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Periodic Boundary Value Problems for Second-Order Functional Differential Equations
Journal of Inequalities and Applications volume 2010, Article number: 598405 (2010)
Abstract
This note considers a periodic boundary value problem for a second-order functional differential equation. We extend the concept of lower and upper solutions and obtain the existence of extreme solutions by using upper and lower solution method.
1. Introduction
Upper and lower solution method plays an important role in studying boundary value problems for nonlinear differential equations; see [1] and the references therein. Recently, many authors are devoted to extend its applications to boundary value problems of functional differential equations [2–5]. Suppose is one upper solution or lower solution of periodic boundary value problems for second-order differential equation; the condition
is required. A neutral problem is that whether we can define upper and lower solution without assuming
. The aim of the present paper is to discuss the following second order functional differential equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ1_HTML.gif)
where
In this paper, we extended the concept of lower and upper solutions for (1.1). By using the method of upper and lower solutions and monotone iterative technique, we obtained the existence of extreme solutions for the boundary value problem (1.1).
Through this paper, we assume that and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ2_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ3_HTML.gif)
Definition 1.1.
Functions and
are called lower solution and upper solution of the boundary value problem (1.1), respectively if
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ4_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ5_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ6_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ7_HTML.gif)
and
Remark 1.2.
Clearly, and
. For example,
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ8_HTML.gif)
2. Comparison Results
We now present the main results of this section.
Lemma 2.1.
Assume that satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ9_HTML.gif)
then for all
.
Proof.
Suppose, to the contrary, that for some
. We consider the following two cases.
Case 1.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_IEq17_HTML.gif)
on . It is easy to obtain that
on
. Thus
from
. Consequently, we obtain that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ10_HTML.gif)
which contradicts .
Case 2.
There exist such that
and
. Hence, two cases are possible.
Subcase 1.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_IEq27_HTML.gif)
. There exists a such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ11_HTML.gif)
Let Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ12_HTML.gif)
Integrating the above inequality from to
, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ13_HTML.gif)
and then integrate from to
to obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ14_HTML.gif)
that implies This is a contradiction.
Subcase 2.
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_IEq35_HTML.gif)
. There exists a such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ15_HTML.gif)
If , then
If
or
, then
. So
.
Let Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ16_HTML.gif)
When , same as Subcase 1, we obtain that
when , integrating the inequality (2.8) from
to
, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ17_HTML.gif)
and then integrate from to
to obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ18_HTML.gif)
that implies This is a contradiction. The proof is complete.
Lemma 2.2.
Assume that satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ19_HTML.gif)
Then for all
.
Proof.
Put
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ20_HTML.gif)
then and
for all
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ21_HTML.gif)
We have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ22_HTML.gif)
Hence by Lemma 2.1, for all
, which implies that
for
. This completes the proof.
Lemma 2.3.
Assume that satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ23_HTML.gif)
Then for all
.
The proof of Lemma 2.3 is similar to that of Lemma 2.2, here we omit it.
Lemma 2.4.
Assume that satisfies
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ24_HTML.gif)
Then for all
.
Proof.
Suppose that for some
. Then from boundary conditions, we have that there exists a
such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ25_HTML.gif)
Suppose that for
. It is easy to see that
and
for
. From
and
for
, we obtain that
. Therefore,
. It follows that
, a contradiction.
Suppose that there exist such that
and
. Let
be such that
. From the first inequality of (2.16), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ26_HTML.gif)
Integrating the above inequality from to
, we obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ27_HTML.gif)
and then integrate from to
to obtain
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ28_HTML.gif)
Hence, a contradiction. The proof is complete.
3. Linear Problem
In this section, we consider the boundary value problem
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ29_HTML.gif)
where .
Theorem 3.1.
Assume that there exist which are lower and upper solutions of (3.1) and
on
. Then there exists one unique solution
to problem (3.1) and
on
.
Proof.
We first show that the solution of (3.1) is unique. Let and
be solutions of (3.1) and set
. Thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ30_HTML.gif)
By Lemma 2.1, we have that for
, that is,
on
. Similarly, one can obtain that
on
. Hence
.
Next, we prove that if is a solution of (3.1), then
.
Let . If
, then
on
. So we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ31_HTML.gif)
By Lemma 2.1, we have that on
.
If , then
. Thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ32_HTML.gif)
It is easy to see that By Lemma 2.2, we have that
on
. Analogously,
on
.
If , then
. Thus
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ33_HTML.gif)
It is easy to see that By Lemma 2.2, we have that
on
. Analogously,
on
.
Finally, we show that (3.1) has a solution by several steps.
Step 1.
Consider the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ34_HTML.gif)
where and
are defined in (3.1). For any
, we show that (3.6) has a unique solution
.
It is easy to check that (3.6) is equivalent to the integral equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ35_HTML.gif)
where
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ36_HTML.gif)
Define a mapping by
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ37_HTML.gif)
Obviously is a Banach space with norm
. For any
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ38_HTML.gif)
Noting that , condition (1.2) implies that
is a contraction mapping. There exists unique
such that
. Thus (3.6) has a unique solution
Moreover,
.
Step 2.
We show that for any ,
and
are continuous in
, where
is a unique solution of the problem (3.6). Let
be the solution of
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ39_HTML.gif)
Then
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ40_HTML.gif)
From (3.12), we have that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ41_HTML.gif)
Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ42_HTML.gif)
Since ,
exists for any
and
. From (3.12) and (3.14), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ43_HTML.gif)
Step 3.
We show that there exists one such that
, where
is the unique solution of the problem (3.6).
Put
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ44_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ45_HTML.gif)
then for any
and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ46_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ47_HTML.gif)
Using (3.18) and (3.19), one easily obtains that for any
.
Put then
Using Lemma 2.3, we easily obtain that
on
. Hence
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ48_HTML.gif)
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ49_HTML.gif)
Define a function
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ50_HTML.gif)
where is the unique solution of the problem (3.6). Since
is continuous and
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ51_HTML.gif)
there exists one such that
that is,
. Obviously,
is a unique solution of the problem (3.1). This completes the proof.
4. Main Result
Theorem 4.1.
Let the following conditions hold.
The functions ,
are lower and upper solutions of (1.1), respectively, and
on
.
The constants in definition of upper and lower solutions satisfy
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ52_HTML.gif)
for .
Then, there exist monotone sequences with
such that
uniformly on
, and
are the minimal and the maximal solutions of (1.1), respectively, such that
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ53_HTML.gif)
on , where
is any solution of (1.1) such that
on
.
Proof.
Let . For any
, we consider the equation
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ54_HTML.gif)
Theorem 3.1 implies that the problem (4.3) has a unique solution . We define an operator
by
, then
is an operator from
to
.
We shall show that
(a)
(b) is nondecreasing in
.
From and
, we have that (a) holds. To prove (b), we show that
if
.
Let and
then by
, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ55_HTML.gif)
and By Lemma 2.1,
, which implies
.
Define the sequence with
such that
for
From (a) and (b), we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ56_HTML.gif)
on . Therefore, there exist
such that
,
uniformly on
. Clearly,
are solutions of (1.1).
Finally, we prove that if is one solution of (1.1), then
on
. Since
and
, by property (b), we obtain that
for
. Using property (b) repeatedly, we have
![](http://media.springernature.com/full/springer-static/image/art%3A10.1155%2F2010%2F598405/MediaObjects/13660_2009_Article_2200_Equ57_HTML.gif)
for all . Passing to the limit as
, we obtain
This completes the proof.
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Acknowledgment
This work is supported by Scientific Research Fund of Hunan Provincial Education Department (09B033) and the NNSF of China (10871062).
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Fu, X., Wang, W. Periodic Boundary Value Problems for Second-Order Functional Differential Equations. J Inequal Appl 2010, 598405 (2010). https://doi.org/10.1155/2010/598405
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DOI: https://doi.org/10.1155/2010/598405