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Honghua Bin, "Periodic Solutions of Second-Order Difference Problem with Potential Indefinite in Sign", Abstract and Applied Analysis, vol. 2013, Article ID 517036, 8 pages, 2013. https://doi.org/10.1155/2013/517036
Periodic Solutions of Second-Order Difference Problem with Potential Indefinite in Sign
We investigate the periodic solutions of second-order difference problem with potential indefinite in sign. We consider the compactness condition of variational functional and local linking at 0 by introducing new number . By using Morse theory, we obtain some new results concerning the existence of nontrivial periodic solution.
We consider the second-order discrete Hamiltonian systems where is a given integer, , , , stands for the gradient of with respect to the second variable. is -periodic in the first variable and has the form , where for some , , stands for the greatest-integer function. For integers , the discrete interval is denoted by .
In this paper we consider that is sign changing, that is, , are two nonempty subsets of , where , is a -periodic real function, , and .
Consider the second-order Hamiltonian system where is -periodic in , . Here is a continuous, -periodic matrix-value function.
Systems (1) and (3) have been investigated by many authors using various methods, see [1–5]. The dynamical behavior of differential and difference equations was studied by using various methods, and many interesting results have obtained, see [6–10] and references therein. The critical point theory [11–14] is a useful tool to investigate differential equations. Morse theory [15–19] has also been used to solve the asymptotically linear problem. By minimax methods in critical point theory, Tang and Wu , Antonacci [20, 21] considered the problem (3) with potential indefinite in sign, where is superquadratic at zero and infinity. By using Morse theory, Zou and Li  study the existence of -periodic solution of (3), where is asymptotically superquadratic and sign changing. Moroz  studies system (3) where is asymptotically subquadratic and sign changing. Motivated by [5, 10, 19], we investigate periodic solutions for asymptotically superquadratic or subquadratic discrete system (1).
By expression of , system (1) possesses a trivial solution . Here we are interested in finding the nonzero -periodic solution of (1), and we divide the problem into two cases: and . For , one can refer to .
Case 1 (asymptotically superquadratic case: ). In this case, we replace with in (2). Letting , we rewrite (1) as Furthermore, for all , we assume that satisfies (A1) as uniformly in , (A2) as uniformly in .
Before stating the main results, we introduce space , where . For any , we define . Then is a linear space. Let , , , where and are the usual inner product and norm in , respectively. Obviously, is a Hilbert space with dimension and homeomorphism to . For , let . Moreover, for simplicity, we write and instead of and , respectively.
Theorem 2. If , then (4) has a nonzero -periodic solution.
Theorem 3. If , then (5) has a nonzero -periodic solution.
2.1. Variational Functional and (PS) Condition
We say that a -functional on Hilbert space satisfies the Palais-Smale (PS) condition if every sequence in , such that , is bounded and as contains a convergent subsequence.
Lemma 4. Functional satisfies (PS) condition if .
Proof. Let be the (PS) sequence for functional , such that is bounded, and as . Hence, for any , there exist and constant , such that
To prove that satisfies (PS) condition, it suffices to show that is bounded in . Suppose not that there exists a subsequence , as . For simplicity, we write as instead of . Without loss of generality, we assume that there exists , such that Therefore for all , by assumption (A1), there exists such that for large . By the previous argument, it follows that
By (7), we have In terms of (9) and (11), for large , it follows that Set . Dividing by in the previous formula, it follows that for large . Therefore, by being chosen arbitrarily, there is a subsequence that converges to such that
On the other hand, we have Then, by (9) and (11), for large , we get By dividing by in the previous formula, then by , we have as , that is, . By the definition of , see (6), we have . This contradicts with (16) and assumption . The proof is completed.
Lemma 5. Functional satisfies (PS) condition if .
The proof is similar to that of Lemma 4 and is omitted.
2.2. Eigenvalue Problem
Consider eigenvalue problem: that is, , . By the periodicity, the difference system has complexity solution for , where , . Moreover, . Let denote the real eigenvector corresponding to the eigenvalues , where and . Since for some , we can split space as follows: where
By means of eigenvalue problem, we have . Let Then for .
On the other hand, associating to numbers and (see (6)), we set where is the real eigenvector corresponding to eigenvalue . , where denotes the transpose of a vector or a matrix. Moreover, letting , we have , . Therefore, by definition of , if then .
However, by assumption for some , thus . That is to say the equality cannot hold. Therefore our discussion will be distinguished in two cases: and .
Let be a Hilbert space, and let be a functional satisfying the (PS) condition. Write for the set of critical points of functional and for the level set. Denote by the th singular relative homology group with integer coefficients. Let be an isolated critical point with value , the group , and is called the th critical group of at , where is a closed neighbourhood of . Due to the excision of homology , is dependent .
Suppose that is strictly bounded from below by , then the critical groups of at infinity are formally defined  as , .
Proposition 6 (Proposition 2.3, ). Assume that -functional satisfying (PS) condition has a local linking at with respect to ; that is, there exists such that Then , .
By Propostion 6, one proves the following lemmas with respect to .
Lemma 7. If , then , , where as , as . is defined by (23).
Proof. We first consider the following.
Case and , ). By , as , then there exists suitably small, such that as , where see (22) and . By assumption (A2) and , for any given , there exists , such that as , . Thus Let . For , it follows that
We need to prove that for , . We first claim that Indeed, by contradiction, assume that , for some , . Since , where , and is arcwise connected, then there exists a , such that . Thus by the definition of . On the other hand, by the definition of , we have . This is a contradiction with assumption . So the claim (27) holds.
There exists by (27), such that for all , where . For , , sufficiently small, we have Since and satisfies (PS) condition by Lemma 4, so by Proposition 6, we obtain that for .
Case 2 . It is easy to see that by and , where and . We need to claim that , for , .
Suppose not that there exists a sequence such that for large . For , by Lemma 1, we get Set . Then by (29) and the previous formula, we have
On the other hand, by the definition of . Hence by , there exists a subsequence converges to , such that , that is and . Since for , it follows from that Dividing by in the previous inequality, then .
Since , and is arcwise connected, then there exists a such that . Thus by the definition of . On the other hand, by the definition of . This is a contradiction with assumption . That is to say, the claim is valid.
By Proposition 6, we obtain , . The proof is completed.
Lemma 8. If , then for , where as , as .
The proof is similar to that of Lemma 7 and is omitted.
3. Proof of Theorem 2
Lemma 9. Let . If there exists such that for any , , then one has , and .
Proof. We first claim that is sufficiently large, if satisfies condition of Lemma 9. Suppose not there exists such that . So there exists , , such that as . Since for any , we have , thus . It is a contradiction with .
If is large enough, then we can assume that is large enough for and are bounded for . Therefore, by assumption (A1), for any given , there exists such that We claim that . Suppose not that, for , there exists such that By , (33) and (34), we have Set and divided by in the previous inequality. Since can be small enough, then there exists a subsequence that converges to , such that , . Moreover, by (33) and (34), we get Since and , divided by in the previous inequality, we have , that is, , which deduce a contradiction. So the claim holds.
Next we prove that holds. By contradiction, there exists a sequence such that, for , Then, by (7), we get and by (37) and , it follows that
Set and divided by in the previous formula; since and can be small enough, then there exists a subsequence converges to such that , . Moreover, by (37) and the first claim, we get Divided by in the previous formula, and by , it follows that . This is a contradiction with the definition of and condition . So the second claim holds. The proof is completed.
Lemma 10. If , then there exists , such that for any , is a strong deformation retraction of . Moreover, and are homotopy equivalent.
Proof. Now we prove that is a strong deformation retraction of .
By Lemma 9, we have . Let . By Lemma 9, there exists a unique such that . By applying Implicit Function Theorem, is a continuous function in . Let and define , then is a strong deformation retraction. Thus is a strong deformation retraction of .
We next claim that is a strong deformation retraction of . Clearly, in terms of the notations, we have be a function such that Define where is a projection operator. Then is a deformation retraction. Indeed, For , if , then where , that is, . If , it follows that We claim that the equality of the previous formula cannot hold. Otherwise, , for , which implies that . Hence in , which contradicts with the fact . So , that is, as . Therefore, is a deformation retraction from onto , and this completes the proof.
Proof of Theorem 2. Since is well known to be contractile in itself, and by Lemma 10, it follows that is homotopically equivalent to for large enough, then the Betti numbers (cf. [11, 13]) are
Now we suppose that system (4) has only trivial solution; that is, has only critical point , then we have the Morse-type numbers for (cf. ). Moreover, by Lemma 7, for or . Since satisfies (PS) condition by Lemma 4, then using Morse Relation, we have the following. which is a contradiction. Therefore, has at least one critical point and system (4) has at least a nonzero -periodic solution.
4. Proof of Theorem 3
For convenience, we introduce the following notations:
Lemma 11. Let . Then there exists such that for any , where stands for the closed ball in of radius with the center at zero.
Lemma 12. Let . Then there exists such that is a retract of , and is a strong deformation retraction of .
Proof of Theorem 3. We first prove that is contractible in itself. In fact, it is sufficient to show that is starshaped with respect to the origin; that is, implies that for all .
Assume, by a contradiction, that there exists and , such that . It follows from Lemma 11 that . By the monotonicity arguments, this implies that This contradicts the assumption , which implies .
On the other hand, since is contractible in itself, and is starshaped with respect to the origin, then is contractible in itself. The retract of the set which is contractible in itself is also contractible (cf. ); it follows that the set is contractible by Lemma 12.
Combining the previous argument, and are contractible in themselves. By Lemma 8, for or . Therefore, by Morse Relation and the same methods in proof of Theorem 2, it follows that has at least one critical point and system (5) has at least a nonzero -periodic solution.
This research is supported by the National Natural Science Foundation of China under Grants (11101187) and NCETFJ (JA11144), the Excellent Youth Foundation of Fujian Province (2012J06001), and the Foundation of Education of Fujian Province (JA09152).
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