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Abstract and Applied Analysis
Volume 2011 (2011), Article ID 969276, 29 pages
http://dx.doi.org/10.1155/2011/969276
Research Article

Neutral Operator and Neutral Differential Equation

1Department of Mathematics, Zhengzhou University, Zhengzhou 450001, China
2Department of Mathematics, Dresden University of Technology, 01062 Dresden, Germany

Received 22 March 2011; Accepted 15 June 2011

Academic Editor: Allan C. Peterson

Copyright © 2011 Jingli Ren 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.

Abstract

In this paper, we discuss the properties of the neutral operator , and by applying coincidence degree theory and fixed point index theory, we obtain sufficient conditions for the existence, multiplicity, and nonexistence of (positive) periodic solutions to two kinds of second-order differential equations with the prescribed neutral operator.

1. Introduction

In [1], Zhang discussed the properties of the neutral operator , which became an effective tool for the research on differential equations with this prescribed neutral operator, see, for example, [25]. Lu and Ge [6] investigated an extension of , namely, the neutral operator and obtained the existence of periodic solutions for a corresponding neutral differential equation.

In this paper, we consider the neutral operator , where is constant and , , and is an -periodic function for some . Although is a natural generalization of the operator , the class of neutral differential equation with typically possesses a more complicated nonlinearity than neutral differential equation with or . For example, the neutral operators and are homogeneous in the following sense for , whereas the neutral operator in general is inhomogeneous. As a consequence many of the new results for differential equations with the neutral operator will not be a direct extension of known theorems for neutral differential equations.

The paper is organized as follows: in Section 2, we first analyze qualitative properties of the neutral operator which will be helpful for further studies of differential equations with this neutral operator; in Section 3, by Mawhin's continuation theorem, we obtain the existence of periodic solutions for a second-order Rayleigh-type neutral differential equation; in Section 4, by an application of the fixed point index theorem we obtain sufficient conditions for the existence, multiplicity, and nonexistence of positive periodic solutions to second-order neutral differential equation. Several examples are also given to illustrate our results. Our results improve and extend the results in [1, 2, 4, 7].

2. Analysis of the Generalized Neutral Operator

Let with norm . Then is a Banach space. A cone in is defined by , where is a fixed positive number with . Moreover, define operators by

Lemma 2.1. If , then the operator has a continuous inverse on , satisfying
(1)(2). (3).

Proof. We have the following cases
Case 1 (). Let and , . Therefore, Since , we get from that has a continuous inverse with where . Then and consequently Moreover,
Case 2 (). Let By definition of the linear operator , we have where is defined as in Case 1. Summing over yields Since , we obtain that the operator has a bounded inverse , and we get On the other hand, from , we have that is, Let be arbitrary. We are looking for such that that is, Therefore, and hence proving that exists and satisfies Statements (1) and (2) are proved. From the above proof, (3) can easily be deduced.

Lemma 2.2. If and , one has for that

Proof. Since and , by Lemma 2.1, we have for that

Lemma 2.3. If and then for one has

Proof. Since and , , by Lemma 2.1, we have for that

3. Periodic Solutions for Neutral Differential Equation

In this section, we consider the second-order neutral differential equation where and ; and are continuous functions defined on and periodic in with , , , or for all .

We first recall Mawhin's continuation theorem which our study is based upon. Let and be real Banach spaces and a Fredholm operator with index zero, where denotes the domain of . This means that is closed in and . Consider supplementary subspaces , , of , respectively, such that , , and let and denote the natural projections. Clearly, , thus the restriction is invertible. Let denote the inverse of .

Let be an open bounded subset of with . A map is said to be -compact in if is bounded and the operator is compact.

Lemma 3.1 (Gaines and Mawhin [8]). Suppose that and are two Banach spaces and is a Fredholm operator with index zero. Furthermore, is an open bounded set, and is -compact on . Assume that the following conditions hold:
(1);(2);(3), where is an isomorphism.
Then the equation has a solution in .

In order to use Mawhin's continuation theorem to study the existence of -periodic solutions for (3.1), we rewrite (3.1) in the following form: Clearly, if is an -periodic solution to (3.2), then must be an -periodic solution to (3.1). Thus, the problem of finding an -periodic solution for (3.1) reduces to finding one for (3.2).

Recall that with norm . Define with norm . Clearly, and are Banach spaces. Moreover, define by and by Then (3.2) can be converted to the abstract equation . From the definition of , one can easily see that So is a Fredholm operator with index zero. Let and be defined by then =, =. Setting and Im denotes the inverse of , then From (3.5) and (3.8), it is clear that and are continuous and is bounded, and then is compact for any open bounded which means is -compact on .

Now we give our main results on periodic solutions for (3.1).

Theorem 3.2. Suppose there exist positive constants with such that:
(H1), for ;
(H2) sgn, for ;
(H3), for and .
Then (3.1) has at least one solution with period if , where .

Proof. By construction (3.2) has an -periodic solution if and only if the following operator equation has an -periodic solution. From (3.8), we see that is -compact on , where is any open, bounded subset of . For define Then satisfies We first claim that there is a constant such that In view of , we know that there exist two constants such that . From the first equation of (3.11), we have , so Let be, respectively, a global maximum and minimum point of . Clearly, we have Since or , w.l.o.g., suppose , for . Then From we see that Similarly, we have and again by ,
Case 1. If , define , obviously .
Case 2. If , from (3.18) and the fact that is a continuous function in , there exists a constant between and such that . This proves (3.12).
Choose an integer and a constant such that , then . Hence Substituting into the second equation of (3.11) yields that is, Integrating both sides of (3.21) over , we have On the other hand, multiplying both sides of (3.21) by and integrating over , we get Using , we have Besides, we can assert that there exists some positive constant such that In fact, in view of condition and (3.22) we have Define With these sets we get which yields That is, where , proving (3.25).
Substituting (3.25) into (3.24) and recalling (3.19), we get where . Since , we have By applying Lemma 2.1, we have where . Since , then , so we get Applying the inequality for , , it follows from (3.31) and (3.34) that Since , it is easy to see that there exists a constant (independent of ) such that It follows from (3.19) that
By the first equation of (3.11) we have , which implies that there is a constant such that , hence . By the second equation of (3.11) we obtain So, from and (3.25), we have Let , then If , then or . But if , we know that is, . From assumption , we know , which yields a contradiction, one can argue similarly if . We also have , that is, , so conditions (1) and (2) of Lemma 3.1 are both satisfied. Define the isomorphism as follows: Let , then, , We have . So, we can get From , it is obvious that , . Hence So condition (3) of Lemma 3.1 is satisfied. By applying Lemma 3.1, we conclude that equation has a solution on , that is, (3.1) has an -periodic solution .

By using a similar argument, we can obtain the following theorem.

Theorem 3.3. Suppose there exist positive constants with such that:
(H1), for ;(H2) sgn, for ,(H3), for and ,
then (3.1) has at least one solution with period if .

Remark 3.4. If and , the problem of existence of -periodic solutions to (3.1) can be converted to the existence of -periodic solutions to where , and . Clearly, and , and (3.46) can be discussed by using Theorem 3.2 (or Theorem 3.3).

4. Positive Periodic Solutions for Neutral Equations

Consider the following second-order neutral functional differential equation: where is a positive parameter; , and for ; with , , , , , and are -periodic functions.

Define the Banach space as in Section 2, and let . Denote It is easy to see that .

Now we consider (4.1). First let and denote It is clear that . We will show that (4.1) has or positive -periodic solutions for sufficiently large or small , respectively.

In the following we discuss (4.1) in two cases, namely, the case where and (note that implies ; implies ) and the case where and (note that implies ; implies ). Obviously, we have which makes Lemma 2.1 applicable for both cases and also Lemmas 2.2 or 2.3, respectively.

Let denote the cone in as defined in Section 2, where is just as defined above. We also use and .

Let , then from Lemma 2.1 we have . Hence (4.1) can be transformed into which can be further rewritten as where .

Now we discuss the two cases separately.

4.1. Case  I

Assume and .

Lemma 4.1 (see [7]). The equation has a unique -periodic solution where

Lemma 4.2 (see [7]). One has. Furthermore, if , then for all and .

Now we consider and define operators by Clearly are completely continuous for and .

By Lemma 4.1, the solution of (4.10) can be written in the form In view of and , we have and hence Define an operator by Obviously, for any , if , is the unique positive -periodic solution of (4.10).

Lemma 4.3. is completely continuous and

Proof. By the Neumann expansion of , we have Since and are completely continuous, so is . Moreover, by (4.17), and recalling that , we get

Define an operator by

Lemma 4.4. One has .

Proof. From the definition of , it is easy to verify that . For , we have from Lemma 4.3 that On the other hand,