Abstract
By applying Clark’s theorem as altered by Liu and Wang and the truncation method, we obtain a sequence of solutions for a Schrödinger–Poisson system with negative energy. A similar result is also obtained for the Schrödinger-Kirchhoff equation as follows:.
MSC:
35J47; 35J50; 35Q55
1. Introduction
In this paper, we consider stationary Schrödinger–Poisson systems of the form
We assume the following conditions for the nonlinearity f and the potential V:
- is odd, for some and ,
- is continuous and .
The condition (2) means that the nonlinearity f is sublinear at the origin and subcritical at infinity. Under these mild conditions, we have the following multiplicity results.
Theorem 1.
Suppose and are satisfied, then the problem (1) has a sequence of weak solutions .
Example 1.
Let and be defined by
Then, f and V satisfy our assumptions and with , respectively.
Remark 1.
Note that in the assumption , we have assumed that V is positive. This condition can be replaced by the following weaker one which allows V to be negative somewhere:
- , in and for some .
To see this, set , with
Then, and satisfy the assumptions and , and hence we can apply Theorem 1 to the equivalent problem
The problem (1) is variational. Let be ths solution of the second equation in (1). As proposed by Benci et al. [1,2], it is well known that if u is a critical point of
being , then is a solution of (1). Using this idea, many results on (1) have been obtained assuming that
in the last two decades; see [3,4,5] for a case where the Schrödinger operator is positive and [6] for a case where the Schrödinger operator S is indefinite.
In all these papers, and many papers on Schrödinger–Poisson systems, some conditions on the nonlinearity f, like the Ambrosetti–Rabinowitz condition, are needed to ensure that the variational functional satisfies the Palais–Smale condition.
Unlike all these papers, our assumption on the nonlinearity f implies that the limit in (4) is infinity, and our assumptions on f are not sufficient for ensuring the boundedness of sequences. Motivated by He and Wu [7], who studied the semilinear elliptic boundary value problem
on a bounded domain , we apply the truncation method and a version of Clark’s theorem created by Liu and Wang [8] to overcome this difficulty.
Using the same idea, we obtained a similar result for Schrödinger–Kirchhoff equations of the form ()
We make the following assumptions:
- is odd, for some and ,
- is continuous and .
Note that is the critical Sobolev exponent. If , then and and reduce to and , respectively.
Theorem 2.
Suppose and are satisfied, then the problem (5) has a sequence of weak solutions .
For problem (5), we make a remark similar to Remark 1. This paper is organized as follows. In Section 2, we present the functional spaces as our frameworks for studying problems (1) and (5), and Clark’s theorem as altered by Liu and Wang [8] is also recalled in this section. The proofs of Theorems 1 and 2 will be given in Section 3 and Section 4, respectively.
2. Preliminaries
Equip the subspace ()
with the norm and the corresponding inner product given by
Then, is a Hilbert space. Since , to prove Theorem 1, it suffices to find a sequence of critical points of the -functional given in (3). Note that by the definition of ,
Obviously, the embedding is continuous. According to [9] [Lemma 2.1], we obtain the following result.
Proposition 1.
Under the assumption , can be continuously embedded into for and the embedding is compact for .
Corollary 1.
Under the assumptions and , if in , then up to a subsequence
Proof.
Given , there is such that
Thus, by Hölder inequality
Because is bounded in and in by Proposition 1, we obtain
for some , which implies (6). □
To find the critical points of , the properties of and the functional
are crucial. Similar to [10] [Lemma 2.2 (1)], we have the following proposition.
Proposition 2.
Let be the unique solution of for . Then, there is a constant such that
To conclude this section, we recall Clark’s theorem, which will be needed for proving our main results.
Proposition 3
([8] [Theorem 1.1]). Let X be a Banach space and be an even coercive functional satisfying the condition for and . If for any there is a k-dimensional subspace and such that
where for , , then Ψ has a sequence of critical points such that , .
3. Proof of Theorem 1
Let be a decreasing -function such that ,
We consider the truncated functional ,
The derivative of is given by
for .
Lemma 1.
The functional Ψ is coercive and satisfies for .
Proof.
Let be a sequence such that . Then, for n large, we have and . Thus
Hence is coercive.
For , let be a sequence of . That is , . Then for n large we have
We claim that
We consider two cases:
which implies (11) because .
Now we are ready to prove Theorem 1.
Proof of Theorem 1.
Since for (the unit ball in ), it suffices to show that has a sequence of critical points satisfying . For this purpose, we shall apply Proposition 3.
Firstly, we remark that by condition , there is such that
Let and be a k-dimensional subspace of . Since all norms on are equivalent, there is such that implies ; thus
for . Hence, for , using Proposition 2 we obtain
being for some constant . Because it is clear that there is such that . Hence,
As we have seen, the -functional is even, coercive and satisfies for . Since is trivially true, by Proposition 3 has a sequence of critical points converging to the zero function. This completes the proof of Theorem 1. □
4. Proof of Theorem 2
Proof of Theorem 2.
Now we will prove Theorem 2 using similar method. The solutions of (5) can be found as critical points of given by
As before, to overcome the difficulty that sequences may be unbounded, we consider the truncated functional ,
where is the function given at the beginning of §3. Then,
Similar to the proof of Lemma 1, the functional is coercive.
Given , let be a sequence. Then is bounded, so
for some , and we have an analog of (11):
However, unlike in the proof of Lemma 1, we could not apply Proposition 2 to obtain the analog of (12):
To go around this difficulty, by the boundedness of in we can observe that the left-hand side of (17) is bounded, and hence up to a subsequence we may assume
Now, by analyzing using (16)
and
(see Corollary 1) we obtain
Thus,
which implies that , and we still obtain in . This completes the proof that satisfies for .
Funding
This research received no external funding.
Data Availability Statement
The manuscript has no associated data.
Acknowledgments
The author would like to thank the referees for their invaluable comments on the previous version of this paper.
Conflicts of Interest
The author declare no conflicts of interest.
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