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Discrete Dynamics in Nature and Society
Volume 2013 (2013), Article ID 786548, 8 pages
Weak Solutions for a -Laplacian Antiperiodic Boundary Value Problem with Impulsive Effects
1Department of Mathematics, Qilu Normal University, Jinan, Shandong 250013, China
2School of Mathematics, Shandong University, Jinan, Shandong 250100, China
3Department of Mathematics, Hebei University of Engineering, Handan, Hebei 056038, China
Received 7 March 2013; Accepted 19 May 2013
Academic Editor: Beatrice Paternoster
Copyright © 2013 Keyu Zhang 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.
By virtue of variational method and critical point theory, we will investigate the existence of weak solutions for a -Laplacian impulsive differential equation with antiperiodic boundary conditions.
In this work, we will study the existence of weak solutions for the -Laplacian impulsive differential equation with antiperiodic boundary conditions where , , , , , and , . Moreover, , where and denote the right and left limits, respectively, of at , for .
Impulsive differential equations have become more important in recent years in some mathematical models of real processes and phenomena studied in physics, chemical technology, population dynamics, biotechnology, and economics. We mention, for instance, the books [1–3] dealing with impulsive differential equations.
Recently, many researchers pay their attention to impulsive differential equations by variational method and critical point theory, to the best our knowledge; we refer the reader to [4–8] and references cited therein. Meanwhile, some people begin to study -Laplacian differential equations with impulsive effects; for example, see [9–14].
Chen and Tang  adopted the least action principle and the saddle point theorem to obtain some existence theorems for a second-order -Laplacian system with or without impulsive effects under weak sublinear growth conditions. In , they also considered that a second-order impulsive differential equation with Dirichlet problems has one or infinitely many solutions under more relaxed assumptions on their nonlinearity , which satisfies a kind of new superquadratic and subquadratic conditions.
The problem of finding infinitely many large energy solutions is a very classical problem; there is an extensive literature concerning the existence of infinitely many large energy solutions of a plethora of problems via the symmetric mountain pass theorem and fountain theorem; for instance, see [15–22].
Motivated by the works cited above, in this paper, we shall discuss the problem (1). Firstly, we adopt Browder theorem, introduced in , to prove that the problem (1) has only one weak solution. Secondly, we shall utilize Fountain theorem under Cerami condition (C), which is introduced in , to prove that the problem (1) has infinitely many weak solutions. The results obtained here improve some existing results in the literature.
2. Variational Structure
We first introduce the Banach space , endowed with the norm Then we see that is a reflexive Banach space. We denote by the norm in and in .
In what follows, we shall convert the problem (1) into an integral equation. For all , multiply (1) by and integrate between and to obtain Noting the impulsive effects, for the left integral of (3), we have Substituting (4) into (3), we easily find that
Definition 1. If for each there is a such that (5) holds, then is called a weak solution for (1).
From (5), we can obtain the weak solutions for (1) that coincide with critical points of the energy functional where . Clearly, is class of and its derivative is
Lemma 2. Consider and .
Proof. For , by mean value theorem, and . It follows from Hölder inequality that Consequently, we have Furthermore, for , we have , Hence every bounded set in is equicontinuous in . By Arzela-Ascoli theorem, we claim that . By the same manner, we can prove that . Furthermore, we can obtain by (9) This completes the proof.
Definition 3 (see [23, page 303]). Let be a reflexive real Banach space and its dual. The operator is said to be demicontinuous if maps strongly convergent sequences in to weakly convergent sequences in .
Lemma 4 (Browder theorem, see [23, Theorem 5.3.22]). Let be a reflexive real Banach space. Moreover, let be an operator satisfying the following conditions:(i) is bounded and demicontinuous;(ii) is coercive, that is, ;(iii) is monotone on the space ; that is, for all , one has
Then the equation has at least one solution for every . If, moreover, the inequality (12) is strict for all , , then the equation has precisely one solution for all .
In what follows, we shall introduce Fountain theorem under Cerami condition (C). We first give the definition of Cerami condition (C).
Definition 5 (see [24, Definition 1.1]). Assume that is a Banach space with norm ; we say that satisfies Cerami condition (C), if for all ,(i)any bounded sequence satisfying , possesses a convergent subsequence;(ii)there exist such that for any with , .
As is a reflexive Banach space, there exist (see [25, Section 17]) and such that , and . For , denote , , and . Clearly, with for all . Denote .
Lemma 6 (see [24, Proposition 1.2]). Let , , be defined as above. Assume that satisfies condition (C), and . For each , there exist such that(i), ,(ii).
Then has a sequence of critical points , such that as .
3. Main Results
Now, we list our assumptions on and , .(H1) is a decreased function about , uniformly in , and are increased functions with .(H2) There exist and such that , for all and .(H3) There exist such that , for all , .(H4) There is a positive constant such that , uniformly in .(H5), , for all , .(H6), uniformly on .(H7) is an even function about and are odd functions about , for all .
Theorem 7. Let and (H1)–(H3) hold. Then (1) has precisely a weak solution.
Proof. For , we define Let Then, to find a weak solution of (1) is equivalent to finding a solution for the operator equation . In what follows, we shall sketch the properties of operators . By Hölder inequality, we have Consequently, is bounded: The last integral tends to zero as due to the continuity of the Nemytskii operator . Hence, is continuous. We assume that in ; by Lemma 2, we know that there is a subsequence, still denoted by , which strongly converges to in . Therefore, utilizing the continuity of , we have So, is continuous. In the same methods, we see that is also continuous. On the other hand, by (9) and (11), for , it follows from (H2) and (H3) that Therefore, and are bounded. Up to now, we have proved that is bounded and continuous, so (i) of Lemma 4 holds. Finally, we shall show the monotonicity and coercivity of . Indeed, (H1) implies that for and , (H2) and (H3) enable us to get Therefore, by the span of , we arrive at . As a result, (ii) and (iii) of Lemma 4 hold. Hence, Lemma 4 implies that (1) has precisely a weak solution. This completes the proof.
Lemma 8. Let (H3)–(H5) hold. Then satisfies Cerami condition (C).
Proof. For all , we assume that is bounded and
Going, if necessary, to a subsequence, we can assume that weakly in , and then
By Lemma 2, enables us to obtain that
It follows from weakly in and that
Noting (19), we have
and thus as . Hence, (i) of Definition 5 is satisfied. Next, we prove (ii) of Definition 5; if not, there exists a sequence such that
By (26), there exists a constant such that
On the other hand, (H4) implies that there is a such that , for all and . Furthermore, is bounded for and . Therefore, there exists such that , for all , . This, together with (H5), yields
which implies . Therefore, there is a such that .
It follows from (H3) that there are such that By this and (H5), we can find and thus if (27) holds, which contradicts in (26). This proves that satisfies condition (C).
Theorem 9. Suppose that (H2)–(H7) hold, and then (1) has infinitely many weak solutions.
Proof. By (H7), we know that is even. Denote , and by the compactness of the embedding , we see that as (see [26, Lemma 3.8]). Noting (30), we have by (H5) and Hölder inequality, for any and ,
We easily have as and as . Hence,
On the other hand, by (H6), we find that there are such that Since all the norms of a finite dimensional normed space are equivalent, note that is a norm of , so there exists a such that Noting (9), we have Note that we can choose a large enough such that by (H6) and by (H2), and then there exists positive constants such that By this and (33), we can take , and thus . Up until now, we have proved that the functional satisfies all the conditions of Lemma 6, and then has infinitely many critical points. Equivalently, (1) has infinitely many weak solutions. This completes the proof.
Let and . Consider the problem where and , where and is defined by (35).
For , we can easily have (H2) and (H7) hold. By computation, , , for all, and hence we obtain that (H5) is satisfied.
For and , we see , . Therefore, (H3) holds.
It is obvious that , so (H7) holds; furthermore, As a result of this, (39) (i.e., (H6)) is true.
In , the condition is As known to all, this condition is originally introduced in  and is still present in many works which is used to guarantee the boundedness of (P.S.) sequences of the corresponding functional.
In the problem (40), note that there exists such that and This is impossible if and is an adequately large positive number. On the other hand, if (H4) holds, then there is a sufficiently large such that Therefore, for all and , which is also weaker than the condition if . This means that our results generalize the corresponding results in .
This research is supported by the NNSF China (10971046), Shandong and Hebei Provincial Natural Science Foundation (ZR2012AQ007, A2012402036), GIIFSDU (yzc12063), IIFSDU (2012TS020), and the Project of Shandong Province Higher Educational Science and Technology Program (J09LA55).
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