Research Article | Open Access
Lifang Niu, Kaimin Teng, "Antiperiodic Solutions for -Laplacian Systems via Variational Approach", Journal of Applied Mathematics, vol. 2014, Article ID 324518, 7 pages, 2014. https://doi.org/10.1155/2014/324518
Antiperiodic Solutions for -Laplacian Systems via Variational Approach
We establish the existence of solutions for -Laplacian systems with antiperiodic boundary conditions through using variational methods.
In this paper we investigate the existence of solutions of the following second order -Laplacian systems with antiperiodic boundary condition: where , () satisfies the following fundamental assumption: is measurable in for each , continuous differentiable in for almost every , and there exists and such that for all and a.e. .
In the last few decades, the following second order systems involving periodic boundary condition: have acted as one of the mainstream research problems in the field of differential equation. Under various assumptions of the potential , there have been lots of existence and multiplicity of results in the literatures by using the tool of nonlinear analysis, such as degree theory, minimax methods, and Morse theory. Here we do not even try to review the huge bibliography, but we only list some references for our purpose; for example, we refer the readers to see [1–6] and the references therein.
Comparing problem (1) with problem (3), we observe that the only difference is the boundary conditions. In order to use the variational methods, one of the main difficulties is the variational principle. For this matter, we try to modify some work space such that the variational principle can be established. Thanks to the work of Tian and Henderson , we borrow their ideas to give the variational principle for problem (1).
Note that the study of antiperiodic solutions for nonlinear differential systems of the form is closely related to the study of its periodic solutions. Indeed, if we assume that and , then let and we get that is a solution of systems (4) with conditions and . Since is -periodic in , can be extended to be -periodic over , and hence is a -periodic solution of systems (4).
In this paper, we will establish the existence of solutions for problem (1) by variational method. As far as we know, there are few papers studying the second order systems with antiperiodic boundary conditions by variational methods.
2. Variational Principle
In the sequel, we denote by the -norm (). Let be the space of infinitely differentiable functions from into . Let ; obviously, and () belong to , and thus .
The following fundamental lemma proved in  is essential to establish the variational principle.
Lemma 1. Let . If, for every , where denotes the inner product on , then and there exists such that
Remark 2. From Lemma 1, we have the following facts.(i)A function satisfying (6) is called a weak derivative of . By a Fourier series argument, the weak derivative, if it exists, is unique. We denote by the weak derivative of .(ii)We will identify the equivalence class and its continuous representation (iii)Equations (7) and (8) imply that . For this matter, we only show that . Indeed, using integration by parts, we have By (9), we have (iv)If is continuous on , then by (9) is the classical derivative of .(v)It follows from (9) and Rademacher theorem that is the classical derivative of a.e. on .
The Sobolev space is the space of functions having a weak derivative . Obviously, if , and . The norm over is defined by It is easy to see that is a reflexive Banach space and .
Lemma 3. Let . Then(i)the embedding is compact;(ii)there exits a constant such that Moreover, the embedding is compact.
Proof. From Theorem 8.2 in , for , and . By Hölder’s inequality, we have On the other hand, letting be a unit ball in , for , we have It follows from the Ascoli-Arzelà theorem that has a compact closure in .
Remark 4. There is some differences between antiperiodic boundary value problem and periodic boundary value problem. One is the norm on the work space, and the other is the decomposition of space not like , where . And it is well known that the norms and are equivalent to .
The energy functional corresponding to problem (1) is defined by Under the assumption of , we have the following.
Proof. Similarly as the proof of Theorem 1.4 in , we obtain that and (28) holds. If for all and hence for all , by Lemma 1, there exists a constant such that From Remark 2, we know that has a weak derivative By Remark 2, we get By (24), if , clearly, we have . If not, we see that . Hence, by calculation, we get This implies that and hence the conclusion is proved.
Lemma 6. The functional is weakly lower semicontinuous.
Proof. Assuming in , then by (ii) of Lemma 3, we have in . By hypothesis , we have where is a constant such that for every , all , and . It follows from the weak lower semicontinuity of the norm function in Banach space that Accordingly, the conclusion is completed.
Next, we study the eigenvalue problem
Definition 7. One says is an eigenvalue of problem (28) if there exists , , such that
Lemma 8. The eigenvalue problem (28) possesses the following properties:(a)all the eigenvalues are positive real numbers;(b)for each , one has , where .
Proof. (a) Letting be the eigenfunction corresponding to the eigenvalue , we have
Hence, and if and only if , implying that , a.e. on , where is a constant. Since (), we see that , for all ; this contradicts with the definition of eigenfunction.
(b) By standard minimization arguments, we can prove the conclusion. We omit it.
Remark 9. Note that the eigenvalue of one dimensional vector -Laplacian operator under antiperiodic boundary condition possesses some similar properties as the -Laplacian operator under Dirichlet boundary condition on bounded domain.
3. Main Results and Proof
In this section, we will give some existence and multiplicity of results for problem (1).
Theorem 10. Assume that satisfies hypothesis and the following conditions.There exist and such that with . Then problem (1) has at least one solution in .
Proof. First we show that is coercive on . In fact, from and Lemma 3, we have Therefore, as . As a result, we get a bounded minimizing sequence in . Combining Lemma 6, by Theorem 1.1 and Corollary 1.1 in , problem (1) has at least one solution which minimizes on .
Remark 11. By hypothesis , we see that is summable over in the neighborhood of zero, and thus the condition can be weaken to hold for large.
If , we may assume the function in satisfies that as the same proof of Theorem 10, and we can get the following theorem.
Theorem 12. Under the above assumptions of , then problem (1) has at least one solution in .
Theorem 13. Assume that satisfies hypothesis and the following conditions:, and uniformly for a.e. ;there exist constants and such that uniformly for a.e. , and uniformly for a.e. .
Then problem (1) possesses at least one nontrivial solution in .
In order to prove Theorem 13, we need the following results.
Lemma 14. Suppose , , and hold. Then functional satisfies the -condition; that is, for every sequence , has a convergent subsequence if is bounded and as .
Proof. Suppose , is bounded, and as , and then there exists a constant such that for every . On one hand, by , there exist constants and such that It follows from and (35) that Hence, by (34), (36), and Hölder inequality, we have On the other hand, by , there are constants and such that By , one has where . Hence, from (38) and (39), we get for all and a.e. . Hence, by (34) and (40), one has So is bounded in . If , by (37) and Hölder inequality, it is easy to obtain that is bounded in . If , by Lemma 3, we have Hence, by (37) and , we obtain is bounded in too. Thus, is bounded in . Since is a reflexive Banach space, by Lemma 3, there exists a subsequence, still denoted by , such that Next, we will show that in . Indeed, from (43) and hypothesis , it is easy to obtain that Hence, by (44), we get Therefore, it follows from (45) and Lemmas 3.2 and 3.3 in  that in . The proof of Lemma 14 is completed.
Proof of Theorem 13. By , for every , there exists such that
Combined with (35) and (46), we have
Thus, by Lemma 3, for , one has
Since , then there exist and such that
On the other hand, by , for any , there exists such that
It follows from that
Therefore, we obtain
for all and a.e. .
Now, we choose being the eigenfunction corresponding to which is defined in Lemma 8. For , from (52), we have We choose , and the above inequality implies that In view of Lemma 14, (49), and (54), noting that and applying the mountain pass theorem under the -condition, there exists a critical point of , such that . Hence is a nontrivial solution of problem (1), and this completes the proof.
We can weaken the condition in the following condition : uniformly for a.e. .
Theorem 15. Suppose that satisfies hypotheses , , and . Then problem (1) has at least one nontrivial solution in .
Proof. Checking the proof of Theorem 13, we only need to verify that (49) and (54) hold. In fact, by , there exist two constants and such that
From (35) and (56), we have
Thus, for , one has
Since and , then (49) holds.
By , there exist two constants and such that Hence, by and (59), we get for all and a.e. . Therefore, (60) and imply that Hence, we complete the proof.
Next, we consider the asymptotically quadratic case. For this purpose, we suppose satisfies the following conditions: uniformly for a.e. ;there exists such that for all and a.e. and uniformly for a.e. .
Theorem 16. Suppose , , , and hold. Then problem (1) possesses at least one nontrivial solution in .
Proof. Paralleling to the proof of Theorem 13, we only need to verify that satisfies the -condition. Suppose , is bounded, and as . Hence, it is easy to get that
We next show that is bounded in . If not, without loss of generality, we can assume that as . Letting , then , and so going to a sequence if necessary, we assume that in and in .
By , there exist two constants and such that From and (64), we get for all and a.e. . Hence, by (65), one has which implies that and thus . Therefore, there exists a subset with such that on .
By , from Lemma 2 in , then for , there exists a subset with such that uniformly for . Obviously, . If not, we assume . Since , thus we have which leads to a contradiction. Hence, we have proved that
From , we have By (68) and (70), we get as . This contradicts with (63). The proof is completed.
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
Kaimin Teng is supported by the NSFC under Grant 11226117 and the Shanxi Province Science Foundation for Youths under Grant 2013021001-3.
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