A Nonlinear Schrödinger Equation Resonating at an Essential Spectrum

Chen, Shaowei; Zhou, Haijun

doi:https://doi.org/10.1155/2016/3042493

Advances in Mathematical Physics

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Research Article | Open Access

Volume 2016 | Article ID 3042493 | https://doi.org/10.1155/2016/3042493

A Nonlinear Schrödinger Equation Resonating at an Essential Spectrum

Shaowei Chen¹and Haijun Zhou¹

Academic Editor: Jacopo Bellazzini

Received12 Dec 2015

Accepted14 Feb 2016

Published02 Mar 2016

Abstract

We consider the nonlinear Schrödinger equation . The potential function satisfies that the essential spectrum of the Schrödinger operator is and this Schrödinger operator has infinitely many negative eigenvalues accumulating at zero. The nonlinearity satisfies the resonance type condition . Under some additional conditions on and , we prove that this equation has infinitely many solutions.

1. Introduction and Statement of Results

In this paper, the following nonlinear Schrödinger problem in () is considered: The nonlinearity satisfies the resonance type condition: Here, denotes the essential spectrum (see [1, Chapter 7] or [2, Chapter 7.4]) of the Schrödinger operator defined as follows:

Equation (1) arises in quantum mechanics and is related to the study of the nonlinear Schrödinger equation for a particle in an electromagnetic field and it has attracted considerable attention from researchers in recent decades. One can see [3–6] and the references therein. However, there are very few results on (1) with the resonance type condition (2).

The following nonlinear elliptic problem with resonance type conditions in bounded domain has been studied by many authors and numerous existence and multiplicity results have been obtained in the past forty years: Here, is a bounded domain in , and satisfies the resonance type condition: is the th eigenvalue of with -Dirichlet boundary condition on . We refer to [7–10] and references therein for more detailed discussions of some historical results. This problem has deeply inspired developments in critical point theory in the last forty years, such as the Landesman-Lazer type conditions (e.g., [8, 10]), Morse theory (e.g., [9]), and the variational reduction method (e.g., [11]). The difficulty of this problem lies in the proof of the boundedness of the Palais-Smale sequence or the sequence (see Definition 5 in Section 2) of its corresponding functional.

The resonant Schrödinger problem (1) is much more difficult than the bounded domain case and it has fewer studies. Unlike the case of bounded domains for which the linear operators in bounded domain are compact, there are continuous spectra of the linear operator . Moreover, the proofs of boundedness and compactness of the sequence of the corresponding functional of (4) are greatly different from and more difficult than the case of bounded domain (see [4, 5, 12]).

More precisely, this paper considers the nonlinear Schrödinger equation: Here, and is the Sobolev space:

For , the following is assumed:(V₁)Consider , in , , with if and , for some if , and there exist , , and such that, for ,(V₂)For any , and there exists such that

For , the following is assumed:(f₁) is a continuous function:(f₂)There exists such that, for any , Here, For , let(f₃)For any , and there exists such that if , then(f₄)For any , .

The main result of this paper is as follows.

Theorem 1. Suppose that and are satisfied; then, problem (6) has infinitely many solutions.

Remark 2. (i) Under the assumptions and , the essential spectrum of defined by (3) is (see Lemma 3 in Section 2). Therefore, condition (11) indicates that satisfies a resonance type condition at the essential spectrum of .
(ii) A typical example for which satisfies is Here, , , , , and in .
(iii) There are many functions satisfying . Let . Using we can construct a function satisfying from a given . For example, let where Then, by (19), for , and, for , It follows that and for any . Because the even function is strictly increasing in , as , it can be deduced that satisfies , . Moreover, it is easy to verify that as , as , as , and as . Because the even function is strictly increasing in , as , and as , it can be deduced that, for sufficiently small , if , then Therefore, satisfies .

Let The inner product and the norm in are respectively. Here, is the constant in (10).

Let Then, for any , the Gateaux derivative of is as follows: It follows that is a solution of (6) if and only if is a critical point of . In Section 2, it is shown that satisfies the Cerami condition. In Section 3, an abstract critical point theorem from [13] is used to show that has infinitely many critical points.

Notation. Let be Banach space. We denote the dual space of by and denote strong and weak convergence in by and , respectively. For , we denote the Fréchet derivative of at by . The Gateaux derivative of is denoted by . denotes the standard space . We use and to mean and , respectively.

2. The Cerami Condition for

Lemma 3. If satisfies the assumptions , then the essential spectrum of the Schrödinger operator defined by (3) is and has infinitely many negative eigenvalues accumulating at zero.

Proof. Because satisfies , by [14, Theorem XIII.15(b)] or [2, Theorem 10.29(b)], the essential spectrum of the Schrödinger operator defined by (3) is . The second result of this lemma comes from [2, Theorem 10.31] or [14, Theorem XIII.6(a)].

Lemma 4. If satisfies the assumptions and , then has no eigenvalue in .

Proof. Because satisfies the assumptions and , by [15, Corollary ], to prove this lemma, it suffices to prove that satisfies the conditions I and II in [15, Corollary ]. For this, it suffices to prove that, for , the functions are bounded functions in . Because and , there exists such that, for , Because , for , the following is true: Because , the following holds: Combining (29) and (30) yields that is a bounded function in . From , we have , . It follows that is also a bounded function in .

Definition 5 (see [16]). Let be a Banach space. Let and . One can call the fact that satisfies the Cerami condition at , denoted by condition, if for any satisfying there exists such that, up to a subsequence, as . If satisfies the condition for every , then is said to satisfy the Cerami condition. A sequence satisfying (32) is called a sequence of .

It was shown in [13] that the Cerami condition actually suffices to get a deformation theorem (see [8]) and then, by standard minimax arguments, it allows rather general minimax results.

Lemma 6. Suppose that satisfies that and . If in , then

Proof. Because in , is bounded. It follows that From , we deduce that, for any , there exists such that It follows that For and , let Because and in , by Lemma of [6], we have Combining (36) and (38) yields (33).

Lemma 7. Let be a sequence of . Then, up to a subsequence, is bounded.

Proof. By Definition 5, we have Arguing indirectly, assume that and set .
Case 1. is nonvanishing; that is, there exist and such that Set . Then, by (40), in
By (39) and (27), for any , we have, as , It follows that, as , where . If , sending in (42), by (11), and the fact that, for , , we get that This is impossible, because . If has a subsequence bounded in , up to a subsequence, it can be assumed that , and sending in (42), we get that It follows that This implies that is a nonzero eigenfunction of corresponding to eigenvalue . It contradicts Lemma 4.
Case 2. is vanishing; that is, The Lions lemma (see, e.g., [6, Lemma ]) shows that, for any , Here, denotes norm. Equations (39) and (27) show that, as , where is defined in (15). It follows that By Lemma 6 and in , we have Together with (49), this implies By (39), Let By , (17), and (52), the following is true: Here, denotes the Lebesgue measure of . It follows that Let Because (see ), we have Under the assumption , there exists such that It follows that for some constant . Then, by (55) and (47), for , Combining (57) and (59) yields This contradicts (51).
From the two cases given above, it can be deduced that is bounded in .

Lemma 8. The functional satisfies the condition for every .

Proof. Let be a sequence of . By Lemma 7, is bounded. It follows that there exists such that, up to a subsequence, in . Because the embedding is compact, we get that, up to a subsequence, As the proof of Lemma of [6], we have . It follows that We are going to prove that, up to a subsequence, as .
By (39), as , By Lemma 6, Combining (62)–(64) yields Because in , by [17, Theorem ], we have Because for any (see (12)), by (61) and the Fatou lemma, we have Using (65)–(67), it can be deduced that, up to a subsequence, By (13), is a sequence of nonnegative functions. Because by the Fatou lemma and (69), the following is true: It follows that By (12), we deduce that in . Then, by and the Fatou lemma, the following holds: Combining (72) and (74) leads to By (10) and (11), it can be deduced that there exists such that Then, by the mean value theorem, the Hölder inequality, and (76), we get that, for any and for any sufficiently small , there exists such that Because (61), (75), and (77) hold and is bounded, Theorem of [18] shows that It follows that, for every sufficiently small , Here, Equations (10) and (12) show that Combining (81) and (82) yields that, for any sufficiently small , there exists such that By (79) and (83), we get that, for every sufficiently small , From we get that By (86), the following holds: where and Combining (84) and (87) leads to Together with (68), this implies in . This completes the proof.

3. Proof of Theorem 1

For the proof of Theorem 1, the following abstract theorem, which is a corollary of [13, Theorem ], is used.

Theorem 9. Let be a real Hilbert space with norm . Suppose that is a -functional defined in and satisfies the following conditions:(A₁) satisfies the condition for every , and .(A₂) is an even functional; that is, for every .(A₃)There exist and a closed subspace of with finite codimension such that there exists such that Here, (A₄)For any , there exists a subspace with dimension larger than or equal to such that Then, has infinitely many pairs of critical points.

Proof. Arguing indirectly, assume that has only a finite number of critical points, say . The codimension of is here denoted by By , there exists a subspace with dimension larger than or equal to such that This indicates that there exists such that Then, by [13, Theorem ], has at least critical points. This contradicts the assumption that has only critical points.

Let be the eigenvalues of below the essential spectrum of . Each has been repeated in the sequence according to its finite multiplicity. Then, by Lemma 3, the following is true:Let Here, is the constant in (10). From this definition, Let denote the spectral family of the operator defined by (3).

Let Here, is the identity map and It is easy to see that the codimension of equals .

Lemma 10. There exists such that

Proof. From (96) and (98), the following holds: Here, It follows that If , then there exists satisfying , , and Up to a subsequence, it can be assumed that in and a.e. in . Since is a closed subspace of , by in , we have . Because in , by [17, Theorem ], we have Because a.e. in , by the Fatou lemma, the following holds: Moreover, by Lemma 6, Combining (104)–(107) yields Because , by (101) and (108), we have . Then, by (104) and (107), This contradicts , . Therefore, .

Proof of Theorem 1. We are going to prove that the functional satisfies the conditions in Theorem 9.
By Lemma 8, the functional satisfies the condition . By , satisfies the condition .
By (10) and (11), it can be deduced that, for any , there exists such thatHere, Let be the space defined by (98). It has a finite codimension. By Lemma 10 and (110), we have, for any , where the positive constant comes from the Sobolev inequality , . Choosing and , by (112), the following holds: It follows that satisfies the assumption .
Let such that and . Then, for the function , and . Then, by (8), it follows that, for ,where is a positive constant independent of . Because , it follows that there exists such that Let , . These functions have mutually disjoint compact supports. Let Then, is an -dimensional subspace of . By (10) and (11), it can be deduced that, for any , there exist and such that Because have mutually disjoint compact supports, by (115), we have, for , where Let Then, is a bounded set in , since every has a compact support. By (117), where Choosing by (118) and (121), we get that Since , for every , (124) implies Therefore, satisfies the condition of Theorem 9.
Because satisfies the conditions of Theorem 9, by Theorem 9, it has infinitely many critical points. It follows that (6) has infinitely many solutions.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

Shaowei Chen was supported by Science Foundation of Huaqiao University (13BS208) and Promotion Program for Young and Middle-Aged Teacher in Science and Technology Research of Huaqiao University (ZQN-PY119).

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Copyright

Copyright © 2016 Shaowei Chen and Haijun Zhou. 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.

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