1. Introduction
In this paper, we consider the Schrödinger–Poisson equation with critical growth:
The Schrödinger–Poisson equation arises while looking for standing wave solutions of a Schrödinger equation interacting with an electrostatic field. In recent years, many researchers are interested in semiclassical states of
which can be used to describe the transition from quantum to classical mechanics.
When
, problem (
2) reduces to the singularly perturbed problem
In the past decade, there is a lot of results on problem (
3). By using variational methods, Rabinowitz [
1] first obtained the existence of solutions of (
3) under the assumption
In [
2], Wang proved the concentration behavior of solutions of (
3) as
. In [
3], del Pino and Felmer introduced a penalization approach and obtained a localized version of the results in [
1,
2]. In [
4], Jeanjean and Tanaka extended the results of [
3] to a more general case. For other related results, see [
5,
6,
7,
8] and the reference therein.
When
, a lot of research focus on the case
,
. By using the Lyapunov–Schmidt reduction method, the authors in [
9,
10] obtained positive bound state solutions and multi-bump solutions concentrating around a local minimum of the potential
V. In [
11,
12], the authors proved the existence of radically symmetric solutions concentrating on the spheres. It should be pointed it out that, the Lyapunov–Schmidt reduction method is based on the uniqueness or non-degeneracy of solutions of the corresponding limiting equation. Recently, by using variational methods, he [
13] considered the subcritical problem
Under the assumption (
4), he related the number of solutions with the topology of the set where
V attains its minimum and obtained the multiplicity of positive solutions. Subsequently, the authors in [
14] studied the problem
Under suitable assumptions on
,
V,
b and
f, they proved the existence and concentration behavior of positive ground state solutions. For the critical case, He and Zou [
15] studied the Schrödinger–Poisson equation
By using (
4), they obtained the ground state solution concentrating around the global minimum of the potential
V. Furthermore, in [
16,
17], the authors considered the existence, multiplicity and concentration behavior of the critical Schrödinger–Poisson equation
Motivated by the above results, in this paper, we study the multiplicity and concentration behavior of positive solutions of (
1). Before stating the results, we introduce the following conditions:
, and . Moreover, there exist k points , , … in such that each point is the strict global maximum of h.
and .
, , 2, … k and .
and . Moreover, , , 2, … k and .
and . Moreover, is increasing for and .
Theorem 1. Assume that , –, and hold. Then there exists such that problem (1) has at least k different positive solutions , , 2, … k for . Moreover, possesses a maximum point satisfying as . Besides, there exist , such that for , Remark 1. In Theorem 1, we obtain the existence of spikes (multiple solutions concentrating at a single point) on the strict global maximum of h. The behavior of solutions of (1) describes the transition between quantum mechanics and classical mechanics in some sense. Remark 2. Some ideas to prove Theorem 1 come from [18], where the authors studied the subcritical problem In [18], the authors imposed the conditionwhich plays an important role in proving the compactness of the Palais–Smale sequences. This type condition is first introduced by Rabinowitz in [1]. We pointed out that, when we seek multiplicity of solutions, it is crucial to prove the compactness of the Palais–Smale sequence. Many authors solved the problem by imposing the Rabinowitz type assumption, which is restrictive. Similar results can be found in [13,15,16] and the reference therein. In this paper, by estimating the Palais–Smale sequences delicately, we remove this technical condition. In fact, we use a different argument. Compared with the existing results, in this paper, we also need to study the influence of the variable coefficient of the critical term on the problem. Inspired by Theorem 1, a natural question is whether (
1) has multiple solutions without any restriction on
. In this paper, by using the Lusternik–Schnirelman category, we obtain a new result. We assume the following conditions:
is positive and .
.
and . Moreover, there exists such that for .
Theorem 2. Let . Assume that , , , and hold with . Then there exists such that for , problem (1) has two positive solutions , , 2. When
, there are an enormous amount of papers studying problem (
1) or the more general form
Many papers, see for example, [
19,
20,
21,
22,
23,
24,
25,
26,
27], focus on the case
V and
K being positive constant or radially symmetric functions,
or
. If
V is non-radial,
,
, the authors in [
20,
28] obtained the existence of ground state solutions of (
10) for
. If
,
is non-radial, by requiring suitable assumptions on
K and
a, the authors in [
29] obtained ground state and bound state solutions. For other related results, see [
30,
31,
32,
33] and the reference therein. Usually, in order to ensure the boundedness of the Palais–Smale sequences, the Ambrosetti-Rabinowitz condition or some monotonicity condition on
f is needed. It is natural to ask whether we can prove the boundedness of the Palais–Smale sequences without the above restrict conditions. When we study (
1), we solve the problem under mild conditions and get an interesting result. Instead of
and
, we assume the following conditions.
and . Moreover, and there exists such that for .
and for . Moreover, and .
Theorem 3. Let . Assume that , , , and hold. Then problem (1) has a positive solution. The outline of this paper is as follows: in
Section 2, we give some important lemmas; in
Section 3, we prove Theorems 1; in
Section 4, we prove Theorem 2 and 3; in
Section 5, we make the conclusions.
Notations:
, .
denotes the Hilbert space equipped with the inner product and the norm , denotes the Sobolev space equipped with the norm .
denotes the best Sobolev constant.
C denotes a positive constant (possibly different).
3. Proof of Theorem 1
By
, we derive that
By
, we have
,
, 2, …
k. We consider the equation:
Let
. Then the functional of (
14) is
Let
, where
. It is well known that
is attained by
w. Moreover,
and
. Let
. Define the functional on
by
Let
, where
. For any
, by
, we have
. Moreover, there exists a unique
satisfying
. Then by
,
,
, we get
So . Similarly, we have .
For , denote the hypercube centered at , , 2, … k. Denote and the closure and the boundary of , respectively. By , we have . Moreover, there exist , such that , , 2, … k are disjoint, for and .
Let . Define such that for , for , and . Let , , 2, … k. Let . Define such that for , for . For , define . Then is continuous in .
For simplicity, denote . Set , , , 2, … k. Let , , , 2, … k.
Lemma 3. Let , 2, … k. Then for any , there exists such that for .
Proof of Lemma 3. It is clear that
. Let
, where
. By
, we derive that
admits a unique critical point
corresponding to its maximum. Then
,
. So
. Note that
. By the Lebesgue dominated convergence theorem,
Then
for
small. Since
, by the definition of
, we get
. So we just prove
for
small. By
and Lemma 1
, we derive that
By
,
,
and the Lebesgue dominated convergence theorem, we get
By
, we have
From (
15)–(
17), we obtain that
is bounded. Furthermore, by (
17),
,
,
, for
, there exists
satisfying
Then
in view of (
15) and (
16). By (
15)–(
17),
that is,
. By
, there exists a unique
satisfying
. Since
, we have
. Then by (
15) and (
16),
So there exists such that for . □
Lemma 4. Let , 2, … k. Then there exist δ, such that for .
Proof of Lemma 4. Assume to the contrary that there exists
such that
. Then there exists
satisfying
. By
, there exists a unique
such that
. Then
from which we get
. Moreover,
Since
, by the Ekeland’s variational principle, there exist
,
satisfying
,
,
, where
. By the standard argument, we derive that
as
. Then
By (
13), we get
is bounded. By the Lions Lemma,
for any
, or there exists
such that
weakly in
. If
for any
, then
. Let
. Then
. Since
, we have
. By
, we get
. Thus,
, a contradiction with
. So
weakly in
. By (
19), we have
,
. Then
. Moreover,
So
in
. Note that
. By
, we get
. Thus,
If
, by
in
, we get
, a contradiction with
and
. So
is bounded. Assume that
as
. If
, by (
20), we get
, a contradiction. If
, we have
. Then
. By
,
in
,
Then by (
18), we get
, a contradiction with
. □
By Lemmas 3 and 4, there exists
such that
for
. Furthermore,
as
. By
, for
, there exists
such that
Note that
. We choose
small such that for
,
Following the ideas of [
18,
35], we can use the implicit function theorem to get the following result. Since the proof is standard, we omit it here.
Lemma 5. Let , , 2, … k. Then for any , there exist and a differential function , where and , satisfying , . Moreover, for any ,that is, , where . Lemma 6. Let , , 2, … k. Then there exists satisfying , . Moreover, converges strongly in up to a subsequence.
Proof of Lemma 6. By the definition of
, there is
satisfying
. Then
is bounded in view of (
13). By the Ekeland’s variational principle,
,
for any
. By Lemma 5, there exist
and
satisfying
for any
with
. Let
, where
,
. For
small,
where
. Dividing by
t and let
,
. By Lemma 5,
. By (
13),
So . If , we get , a contradiction with . Then . Since is bounded, we know is bounded. So . Assume weakly in .
Case 1. weakly in .
Define the functional on
by
By
,
, we get
,
. By the Lions Lemma,
for any
, or there exists
with
satisfying
weakly in
. If
for any
, by Lemma 1
, we get
. Moreover,
. Then
Since
, we assume
. By
, we get
. So
, a contradiction. Thus,
weakly in
with
. Define
By
weakly in
and
, we get
Then
. Let
. By Lemma
in [
36], we have
By the Brezis–Lieb Lemma in [
34],
. By the Lebesgue dominated convergence theorem,
. Then
Combining (24) and (25) and Lemma 1
, we get
If
in
, that is,
in
, by
,
a contradiction with
. So
converges weakly and not strongly to 0 in
. By Lemma 8.9 in [
34],
for any
. By the Lebesgue dominated convergence theorem,
for any
. Then
Similar to Lemma 8.1 in [
34], we derive that
Together with Lemma 1
,
,
, we get
. Thus,
By (
21), (
27) and
, we derive that
So
. Since
converges weakly and not strongly to 0 in
, we get
. Since
, we have
. Then by (26) and (27),
a contradiction with (
22).
Case 2. weakly in .
By
weakly in
, we have
. By Lemma 1.3 in [
36],
where
. Together with the Brezis–Lieb Lemma in [
34] and Lemma 1
, we get
. By Lemma 8.9 in [
34], for any
,
. Similar to Lemma 8.1 in [
34], for any
,
. Together with Lemma 1
, we get
. Thus,
We claim
in
. Otherwise,
converges weakly and not strongly to 0 in
. By
,
, we have
By (
21),
and
,
By the Sobolev embedding theorem, we get
. Since
, we have
. Then by (28),
a contradiction with (
22). So
in
, that is,
in
. □
Lemma 7. Let , Then problem (11) has at least k different positive solutions , , 2, … k. Proof of Lemma 7. Let , 2, … k. By Lemma 6, we have , , . Moreover, in . Then , , . By , we have . Since , where , , 2, … k are disjoint, we derive that , , 2, … k are different. Obviously, is non-negative. By the maximum principle, is positive. □
Now we study the behavior of as .
Lemma 8. Let , 2, … k. Then there exist , , , satisfying for .
Proof of Lemma 8. Otherwise, there exists
such that for any
,
By the Lions Lemma,
for any
. Since
,
, similar to the argument of (
19), we get
, a contradiction. □
Lemma 9. as .
Proof of Lemma 9. We first prove
is bounded. Assume to the contrary that there exists
satisfying
. By
,
, we derive that
is bounded. Let
. By Lemma 8, we get
weakly in
. Let
Then
,
. Furthermore,
By
, we derive that
that is,
. By
,
,
Then by Fatou’s Lemma and
,
Since
, we have
. Combining (29) and (30), we get
in
. Note that
Then by , we get , a contradiction with .
Now we prove
as
. Since
is bounded, we assume
as
. Let
. By Lemma 8, we get
weakly in
. Then by
, we derive that
Then
. By
,
,
Since
, by
, we have
. Moreover, there exists a unique
satisfying
. Then
Combining (31)–(33), we get
and
in
. Then
. Note that
If , then , a contradiction with . So , from which we derive that . Since , , we get as . □
Proof of Theorem 1. By Lemma 7, problem (
11) has at least
k different positive solutions
,
, 2, …
k. Let
. Then
By Lemma 9,
in
,
. By the argument of Lemmas 3.8 and 3.11 in [
15],
uniformly for
and there exists
independent of
satisfying
. Furthermore, there exist
,
such that
uniformly for
.
We claim there exists such that uniformly for . Otherwise, as . By (34) and , we get . Then , a contradiction with in . Let be the maximum point of . Then . By uniformly for , we derive that there exists such that .
Since
, we know
is the maximum point of
. By
as
and
, we get
as
. Moreover, for
,
Let
,
. Then
is the positive solution of (
1). Furthermore,
is the maximum point of
,
and there exist
,
such that
for
. □
4. Proof of Theorems 2 and 3
For simplicity, let
. Denote
the Hilbert space with the norm
. Let
Let , where .
We first prove Theorem 3. Let X be the Banach space. Recall that is a sequence for the functional I if and as .
Theorem 4 ([
37])
. Let X be the Banach space and satisfyingfor some and with . Let , where . Then there exists a sequence for the functional I satisfying . For any with , by , there exist , such that for . Let . Define , where such that for , , .
Lemma 10. Assume that , , and hold. Then there exists independent of with such that for any , Proof of Lemma 10. From Lemma 1
, there exists
such that
By Lemma 2 and
, there exists
independent of
y such that for
,
Then there exist a small
and a large
independent of
satisfying
By Lemmas 1 and 2, we get
By
, we have
. Then for
, there exists
satisfying
for
. Let
. Note that
for
. Let
and
. Then for
and
, we have
. Since
, we derive that
Thus, for
,
Combining (35) and (36), we get Lemma 10. □
Define the functional on
H by
Lemma 11. Assume that , , , and hold. If such that weakly in H, and , then and weakly in H. Moreover, if , then in H.
Proof of Lemma 11. By
weakly in
H, we have
. Let
. By Lemma 1.3 in [
36], we obtain that
. By
and
, for any
, there exists
such that
. Then for any
,
Since
is bounded, by
, for any
, there exists
such that
By
weakly in
H, we have
. Thus,
By the Brezis–Lieb Lemma in [
34], we get
and
. Together with Lemma 1
, we have
By Lemma 8.9 in [
34], we have
for any
. Similar to Lemma 8.1 in [
34], we derive for any
, there holds
. Together with Lemma 1 , we get . So . Similar to (37), we get . Thus, we have .
Assume to the contrary that weakly in H. Then . Assume
. Then . If , by , we get . So . Since , we get , a contradiction. So . By , we get in H, a contradiction with . So weakly in H.
If , similar to the above argument, we can derive that in H. We omit the proof here. □
Proof of Theorem 3. From
,
, for
, there exists
such that
. By the Sobolev embedding theorem,
Then there exist
,
such that
for
. Furthermore,
. By the argument of Lemma 4.1,
. So by Lemma 10 and Theorem 4, there is
such that
If
is bounded, by Lemma 11, we have
weakly in
H and
, that is, (
1) has a positive solution. So we just need to prove that
is bounded.
Otherwise, we have . Let . Then weakly in H.
Case 1.
a.e.
. Let
and
. By
,
and
, for
, there exists
such that
From (39), we derive that
Similar to (37), we have . Then by , and , we get , a contradiction.
Case 2.
. Let
. Then the measure of
is positive. For
, by
, we get
. Let
. By
,
and Fatou’s Lemma,
Furthermore, for
, there exists
such that
. Then
from which we get
Combining (40) and (41), we derive that
On the other hand, by Lemma 1,
,
So , a contradiction with (42). □
Now we prove Theorem 2. By the Lagrange multipliers Theorem, we can derive the following result. Since the proof is standard, we omit it here.
Lemma 12. Assume that , , and hold. Let such that and . Then . Moreover, is bounded.
Lemma 13. Assume that , and hold. Then there exists such that for with , where .
Proof of Lemma 13. Assume to the contrary that there exists
such that
and
. Then
. By
, we get
. So
from which we derive that
,
Let . We get , ,
. By Theorem 1.41 in [
34], there exist
and
such that
So
,
, that is,
S is attained by
. By [
38],
, where
,
,
. Thus,
By , there exist and such that for .
Case 1. as .
Let . By , we get . Then . By (44), we have a.e. By Fatou’s Lemma, we get , a contradiction.
Case 2. as .
Similar to Case 1, we get a contradiction.
Case 3. as and for large n.
Assume
. Then
. By (43), we have
. Then by (44),
By the Lebesgue dominated convergence theorem, we derive that , a contradiction.
Case 4.
as
and there exists a subsequence of
(still denoted by
) satisfying
. By
and (45),
By
and
, we have
for large
n. For any
x,
,
From (46)–(48), we get
a contradiction. □
Lemma 14. Assume that , , and hold. Then there exists such that for and with , there holds .
Proof of Lemma 14. By
, for any
, there exists a unique
such that
. Then
If
, then
that is,
, a contradiction. So
. By (49),
Since
, by
,
, for
, there exists
such that
. Then for
,
Since
, we obtain that there exists
such that
. By
with
, we get
. Then
. So
. By (50), there exists
such that for
,
Then
. Together with (49) and
, we derive that there exists
such that
. By
, we obtain that for
with
,
By and , there exists such that for . Since , by Lemma 13, we get , that is, . □
We introduce the Lusternik–Schnirelman category.
Definition 1. For a topological space X, a nonempty, closed subset is contractible to a point y in X if and only if there exists a continuous mapping such that for and for .
Definition 2. In particular, if there does not exist finitely many closed subsets such that and is contractible to a point in X for all i, denote .
The following two lemmas are introduced to prove Theorem 2.
Lemma 15 (Lemma 2.5 in [
39])
. Let X be a topological space. Assume there exist two continuous mappingsuch that is homotopic to identity, that is, there is a continuous mapping such that for and for . Then .
Lemma 16 (Proposition 2.4 in [
39])
. Let M be a Hilbert manifold and . If there exist and such that satisfies the Palais–Smale condition for and , then admits at least k critical points in . Proof of Theorem 2. We note that
. By the proof of Theorem 3, there exist
,
such that
By the argument of Lemma 10,
. Then
attained its maximum at a
. So
. We note that
By
, we get
is unique. Moreover,
. By Lemma 10, for
,
Define by . Then P is continuous and
. Define
by
. By Lemma 14, we know
Q is continuous. Define
such that
for
,
for
and
for
. By the argument of Case 4 in Lemma 13, we have
By a direct calculation, for
,
By a direct calculation, for
,
Thus, we have
. By Lemma 2,
Then by the continuity of
Q, we obtain that
,
for
and
for
. By Lemma 15, we have
By Lemmas 11 and 12, we know I satisfies the condition for . Then by Lemma 16, we obtain that I has two nonnegative critical points , , 2. By the maximum principle, is positive. □