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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
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
Definition 1.1.
Functions and are called lower solution and upper solution of the boundary value problem (1.1), respectively if
where
and
Remark 1.2.
Clearly, and . For example,
2. Comparison Results
We now present the main results of this section.
Lemma 2.1.
Assume that satisfies
then for all .
Proof.
Suppose, to the contrary, that for some . We consider the following two cases.
Case 1.
on . It is easy to obtain that on . Thus from . Consequently, we obtain that
which contradicts .
Case 2.
There exist such that and . Hence, two cases are possible.
Subcase 1.
. There exists a such that
Let Then
Integrating the above inequality from to , we obtain
and then integrate from to to obtain
that implies This is a contradiction.
Subcase 2.
. There exists a such that
If , then If or , then . So .
Let Then
When , same as Subcase 1, we obtain that
when , integrating the inequality (2.8) from to , we obtain
and then integrate from to to obtain
that implies This is a contradiction. The proof is complete.
Lemma 2.2.
Assume that satisfies
Then for all .
Proof.
Put
then and for all and
We have
Hence by Lemma 2.1, for all , which implies that for . This completes the proof.
Lemma 2.3.
Assume that satisfies
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
Then for all .
Proof.
Suppose that for some . Then from boundary conditions, we have that there exists a such that
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
Integrating the above inequality from to , we obtain
and then integrate from to to obtain
Hence, a contradiction. The proof is complete.
3. Linear Problem
In this section, we consider the boundary value problem
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
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
By Lemma 2.1, we have that on .
If , then . Thus
It is easy to see that By Lemma 2.2, we have that on . Analogously, on .
If , then . Thus
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
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
where
Define a mapping by
Obviously is a Banach space with norm . For any , we have
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
Then
From (3.12), we have that
Hence
Since , exists for any and . From (3.12) and (3.14), we have
Step 3.
We show that there exists one such that , where is the unique solution of the problem (3.6).
Put
then for any and
Using (3.18) and (3.19), one easily obtains that for any .
Put then Using Lemma 2.3, we easily obtain that on . Hence
Define a function
where is the unique solution of the problem (3.6). Since is continuous and
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
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
on , where is any solution of (1.1) such that on .
Proof.
Let . For any , we consider the equation
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
and By Lemma 2.1, , which implies .
Define the sequence with such that for From (a) and (b), we have
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
for all . Passing to the limit as , we obtain This completes the proof.
References
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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