Abstract
By the virtue of variational method and critical point theory, we give some existence results of weak solutions for a -Laplacian impulsive differential equation with Dirichlet boundary conditions.
1. Introduction
In this paper, we shall consider the following problem: where , , and , are continuous.
Many evolution processes are characterized by the fact that, at certain moments of time, they experience a change of state abruptly. These processes are subject to short-term perturbations whose duration is negligible in comparison with the duration of the process. Consequently, it is natural to assume that these perturbations act instantaneously, that is, in the form of impulses. Thus impulsive differential equations appear as a natural description of observed evolution phenomena of several real world problems.
Recently, there have been many papers to study impulsive problems by variational method and critical point theory, such as [1–11] and the references therein.
In [7], Nieto and O’Regan studied the linear Dirichlet impulsive problem and the nonlinear Dirichlet impulsive problem In the paper, they have shown that the impulsive problem minimizes some (energy) functional, and the critical points of that functional are indeed solutions of the impulsive problem.
In [3, 4], Sun et al. utilized some variant fountain theorems by [12] to consider the existence of infinitely many solutions for the following two impulsive problems: Admittedly, they obtained many perfect results.
We also note that some people begin to study -Laplacian differential equations with impulsive effects; for example, see [1, 2, 8–11].
In [1], Chen and Tang considered the -Laplacian impulsive problem They established some existence theorems for one or infinitely many solutions under more relaxed assumptions on their nonlinearity , which satisfies a kind of new superquadratic and subquadratic condition.
In [8], Bogun discussed the existence of weak solutions for the -Laplacian problem with superlinear impulses by the virtue of mountain pass theorem and symmetric mountain pass theorem
In [9], Xu et al. reconsidered the previous problem by topological degree theory and Fountain theorem under Cerami condition.
In [11], by the virtue of three critical points theorem obtained by Bai and Dai is studied the existence of at least three solutions for the following -Laplacian boundary value problem:
Motivated by the previous facts, in this paper, our aim is to study the existence and multiplicity of weak solutions for impulsive problem (1) by using variational method and critical point theory. It is well known that the Ambrosetti-Rabinowitz type condition is to ensure the boundedness of all (PS) sequences of the corresponding functional. However, without it, it will become more complicated. Therefore, we will use new variant fountain theorems due to Zou [12] to overcome this difficulty and obtain infinitely many weak solutions for (1). On the other hand, for the superlinear at and asymptotically linear at , we obtain a weak solution for (1) by the mountain pass theorem. The results obtained here improve some existing results in the literature.
2. Preliminaries
In this section, we recall some fundamental facts of critical point theory which will be used in the proofs of our main results. Let be the Sobolev space with the usual norm It is clear that is a reflexive Banach space. Next, we make a finite dimensional decomposition for . In order to do this, we first need to consider the eigenvalue problem It is well known that the set of all eigenvalues of the problem (9) is given by the sequence of positive numbers (see [1, 13–15]) We denote by the corresponding eigenfunctions associated with for all , and . Moreover, the first eigenvalue is simple and isolated, and is positive in . Furthermore, the Poincaré inequality holds. Note that we can normalize such that Fix any define and where By [16, Section 5], the conclusions are Note the definitions of and ; by (9), we have For each , multiply by on both sides of (16) to obtain In particular, choosing , we see for all .
We denote the norms in and as follows: By the Sobolev embedding theorem, the embeddings and are compact. Consequently, we also find that there are two constants and such that
For , we have that and are both absolutely continuous. Hence, for any . If , then is absolutely continuous. In this case, the one-sided derivatives , may not exist. It leads to the impulsive effects. As a result, we need to introduce a different concept of solution. Suppose that satisfies the Dirichlet condition . Assume that, for every , and . Let .
Take and multiply the two sides of the equality by and integrate from to : For the left term, in view of impulsive effects, we find Consequently, Considering the previous, we introduce the following concept for the solution for (1).
Definition 1. One says that a function is a weak solution for (1) if the identity
Consider the functional defined by
where . Note that for the continuity of and , we see . Furthermore, the derivative of is
Thus, we easily know that weak solutions of (1) coincide with the critical points of the -functional .
For the reader’s convenience, we now present some critical point theorems; one can refer to [12, 17–22] for more details.
Definition 2. Let be a real Banach space and . For any sequence , if is bounded and as possesses a convergent subsequence, then we say that satisfies the Palais-Smale condition (PS condition for short).
Definition 3. One says that satisfies condition if the existence of a sequence such that and as implies that has a convergent subsequence.
Lemma 4 (see [17]). Let satisfy (PS) condition. Suppose that(i),(ii)there exist and such that for all , ,(iii)there exists with such that .
Then has a critical value . Moreover, can be characterized as , where .
Let be a Banach space equipped with the norm and , where for any . Set and . In the following, one will introduce variant fountain theorems by Zou [12]. Let and the subspaces and be defined as previously. Consider the following -functional defined by
Lemma 5. If the functional satisfies the following:(T1) maps bounded sets to bounded sets uniformly for . Moreover, for all ,(T2) for all ; moreover, or as ,(T3) there exists such that then where and . Moreover, for a.e. , there exists a sequence such that
Now, we list our assumptions on and .(H1) There exist such that , ,(H2) uniformly for ,(H3) uniformly for ,(H4) There exist and such that (H5) for all and , uniformly on ,(H6) there is a positive constant such that , uniformly on ,(H7), , ,(H8), , ,(H9) There exist and such that , and ,(H10) and are odd functions about , for all .
Remark 6. (1) As known to all, (H4) implies that ; that is, is superlinear at with respect to . In view of (H3) and (H4), we see that is superlinear at and asymptotically linear at ; this is a new case. However, the nonlinearity in [9] is asymptotically linear at .
(2) Condition (H6) is weaker than the well-known Ambrosetti-Rabinowitz condition (H4); also see condition in [8]. Indeed, by (H6), there is a such that
Consequently, for all and , which is weaker than condition (H4).
3. Main Results
Theorem 7. Suppose that (H1)–(H4), (H7), and (H9) hold, , and with for all . Then (1) has a weak solution.
Proof. From (H1)–(H4), for all , there exist such that
Choose such that , together with (33), (11), (19), and (H7); we obtain
If is small enough, (ii) of Lemma 4 can be proved.
On the other hand, we can take such that ; by (34), (19), and (H9), noting that , we find
Therefore, (iii) of Lemma 4 is also proved, as required.
Now, we only claim that satisfies (PS) condition. Supposing that is a (PS) sequence, for all , we have
where is a constant and as . Next, we will show that is a bounded sequence in . If not, there is a subsequence of , still denoted by , such that
Define , then , and thus it has a subsequence, still denoted , such that weakly in , strongly in , , and , , where , .
Divide (38) by to get
Passing to the limit in (40), we see
with the fact that
Now, we claim that , . Indeed, in (41), taking , we arrive at
where . However, by (H3) and (H4), there is such that
In addition, note that and the super-linearity of we have
Consequently, if , by the Fatou theorem, we get
which contradicts the fact of (43).
Obviously, . From (H2) and (H3), there exists such that . By (41), Lebesgue’s dominated convergence theorem enables us to see
This contradicts for all . Therefore, is bounded, as required. Going, if necessary, to a subsequence, we can assume that weakly in ; then
enables us to obtain that
It follows from weakly in and that
Note that
and thus as . So, satisfies (PS) condition. This completes the proof.
In what follows, we will utilize Lemma 5 to study (1). Now, we define a class of functionals on by It is easy to know that for all and the critical points of correspond to the weak solutions of problem (1). Note that , where is the functional defined in (25).
Theorem 8. Assume that (H1) and (H5)–(H10) hold. Then (1) possesses infinitely many weak solutions.
Proof. We first prove that there is a positive integer and two sequences as such that
where and .
Step 1. We will show that (53) holds true.
From (H1), we see that there exist , such that
Consequently, by (H5) and (H7), we obtain
Let , . Then by [19, Lemma 3.8], as . Since , we find
Let as . Then there exists such that , . Therefore,
Step 2. We will show that (54) holds true.
We first prove that there exists such that
There exists otherwise a sequence such that
For each , let and
Passing to a subsequence if necessary, we may assume in for some since is of finite dimension. We easily find . Consequently, there exists a constant such that
Indeed, if not, then we have
which implies
This leads to , contradicting to . In view of and the equivalence of any two norms on , we have
For every , denote
and , where is defined by (62). Then for large enough, by (62), we see
Consequently, for large enough, we arrive immediately at
This contradicts (65). Therefore, (59) holds. For the given in (59), we let
Then by (59), we find
By (H5), for any , there is a constant such that
where is determined in (59). Therefore,
Now for any , if we take , noting that and and large enough, we have
Step 3. The continuity and and imply that maps bounded sets to bounded sets uniformly for . In view of (H10), for all . Thus condition (T1) of Lemma 5 holds. Besides, by (H7), we get
and since . Thus the condition (T2) of Lemma 5 holds. For step 1 and step 2, the condition (T3) of Lemma 5 also holds for all . Consequently, Lemma 5 implies that, for any and a.e. , there exists a sequence such that
where , and , .
Furthermore, we easily have
where and as .
Claim 1. possesses a strong convergent subsequence in , for all and . In fact, by the boundedness of , passing to a subsequence, as , we may assume in . By the method of Theorem 7, we easily prove that strongly in .
Thus, for each , we can choose such that the sequence obtained a convergent subsequence, and passing again to a subsequence, we may assume
Thus we obtain
Claim 2. is bounded in and has a convergent subsequence with the limit for all . For convenience, we set for all . If not, is unbounded in ; that is, . By (H6), there is such that
Combining this and (H8), we have
This implies that
On the other hand, by (H1) and (H9), we see
Consequently, noting that and , we have
This is a contradiction. Therefore, is bounded in . By claim 1, we see that has a convergent subsequence, which converges to an element for all .
Hence, passing to the limit in (78), we see
Since as , we get infinitely many nontrivial critical points of . Therefore (1) possesses infinitely many nontrivial solutions by Lemma 5. This completes the proof.
Remark 9. (1) Let
where , , with avoiding for all . Then (H1)–(H4) hold true with and with for all .
(2) Let , , , and for all and . Then , . Clearly, (H1), (H7), (H9), and (H10) are satisfied.(1), uniformly on , and for all . So, (H5) holds.(2), uniformly on , and hence (H6) holds.(3), , . Therefore, (H8) holds.
Acknowledgments
This research is supported by the NNSF, China (10971046), Shandong and Hebei Provincial Natural Science Foundation (ZR2012AQ007, A2012402036), GIIFSDU (yzc12063), and IIFSDU (2012TS020).