Existence and Lyapunov Stability of Periodic Solutions for Generalized Higher-Order Neutral Differential Equations

Abstract

Existence and Lyapunov stability of periodic solutions for a generalized higher-order 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

(1.1)

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:

(1.2)

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 lower-order neutral functional differential equations, while studies on higher-order neutral functional differential equations are rather infrequent, especially on higher-order -Laplacian neutral functional differential equations. In this paper, we consider the following generalized higher-order neutral functional differential equation:

(1.3)

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 first-order 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

(2.1)

where is a fixed real number with and

(2.2)

For the sake of convenience, throughout this paper we denote by a positive real number, and for any continuous function , we write

(2.3)

Let be the operator on given by

(2.4)

Lemma 2.3.

The operator has a continuous inverse on satisfying the following:

(2.5)

(2.6)

(2.7)

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

(2.8)

Clearly, and . Note also that . Therefore, has a continuous inverse with ; here . Hence,

(2.9)

and so

(2.10)

Case 2 ().

Define operators

(2.11)

From the definition of the linear operator , we have

(2.12)

Since , the operator has a bounded inverse with

(2.13)

and so, for any ,

(2.14)

On the other hand, from , we have

(2.15)

That is,

(2.16)

Now, for any , if satisfies

(2.17)

then we have

(2.18)

or

(2.19)

So, we have

(2.20)

So, exists and satisfies

(2.21)

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

(2.22)

Lemma 2.5 (see [4]).

If , then

(2.23)

Now we consider (2.22). Define the conjugate index by . Introducing new variables

(2.24)

Using the fact that and by Lemma 2.3, (1.3) can be rewritten as

(2.25)

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

(2.26)

with norm . Obviously, and are Banach spaces. Define

(2.27)

by

(2.28)

Moreover, define

(2.29)

by

(2.30)

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

(2.31)

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

(2.32)

where

(2.33)

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

(2.34)

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

(2.35)

By Lemma 2.5 and Lemma 2.2, we have

(2.36)

Combining (2.35) and (2.36), we get

(2.37)

Similarly, we get

(2.38)

This completes the proof of Lemma 2.6.

Remark 2.7.

In particular, if we take , then and

(2.39)

In this case, (2.34) is transformed into

(2.40)

3. Main Results

For the sake of convenience, we list the following assumptions which will be used repeatedly in the sequel.

(H₁)There exists a constant such that

(3.1)

H₂There exists a constant such that

(3.2)

H₃ There exist nonnegative constants such that

(3.3)

H₄ There exist nonnegative constants such that

(3.4)

for all .

Theorem 3.1.

If and hold, then (1.3) has at least one nonconstant -periodic solution.

Proof.

Consider the equation

(3.5)

Let . If , then

(3.6)

We first claim that there exists a constant such that

(3.7)

Integrating the last equation of (3.6) over , we have

(3.8)

By the continuity of , there exists such that

(3.9)

From assumption , we get (3.7). As a consequence, we have

(3.10)

On the other hand, multiplying both sides of the last equation of (3.6) by and integrating over , using assumption we have

(3.11)

It is easy to see that there exists a constant (independent of ) such that

(3.12)

From , there exists a point such that . By Hölder's inequality, we have

(3.13)

From , there exists a point such that , and we have

(3.14)

Continuing this way for , we get

(3.15)

Hence,

(3.16)

Meanwhile, from (3.10), we get

(3.17)

Let . Then, obviously , and .

Let . If , then , which means that and . We see that

(3.18)

So,

(3.19)

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:

(3.20)

Let , . Then, for all ,

(3.21)

From , it is obvious that for all . Therefore,

(3.22)

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

(3.23)

(3.24)

We claim that is bounded.

Multiplying both sides of (3.23) by and integrating over , by using assumption we have

(3.25)

Applying Hölder's inequality, we have

(3.26)

Applying Lemma 2.6 and (3.26), we have

(3.27)

Case 1.

If and , then it is easy to see that there exists a constant (independent of ) such that

(3.28)

Case 2.

If , then it is easy to see that there exists a constant (independent of ) such that

(3.29)

From , there exists a point such that . By Hölder's inequality, we have

(3.30)

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

(3.31)

where and , then the results of Theorems 3.1 and 3.2 still hold.

Remark 3.4.

If , then (1.3) is transformed into

(3.32)

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

(3.33)

Then, system (3.32) is transformed into

(3.34)

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

(3.35)

and so

(3.36)

Let . Then,

(3.37)

Take , and define a function by

(3.38)

Let . It is obvious that and . From and Lemma 2.3, we get

(3.39)

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

(3.40)

Here is a constant with . Comparing with (1.3), we have and

(3.41)

Observe that has period and satisfies

(3.42)

Pick . Then,

(3.43)

for all with . Hence, holds. On the other hand, since

(3.44)

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

(3.45)

So by (2) of Theorem 3.2, (3.40) has at least one nonconstant -periodic solution.