Existence and Lyapunov stability of periodic solutions for a generalized higherorder neutral differential equation are established.
1. Introduction
In recent years, there is a good amount of work on periodic solutions for neutral differential equations (see [1–11] and the references cited therein). For example, the following neutral differential equations
have been studied in [1, 3, 8], respectively, and existence criteria of periodic solutions were established for these equations. Afterwards, along with intensive research on the Laplacian, some authors [4, 11] start to consider the following Laplacian neutral functional differential equations:
and by using topological degree theory and some analysis skills, existence results of periodic solutions for (1.2) have been presented.
In general, most of the existing results are concentrated on lowerorder neutral functional differential equations, while studies on higherorder neutral functional differential equations are rather infrequent, especially on higherorder Laplacian neutral functional differential equations. In this paper, we consider the following generalized higherorder neutral functional differential equation:
where is given by with being a constant, is a continuous function defined on and is periodic with respect to with period , that is, for all , and , are constants.
Since the neutral operator is divided into two cases and , it is natural to study the neutral differential equation separately according to these two cases. The case has been studied in [5]. Now we consider (1.3) for the case . So throughout this paper, we always assume that , and the paper is organized as follows. We first transform (1.3) into a system of firstorder differential equations, and then by applying Mawhin's continuation theory and some new inequalities, we obtain sufficient conditions for the existence of periodic solutions for (1.3). The Lyapunov stability of periodic solutions for the equation will then be established. Finally, an example is given to illustrate our results.
2. Preparation
First, we recall two lemmas. Let and be real Banach spaces and let be a Fredholm operator with index zero; here denotes the domain of . This means that is closed in and . Consider supplementary subspaces , of , , respectively, such that , . Let and denote the natural projections. Clearly, and so 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 2.1 (see [12]).
Suppose that and are two Banach spaces, and suppose that is a Fredholm operator with index zero. Let be an open bounded set and let be compact on . Assume that the following conditions hold:
(1)
(2)
(3), where is an isomorphism.
Then, the equation has a solution in .
Lemma 2.2 (see [13]).
If and , then
where is a fixed real number with and
For the sake of convenience, throughout this paper we denote by a positive real number, and for any continuous function , we write
Let be the operator on given by
Lemma 2.3.
The operator has a continuous inverse on satisfying the following:
Remark 2.4.
This lemma is basically proved in [3, 10]. For the convenience of the readers, we present a detailed proof here as follows.
Proof.
We split it into the following two cases.
Case 1 ().
Define an operator by
Clearly, and . Note also that . Therefore, has a continuous inverse with ; here . Hence,
and so
Case 2 ().
Define operators
From the definition of the linear operator , we have
Since , the operator has a bounded inverse with
and so, for any ,
On the other hand, from , we have
That is,
Now, for any , if satisfies
then we have
or
So, we have
So, exists and satisfies
This proves (1) and (2) of Lemma 2.3. Finally, (3) is easily verified.
By Hale's terminology [14], a solution of (1.3) is that such that and (1.3) is satisfied on . In general, does not belong to But we can see easily from that a solution of (1.3) must belong to . Equation (1.3) is transformed into
Lemma 2.5 (see [4]).
If , then
Now we consider (2.22). Define the conjugate index by . Introducing new variables
Using the fact that and by Lemma 2.3, (1.3) can be rewritten as
It is clear that, if is a periodic solution to (2.25), then must be a periodic solution to (1.3). Thus, the problem of finding a periodic solution for (1.3) reduces to finding one for (2.25).
Define the linear spaces
with norm . Obviously, and are Banach spaces. Define
by
Moreover, define
by
Then, (2.25) can be rewritten as the abstract equation . From the definition of , one can easily see that and . So, is a Fredholm operator with index zero. Let and be defined by
It is easy to see that . Moreover, for all , if we write , we have and so . This is to say and So, is a Fredholm operator with index zero. Let denote the inverse of , then we have
where
From (2.30) and (2.33), it is clear that and are continuous, and is bounded, and so is compact for any open bounded . Hence, is compact on . For the function defined as (2.24), we have the following.
Lemma 2.6.
If and , then
where
Proof.
From , there is a point such that . Let . Then, . From , there is a point such that . Let . Then, Continuing this way, we get from a point such that . Let . Then, From , we have , so there is a point such that ; hence, we have . Let . Then, Continuing this way, we get from that there is a point such that . Let . Then, By Lemma 2.2, we have
By Lemma 2.5 and Lemma 2.2, we have
Combining (2.35) and (2.36), we get
Similarly, we get
This completes the proof of Lemma 2.6.
Remark 2.7.
In particular, if we take , then and
In this case, (2.34) is transformed into
3. Main Results
For the sake of convenience, we list the following assumptions which will be used repeatedly in the sequel.
There exists a constant such that
There exists a constant such that
There exist nonnegative constants such that
There exist nonnegative constants such that
for all .
Theorem 3.1.
If and hold, then (1.3) has at least one nonconstant periodic solution.
Proof.
Consider the equation
Let . If , then
We first claim that there exists a constant such that
Integrating the last equation of (3.6) over , we have
By the continuity of , there exists such that
From assumption , we get (3.7). As a consequence, we have
On the other hand, multiplying both sides of the last equation of (3.6) by and integrating over , using assumption we have
It is easy to see that there exists a constant (independent of ) such that
From , there exists a point such that . By Hölder's inequality, we have
From , there exists a point such that , and we have
Continuing this way for , we get
Hence,
Meanwhile, from (3.10), we get
Let . Then, obviously , and .
Let . If , then , which means that and . We see that
So,
Now take . By the analysis above, it is easy to see that , , and conditions (1) and (2) of Lemma 2.1 are satisfied.
Next we show that condition (3) of Lemma 2.1 is also satisfied. Define an isomorphism as follows:
Let , . Then, for all ,
From , it is obvious that for all . Therefore,
which means that condition (3) of Lemma 2.1 is also satisfied. By applying Lemma 2.1, we conclude that equation has a solution on ; that is, (1.3) has a periodic solution with .
Finally, observe that is not constant. For, if (constant), then from (1.3) we have , which contradicts the assumption that . The proof is complete.
Theorem 3.2.
If and hold, then (1.3) has at least one nonconstant periodic solution if one of the following conditions holds:
(1),
(2) and
Proof.
Let be defined as in Theorem 3.1. If then from the proof of Theorem 3.1 we have
We claim that is bounded.
Multiplying both sides of (3.23) by and integrating over , by using assumption we have
Applying Hölder's inequality, we have
Applying Lemma 2.6 and (3.26), we have
Case 1.
If and , then it is easy to see that there exists a constant (independent of ) such that
Case 2.
If , then it is easy to see that there exists a constant (independent of ) such that
From , there exists a point such that . By Hölder's inequality, we have
This proves the claim, and the rest of the proof of the theorem is identical to that of Theorem 3.1.
Remark 3.3.
If (1.3) takes the form
where and , then the results of Theorems 3.1 and 3.2 still hold.
Remark 3.4.
If , then (1.3) is transformed into
and the results of Theorems 3.1 and 3.2 still hold.
Next, we study the Lyapunov stability of the periodic solutions of (3.32).
Theorem 3.5.
Assume that holds. Then every periodic solution of (3.32) is Lyapunov stable.
Proof.
Let
Then, system (3.32) is transformed into
Suppose now that is a periodic solution of (3.34). Let be any arbitrary solution of (3.34). For any , write . Then, it follows from (3.34) that
and so
Let . Then,
Take , and define a function by
Let . It is obvious that and . From and Lemma 2.3, we get
Hence, is a Lyapunov function for nonautonomous (3.32) (see [15, page 50]), and so the periodic solution of (3.32) is Lyapunov stable.
Finally, we present an example to illustrate our result.
Example 3.6.
Consider the order delay differential equation
Here is a constant with . Comparing with (1.3), we have and
Observe that has period and satisfies
Pick . Then,
for all with . Hence, holds. On the other hand, since
assumption holds with .
Case 1.
If , then by (1) of Theorem 3.2, (3.40) has at least one nonconstant periodic solution.
Case 2.
If , then
So by (2) of Theorem 3.2, (3.40) has at least one nonconstant periodic solution.
Acknowledgments
This paper is partially supported by the National Natural Science Foundation of China (10971202), and the Research Grant Council of Hong Kong SAR, China (project no. HKU7016/07P).
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