Nontrivial solutions for a nonlinear Schrödinger equation with nonsymmetric coefficients

Nonlinear Analysis 195 (2020) 111755

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Nontrivial solutions for a nonlinear Schrödinger equation with nonsymmetric coefficients✩ Xiaoping Wang, Fangfang Liao ∗ Department of Mathematics, Xiangnan University, Chenzhou, Hunan 423000, PR China

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abstract

Article history: Received 22 December 2019 Accepted 12 January 2020 Communicated by Vicentiu D. Radulescu

This paper is dedicated to studying the nonlinear Schrödinger equations of the form { −△u + λu = a(x)f (u), x ∈ RN ; u ∈ H 1 (RN ),

MSC: 35J20 35J65

where a ∈ C 1 (RN , [0, ∞)) is nonsymmetric and satisfies a(x) ≤ (̸≡) a∞ := lim|y|→∞ a(y), and f ∈ C(R, R). By using concentration compactness arguments and projections on a general Pohozaev type manifold, we prove that the above problem has a nontrivial solution under weaker assumptions on a and f than the existing results in literature. © 2020 Elsevier Ltd. All rights reserved.

Keywords: Schrödinger equation Pohozaev identity Concentration compactness Cerami sequence Barycenter

1. Introduction In this paper, we consider the nonlinear Schr¨ odinger equations of the form: { −△u + λu = a(x)f (u), x ∈ RN ; u ∈ H 1 (RN ),

(1.1)

where N ≥ 3, λ > 0, a : RN → R and f : R → R satisfy the following basic assumptions: (A1) a ∈ C(RN , [0, ∞)), with a0 := inf x∈RN a(x) > 0; (A2) a∞ := lim|y|→∞ a(y) < ∞; (F1) f ∈ C(R, R), tf (t) ≥ 0, and there exist constant C0 > 0 and p ∈ (2, 2∗ ) such that ( ) p−1 |f (t)| ≤ C0 1 + |t| , ∀ t ∈ R, where 2∗ = 2N/(N − 2); (F2) f (t) = o(t) as t → 0. ✩ This work was supported by the NNSF (11701487, 11626202), Hunan Provincial Natural Science Foundation of China (2016JJ6137), Scientific Research Fund of Hunan Provincial Education Department (19C1700). ∗ Corresponding author. E-mail addresses: [email protected] (X.P. Wang), [email protected] (F.F. Liao).

https://doi.org/10.1016/j.na.2020.111755 0362-546X/© 2020 Elsevier Ltd. All rights reserved.

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X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

Clearly, under (A1), (A2), (F1) and (F2), the weak solutions of (1.1) correspond to the critical points of the energy functional defined in H 1 (RN ) by ∫ ( ∫ ) 1 2 2 I(u) = |∇u| + λu dx − a(x)F (u)dx, (1.2) 2 RN RN ∫t where F (t) := 0 f (s)ds. If a(x) ≡ a∞ , then (1.1) reduces to the following autonomous form: { −△u + λu = a∞ f (u), x ∈ RN ; (1.3) u ∈ H 1 (RN ), its energy functional is as follows: I ∞ (u) =

1 2

∫

(

∫ ) 2 |∇u| + λu2 dx − a∞

RN

F (u)dx.

(1.4)

RN

When a is radially symmetric, the existence of solution for problem (1.1) was obtained by Strauss [23], Berestycki and Lions [5] if f is superlinear at infinity, and by [1,5,24] if f is asymptotically linear at infinity. In these papers, the radial symmetry plays a crucial role since which can restore the compactness of the (PS)-sequence for the energy functional I. When a is not radial symmetry, problem (1.1) with a(x) > a∞ > 0 was also solved in [17] by constrained minimization and concentration-compactness arguments. They take advantage of the role played by the inequality a(x) > a∞ in restoring compactness in RN . More related results can be found in [9,11–14,20,21,25– 29]. However, in case a(x) ≤ (̸≡) a∞ and f is superlinear at infinity, nonsymmetric problem (1.1) cannot be solved by minimization [6]. In order to find a positive solution, Benci and Cerami [4] gave a description of the behavior of the Palais–Smale sequences together with some energy estimates, which allow to locate some interval of values in which the compactness is obtained [6]. Under this type of nonlinearity and some decay assumption of a(x) to limit value a∞ , Bahri and Lions [2] were able to prove that (1.1) has a positive solution. On the other hand, if f is asymptotically linear at infinity, the existence of a positive solution is exploited in [1,5,24] and their references. Let us point out that the main obstacle in finding a solution of (1.1) in the case a(x) ≤ (̸≡) a∞ is that the geometrical hypotheses on a(x) do not allow to use concentration compactness arguments as in [17] (see also [6]). There is a problem of lack of compactness due to the fact that the problem is studied in an unbounded domain RN . In general, this difficulty is circumvent by assuming symmetry properties of a, for example that a is radially symmetric as in [1,5] and [23] among many others. In this paper, a is not assumed to be symmetric, inspired by [2,4,15,16], we shall prove the nonexistence and the existence of nontrivial solutions for (1.1) under assumptions that a(x) ≤ (̸≡)a∞ and lim|x|→∞ a(x) = a∞ . In order to state clearly the results we obtain. We always assume that (1.3) admits a unique positive solution and denote by w the unique positive, radially symmetric, ground state solution and we set m∞ = I ∞ (w). Moreover, we introduce the following further assumptions on a and f : (A3) a ∈ C 1 (RN , R) and t ↦→ a(tx) + N1 ∇a(tx) · (tx) is nondecreasing on (0, ∞) for all x ∈ RN ; (A3′ ) a ∈ C 1 (RN , R), ∇a(x) · x ≥ 0 and a(x) + N1 ∇a(x) · x ≤ (̸≡) a∞ , for all x ∈ RN ; (∫ )−1 (A4) supRN |a∞ − a(x)| ≤ β0 RN F (w)dx , where β0 is the unique positive root of the equation )/N (N − 2)t2/N + 2m(2−N t = (N − 2)m2/N ∞ ∞ ;

(F3) there exists s0 > 0 such that a∞ F (s0 ) > λ2 s20 ; (F4) lim|t|→∞ Ft(t) 2 = ∞;

(1.5)

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

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(F5) F(t) := f (t)t − 2F (t) ≥ 0, and there exist constants c0 > 0 and κ > max{1, N/2} such that 3λ f (t) κ κ ≥ ⇒ |f (t)| ≤ c0 |t| F(t). t 4a∞ Our main results read as follows. Theorem 1.1. Assume that a and f satisfy (A1), (A2), (A3′ ) and (F1)–(F3). Then m := inf u∈M I(u) is not a critical level for the functional I. In particular, the infimum m is not achieved. Theorem 1.2. Assume that a and f satisfy (A1)–(A4), (F1), (F2), (F4) and (F5), and (1.3) admits a unique positive solution. Then problem (1.1) has a solution u ¯ ∈ H 1 (RN ) \ {0}. Remark 1.3. In [15], instead of (A3′ ) (or (A3)), the authors used the following stronger assumptions: (A5) a ∈ C 2 (RN , R), ∇a(x) · x ≥ 0 and a(x) + N1 ∇a(x) · x < a∞ , for all x ∈ RN ; (A6) ∇a(x) · x + x·H·x ≥ 0, for all x ∈ RN , where H represents the Hessian matrix of the function a. N In our assumptions on f , f (t)/t is not necessarily an increasing function. This kind of nonlinearity imposes another difficulty to the problem. Projections on Nehari manifold of the functional associated with the problem (1.1) are not possible in general. Therefore, one is motivated to use the more suitable projections on the set of points which satisfy the Pohozaev identity [19], the so-called the Pohozaev manifold of (1.1), instead. This idea is performed in [1,15,16,18,22] and references therein. As in [15,16], by using concentration compactness arguments and projections on a general Pohozaev type manifold and some new tricks, we can prove our results. Throughout the paper we make use of the following notations: ♠ H 1 (RN ) denotes the usual Sobolev space equipped with the inner product and norm ∫ (u, v) = (∇u · ∇v + uv)dx, ∥u∥ = (u, u)1/2 , ∀ u, v ∈ H 1 (RN ); RN

(∫ )1/s s ♠ Ls (RN )(1 ≤ s < ∞) denotes the Lebesgue space with the norm ∥u∥s = RN |u| dx ; 1 N ♠ For any u ∈ H (R ) \ {0}, ut (x) := u(x/t) for t > 0; ♠ For any x ∈ RN and r > 0, Br (x) := {y ∈ RN : |y − x| < r}; ♠ C1 , C2 , . . . denote positive constants possibly different in different places. The rest of the paper is organized as follows. In Section 2, we give some preliminaries. In Section 3, we complete the proof of Theorem 1.1. Section 4 is devoted to finding a nontrivial solution for (1.1) and Theorem 1.2 will be proved in this section. 2. Notation and preliminaries In this section, we first give some useful lemmas. The main strategy comes from [10]. Since a(x) ≡ a∞ satisfies (A1)–(A3), thus all conclusions on I are also true for I ∞ . First, we define two functionals: ∫ λN N −2 ∥∇u∥22 + ∥u∥22 − [N a(x) + ∇a(x) · x]F (u)dx (2.1) P(u) := 2 2 RN

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

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and P ∞ (u) :=

N −2 Nλ ∥∇u∥22 + ∥u∥22 − N a∞ 2 2

∫ F (u)dx

(2.2)

RN

which is the Poho˘zaev functional associated with (1.1) and (1.3), respectively (see [14,19]). Let { } M := u ∈ H 1 (RN ) \ {0} : P(u) = 0

(2.3)

{ } M∞ := u ∈ H 1 (RN ) \ {0} : P ∞ (u) = 0 .

(2.4)

and

By a simple calculation, we can verify Lemma 2.1. Lemma 2.1. The following inequality holds: g(t) := 2 − N tN −2 + (N − 2)tN > g(1) = 0,

∀ t ∈ [0, 1) ∪ (1, +∞).

(2.5)

∀ t ≥ 0, x ∈ RN .

(2.6)

Moreover (A3) implies the following inequality holds ( ) N tN [a(x) − a(tx)] + tN − 1 ∇a(x) · x ≤ 0, Further, if (A2) holds also, then a(x) +

1 ∇a(x) · x ≤ a∞ , N

∀ x ∈ RN

(2.7)

and ∇a(x) · x ≥ 0,

∀ x ∈ RN .

(2.8)

Lemma 2.2. Assume that (A1), (A3), (F1) and (F2) hold. Then 1 − tN 2 − N tN −2 + (N − 2)tN P(u) + ∥∇u∥22 , N 2N ∀ u ∈ H 1 (RN ), t > 0.

I(u) ≥ I(ut ) +

Proof . Note that I(ut ) =

tN −2 λtN ∥∇u∥22 + 2 2

∫ RN

u2 dx − tN

∫ a(tx)F (u)dx. RN

Thus, by (1.2), (2.1), (2.5), (2.6) and (2.10), one has I(u) − I(ut ) ∫ [ ] 1 − tN −2 λ(1 − tN ) 2 2 = ∥∇u∥2 + ∥u∥2 − a(x) − tN a(tx) F (u)dx 2 2 N { } ∫R 1 − tN N − 2 λN 2 2 = ∥∇u∥2 + ∥u∥2 − [N a(x) + ∇a(x) · x]F (u)dx N 2 2 RN 2 − N tN −2 + (N − 2)tN + ∥∇u∥22 2N ∫ { N } 1 − N t [a(x) − a(tx)] − (1 − tN )∇a(x) · x F (u)dx N RN 1 − tN 2 − N tN −2 + (N − 2)tN ≥ P(u) + ∥∇u∥22 , ∀ u ∈ H 1 (RN ), t > 0. N 2N This shows that (2.9) holds.

□

(2.9)

(2.10)

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

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From Lemma 2.2, we have the following two corollaries. Corollary 2.3. Assume that (F1) and (F2) hold. Then 2 − N tN −2 + (N − 2)tN 1 − tN ∞ P (u) + ∥∇u∥22 , N 2N ∀ u ∈ H 1 (RN ), t > 0.

I ∞ (u) = I ∞ (ut ) +

(2.11)

Corollary 2.4. Assume that (A1), (A3), (F1) and (F2) hold. Then for u ∈ M I(u) = max I(ut ). t>0

To show M = ̸ ∅, we define a set Λ as follows: { ∫ Λ = u ∈ H 1 (RN ) : RN

[

] } λ 2 u − a∞ F (u) dx < 0 . 2

(2.12)

(2.13)

Lemma 2.5. Assume that (A1), (A2), (A3′ ) and (F1)–(F3) hold. Then Λ ̸= ∅ and {

} u ∈ H 1 (RN ) \ {0} : P ∞ (u) ≤ 0 or P(u) ≤ 0 ⊂ Λ.

(2.14)

Proof . In view of the proof of [5, Theorem 2], (F3) implies Λ ̸= ∅. Next, we have two cases to distinguish: (1). u ∈ H 1 (RN ) \ {0} and P ∞ (u) ≤ 0, then (2.2) implies u ∈ Λ. (2). u ∈ H 1 (RN ) \ {0} and P(u) ≤ 0, then it follows from (2.1) and (A3′ ) that ] λ 2 u − a∞ F (u) dx RN 2 ] ∫ [ N −2 ∇a(x) · x 2 = P(u) − ∥∇u∥2 + N (a(x) − a∞ ) + F (u)dx 2 N RN N −2 ≤− ∥∇u∥22 < 0, 2 ∫

[

N

which implies u ∈ Λ. □ Lemma 2.6. Assume that (A1), (A2), (F1) and (F2) hold. Then for any u ∈ Λ, there exists a tu > 0 such that utu ∈ M. Furthermore, if (A3) holds, then tu is unique. Proof . Let u ∈ Λ be fixed and define a function ζ(t) := I(ut ) on (0, ∞). Clearly, by (2.1) and (2.10), we have ζ ′ (t) = 0 ∫ ∫ N − 2 N −2 tN ⇔ t ∥∇u∥22 + [N a(tx) + ∇a(tx) · (tx)]u2 dx − N tN F (u)dx = 0 2 2 RN RN ⇔ P(ut ) = 0 ⇔ ut ∈ M.

(2.15)

It is easy to verify, using (A1), (A2), (F1), (2.10) and the definition of Λ, that limt→0 ζ(t) = 0, ζ(t) > 0 for t > 0 small and ζ(t) < 0 for t large. Therefore maxt∈[0,∞) ζ(t) is achieved at some tu > 0 so that ζ ′ (tu ) = 0 and utu ∈ M.

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

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Next we assume that (A3) holds. We claim that tu is unique for any u ∈ Λ. In fact, for any given u ∈ Λ, let t1 , t2 > 0 such that ut1 , ut2 ∈ M. Then P (ut1 ) = P (ut2 ) = 0. Jointly with (2.9), we have N 2tN − N t21 t2N −2 + (N − 2)tN tN 2 1 − t2 P (ut1 ) + 1 ∥∇ut1 ∥22 N N t1 2N tN 1 −2 2tN − N t21 tN + (N − 2)tN 2 2 ∥∇u∥22 = I (ut2 ) + 1 2N t21

I (ut1 ) ≥ I (ut2 ) +

(2.16)

and 2 N −2 N 2tN + (N − 2)tN tN 2 − N t2 t1 1 2 − t1 P (u ∥∇ut2 ∥22 ) + t 2 N tN 2N tN 2 2 −2 2tN − N t22 tN + (N − 2)tN 1 1 ∥∇u∥22 . = I (ut1 ) + 2 2N t22

I (ut2 ) ≥ I (ut1 ) +

(2.17)

(2.16) and (2.17) imply t1 = t2 . Therefore, tu > 0 is unique for any u ∈ Λ. □ Corollary 2.7. Assume that (F1) and (F2) hold. Then for any u ∈ Λ, there exists a unique tu > 0 such that utu ∈ M∞ . From Corollary 2.4, Lemmas 2.5 and 2.6, we have M = ̸ ∅ and the following lemma. Lemma 2.8. Assume that (A1)–(A3) and (F1)–(F3) hold. Then inf I(u) := m = inf max I(ut ).

u∈M

u∈Λ t>0

By a standard argument, we can prove the following lemma. Lemma 2.9. Assume that (A1), (A2), (F1) and (F2) hold. If un ⇀ u ¯ in H 1 (RN ), then I(un ) = I(¯ u) + I(un − u ¯) + o(1)

(2.18)

P(un ) = P(¯ u) + P(un − u ¯) + o(1).

(2.19)

and

3. Nonexistence result for (1.1) In this section, we give the proof of Theorem 1.1. In the rest of two sections, we always assume that a(x) ̸≡ a∞ in (A2) (if a(x) ≡ a∞ , we recall that Theorem 1.1 is contained in Theorem 1.2). Lemma 3.1. Assume that (A1), (A2), (A3′ ) and (F1)–(F3) hold. If u ∈ M∞ , then there exists tu ≥ 1 such that utu ∈ M. Proof . Since u ∈ M∞ , then it follows from (2.2) that N −2 Nλ ∥∇u∥22 + ∥u∥22 − N a∞ 2 2

∫ F (u)dx = 0. RN

(3.1)

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

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In view of Lemmas 2.5 and 2.6, there exists tu > 0 such that utu ∈ M. From (A3′ ), (F1), (2.1) and (3.1), one has 0 = P (utu ) ∫ N − 2 N −2 λN tN u tu ∥∇u∥22 + ∥u∥22 − tN [N a(tu x) + ∇a(tu x) · (tu x)]F (u)dx = u 2 2 RN } { ∫ (N − 2) N −2 2 2 2 (1 − tu )∥∇u∥2 + tu [N (a∞ − a(tu x)) − ∇a(tu x) · (tu x)]F (u)dx = tu 2 RN (N − 2) N −2 ≥ tu (1 − t2u )∥∇u∥22 , 2

(3.2)

which implies tu ≥ 1. □ Similarly to Lemma 3.1, we can prove the following lemma. Lemma 3.2. Assume that (A1), (A2), (A3′ ) and (F1)–(F3) hold. If u ∈ M, then there exists tu ∈ (0, 1] such that utu ∈ M∞ . Lemma 3.3. Assume that (A1), (A2), (A3′ ) and (F1)–(F3) hold. If u ∈ M∞ , then u(· − y) ∈ M∞ for all y ∈ RN . Moreover, for every y ∈ RN , there exists θy ≥ 1 such that uθy (· − y) ∈ M and lim θy = 1.

|y|→∞

(3.3)

Proof . Let u ∈ M∞ . Then it follows from the translation invariance of I ∞ that u(· − y) ∈ M∞ , for all y ∈ RN . Furthermore, from Lemma 3.1 there exists θy ≥ 1 such that uθy (· − y) ∈ M. By (A2), (A3′ ), (2.2), (2.1) and the Lebesgue Dominated Convergence Theorem, we have ( ) 0 = lim inf θy−N P uθy (· − y) |y|→∞ [ N − 2 −2 θy ∥∇u∥22 = lim inf 2 |y|→∞ ] ∫ λN 2 + ∥u∥2 − [N a(θy x + y) + ∇a(θy x + y) · (θy x + y)]F (u)dx 2 RN { (N − 2)(θy−2 − 1) = lim inf ∥∇u∥22 2 |y|→∞ } ∫ + [N (a∞ − a(θy x + y)) − ∇a(θy x + y) · (θy x + y)]F (u)dx RN ( ) N −2 −2 = lim inf θy − 1 ∥∇u∥22 , 2 |y|→∞ which implies lim sup|y|→∞ θy = 1, and so lim|y|→∞ θy = 1. □ Lemma 3.4. Assume that (A1), (A2), (A3′ ) and (F1)–(F3) hold. Then (i) there exists ρ0 > 0 such that ∥u∥ ≥ ρ0 , ∀ u ∈ M; (ii) m = inf u∈M I(u) > 0.

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Proof . (i). Since P(u) = 0, ∀ u ∈ M, by (A3′ ), (F1), (F2), (2.1) and Sobolev embedding theorem, one has ∫ λN N −2 2 2 ∥∇u∥2 + ∥u∥2 = [N a(x) + ∇a(x) · x]F (u)dx 2 2 RN∫ ≤ a∞ F (u)dx RN ∗ λN ≤ ∥u∥22 + C1 ∥u∥22∗ 2 ∗ λN ≤ ∥u∥22 + C2 ∥∇u∥22 , 2

(3.4)

which implies ( ∥∇u∥2 ≥ ρ0 :=

N −2 2C2

)(N −2)/4 ,

∀ u ∈ M.

(ii). For u ∈ M, from (A3′ ), (F1), (1.2), (2.1) and (3.5), we have ∫ 1 ρ2 1 ∇a(x) · xF (u)dx ≥ 0 , I(u) = ∥∇u∥22 + N N RN N

(3.5)

∀ u ∈ M.

This shows that m = inf u∈M I(u) > 0. □ Similarly to [15, Lemma 3.13], we can prove the following lemma. Lemma 3.5. Assume that (A1), (A2), (A3′ ) and (F1)–(F3) hold. Then m = m∞ . Proof of Theorem 1.1. Suppose, by contradiction, that there exists u ¯ ∈ M such that I(¯ u) = m. By ∞ ′ Lemma 3.2, there exists tu¯ ∈ (0, 1] such that u ¯tu¯ ∈ M . Then it follows from (A3 ), (F1), (1.2), (1.4), (2.2), (2.1) and Lemma 3.5 that m = I(¯ u) ∫ 1 1 ∥∇¯ u∥22 + ∇a(x) · xF (¯ u)dx = N N RN ∫ tN −2 1 ≥ u¯ ∥∇¯ u∥22 + ∇a(x) · xF (¯ u)dx N N RN ∫ 1 ∇a(x) · xF (¯ u)dx = I ∞ (¯ utu¯ ) + N RN ≥ m∞ = m.

(3.6)

This shows that tu¯ = 1, u ¯ ∈ M∞ and I ∞ (¯ u) = m∞ . In view of [10, Lemma 2.14], (I ∞ )′ (¯ u) = 0. We can show that u ¯ > 0 by (F1) and a standard argument. Hence, it follows from (A3′ ), (1.2), (1.4), (2.2), (2.1) and Lemma 3.5 that ∫ 1 ∞ m = I(¯ u) = I (¯ u) + ∇a(x) · xF (¯ u)dx > m∞ = m. (3.7) N RN This contradiction shows the conclusion of the theorem holds. 4. Existence of a positive solution for (1.1) In this section, we give the proof of Theorem 1.2. Lemma 4.1. Assume that (A1)–(A3), (F1), (F2), (F4) and (F5) hold. Let {un } ⊂ H 1 (RN ) be a Cerami sequence (Ce)d with d > 0. Then {un } is bounded.

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

Proof . Since {un } ⊂ H 1 (RN ) is a Cerami sequence (Ce)d , then we have ∫ 1 a(x)F(un )dx. d + o(1) = I(un ) − ⟨I ′ (un ), un ⟩ = 2 RN

9

(4.1)

To prove the boundedness of {un }, arguing by contradiction, suppose that ∥un ∥ → ∞ as n → ∞. Let vn = un /∥un ∥. Then 1 = ∥vn ∥2 . Passing to a subsequence, we have vn ⇀ v¯ in H 1 (RN ). ∫ 2 If δ := lim supn→∞ supy∈RN B (y) |vn | dx = 0, then by Lions’ concentration compactness principle [30, 1 Lemma 1.21], vn → 0 in Ls (RN ) for 2 < s < 2∗ . Set { } 3λ f (un ) Ωn := x ∈ RN : ≤ . (4.2) un 4a∞ Using (2.7), (2.8) and λ∥vn ∥22 ≤ ∥vn ∥2 = 1, one has ∫ Ωn

a(x)f (un ) 2 3λ 3 vn dx ≤ ∥vn ∥22 ≤ . un 4 4

(4.3)

Moreover, by virtue of (F5), (4.1) and the H¨ older inequality, one can get ∫ RN \Ωn

a(x)f (un ) 2 vn dx ≤ un

]1/κ ⏐ ⏐ ⏐ f (un ) ⏐κ ⏐ ⏐ ∥vn ∥22κ′ ⏐ un ⏐ dx N R \Ωn [∫ ]1/κ

[∫

≤ C1

a(x)F(un )dx RN \Ωn

∥vn ∥22κ′ = o(1).

(4.4)

From (1.2), (4.3) and (4.4), we have ∫ a(x)f (un ) 2 ∥un ∥2 − ⟨I ′ (un ), un ⟩ = vn dx 1 + o(1) = ∥un ∥2 un N ∫ ∫ R a(x)f (un ) 2 a(x)f (un ) 2 = vn dx + vn dx u un N n Ωn R \Ωn 3 ≤ + o(1). 4 This contradiction shows that δ > 0. By a standard argument, we can prove that {un } is bounded in H 1 (RN ). □ Lemma 4.2 ([14,15]). Assume that (A1), (A2), (F1) and (F2) hold. Let {un } be a bounded Cerami sequence (Ce)d with d > 0 for I. Then there exists a subsequence of {un }, still denoted by {un }, an integer k ∈ N ∪ {0}, a sequence {yni } and wi ∈ H 1 (R3 ) for 1 ≤ i ≤ k, such that un ⇀ u0 with I ′ (u0 ) = 0; wi ̸= 0 and (I ∞ )′ (wi ) = 0 for   1 ≤ i ≤ k; ∑k  i i  un − u0 − i=1 w (· − yn ) → 0; ∑k (iv) I(un ) → I(u0 ) + i=1 I ∞ (wi );

(i) (ii) (iii)

where we agree that in the case k = 0 the above holds without wi . Lemma 4.3. Assume that (A1)–(A3) and (F1)–(F3) hold. Let {un } ⊂ M be a minimizing sequence such that I(un ) → m. Then either un → u ¯ ̸= 0 in H 1 (RN ) or there exist w ∈ H 1 (RN ) \ {0} and a sequence {yn }

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with |yn | → ∞ such that (i) I ∞ (w) = m and (I ∞ )′ (w) = 0; (ii) ∥un − w(· − yn )∥ → 0. Proof . Since P(un ) = 0, then it follows from (2.9) with t → 0 that m + o(1) = I(un ) ≥

1 ∥∇un ∥22 . N

(4.5)

This shows that {∥∇un ∥2 } is bounded. Next, we prove that {∥un ∥} is also bounded. By (F1), (F2), (2.1), (2.7) and the Sobolev embedding theorem, one has ∫ N −2 λN ∥∇un ∥22 + ∥un ∥22 = [N a(x) + ∇a(x) · x]F (un )dx 2 2 RN ∗ λN ≤ ∥un ∥22 + C4 ∥un ∥22∗ 4 ∗ ∗ λN ∥un ∥22 + C4 S −2 /2 ∥∇un ∥22 . (4.6) ≤ 4 This shows that {un } is bounded in H 1 (RN ). Passing to a subsequence, we have un ⇀ u ¯ in H 1 (RN ). Then s N ∗ N un → u ¯ in Lloc (R ) for 2 ≤ s < 2 and un → u ¯ a.e. in R . There are two possible cases: (i). u ¯ = 0 and (ii). u ¯ ̸= 0. Case (i). u ¯ = 0, i.e. un ⇀ 0 in H 1 (RN ). Then un → 0 in Lsloc (RN ) for 2 ≤ s < 2∗ and un → 0 a.e. in RN . By (A2), (F1), (F2), (2.7) and (2.8), it is easy to show that ∫ ∫ ∇a(x) · xF (un )dx = 0. (4.7) [a∞ − a(x)]F (un )dx = lim lim n→∞

n→∞

RN

RN

From (1.2), (1.4), (2.2), (2.1) and (4.7), one can get I ∞ (un ) → m,

P ∞ (un ) → 0.

(4.8)

From Lemma 4.2(ii), (2.2), (4.8), (F1), (F2) and Lions’ concentration compactness principle [30, Lemma 1.21], we can prove that there exist δ > 0 and a sequence {yn } ⊂ RN with |yn | → ∞ such that ∫ 2 |un | dx > δ. Let u ˆn (x) = un (x + yn ). Then we have ∥ˆ un ∥ = ∥un ∥ and B1 (yn ) ∫ 2 P ∞ (ˆ un ) = o(1), I ∞ (ˆ un ) → m, |ˆ un | dx > δ. (4.9) B1 (0)

Therefore, there exists u ˆ ∈ H 1 (RN ) \ {0} such that, passing to a subsequence, ⎧ ˆn ⇀ u ˆ, in H 1 (RN ); ⎪ ⎨ u u ˆn → u ˆ, in Lsloc (RN ), ∀ s ∈ [1, 2∗ ); ⎪ ⎩ u ˆn → u ˆ, a.e. on RN .

(4.10)

Let wn = u ˆn − u ˆ. Then (4.10) and Lemma 2.9 yield I ∞ (ˆ un ) = I ∞ (ˆ u) + I ∞ (wn ) + o(1)

(4.11)

P ∞ (ˆ un ) = P ∞ (ˆ u) + P ∞ (wn ) + o(1).

(4.12)

and Moreover,

1 1 ∥∇wn ∥22 = m − ∥∇ˆ u∥22 + o(1), N N

P ∞ (wn ) = −P ∞ (ˆ u) + o(1).

(4.13)

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If there exists a subsequence {wni } of {wn } such that wni = 0, then going to this subsequence, we have I ∞ (ˆ u) = m,

P ∞ (ˆ u) = 0.

(4.14)

Next, we assume that wn ̸= 0. We claim that P ∞ (ˆ u) ≤ 0. Otherwise, if P ∞ (ˆ u) > 0, then (4.13) implies ∞ P (wn ) < 0 for large n. In view of Lemma 2.6 and Corollary 2.7, there exists tn > 0 such that (wn )tn ∈ M∞ . From (1.4), (2.2), (2.11) and (4.13), we obtain m−

1 1 ∥∇ˆ u∥22 + o(1) = ∥∇wn ∥22 N N 1 ∞ P (wn ) N ( ) tN ≥ I ∞ (wn )tn − n P ∞ (wn ) N tN n ∞ ∞ ≥m − P (wn ) ≥ m∞ , N = I ∞ (wn ) −

which implies P ∞ (ˆ u) ≤ 0 due to ∥∇ˆ u∥2 > 0. Since u ˆ ̸= 0 and P ∞ (ˆ u) ≤ 0, in view of Lemma 2.6 and ∞ ˆ Corollary 2.7, there exists t > 0 such that u ˆtˆ ∈ M . From (1.4), (2.2), (2.11), (4.13) and the weak semicontinuity of norm, one has ] [ 1 un ) m = lim I ∞ (ˆ un ) − P ∞ (ˆ n→∞ N 1 1 = lim ∥∇ˆ un ∥22 ≥ ∥∇ˆ u∥22 N n→∞ N 1 tˆN ∞ = I ∞ (ˆ u) − P ∞ (ˆ u) ≥ I ∞ (ˆ utˆ) − P (ˆ u) N N tˆN ∞ P (ˆ u) ≥ m∞ − N tˆN ∞ ≥ m− P (ˆ u) ≥ m, N which implies (4.14) holds also. From (4.11), (4.12) and (4.14), we have I ∞ (wn ) = o(1),

P ∞ (wn ) = o(1),

(4.15)

which, together with (F1), (F2), (1.4) and (2.2), implies that ∥wn ∥ → 0 as n → ∞. Case (ii). u ¯ ̸= 0. Let vn = un − u ¯. Then Lemma 2.9 yields I(un ) = I(¯ u) + I(vn ) + o(1)

(4.16)

P(un ) = P(¯ u) + P(vn ) + o(1).

(4.17)

and Set Ψ (u) =

1 1 ∥∇u∥22 + N N

∫ ∇a(x) · xF (u)dx.

(4.18)

RN

Then it follows from (2.8) and (4.18) that Ψ (u) ≥

1 ∥∇u∥22 , N

∀ u ∈ H 1 (RN ).

(4.19)

Since I(un ) → m and P(un ) = 0, then it follows from (1.2), (2.1), (4.16)–(4.18) that Ψ (vn ) = m − Ψ (¯ u) + o(1),

P(vn ) = −P(¯ u) + o(1).

(4.20)

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If there exists a subsequence {vni } of {vn } such that vni = 0, then going to this subsequence, we have I(¯ u) = m,

P(¯ u) = 0,

(4.21)

which implies the conclusion of Lemma 3.2 holds. Next, we assume that vn ̸= 0. We claim that P(¯ u) ≤ 0. Otherwise P(¯ u) > 0, then (4.20) implies P(vn ) < 0 for large n. In view of Lemmas 2.5 and 2.6, there exists tn > 0 such that (vn )tn ∈ M. From (1.2), (2.1), (2.9) and (4.20), we obtain m − Ψ (¯ u) + o(1) = Ψ (vn ) 1 P(vn ) N ( ) tN ≥ I (vn )tn − n P(vn ) N tN ≥ m − n P(vn ) ≥ m, N = I(vn ) −

which implies P(¯ u) ≤ 0 due to Ψ (¯ u) > 0. Since u ¯ ̸= 0 and P(¯ u) ≤ 0, in view of Lemmas 2.5 and 2.6, there exists t¯ > 0 such that u ¯t¯ ∈ M. From (1.2), (2.1), (2.9), (4.18), (4.19) and the weak semicontinuity of norm, one has ] [ 1 m = lim I(un ) − P(un ) n→∞ N = lim Ψ (un ) ≥ Ψ (¯ u) n→∞

= I(¯ u) − ≥ m−

1 t¯N P(¯ u) ≥ I (¯ ut¯) − P(¯ u) N N

t¯N P(¯ u) ≥ m, N

which implies (4.21) also holds. Similar to Case (1), we can show that ∥vn ∥ → 0 as n → ∞ from (A2), (A3), (F1), (F2), (4.16), (4.17) and (4.21). □ Similar to the proof of [8, Lemma 6.2], we can prove the following lemma. Lemma 4.4. Assume that (A1)–(A3), (F1), (F2), (F4) and (F5) hold. Then I satisfies condition (Ce) at level d ∈ (m∞ , 2m∞ ). Inspired by [15], let w ∈ H 1 (RN ) be the positive, radially symmetric, ground state solution of (1.3). We define the operator Π : RN → M by Π [y](x) = wty (x − y). Lemma 4.5. Assume that (A1)–(A4), (F1) and (F2) hold. Then I(Π [y]) < 2m∞ , Proof . Since w ∈ M∞ and m∞ = I ∞ (w), then m∞ =

∀ y ∈ RN . 1 2 N ∥∇w∥2 ,

N −2 λN ∥∇w∥22 + ∥w∥22 − N a∞ 2 2 [

and

∫ F (w)dx = 0.

(4.23)

RN

Set t∗ :=

(4.22)

(N − 2)m∞ (N − 2)m∞ − 2β0

]1/2 .

(4.24)

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Since β0 is the unique positive root of (1.5), then 1 < t∗ < ∞. Hence, it follows from (A4), (1.2), (1.5), (4.27) and (4.24) that ∫ −2 tN λtN y y a(ty x + y)F (w)dx I(Π [y]) = ∥∇w∥22 + ∥w∥22 − tN y 2 2 RN ∫ −2 N tN − (N − 2)tN y y = ∥∇w∥22 + tN [a∞ − a(ty x + y)] F (w)dx y 2N RN [ N −2 ] ∫ m∞ N ty − (N − 2)tN y = + tN [a∞ − a(ty x + y)] F (w)dx y 2 N R ] [ −2 m∞ N tN − (N − 2)tN y y + β0 t N ≤ y 2 [ N −2 ] N m∞ N t∗ − (N − 2)t∗ ≤ + β0 t N ∗ 2 ]N/2 [ (N − 2)m∞ < m∞ + β 0 (N − 2)m∞ − 2β0 = 2m∞ . □ As in [7], next we introduce the barycenter function, which is going to be crucial for proving the existence of a solution of problem (1.1). Definition 4.6. Define the barycenter function of a given function u ∈ H 1 (RN ) \ {0} as follows: let ∫ 1 |u(y)|dy, µ(u)(x) = |B1 | B1 (x) with µ(u) ∈ L∞ (RN ) and µ is a continuous function. Subsequently, take [ ]+ 1 u ˆ(x) = µ(u)(x) − max µ(u) . 2 It follows that u ˆ ∈ C0 (RN ). Now define the barycenter of u by ∫ 1 xˆ u(x)dx ∈ RN . β(u)(x) = ∥ˆ u∥1 RN Since u ˆ has compact support, by definition, β(u) is well defined, and it has the following properties. Lemma 4.7 ([15]). The function β satisfies the following properties: (a) β is a continuous function in H 1 (RN ) \ {0}; (b) If u is radial, then β(u) = 0; (c) Given y ∈ RN , β(u(· − y)) = β(u) + y. Now we define b := inf{I(u) : u ∈ M, β(u) = 0}.

(4.25)

It is clear that b ≥ m∞ . Moreover, the following result is true: Lemma 4.8. Assume that (A1)–(A4), (F1), (F2) and (F5) hold. Then b > m∞ . Proof . Suppose, by contradiction, that b = m∞ . By the definition of b, there exists a (minimizing) sequence {un } ∈ {u ∈ M, β(u) = 0} such that I(un ) → m∞ = m. By Lemma 4.3, there are two possible cases: Case (1) un → u ¯ ∈ M in H 1 (RN ); Case (2) there exist w ∈ H 1 (RN ) \ {0} and a sequence {yn } with |yn | → ∞

X.P. Wang and F.F. Liao / Nonlinear Analysis 195 (2020) 111755

14

such that (i) I ∞ (w) = m and (I ∞ )′ (w) = 0; (ii) ∥un − w(· − yn )∥ → 0. Case (1). In this case, one has I(¯ u) = inf M I, which contradicts with Theorem 1.1. Case (2). In this case, one has un (· + yn ) = w + on (1). Calculating the barycenter function on both sides, we have β(un (· + yn )) = β(un ) − yn = −yn and β(w + on (1)) = β(w) + o(1), since β is continuous. Since |yn | → ∞, so we arrive at a contradiction. Therefore, we must have b > m∞ . □ Similarly to [15, Lemmas 4.13, 4.14], we can prove the following two lemmas. Lemma 4.9. Assume that (A1)–(A4), (F1), (F2) and (F5) hold. Then β(Π [y](x)) = y,

∀ y ∈ RN .

(4.26)

Lemma 4.10. Assume that (A1)–(A4), (F1), (F2) and (F5) hold. Then I(Π [y]) → c∞ ,

as |y| → ∞.

(4.27)

We will need a version of the Linking Theorem with Cerami condition by P. Bartolo, V. Benci and D. Fortunato in [3] (see Theorem 2.3), which we state here for the sake of completeness. Definition 4.11. Let S be a closed subset of a Banach space X, and Q a submanifold of X with relative boundary ∂Q. We say that S and ∂Q “link” if: (1) S ∩ ∂Q = ∅; (2) for any h ∈ C 0 (X, X) such that h|∂Q = id, then S ∩ h(Q) ̸= ∅. Moreover, if S and Q are as above and B is a subset of C 0 (X, X), then S and ∂Q “link” with respect to B if (1) and (2) hold for any h ∈ B. Lemma 4.12 (Linking). Suppose that I ∈ C 1 (X, R) is a functional satisfying (Ce) condition. Consider a closed subset S ⊂ X and a submanifold Q ⊂ X with relative boundary ∂Q; suppose also that: (1) S and ∂Q “link”; (2) α := inf u∈S I(u) > supu∈∂Q I(u) := α0 ; (3) supu∈∂Q I(u) < ∞. If B = {h ∈ C 0 (X, X); h|∂Q = id}, then the real number τ = inf h∈B supu∈Q I(h(u)) defines a critical value of I, with τ ≥ α. Proof of Theorem 1.2. Since P(Π [y]) = 0 for any y ∈ RN , then it follows from (1.2), (2.1), (2.8) and Lemma 3.5 that ∫ 1 1 ∥∇Π [y]∥22 + ∇a(x) · xF (Π [y])dx I(Π [y]) = N N RN ∫ −2 tN tN y y = ∥∇ω∥22 + ∇a(x) · xF (w)dx N N RN 1 > ∥∇ω∥22 = m∞ , ∀ y ∈ RN . (4.28) N

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In view of Lemmas 4.8 and 4.10, we have b > m∞ = lim|y|→∞ I(Π [y]). Then there exists ρ¯ > 0 such that for every ρ ≥ ρ¯, m∞ < max I(Π [y]) < b. (4.29) |y|=ρ

In order to apply the Linking Theorem, we take Q := Π (Bρ¯(0)) and S := {u ∈ M : β(u) = 0}. We can prove that S and ∂Q “link” by the same argument as in [15]. Furthermore, from the definitions of b and Q and (4.29), we may write b = inf I > max I. S

∂Q

Let us define H := {h ∈ C(Q, M) : h|∂Q = id}, ∞

d := inf max I(h(u)). h∈H u∈Q

∞

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