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Abstract and Applied Analysis
Volume 2013 (2013), Article ID 504573, 10 pages
Positive Periodic Solution of Second-Order Coupled Systems with Singularities
School of Mathematical Sciences, Capital Normal University, Beijing 100048, China
Received 1 March 2013; Revised 11 June 2013; Accepted 13 June 2013
Academic Editor: Chuanzhi Bai
Copyright © 2013 Tiantian Ma. 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.
This paper establishes the existence of periodic solution for a kind of second-order singular nonautonomous coupled systems. Our approach is based on fixed point theorem in cones. Examples are given to illustrate the main result.
We are concerned with the existence of positive -periodic solution for the second-order nonautonomous singular coupled systems: where , , , are Lebesgue integrable, may be singular at , and can have finitely many singularities.
Singular differential equations or systems arise from many branches of applied mathematics and physics such as gas dynamics, Newtonian fluid mechanics, and nuclear physics, which have been widely studied by many authors (see [1–7] and references therein). Some classical-tools have been used to study the positive solutions for two point nonperiodic boundary value problems of coupled systems [8, 9]. However, there are few works on periodic solutions of second order nonautonomous singular coupled systems of type (1).
In the recent papers [10, 11], the periodic solutions of singular coupled systems were proved by using some fixed point theorems in cones for completely continuous operators, where , . When the Green's function (), associated with the periodic boundary problem is nonnegative for all and () satisfies weak singularities where , , , , and , are strictly positive on some positive measure subsets of , some sufficient conditions for the existence of periodic solutions of (2) were obtained in [10, 11].
Motivated by the papers [9–11], we consider the existence of positive -periodic solution of (1). Owing to the disappearing of the terms () in (2), the methods in [9, 10] are no longer valid. In present paper, we will deal with the periodic solutions of (1) under new conditions. Let be a constant satisfying . Denote by the Green function of which can be expressed by By a direct computation, we can get Assume () satisfies the following conditions.(H1) For and there exist constants , such that, for any constants , , (H2) and where , , .
Theorem 3. Assume that (H1), (H2) hold. Then (1) has at least one positive -periodic solution.
Lemma 4 (see ). Let be a Banach space, a cone in , , two nonempty bounded open sets in , . is a completely continuous operator. If(i), , ,(ii), , , ,then has a fixed point in .
Lemma 5. If satisfy (H1), then, for , are increasing on and, for any , uniformly with respect to .
Proof. We only deal with . Without loss of generality, let . If , we get . If , let , then . From (8), we get
which means is increasing on .
Assume . It follows from (11) that . Thus From (H1), for any , we obtain Therefore uniformly with respect to .
Let . We know is a Banach space with the norm . Define the sets It is easy to check that , are cones in and . Throughout this paper, we consider the space . It is easy to see is a Banach space with the norm,
We can get the conclusion that , are cones in and .
For any , define the function Then the solution of periodic boundary value problem can be expressed by , and the solution of periodic boundary value problem can be expressed by . Obviously, , . Then From (11), (12), and Lemma 5, we have Then, for any fixed , it follows from that
Thus, we can define the operator , by for . Then, we have the following lemma.
Lemma 6. Assuming that (H1), (H2) hold, then has a fixed point if and only if has one positive -periodic solution.
Lemma 7. Assuming that (H1), (H2) hold, then and is completely continuous.
Proof. For , we have
Then, we can get
which means .
Let be any bounded set. Then there exists a constant such that, for any , From (27), we have Let Thus which implies that is bounded.
Next we prove that is equicontinuous. For any , , we know From (25) and (H2), we have
Using the same method, we can obtain . Therefore, is equicontinuous. According to Ascoli-Arzela theorem, is a relatively compact set.
Next, we prove that is continuous. Suppose , , , that is, , , . We know that there exists a constant such that
We shall prove , , that is,
We first deal with , . Otherwise, there exist , such that . Without loss of generality, we can assume . We know
Next, we show , . In fact, Let Since and is continuous, we know , . From (24), we know It can be inferred from (8) that
Set . Thus, we get From (H2), we know . Using Lebesgue-dominated convergence theorem, we get
From (40) and (47), we obtain which is a contradiction. Thus, we know , . Using the same method, we can obtain , . Then which means is continuous. Therefore is a completely continuous operator.
3. Proof of Theorem 3
We proceed to prove Theorem 3 in two steps.
(1) Let . We can get For , we have the following two cases.
Case I. One has . Under this condition, we can get
Otherwise, there exist , such that . As
then . By a direct computation, we know satisfies
Then, we get
Since , integrating both sides of (54) on , we get Then That is, which contradicts with (H2).
Case II. One has . Under this condition, we can get
Otherwise, there exist , such that . As
then . By a direct computation, we know satisfies
For , using the same method as condition I, we obtain which is also a contradiction.
(2) Choose an interval satisfying . Set . From Lemma 5, there exists , such that
Let . Define . We can get
For , we have the following two cases.
Case I. One has . Under this condition, we know Thus . Furthermore, we obtain from (62) that On the other hand, for , we know . From the choice of , we get .
Case II. One has . Under this condition, we get
Thus . From (62), we get
For , , from the choice of , we know .
Furthermore, we can obtain It implies .
From Lemma 4, we know has a fixed point in . For , we have the following three cases.
Case 1. One has
Case 2. One has
Case 3. One has
Next, we show Cases 1 and 2 are impossible. In Case 1, we have It follows that . By a direct computation, we know satisfies Then, we get Since , integrating both sides of (74) on , we get Then That is, which contradicts with (H2). Using the same method, we can prove that Case 2 is also impossible. Therefore, Case 3 is satisfied and has a fixed point in satisfying Since from Lemma 6, we know satisfy
Let . For we obtain
This means is one positive -periodic solution of (1).
4. Applications of Theorem 3
Finally, we give some examples as the applications of Theorem 3:
Choosing , , , , , , and , , then (H1) is satisfied. Notice (H2) also holds, since
Existence of the positive -periodic solutions is guaranteed from Theorem 3. We can also consider the following examples and the same result can be obtained:
The author is grateful to the referees for valuable comments and useful remarks on the paper. Research supported by China Postdoctoral Science Foundation (2012M510341), Beijing Natural Science Foundation (1112006) and the Grant of Beijing Education Committee Key Project (KZ20130028031).
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