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Abstract and Applied Analysis
Volume 2010, Article ID 152460, 27 pages
http://dx.doi.org/10.1155/2010/152460
Research Article

On the Existence of Solutions for Impulsive Duffing Dynamic Equations on Time Scales with Dirichlet Boundary Conditions

Department of Mathematics, Yunnan University, Yunnan, Kunming 650091, China

Received 4 June 2010; Accepted 16 August 2010

Academic Editor: Douglas Robert Anderson

Copyright © 2010 Yongkun Li and Tianwei Zhang. 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

By using critical point theory, some new sufficient conditions for the existence of solutions of impulsive Duffing dynamic equations on time scales with Dirichlet boundary conditions are obtained. Some examples are also given to illustrate our results.

1. Introduction

Consider the following Duffing dynamic equations on time scales with impulsive effects where , is a regressive constant, , , , is continuous, and , are continuous.

Obviously, system (1.1) covers Duffing equations (when ) The Duffing equation has been used to model the nonlinear dynamics of special types of mechanical and electrical systems. This differential equation has been named after the studies of Duffing in 1918 [1], has a cubic nonlinearity, and describes an oscillator. The main applications have been in electronics, but it can also have applications in mechanics and in biology. For example, the brain is full of oscillators at micro- and macrolevel [2]. There are applications in neurology, ecology, secure communications, chaotic synchronization, and so on. Due to the rich behaviour of these equations, the most general forced forms of the Duffing equation (1.2) have been studied by many researchers [312].

The study of dynamic equations on time scales goes back to its founder Stefan Hilger [13], and is a new area of still fairly theoretical exploration in mathematics. Motivating the subject is the notion that dynamic equations on time scales can build bridges between continuous and discrete equations. Further, the study of time scales has led to several important applications, for example, in the study of insect population models, neural networks, heat transfer, and epidemic models [1416].

Impulsive effects exist widely in many evolution processes in which their states are changed abruptly at certain moments of time. The theory of impulsive differential systems has been developed by numerous mathematicians (see [1724]). Applications of impulsive differential equations with or without delays occur in biology, medicine, mechanics, engineering, chaos theory, and so on (see [2532]).

In addition, system (1.1) also includes which were studied by papers [33, 34], and some existence results were obtained by using some critical point theorems.

Our purpose in this paper is to study the variational structure of problem (1.1) in an appropriate space of functions and the existence of solutions for problem (1.1) by means of some critical point theorems. The organization of this paper is as follows. In Section 2, we make some preparations. In Section 3, we will study the variational structure of problem (1.1) and give some important lemmas which will be used in later section. In Section 4, by applying some critical point theorems, we establish sufficient conditions for the existence of solutions to problem (1.1). Some illustrative examples are also given in Section 4.

2. Preliminaries

In this section, we will first recall some basic definitions and lemmas which are used in what follows.

Definition 2.1 (see [13]). A time scale is an arbitrary nonempty closed subset of the real set with the topology and ordering inherited from . The forward and backward jump operators , and the graininess are defined, respectively, by The point is called left dense, left scattered, right dense, or right scattered if , , , or , respectively. Points that are right dense and left dense at the same time are called dense. If has a left-scattered maximum , defined ; otherwise, set . If has a right-scattered minimum , defined ; otherwise, set .

Definition 2.2 (see [13]). For and , then the delta derivative of at the point is defined to be the number (provided it exists) with the property that for each , there is a neighborhood of such that

Definition 2.3 (see [13]). A function is rd continuous provided it is continuous at each right-dense point in and has a left-sided limit at each left-dense point in . The set of rd-continuous functions will be denoted by .

Definition 2.4 (see [13]). Assume that is an arbitrary time scale. We say that a function is regressive provided

Definition 2.5 (see [13]). Assume that is an arbitrary time scale and a function is regressive, then we define the exponential function on by in which for , here is the principal logarithm function.

A function is called rd continuous provided it is continuous at each right-dense point and its left-sided limit exists (finite) at each left-dense point in . We write . If is differentiable with , we write . If is differentiable on with , we write . Similar to the classical Riemann's definition of integrability, the concept of the Riemann delta integral on time scales is given in [35]. We know that many familiar functions, including monotone continuous, piecewise continuous, right-dense continuous functions, are Riemann delta integrable. In the following lemma, we present some properties of the integral that will be needed later.

Lemma 2.6 (see [35]). Let be two functions and . Then one has the following. Let and be Riemann delta integrable functions on , and . Then is Riemann delta integrable andFundamental Theorem of Calculus. Let be a continuous function on such that is delta differentiable on . If is Riemann delta integrable from to , thenIntegration by Parts. Let and be continuous functions on that are differentiable on . If and are Riemann delta integrable from to , thenIf is Riemann delta integrable on , then so is , and

The construction of the -measure on and the following concepts can be found in [35]. For each , the single-point set is -measurable, and its -measure is given byIf and , thenIf and , then

The Lebesgue integral associated with the measure on is called the Lebesgue delta integral. For a (measurable) set and a function , the corresponding integral of on is denoted by . All theorems of the general Lebesgue integration theory hold also for the Lebesgue delta integral on . Moreover, comparing the Lebesgue delta integral with the Riemann delta integral, we have the following.

Lemma 2.7 (see [35]). Let be a closed bounded interval in and let be a bounded real-valued function defined on . If is Riemann delta integrable from to , then is Lebesgue delta integrable on and where and indicate the Riemann delta integral and Lebesgue delta integral, respectively.

Assume that and . Let be the set of By Lemma in [36], is a complete linear space with the norm defined by

Let denote the linear space of all continuous functions with the maximum norm .

Lemma 2.8 (Hlder inequality [37]). Let and q the conjugate number of . Then

At the end of this section, we recall some notation and known results from critical point theory.

Let be a real normed space and let be a functional from to . is called weakly continuous if is called lower semicontinuous if is called weakly lower semicontinuous if

Let be a real Hilbert space, , which means that is a continuously Fréchet-differentiable functional defined on . is said to satisfy the Palais-Smale condition (P.S. condition) if any sequence for which is bounded and as possesses a convergent subsequence in .

Let be the open ball in with radius and centered at and let denote its boundary. The following lemma is taken from [38, 39] and will play an important role in the proof of our main results.

Lemma 2.9 (see [38]). If is a real normed space and is lower semicontinuous and convex, then is weakly lower semicontinuous.

Lemma 2.10 (see [39]). Let be a real Banach space, be weakly lower(upper) semicontinuous, and Then, there exists such that

Lemma 2.11 (Mountain Pass Theorem [40]). Let be a real Hilbert space and let satisfy P.S. condition, and satisfies (J1)There are constants such that for all , where . (J2) and there exists such that . Then possesses a critical value . Moreover, can be characterized as where

3. Variational Structure

In this section, we will establish the corresponding variational framework for problem (1.1).

Let be the function space of the form can be equipped with the inner product inducing the norm

Let be a Cauchy sequence in , that is, as . Since is a complete space, there exists a such that as . Define a function by The function is called a weak derivative of and is also denoted by .

Remark 3.1. If has a weak derivative, then this weak derivative is unique in , that is, if are both weak derivatives of , then . If is delta differential on , then its weak derivative is its delta derivative.

From (3.4), it is clear that and Thus, .

Define the space as It is clear that , and we define the inner product in as inducing the norm where represents the weak derivative of . Moreover, we have the following result.

Lemma 3.2 (see [33]). is a complete space in the norm and inequality holds, that is, .

Also, we consider the inner product inducing the norm

Throughout this paper, it will be assumed that ess , where is the smallest eigenvalue of the problem From [41], may be defined as Applying the integration by parts in Lemma 2.6, we have which implies that

For convenience, we introduce the following notations:

Lemma 3.3. Suppose that (H1). Then one has which implies that the norm and the norm are equivalent, where

Proof
Case 1 (). For every , we have On the other hand, by Lemma 2.6, one has Therefore, the norm and the norm are equivalent.
Case 2 (). For every , we have from (3.15) that Similar to Case 1, it is easy to obtain the desired results. This completes the proof.

Together with (3.15) and Lemma 3.3, it is not difficult to obtain the following.

Lemma 3.4 (Poincaré inequality). For all , one has that is, .

Lemma 3.5. for all , where .

Proof. For every , we have Therefore, This completes the proof.

For , we have that and are both absolutely continuous. Hence, for any .

If , then is absolutely continuous and . In this case, may not hold for some . It leads to the impulsive effects.

Take and multiply the two sides of the equality by and integrate from to ; we have Furthermore, Combining (3.26), we have Considering the above, we introduce the following concept solution for problem (1.1).

Definition 3.6. We say that a function is a weak solution of problem (1.1) if the identity holds for any .

Consider the functional defined by where . Using the continuity of and , , one has that . For any , we have Therefore, the solutions of problem (1.1) are the corresponding critical points of .

Lemma 3.7 (see [33]). The functionals defined by are weakly continuous on .

4. Main Results

Theorem 4.1. Assume that (H1) holds. Suppose further the following.(H2)There exist and such that(H3)There exist and such that Then problem (1.1) has at least one weak solution.

Proof. According to Lemma 3.7, are weakly continuous on . On the other hand, it is clear that is continuous and convex on . By Lemma 2.9, is weakly lower semicontinuous. So is weakly lower semicontinuous on .
From (H2)-(H3), we have for all . This implies that , and is coercive. By Lemma 2.10, has a minimum point on , which is a critical point of . Therefore, problem (1.1) has at least one weak solution. This completes the proof.

Example 4.2. Consider the following impulsive Duffing equations on time scales Then, problem (4.4) has at least one solution.

Proof. The result is easy to obtain from Theorem 4.1 and we should omit it. This completes the proof.

Theorem 4.3. Assume that (H1) holds. Suppose further the following. (H4)There exist and such that(H5)There exist and such that(H6). Then problem (1.1) has at least one weak solution.

Proof. According to Theorem 4.1, we know that is weakly lower semicontinuous on .
From (H4)(H6), we have for all . This implies that , and is coercive. By Lemma 2.10, has a minimum point on , which is a critical point of . Therefore, problem (1.1) has at least one weak solution. This completes the proof.

Example 4.4. Consider the following impulsive Duffing equations on time scales: where Then problem (4.8) has at least one solution.

Proof. By an easy calculation, we obtain It is easy to verify that all the conditions of Theorem 4.3 are satisfied. From Theorem 4.3, problem (4.8) has at least one solution. This completes the proof.

Lemma 4.5. Assume that (H1) and (H3) hold. Suppose further the following. (H7)There exist constants and , such that where Then satisfies the P.S. condition in .

Proof. Let be a sequence in satisfying that is bounded and as . Let for all . By the continuity of and (H7), there exists a constant such that which implies that is bounded in . Taking into account and employing Lemma 3.5, we know that is relatively compact in . Thus possesses a convergent subsequence in , that is, as . since , we know that for all . From the above inequality, it is easy to see that From the continuity of and , , we have hold uniformly for as . Set Clearly, as . We rewrite (3.31) as for any . Therefore, Thus, for any , it follows that Hence, Observing that as , it is easy to see that Together with (4.16)–(4.21), we get In other words, possesses a convergent subsequence in . The P.S. condition is now satisfied. This completes the proof.

Theorem 4.6. Assume that (H1), (H3), and (H7) hold. Suppose further the following.(H8)There exists constant such that(H9), where Then problem (1.1) has at least one weak solution.

Proof. It suffices to show that possesses a nonzero critical point. Now, we need to verify that all assumptions of the Mountain Pass Theorem hold. The P.S. condition follows from Lemma 4.5. Next, we will check conditions (J1) and (J2) of Lemma 2.11. Assume that . In view of (H8), for , we get that Hence, which implies that At the same time, we can obtain that Together (4.27)-(4.28) and the expression of functional , it follows that Setting Then we have