 ≤ |λ1 − λ2 | |δ1 − δ2 |
M N
ut, λ1 , δ1 − ut, λ2 , δ2 max
1
t∈J
1 b2
ut, λ1 , δ1 − ut, λ2 , δ2 1 ab1
Gt, sds
0
m
Lk x − y
k1
≤ |λ1 − λ2 | |δ1 − δ2 | μ
ut, λ1 , δ1 − ut, λ2 , δ2 .
3.24
Hence
ut, λ1 , δ1 − ut, λ2 , δ2 0 ≤
1
|λ1 − λ2 | |δ1 − δ2 |.
1−μ
3.25
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Step 4. We show that
αt ≤ ut, λ, δ ≤ βt
3.26
for any t ∈ J, λ ∈ cαη, cβη, and δ ∈ dαξ, dβξ, where ut, λ, δ is unique solution
of the boundary value problem 3.16.
Let mt αt − ut, λ, δ. From 3.9, 3.11, 3.12, and 3.16, we have that m0 −
am 0 ≤ cmη, m1 bm 1 ≤ dmξ, and
− m t Mmt Nmθt
−α t Mαt Nαθt u t, λ − Mut, λ, δ − Nuθt, λ, δ ≤ σt − σt ≤ 0,
Δm tk ≥ Lk utk .
3.27
By Theorem 2.1, we obtain that m ≤ 0 on J, that is, αt ≤ ut, λ, δ on J. Similarly, ut, λ, δ ≤
βt on J.
Step 5. Let D cαη, cβη × dαξ, dβξ. Define a mapping F : D → R2 by
Fλ, δ u η, λ, δ , uξ, λ, δ ,
3.28
where ut, λ, δ is unique solution of the boundary value problem 3.16. From Step 4, we
have
FD ⊂ D.
3.29
Since D is a compact convex set and F is continuous, it follows by Schauder’s fixed point
theorem that F has a fixed point λ0 , δ0 ∈ D such that
u η, λ0 , δ0 λ0 ,
3.30
uξ, λ0 , δ0 δ0 .
Obviously, ut, λ0 , δ0 is unique solution of the boundary value problem 3.1. This completes
the proof.
4. Main Results
Let M ∈ R, N ∈ R. We first give the following definition.
Definition 4.1. A function α ∈ E is called a lower solution of the boundary value problem 1.2
if
−α t M rcα t Ncα θt ≤ ft, αt, αθt,
Δα tk ≥ Ik αtk Lk cα tk ,
k 1, . . . , m.
t ∈ J, t /
tk ,
4.1
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13
Definition 4.2. A function β ∈ E is called an upper solution of the boundary value problem
1.2 if
−β t − M rc−β t − Nc−β θt ≥ f t, βt, βθt
Δβ tk ≤ Ik βtk − Lk c−β tk ,
t ∈ J, t ∈ J, t /
tk ,
k 1, . . . , m.
4.2
Our main result is the following theorem.
Theorem 4.3. Assume that H holds. If the following conditions are satisfied:
H1 α, β are lower and upper solutions for boundary value problem (1.2) respectively, and
αt ≤ βt on J,
H2 the constants M, N in definition of upper and lower solutions satisfy 2.3, 3.2, and
f t, x, y − f t, x, y ≥ −Mx − x − N y − y ,
Ik x − Ik y ≥ Lk x − y , x ≤ y,
4.3
for αt ≤ x ≤ x ≤ βt, αθt ≤ y ≤ y ≤ βθt, t ∈ J.
Then, there exist monotone sequences {αn }, {βn } with α0 α, β0 β such that
limn → ∞ αn t ρt, limn → ∞ βn t t uniformly on J, and ρ, are the minimal and the maximal
solutions of (1.2), respectively, such that
α0 ≤ α1 ≤ α2 ≤ · · · αn ≤ ρ ≤ x ≤ ≤ βn ≤ · · · ≤ β2 ≤ β1 ≤ β0
4.4
on J, where x is any solution of 1.1 such that αt ≤ xt ≤ βt on J.
Proof. Let α, β {u ∈ E : αt ≤ ut ≤ βt, t ∈ J}. For any γ ∈ α, β, we consider the
boundary value problem
−u t Mut Nuθt f t, γt, γθt Mγt Nγθt,
Δu tk Ik γtk − Lk utk − γtk , k 1, . . . , m.
u1 bu 1 duξ.
u0 − ax 0 cu η ,
t ∈ J,
4.5
Since α is a lower solution of 1.2, from H2 , we have that
− α t Mαt Nαθt
≤ ft, αt, αθt Mαt Nαθt − M rcα t − Ncα θt
≤ f t, γt, γθt Mγt Nγθt − M rcα t − Ncα θt,
Δα tk ≥ Ik αtk Lk cα tk ≥ Ik γtk Lk αtk − Lk γtk Lk cα tk .
4.6
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Similarly, we have that
− β t Mβt Nβθt
≥ f t, γt, γθt Mγt Nγθt M rc−β t Nc−β θt,
Δβ tk ≤ Ik βtk − Lk c−β tk ≤ Ik γtk Lk βtk − Lk γtk − Lk c−β tk .
4.7
By Theorem 3.1, the boundary value problem 4.5 has a unique solution u ∈ α, β.
We define an operator Ψ by u Ψγ, then Ψ is an operator from α, β to α, β.
We will show that
a α ≤ Ψα,
Ψβ ≤ β,
b Ψ is nondecreasing in α, β.
From Ψα ∈ α, β and Ψβ ∈ α, β, we have that a holds. To prove b, we show that
Ψν1 ≤ Ψν2 if α ≤ ν1 ≤ ν2 ≤ β.
Let ν1∗ Ψν1 , ν2∗ Ψν2 ,and p ν1∗ − ν2∗ , then by H2 and boundary conditions, we
have that
− p t Mpt Npθt
ft, ν1 t, ν1 θt Mν1 t Nν1 θt
− ft, ν2 t, ν2 θt − Mν2 t − Nν2 θt ≤ 0,
Δp tk ≥ Lk ptk ,
p0 − ap 0 cp η ,
4.8
p1 pu 1 dpξ.
By Theorem 2.1, pt ≤ 0 on J, which implies that Ψν1 ≤ Ψν2 .
Define the sequences {αn }, {βn } with α0 α, β0 β such that αn1 Ψαn , βn1 Ψβn
for n 0, 1, 2, . . . From a and b, we have
α0 ≤ α1 ≤ α2 ≤ · · · ≤ αn ≤ βn ≤ · · · ≤ β2 ≤ β1 ≤ β0
4.9
on t ∈ J, and each αn , βn ∈ E satisfies
−αn t Mαn t Nαn θt ft, αn−1 t, αn−1 θt Mαn−1 t Nαn−1 θt, t ∈ J, t /
tk ,
Δαn tk Ik αn−1 tk Lk αn tk − αn−1 tk , k 1, 2, . . . , m,
αn 1 bαn 1 dαn ξ,
αn 0 − aαn 0 cαn η ,
−βn t Mβn t Nβn θt f t, βn−1 t, βn−1 θt Mβn−1 t Nβn−1 θt, t ∈ J, t / tk ,
Δβn tk Ik βn−1 tk Lk βn tk − βn−1 tk , k 1, 2, . . . , m,
βn 1 bβn 1 dβn ξ.
βn 0 − aβn 0 cβn η ,
4.10
Mathematical Problems in Engineering
15
Therefore, there exist ρ, such that such that limn → ∞ αn t ρt, limn → ∞ βn t t
uniformly on J. Clearly, ρ, are solutions of 1.1.
Finally, we prove that if x ∈ α0 , β0 is any solution of 1.1, then ρt ≤ xt ≤ t
on J. To this end, we assume, without loss of generality, that αn t ≤ xt ≤ βn t for some n.
Since α0 t ≤ xt ≤ β0 t, from property b, we can obtain
αn1 t ≤ xt ≤ βn1 t,
t ∈ J.
4.11
Hence, we can conclude that
αn t ≤ xt ≤ βn t,
∀n.
4.12
t ∈ J.
4.13
Passing to the limit as n → ∞, we obtain
ρt ≤ xt ≤ t,
This completes the proof.
Acknowledgments
This work is supported by the NNSF of China 10571050;10871062 and Hunan Provincial
Natural Science Foundation of China NO:09JJ3010, and Science Research Fund of Hunan
provincial Education Department No: 06C052 .
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