#### Abstract

Based on a continuation theorem of Mawhin, a unique periodic solution is found for a second-order nonlinear differential equation with piecewise constant argument.

#### 1. Introduction

Qualitative behaviors of first-order delay differential equations with piecewise constant arguments are the subject of many investigations (see, e.g., [1–19]), while those of higher-order equations are not.

However, there are reasons for studying higher-order equations with piecewise constant arguments. Indeed, as mentioned in [10], a potential application of these equations is in the stabilization of hybrid control systems with feedback delay, where a hybrid system is one with a continuous plant and with a discrete (sampled) controller. As an example, suppose that a moving particle with time variable mass is subjected to a restoring controller which acts at sampled time Then Newton's second law asserts that

Since this equation is “similar” to the harmonic oscillator equation

we expect that the well-known qualitative behavior of the later equation may also be found in the former equation, provided appropriate conditions on and are imposed.

In this paper we study a slightly more general second-order delay differential equation with piecewise constant argument:

where is a real continuous function defined on with positive *integer* period for and are continuous function defined on with period for and

By a solution of (1.3) we mean a function which is defined on and which satisfies the following conditions: (i) is continuous on (ii) is differentiable at each point with the possible exception of the points where one-sided derivatives exist, and (iii) substitution of into (1.3) leads to an identity on each interval with integral endpoints.

In this note, existence and uniqueness criteria for periodic solutions of (1.3) will be established. For this purpose, we will make use of a continuation theorem of Mawhin. Let and be two Banach spaces and is a linear mapping and a continuous mapping. The mapping will be called a Fredholm mapping of index zero if , and is closed in If is a Fredholm mapping of index zero, there exist continuous projectors and such that and It follows that has an inverse which will be denoted by . If is an open and bounded subset of , the mapping will be called -compact on if is bounded and is compact. Since is isomorphic to there exists an isomorphism .

Theorem 1 A (Mawhin's continuation theorem [18]). *Let be a Fredholm mapping of index zero, and let be -compact on . Suppose that*(i)*for each , *(ii)*for each and **Then the equation has at least one solution in .*

#### 2. Existence and Uniqueness Criteria

Our main results of this paper are as follows.

Theorem 2.1. *Suppose that there exist constants and such that*(i)* for and ,*(ii)* (or ).**If then (1.3) has an -periodic solution. Furthermore, the -periodic solution is unique if in addition one has the following.*(iii)* is strictly monotonous in and there exists nonnegative constant such that*

Theorem 2.2. *Suppose that there exist constants and such that*()* for and ,*()* (or ).**If then (1.3) has an -periodic solution. Furthermore, the -periodic solution is unique if in addition one has the following.*(iii)* is strictly monotonous in and there exists nonnegative constant such that (2.1) holds.*

We only give the proof of Theorem 2.1, as Theorem 2.2 can be proved similarly.

First we make the simple observation that is an -periodic solution of the following equation:

if, and only if, is an -periodic solution of (1.3). Next, let be the Banach space of all real -periodic continuously differentiable functions of the form which is defined on and endowed with the usual linear structure as well as the norm . Let be the Banach space of all real continuous functions of the form such that where and and endowed with the usual linear structure as well as the norm Let the zero element of and be denoted by and respectively.

Define the mappings and respectively, by

Let

Since and , is a well-defined operator from to Let us define and respectively, by

for and

for

Lemma 2.3. *Let the mapping be defined by (2.3). Then
*

*Proof. *It suffices to show that if is a real -periodic continuously differentiable function which satisfies
then is a constant function. To see this, note that for such a function
Hence by integrating both sides of the above equality from to we see that
Since is positive, continuous, and periodic,
Since is bounded, we may infer from (2.11) that . But then (2.9) implies for The proof is complete.

Lemma 2.4. *Let the mapping be defined by (2.3). Then
*

*Proof. *It suffices to show that for each that satisfies there is a such that
But this is relatively easy, since we may let
Then it may easily be checked that (2.14) holds. The proof is complete.

Lemma 2.5. *The mapping defined by (2.3) is a Fredholm mapping of index zero.*

Indeed, from Lemmas 2.3 and 2.4 and the definition of From (2.13), we see that is closed in Hence is a Fredholm mapping of index zero.

Lemma 2.6. *Let the mapping and be defined by (2.3), (2.6), and (2.7), respectively. Then and *

Indeed, from Lemmas 2.3 and 2.4 and defining conditions (2.6) and (2.7), it is easy to see that and

Lemma 2.7. *Let and be defined by (2.3) and (2.4), respectively. Suppose that is an open and bounded subset of Then is -compact on .*

*Proof. *It is easy to see that for any
so that
These lead us to
where is defined by (2.15). By (2.18), we see that is bounded. Noting that (2.7) holds and is a completely continuous mapping, by means of the Arzela-Ascoli theorem we know that is relatively compact. Thus is -compact on . The proof is complete.

Lemma 2.8. *Suppose that is a real, bounded and continuous function on and exists. Then there is a point such that
*

The above result is only a slight extension of the integral mean value theorem and is easily proved.

Lemma 2.9. *Suppose that condition (i) in Theorem 2.1 holds. Suppose further that satisfies
**
Then there is such that *

*Proof. *From (2.22) and Lemma 2.8**, **we have for such that
In case from the condition (i) in Theorem 2.1 and (2.23), we know that Suppose Our assertion is true if one of has absolute value less than or equal to Otherwise, there should be and among and such that and Since is continuous, in view of the intermediate value theorem, there is such that (here or ). Since is periodic, there is such that The proof is complete.

Now, we consider that following equation:

where

Lemma 2.10. *Suppose that conditions (i) and (ii) of Theorem 2.1 hold. If then there are positive constants and such that for any -periodic solution of (2.24),
*

*Proof. *Let be a -periodic solution of (2.24). By (2.24) and our assumption that we have
By Lemma 2.9, there is such that
Since and are with period thus for any we have
From (2.28), we see that for any
It is easy to see from (2.27) and (2.29) that for any
In view of the condition we know that there is a positive number such that
From condition (ii), we see that there is a such that for and
Let
By (2.32) and (2.33), we have
From (2.34) and (2.36), we have
In view of condition (i), (2.26), (2.37), and (2.38), we get
It follows from (2.37), (2.38), and (2.39) that
Since thus there is a such that In view of (2.24) and the fact that , we conclude that for any
From (2.40) and (2.41), we see that
It follows from (2.30), (2.31), and (2.42) that
where
Let then from (2.43) we have
From (2.42) and (2.45), for any we have
where
The proof is complete.

Lemma 2.11. *Suppose that condition (iii) of Theorem 2.1 is satisfied. Then (1.3) has at most one -periodic solution.*

*Proof. *Suppose that and are two -periodic solutions of (1.3). Set . Then we have
*Case 2 (i). *For all . Without loss of generality, we assume that , that is, for Integrating (2.48) from to , we have
Combining condition (iii) and , either
or
holds. This is contrary to (2.49).*Case 2 (ii). *There exist such that . As in the proof of (2.30) in Lemma 2.10, we have
On the other hand, since thus there is a such that . In view of (2.48), we conclude that for any
By (2.53) and the fact that , we have for any
It follows that for any
We know that for any
From (2.56), we have
By (2.52), we get
It is easy to see from (2.57) and (2.58) that
By condition (iii) of Theorem 2.1, we see that Thus (2.58) leads us to which is contrary to So (1.3) has at most one -periodic solution. The proof is complete.

We now turn to the proof of Theorem 2.1. Suppose Let and be defined by (2.3), (2.4), (2.6), and (2.7), respectively. By Lemma 2.10, there are positive constants and such that for any -periodic solution of (2.24) such that (2.25) holds. Set

where is a fixed number which satisfies It is easy to see that is an open and bounded subset of Furthermore, in view of Lemmas 2.5 and 2.7, is a Fredholm mapping of index zero and is -compact on . Noting that by Lemma 2.10, for each and Next note that a function must be constant: or Hence by (i) and (2.17), Hence by conditions (i), (iii) and (2.17),

so The isomorphism is defined by for and Then

In particular, we see that if then

and if then Consider the mapping

From (2.63) and (2.65), for each and we have

Similarly, from (2.64) and (2.65), for each and we have

By (2.66) and (2.67), is a homotopy. This shows that

By Theorem , we see that equation has at least one solution in In other words, (1.3) has an -periodic solution Furthermore, if (iii) is satisfied, from Lemma 2.11, we know that (1.3) has an -periodic solution only. The proof is complete.

#### 3. Example

Consider the equation

and we can show that it has a nontrivial -periodic solution. Indeed, take

We see that Let and Then condition (i) of Theorem 2.1 is satisfied:

Let and Then conditions (i), (ii) and (iii), of Theorem 2.1 are satisfied. Note further that Therefore (3.1) has exactly one -periodic solution. Furthermore, it is easy to see that any solution of (3.1) must be nontrivial. We have thus shown the existence of a unique nontrivial -periodic solution of (3.1).

#### Acknowledgment

The first author is supported by the Natural Science Foundation of Guangdong Province of China under Grant no. 9151008002000012.