Existence and multiplicity of solutions for Schrödinger equation with inverse square potential and Hardy–Sobolev critical exponent

Nonlinear Analysis: Real World Applications 46 (2019) 525–544

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Nonlinear Analysis: Real World Applications www.elsevier.com/locate/nonrwa

Existence and multiplicity of solutions for Schrödinger equation with inverse square potential and Hardy–Sobolev critical exponent✩ Cong Wang, Yan-Ying Shang ∗ School of Mathematics and Statistics, Southwest University, Chongqing 400715, People’s Republic of China

article

info

abstract In this paper, we study a Schrödinger equation involving Hardy–Sobolev critical exponent in RN , the existence of a ground state solution and multiplicity of solutions are established. Our method relies upon Ekeland’s variational principle, Nehari manifold and Mountain Pass Theorem. © 2018 Elsevier Ltd. All rights reserved.

Article history: Received 27 April 2017 Received in revised form 29 September 2018 Accepted 1 October 2018 Available online 24 October 2018 Keywords: Hardy–Sobolev critical exponent Ekeland’s variational principle Schrödinger equation

1. Introduction and main results In this paper, we consider the following equation − ∆u − µ

2∗ (s)−2

u 2

|x|

△

+ (k + V (x))u =

|u|

s

|x|

u

r−2

+ K(x)|u|

u + f (x),

x ∈ RN ,

(1.1)

2

−s) , 0 ≤ s < 2, 2∗ (s) = 2(N where N ≥ 3, 0 ≤ µ < µ = (N −2) 4 N −2 is the Hardy–Sobolev critical exponent. If s = 0, then 2∗ := 2∗ (0) = N2N −2 is the Sobolev critical exponent. k ≥ 0, V is a given potential, K and r satisfy the following assumption: −s (K): inf x∈RN K(x) > 0 and K is locally bounded in RN \{0}, K(x) = O(|x| ) in the bounded neighborhood −t G of the origin, K(x) = O(|x| ) as |x| → ∞, 0 ≤ s < t < 2, 2∗ (t) < r < 2∗ (s).

✩ Supported by National Natural Science Foundation of China (No. 11471267), the Fundamental Research Funds for the Central Universities, China (No. XDJK2016C119). ∗ Corresponding author. E-mail address: [email protected] (Y. Shang).

https://doi.org/10.1016/j.nonrwa.2018.10.002 1468-1218/© 2018 Elsevier Ltd. All rights reserved.

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C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

f ̸≡ 0 is some given function and satisfies f ∈ H −1 , H −1 represents the dual space of H and we define: { 1 N △ H (R ), k > 0, H= E, k = 0, where H 1 (RN ) is the standard Sobolev space with the usual norm [∫ ∥u∥H 1 (RN ) := We define

{ E :=

(

2

2

|∇u| + u

)

] 21 dx .

RN

u ∈ Hµ (RN ) :

∫

V (x)u2 dx < +∞

}

RN

with the scalar product and norm are given by ∫ 1 u2 ⟨u, v⟩ := (∇u · ∇v − µ 2 + V (x)uv)dx and ∥u∥E = ⟨u, u⟩ 2 |x| RN where Hµ (RN ) is the completion of C0∞ (RN ) in the norm [∫ ∥u∥µ :=

( 2

|∇u| − µ RN

u2 2

|x|

)

] 21 dx

.

The corresponding energy functional of (1.1) is ) ∫ ( ∫ 2 2∗ (s) |u| 1 |u| 1 2 2 |∇u| − µ 2 + (k + V (x))u dx − ∗ dx I(u) = 2 RN 2 (s) RN |x|s |x| ∫ ∫ 1 r − K(x)|u| dx − f (x)udx, r RN RN where u ∈ H. We easily get that the functional I(u) ∈ C 2 (H, R). Hence, it is well known that there possesses a one-toone correspondence between the solutions of problem (1.1) and the critical points of I on H. More precisely, a function u ∈ H is said to be a weak solution to (1.1) if and only if for any v ∈ H −1 , we have ) ∫ ( ∫ 2∗ (s)−2 uv |u| uv ′ dx ∇u · ∇v − µ 2 + (k + V (x))uv dx − ⟨I (u), v⟩ = s |x| |x| RN RN ∫ ∫ r−2 − K(x)|u| uvdx − f (x)vdx, RN

RN

where ⟨ , ⟩ denotes the usual scalar product in H. We first mention that inverse square potential (Hardy term) relies on their criticality. In fact, it has the same homogeneity as critical Sobolev or Hardy–Sobolev critical exponent and does not belong to the Kato class, moreover, Hardy term cannot be regarded as a lower perturbation. There are a lot of difference in regular potentials and singular potentials. We all know that inverse square singular potentials arise in many other physical contexts: molecular physics, quantum cosmology, nuclear physics and linearization of combustion models. For the problem (1.1), if µ = s = 0, k + V (x) ≡ f (x) ≡ 0, K(x) ≡ λ, r ≡ 2 and RN becomes a bounded domain Ω ⊂ RN , then Eq. (1.1) becomes ⎧ ∗ ⎨−∆u = |u|2 −2 u + λu, x ∈ Ω , (1.2) u > 0, x ∈ Ω, ⎩ u = 0, x ∈ ∂Ω ,

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

527

where N ≥ 3. As everyone knows the problem (1.2) comes from celebrated paper [1], if N ≥ 4, Brezis and Nirenberg obtained a solution of problem (1.2) under the case of λ ∈ (0, λ1 ), where λ1 is the first eigenvalue of operator −∆ on H01 (Ω ). However, if N = 3, only when Ω is a ball, then the problem (1.2) has a solution if and only if λ ∈ ( λ41 , λ1 ). Tarantello in [2] investigated the following Dirichlet problem: {

−∆u = |u| u = 0,

2∗ −2

u + f,

x ∈ Ω, x ∈ ∂Ω ,

(1.3)

where Ω is a bounded domain of RN . As we know that the problem (1.3) never has solution if f ≡ 0 (see [1]). However, the case of f ̸≡ 0 and f ∈ (H01 (Ω ))−1 satisfying a suitable condition, Tarantello showed the existence of two solutions by Ekland’s variational principle. For problems of bounded domain, we refer readers to [3–6] and the references therein. For problems in unbounded domain, there are a lot of scholars who studied whole space RN as [7–16]. For problems with critical Sobolev or critical Hardy–Sobolev exponent in RN , which cause the loss of ∗ −s −s compactness of embeddings D1,2 (RN ) ↪→ L2 (s) (RN , |x| )(s = 0, 2), H 1 (RN ) ↪→ Lq (RN , |x| )(2 ≤ q ≤ 2∗ (s), 0 ≤ s ≤ 2). On the one hand, for the problems without inverse square singular potential, that is µ = 0 in (1.1). In [7], Cao and Zhou studied the problem with k + V (x) ≡ 1 and general subcritical nonlinearity f (x, u), they obtained the existence and multiplicity of positive solutions in some different conditions, their method relies upon the proof of Tarantello in [2]. Chen and Peng in [8] investigated the problem (1.1) with k + V (x) ≡ 1 and nonlinearity λ(f (x, u) + h(x)), under certain conditions, they obtained the first positive solution using Ekeland’s variational principle. By barrier method, they proved that there is a constant λ∗ such that the problem has two positive solutions for λ ∈ (0, λ∗ ), no solution for λ > λ∗ , a unique solution for 2 . If k + V (x) ≡ 0, f (x) ≡ 0 λ = λ∗ . On the other hand, many boffins work on problems with 0 < µ < (N −2) 4 in (1.1), Kang and Deng in [9] proved the existence of solutions for the problem with critical Sobolev exponent by the variational method. If k + V (x) ≡ 0, f (x) ≡ λu, using the similar method, Li in [10] proved the existence of nontrivial solutions to the problem with Hardy–Sobolev critical exponent. For the case of ∗ k = f (x) = K(x) = 0, critical term has the form Q(x)u2 −1 and ∂Q/∂r changes sign, Smets in [11] gave two kinds of blow-up possibility for Eq. (1.1), further, he completely characterized positive Palais–Smale sequences for energy functional. What is more, due to the strong singularity in Hardy potential, he found some new phenomenon. When s = 0, k ≡ 0, f (x) = f (x, u) and f (x, u) satisfy some weaker conditions for the problem (1.1), by exploiting suitable blow-up arguments, in [12], the authors proved that Palais–Smale sequences can be represented as sums of scaled critical points of the functional for three limiting equations, then they obtained the existence of positive solutions by Mountain Pass Theorem. If nonlinearity is f (x, u) in (1.1), which is superlinear and subcritical and satisfies a weak monotonicity condition, k = 0, Guo and Mederski in [13] found a ground state solution as a minimizer of the energy functional on a natural constraint when µ > 0 is sufficiently small. Besides, they also found that ground state solution do not exist if µ < 0 and 0 lies below the spectrum of −∆ + V . For more results about related problems, we refer readers to [14–17]. To the best of our knowledge, there are few papers in which the authors investigated the existence of ground state solutions for problems with Hardy potential and Hardy–Sobolev critical exponent in RN . What is more, the problems with Hardy or Hardy–Sobolev critical exponent in RN may have a family of limiting equations, which lead to more difficulty to prove the compactness of energy functional. The following assumptions are needed in the paper: (V1 ): V ∈ C(RN ) ∩ Lq (RN ), N2 ≤ q < ∞. (V2 ): V ∈ L∞ (RN ) and V ∈ F, where F := {g(x) : for any ε > 0, meas {x ∈ B1 (y) : |g(x)| ≥ ε} → 0 as |y| → ∞}

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C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

(V3 ): For k > 0, there exists a constant ξ > 0 such that ) ∫ ∫ ( 4µ 2 2 |∇u| +(k + V (x))u dx ≥ −ξ V (x)u2 dx, u ∈ H 1 (RN ), 1− (N − 2)2 RN RN ) N2 (∫ N N (V4 ): V ∈ C(RN ) ∩ L 2 (RN ) and 1 − Sµ,s (RN ) RN |V − (x)| 2 dx > 0, where Sµ,s (RN ) the best Hardy– Sobolev constant (see (2.2)), V − (x) := − min{V (x), 0}. (f1 ): f ∈ H −1 and satisfies the following: ∫ f udx ≤ φ(t0 ). RN

(f2 ): f ∈ H −1 and satisfies the following: ∫ f udx < φ(t0 ). RN

N

Remark 1.1. The assumption 1 − Sµ,s (R ) ) u2 2 − µ |x| + V (x)u dx > 0, ∀u ̸≡ 0 in E. Since 2 (

∫

2

|∇u| − µ RN

(

∫

2

|∇u| − µ

= RN

(

∫

2

≥

|∇u| − µ RN

( N

≥

u2 2

|x|

u2 2

|x|

u2 2

|x| (∫

(∫

RN

N 2

−

|V (x)| dx

) N2

> 0 in (V4 ) can guarantee

∫

( RN

2

|∇u|

) 2

+ V (x)u

dx ) −

+

2

+ (V (x) − V (x))u

dx

) − V − (x)u2

1 − Sµ,s (R )

dx

) 2 )∫ N |V (x)| dx N 2

−

RN

RN

( 2

|∇u| − µ

u2 |x|

2

) dx

>0 for all u ̸≡ 0 in E, where V + := max{V (x), 0}, V − (x) := − min{V (x), 0}. Moreover, it is easy to see that ∥ · ∥E is equivalent to the norm ∥ · ∥µ . Remark 1.2. Let ∫

) ∫ ∫ 2 2∗ (s) ∗ |u| |u| r 2 |∇u| −µ 2 +(k+V (x))u dx−tr−1 K(x)|u| dx−t2 (s)−1 s . |x| RN RN |x|

(

φ(t) := t RN

2

By careful computation, it is obvious that φ(t) is concave and has a unique maximum at t0 , note that φ(t0 ) in conditions (f1 ) and (f2 ) is maximum of φ. 2

−2) Remark 1.3. If µ = µ, the operator −∆ − (N4|x| is not equivalent to −∆ any more (see [18]). Hence, in 2 this paper, we only consider the case of µ < µ.

The main results of this paper are the following theorems: Theorem 1. Assume N ≥ 3, k > 0, 0 ≤ µ < µ, 0 ≤ s < 2, Suppose that (V3 ), (K), (f1 ), f (x) ̸≡ 0 hold and V satisfies (V1 ) or (V2 ). Then the problem (1.1) has a ground state solution u0 in H 1 (RN ) and u0 ≥ 0 if f ≥ 0. In addition, if (f2 ) holds, then u0 is a local minimum for (1.1).

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

529

Remark 1.4. (1) Assumption (V1 ) or (V2 ) implies that lim|x|→∞ V (x) = 0. (2) Note that k + V may be negative in some bounded domain of RN when k > 0. We give an example: V˜ (x) = −(k + h)e−|x| , x ∈ RN where k > 0, h ∈ (0, h∗ ) and h∗ is a small positive constant. Through careful computation, we easily know that V˜ (x) satisfies (V1 ), (V2 ), what is more, V˜ (x) ∈ C(RN ) and V˜ (x) satisfies the following: (i) V˜ (x) → 0 as |x| → +∞; (ii) −h ≤ k + V˜ (x) and the set {x ∈ RN : −h ≤ k + V˜ (x) ≤ 0} is nonempty and bounded. According to the Theorem 5.1 in [12], it is easy to see that k + V˜ (x) is negative in some bounded domain of RN and V˜ (x) satisfies (V3 ). Theorem 2. Assume N ≥ 7, k = 0, 0 ≤ µ < µ, 0 ≤ s < 2. Suppose that (V4 ), (K), (f2 ) and f (x) ̸≡ 0 hold. Then the problem (1.1) has at least two weak solutions u ˜0 , u ˜1 in E and u ˜0 , u ˜1 ≥ 0 if f ≥ 0. Remark 1.5. (1) In Theorem 1, we only assume that 0 ≤ µ < µ, it is weaker than the assumption 0 ≤ µ < µ − 4 in [3]. (2) In Theorem 1, we only need N ≥ 3, which is weaker than the assumption N ≥ 7 in [3]. (3) The first solution in Theorem 2 is a ground state solution. (4) For f ≥ 0, we know that the problem (1.1) cannot admit positive solution when ∥f ∥H −1 is too large. So our approach necessarily breaks down if ∥f ∥H −1 is large, however, conditions (f1 ), (f2 ) can guarantee the existence of solutions. (5) Our results generalize the theorems of [2], where the author only studied the elliptic equation with critical Sobolev exponent in bounded domain. For critical Hardy–Sobolev exponent in unbounded domain, it is more thorny and complicated. The whole paper is organized as follows. In the forthcoming, we show some preliminaries. In the third section, we mainly give the proof of theorems. 2. Preliminaries The following Hardy inequality (2.1) and the best Hardy–Sobolev constant (2.2) are related to the problem (1.1) ∫ ∫ 2 4 |u| 2 dx ≤ |∇u| dx. (2.1) 2 2 (N − 2) N N R |x| R ) ∫ ( 2 u2 |∇u| − µ dx N 2 R |x| . (2.2) Sµ,s (RN ) = inf 2 ( ) ∗ ∫ u∈H 1 (RN )\{0} 2∗ (s) |u|2 (s) |x|s dx RN According to [19], we know that the best constant Sµ,s (RN ) is achieved by a family of functions: √

( ε Uµ,s (x)

where β =

√

:=

µ ) 2−s

2

2εβ (N −s)

√

µ N −2 ( √ (2−s)β ) 2−s √ µ−β |x| ε + |x| µ

, ∀ε > 0,

ε µ − µ. What is more, the functions Uµ,s (x) solve the equation

−∆u − µ

2∗ (s)−2

u 2

|x|

=

|u|

s

|x|

u

, RN \ {0}

(2.3)

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C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

and satisfy ∫

2

( 2

ε |∇Uµ,s |

RN

−µ

ε |Uµ,s |

|x|

)

2

2∗ (s)

ε |Uµ,s | s |x|

∫ dx = RN

N −s

dx = Sµ,s (RN ) 2−s .

Set • p = 2∗ (s), H(1 := H 1 (RN ), Vk := k)+ V (x). ∫ 2 u2 2 • ∥u∥2 = RN |∇u| − µ |x| dx, k > 0. 2 + Vk u (∫ )1 r • Lr (RN , K(x)) is Banach space with the norm: ∥u∥Lr := RN K(x)|u| dx r . • BR (0) := {x ∈ RN : |x| < R}. • C or Ci (i = 1, 2, 3, · · · ) represent all kinds of positive constants. We define the following N ehari manifold: N = {u ∈ H | ⟨I ′ (u), u⟩ = 0} Remark 2.1. If f = 0, then Theorem 1 remains valid and gives the trivial solution u0 = 0. Moreover, in this case of u0 is a local minimum for I, necessarily: ∫ ∫ p |u0 | r dx − (r − 1) |u0 | dx ≥ 0. ∥u0 ∥2 − (p − 1) s RN RN |x| For any u ∈ N , we have △

J(u) = ⟨I ′ (u), u⟩ = ∥u∥2 −

∫ RN

p

|u| s dx − |x|

∫

r

∫

K(x)|u| dx − RN

f (x)udx = 0.

(2.4)

RN

Moreover, as for J, there are merely three situations: ⟨J ′ (u), u⟩ > 0, ⟨J ′ (u), u⟩ = 0, ⟨J ′ (u), u⟩ < 0. Combining with (2.4), we get ⟨J ′ (u), u⟩ ∫ ∫ p |u| r K(x)|u| dx− f (x)udx = 2∥u∥ −p s dx−r N |x| RN RN ( ) ∫R ∫ ∫ ∫ p p |u| |u| r r 2 2 = 2∥u∥ −p K(x)|u| dx− ∥u∥ − K(x)|u| dx s dx−r s dx− RN |x| RN RN |x| RN ∫ ∫ p |u| r = ∥u∥2 − (p − 1) K(x)|u| dx s dx − (r − 1) |x| N N R R 2

∫

Therefore, we can make the following splitting for N : ∫ ∫ p |u| r N + = {u ∈ N | ∥u∥2 − (p − 1) dx − (r − 1) K(x)|u| dx > 0}. s |x| N N R R ∫ ∫ p |u| r N 0 = {u ∈ N | ∥u∥2 − (p − 1) K(x)|u| dx = 0}. s dx − (r − 1) |x| N N R R ∫ ∫ p |u| r − 2 N = {u ∈ N | ∥u∥ − (p − 1) K(x)|u| dx < 0}. s dx − (r − 1) |x| N N R R c0 := inf I(u). u∈N

(2.5)

Lemma 2.2. Assume (V3 ) holds and V satisfies (V1 ) or (V2 ). Then there exist two positive constants C1 and C2 such that ) ∫ ( 2 |u| 2 2 2 C1 ∥u∥H 1 (RN ) ≤ |∇u| − µ 2 + Vk u dx ≤ C2 ∥u∥2H 1 (RN ) . |x| RN

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

Proof . According to (2.1), we get that ) ∫ ( 2 |u| 2 2 |∇u| − µ 2 + Vk u dx |x| RN ) ) ∫ ∫ (( 4µ 2 2 |∇u| + u dx − (1 − Vk ) u2 dx ≥ 1− (N − 2)2 RN RN ) ) ) ) ∫ (( ∫ (( 1 4µ 4µ 2 2 2 2 = |∇u| + u dx − |∇u| +u dx 1− 1− (N − 2)2 ξ + 1 RN (N − 2)2 RN ) ) ∫ (( ∫ 1 4µ 2 |∇u| +u2 dx − (1 − Vk ) u2 dx. + 1− ξ + 1 RN (N − 2)2 N R It follows from (V3 ) that ∫ ( 1− RN

531

(2.6)

) ∫ ∫ 4µ 2 2 2 |∇u| +u dx ≥ (1 − Vk )u dx −ξ V (x)u2 dx. (N − 2)2 RN RN

Moreover, one has ) ) ∫ (( ∫ 4µ 1 2 2 1− |∇u| +u dx − (1 − Vk ) u2 dx ξ + 1 RN (N − 2)2 RN (∫ ) ∫ ∫ 1 (1 − k − V (x))u2 dx −ξ V (x)u2 dx − (1 − k − V (x)) u2 dx ≥ ξ+1 N N N R R ∫R ξ(k−1) 2 = u dx. ξ +1 RN Combining (2.6) and (2.7), we have ) ∫ ( 2 |u| 2 2 |∇u| − µ 2 + Vk u dx |x| RN ( )∫ (( ) ) ∫ 1 4µ ξ(k−1) 2 2 ≥ 1− 1− u2 dx. |∇u| +u dx + ξ + 1 RN (N − 2)2 ξ +1 RN If k − 1 ≥ 0, by (2.8), we deduce that there exists a positive constant C such that ) ∫ ( 2 |u| 2 |∇u| − µ 2 + Vk u2 dx ≥ C∥u∥2H 1 (RN ) . |x| RN If k − 1 < 0, from (2.8), then there exists a positive constant C such that ) ∫ ( 2 |u| 2 2 |∇u| − µ 2 + Vk u dx |x| RN ) ( )∫ (( ( )2 ) 1 ξ(1 − k) 2 2 ≥ 1− − 1− µ |∇u| +u2 dx ξ+1 ξ+1 N −2 RN ≥ C∥u∥2H 1 (RN ) . Therefore, there exists C1 > 0 such that ) ∫ ( 2 |u| 2 |∇u| − µ 2 + Vk u2 dx ≥ C1 ∥u∥2H 1 (RN ) . |x| RN On the other hand, from (V1 ) or (V2 ) we get that ) ∫ ( ∫ ( 2 ) |u| 2 2 2 |∇u| − µ 2 + Vk u dx ≤ |∇u| + ∥Vk ∥L∞ u2 dx ≤ C2 ∥u∥H 1 (RN ) |x| RN RN

(2.7)

(2.8)

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

532

for any u ∈ H 1 , where ∥Vk ∥L∞ := ess sup|Vk | is the norm of Banach space L∞ (RN ). The proof of the lemma is complete. □ Lemma 2.3. Assume (V3 ), (K) hold and V satisfies (V2 ) or (V2 ). Then the embedding H 1 ↪→ Lr (RN , K(x)) is compact. Proof . Our method is similar with Lemma 2.2 in [9], for reader’s convenience, we give a sketch of the proof. Set 0 ≤ µ < µ, from (2.1), (2.2) and Lemma 2.3, we deduce that there exist C1 (µ) > 0 and C2 (µ) > 0 such that ∫ ∫ 2 p |u| |u| 2 2 ≤ C ∥u∥ , 1 s ≤ C2 ∥u∥ . 2 RN |x| RN |x| Let R1 > 0, R2 > 0 and 0 < R1 < R2 , define ∫ ∫ r r I1 (u) = K(x)|u| dx, I2 (u) = K(x)|u| dx, |x|>R2 ∫|x|
From (2.2), as R1 → 0, we have r

) pr (∫ ) p−r p p |u| 1 dx dx s s |x|
|u| I1 (u) ≤ C s dx ≤ C |x|
r

≤ CR1δ1 ∥u∥ p , where δ1 = (N −s)(p−r) = p (2.2), as R2 → ∞,

N −2 2 (p

− r), since N ≤ 3, r < p, then δ1 > 0. Further, I1 (u) → 0 as R1 → 0. From

r

|u| I2 (u) ≤ C t dx ≤ C |x|>R2 |x| ∫

) pr (∫ ) p−r p p |u| 1 s dx pt−sr dx |x|>R2 |x| |x|>R2 |x| p−r

(∫

≤ C∥u∥

r p

(∫

+∞

R2

(

lN −1 l

pt−sr p−r

) p−r p dl

r

≤ CR2δ2 ∥u∥ p ,

)

p−r where δ2 = N − pt−sr = N − t − (N −2)r , since r > 2∗ (t), then δ2 < 0. Moreover, I2 (u) → 0 as p−r p 2 R2 → ∞. Let {un } ⊂ H 1 be a bounded sequence and Ω := {x ∈ RN , R1 < |x| < R2 }, by the compactness of the embedding H 1 (Ω ) ↪→ Lr (Ω ) and the local boundedness of K(x), we get that {un } has a subsequence, still defined by {un }, such that I3 (un − u) → 0 as n → ∞ for some u ∈ H 1 . Let R1 → 0 and R2 → ∞ we ∫ r have RN K(x)|un − u| → 0. We complete the proof of the lemma. □

Lemma 2.4. Let f ̸≡ 0 satisfies (f2 ). For any u ∈ H 1 , u ̸= 0, there exists a unique t+ := t+ (u) > 0 such ∫ that t+ u ∈ N − . In particular, t+ > t0 and I(t+ u) = maxt≥t0 I(tu). Moreover, if RN f u > 0, then there exists a unique t− : t− (u) > 0 such that t− u ∈ N + . In particular, t− < t0 and I(t− u) ≤ I(tu), ∀ ∈ [0, t+ ]. Proof . According to Remark 1.1, we get that φ(t) has a unique maximum at t0 and ∫ ∫ p |u| r r−1 dx − t K(x)|u| dx. φ(t0 ) = t0 ∥u∥2 − tp−1 s 0 0 |x| N N R R

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

533

∫ ∫ Moreover, if RN f u ≤ 0, then there exists a unique t+ > t0 such that φ(t+ ) = RN f udx and φ(t+ ) < 0, that is t+ u ∈ N − and I(t+ u) ≥ I(tu), ∀ t ≥ t0 . ∫ In the case of RN f udx > 0, by the assumption (f2 ), we have ∫ f udx < φ(t0 ). RN

Therefore, in this case, we have unique 0 < t− < t0 < t+ such that ∫ + φ(t ) = f udx = φ(t− ) RN

and φ′ (t− ) > 0 > φ′ (t+ ). Namely, t+ u ∈ N − and t− u ∈ N + . Further, we have I(t+ u) ≥ I(tu), ∀t ≥ t− and I(t− u) ≤ I(tu), ∀t ∈ [0, t+ ]. □ Lemma 2.5. Assume f satisfies (f2 ), for every u ∈ N , u ̸= 0, we have ∫ ∫ p |u| r dx − (r − 1) K(x)|u| dx ̸= 0, ∥u∥2 − (p − 1) s RN RN |x|

(2.9)

that is N 0 = {0}. Proof . Although the result also holds for f = 0, we will only prove the case of f ̸≡ 0. Proving by contradiction and assume there exists u0 ∈ N , u0 ̸= 0 such that ∫ ∫ p |u0 | r 2 ∥u0 ∥ − (p − 1) K(x)|u0 | dx ≡ 0. (2.10) s dx − (r − 1) |x| N N R R ∫ ∫ p r For φ′ (t) = ∥u0 ∥ − (r − 1)tr−1 RN K(x)|u0 | dx − (p − 1)tp−1 RN |u|x|0 |s dx = 0, we know φ′ (t) has a unique solution. By (2.10), we have φ′ (1) = 0, that is φ(t0 ) = φ(1). According to (f2 ), we know ∫ ∫ ∫ p |u0 | r dx − K(x)|u0 | dx. f u0 dx < ∥u0 ∥2 − s |x| N N N R R R However, the fact of u0 ∈ N implies that ∫ ∫ f u0 dx = ∥u0 ∥2 − RN

RN

p

|u0 | s dx − |x|

∫

(2.11)

r

K(x)|u0 | dx, RN

□

which contradicts (2.11). Therefore, the proof of the lemma is complete.

Lemma 2.6. Let f ̸≡ 0 satisfies (f2 ), given u ∈ N , u ̸= 0, then there exist ε > 0 and a differentiable function t = t(ω) > 0, ω ∈ H 1 , ∥ω∥ < ε satisfying the following t(0) = 1, t(ω)(u − ω) ∈ N , for ∥ω∥ < ε and ′

⟨t (0), ω⟩ = where ⟨u, ω⟩ =

∫

RN

2⟨u, ω⟩ − p ∥u∥2

( uω ∇u∇ω − µ |x| 2

∫

RN

|u|p−2 uω |x|s

−r

∫

RN

|u|p dx RN |x|s

∫

− (p − 1) ) + Vk (x)uω dx.

K(x)|u|

r−2

− (r − 1)

∫

uωdx −

RN

∫

RN r

f ωdx

K(x)|u| dx

,

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Proof . We firstly define the map F : R × H 1 → R ∫ ∫ ∫ p |u−ω| r r−1 F (t, ω) = t∥u−ω∥2 −tp−1 dx−t K(x)|u−ω| dx− f (u−ω)dx. s |x| RN RN RN Since F (1, 0) = 0,

∂F (1, 0) = ∥u∥2 − (p − 1) ∂t

∫ RN

p

|u| s dx − (r − 1) |x|

∫

r

K(x)|u| dx ̸= 0. RN

Applying the implicit theorem at the point (1, 0), we can get the results of this Lemma.

□

Lemma 2.7. Assume (f2 ) holds. Then there exists a minimizing sequence {un } ⊂ N for (2.5) such that I(un ) < c0 +

1 1 and I(un ) ≥ I(un ) − ∥ω − un ∥, ∀ω ∈ N . n n

Moreover, we deduce that ∥I ′ (un )∥(H 1 )−1 → 0 as n → ∞. Proof . We firstly show that I is bounded from below. For u ∈ N we have ∫ ∫ ∫ p 1 1 |u| 1 r I(u) = ∥u∥2 − K(x)|u| dx − f udx s dx − 2 p N |x| r N RN ) R ( ) ∫R ( ) ( ∫ p 1 1 |u| 1 1 1 2 − − ∥u∥ + f udx = s dx − 1 − 2 r r p r N |x| RN ( ) ( )∫ R 1 1 1 ≥ − ∥u∥2 − 1 − f udx 2 r r N ( ) ( ) R 1 1 1 ≥ − ∥u∥2 − 1 − ∥f ∥(H 1 )−1 ∥u∥. 2 r r

(2.12)

For any ϵ > 0, it follows from the Young inequality that ( )2 ( ) 1 − 1r ∥f ∥2(H 1 )−1 1 ϵ2 ∥u∥2 1− ∥f ∥(H 1 )−1 ∥u∥ ≤ + . r 2ϵ2 2 Taking ϵ =

( r−2 ) 12 r

, then ( ) ( ) 1 (r − 1)2 1 1 2 1− ∥f ∥(H 1 )−1 ∥u∥ ≤ ∥f ∥(H 1 )−1 + − ∥u∥2 . r 2r(r − 2) 2 r

(2.13)

From (2.12) and (2.13), we have I(u) ≥ − In particular, c0 ≥ −

(r − 1)2 ∥f ∥2(H 1 )−1 . 2r(r − 2)

(r − 1)2 ∥f ∥2(H 1 )−1 . 2r(r − 2)

In order to find an upper bound for c0 , let v ∈ H 1 be the weak solution of the following equation: { −∆v − µ |x|v 2 + Vk (x)v = f (x), x ∈ RN , v ∈ H 1. Then for f ̸≡ 0, we obtain

∫ RN

f vdx = ∥v∥2 > 0.

(2.14)

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+ From Lemma 2.4, we can get a t1 := t− 1 (v) such that t1 v ∈ N and t1 v ∈ N . Moreover, ∫ ∫ ∫ p tr1 t21 |v| tp1 r 2 dx − I(t1 v) = ∥v∥ − K(x)|v| dx − t1 f vdx 2 p RN |x|s r RN RN ( ) ( ) ∫ ∫ p 1 p |v| 1 r t2 r t dx + 1 − t K(x)|v| dx = − 1 ∥v∥2 + 1 − 2 p 1 RN |x|s r 1 RN ( ) ∫ ∫ p t2 1 |v| r r dx + (r − 1)t ≤ − 1 ∥v∥2 + (p − 1)tp1 K(x)|v| dx s 1 2 r RN |x| RN 2 2 t1 t < − ∥v∥2 + 1 ∥v∥2 2 r < 0.

Hence c0 ≤ I(t1 v) < 0.

(2.15)

Applying the Ekeland’s variational principle to the minimization problem (2.5), we can obtain a minimizing sequence {un } ⊂ N satisfying conditions of this lemma. Taking n large enough, from (2.15), we get ) ( 1 1 1 2 − t ∥v∥2 > c0 + > I(un ) − 2 r 1 n ∫ ∫ ∫ p 1 1 |u| 1 r = ∥un ∥2 − dx − K(x)|u| dx − f un dx 2 p RN |x|s r N RN ( ( ) ( ) ∫R ) ∫ p 1 1 1 1 |un | 1 2 − − ∥un ∥ + f un dx = s dx − 1 − 2 r r p r N |x| RN ) ( )∫ R ( 1 1 1 − ∥un ∥2 − 1 − f un dx ≥ 2 r r RN ( )∫ 1 ≥ − 1− f un dx. (2.16) r RN Therefore,

∫ ∥f ∥(H 1 )−1 ∥un ∥ ≥

f un dx > RN

r−2 2 t ∥v∥2 > 0. 2(r − 1) 1

(2.17)

Hence un ̸= 0. By (2.16) and (2.17), we have ) ( )∫ ( 1 r−1 1 1 − ∥un ∥2 ≤ 1 − f un dx ≤ ∥f ∥(H 1 )−1 ∥un ∥. 2 r r r N R Moreover, ∥un ∥ ≤ It follows from (2.17) that ∥un ∥ ≥

2(r − 1) ∥f ∥(H 1 )−1 . r−2

r−2 2 t ∥v∥2 ∥f ∥−1 . (H 1 )−1 2(r − 1) 1

Thus, r−2 2 2(r − 1) t ∥v∥2 ∥f ∥(H 1 )−1 ≤ ∥un ∥ ≤ ∥f ∥(H 1 )−1 . (2.18) 2(r − 1) 1 r−2 We apply contradiction to prove I ′ (un ) → 0. We assume ∥I ′ (un )∥ > 0 as n large enough. For u = un and δI ′ (un ) ω = ∥I ′ (u )∥ with δ > 0 small enough. According to Lemma 2.6, we can obtain tn (δ) := t(ω) such that n ωδ := tn (δ)(un − ω) ∈ N . From the conditions of this Lemma, one has 1 ∥∇(ωδ − un )∥ ≥ I(un ) − I(ωδ ) n = (1 − tn (δ))⟨I ′ (ωδ ), ωn ⟩δtn (δ)⟨I ′ (ωδ ),

I ′ (un ) ⟩ + o(δ). ∥I ′ (un )∥

536

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

Dividing by δ > 0 and passing to the limit as δ → 0, we have 1 (1 + |t′n (0)| ∥un ∥) ≥ −tn (0)⟨I ′ (un ), un ⟩ + ∥I ′ (un )∥ = ∥I ′ (un )∥, n ′

(un ) where t′n (0) := ⟨t′ (0), ∥II ′ (u ⟩. Finally, we prove that {|tn (0)|} is bounded uniformly in n. According to n )∥ (2.18) and Lemma 2.6, we get that there exists C > 0 such that C . |t′n (0)| ≤ ∫ ∫ p r |u | n ∥un ∥2 −(p − 1) RN |x|s dx−(r−1) RN K(x)|un | dx

Consequently, we only need to verify that ∫ ∫ p |un | r |∥un ∥2 − (p − 1) dx − (r − 1) K(x)|un | dx| ≥ β > 0. s |x| N N R R

(2.19)

Now, we will verify (2.19) by contradiction. We assume there exists a subsequence (still denoted by {un }) such that ∫ ∫ p |un | r ∥un ∥2 − (p − 1) K(x)|un | dx = o(1). (2.20) s dx − (r − 1) RN |x| RN Combining (2.18) and (2.20), then there exists a suitable constant γ > 0 such that ∫ ∫ p |un | r (p − 1) K(x)|un | dx ≥ γ for n large enough. (2.21) s dx + (r − 1) |x| N N R R The fact of un ∈ N and (2.20) imply that ∫ ∫ f un = (p − 2) RN

RN

p

|un | s dx + (r − 2) |x|

∫

r

K(x)|un | dx + o(1).

(2.22)

RN

We will apply the following elementary inequality: ′

′

′

(a + b)γ ≥ aγ + γ ′ aγ −1 b, a, b ≥ 0, γ ′ > 1. (2.23) ∫ ∫ p r n| ′ For φ′ (t) = ∥un ∥ − (r − 1)tr−2 RN K(x)|un | dx − (p − 1)tp−2 RN |u|x| s dx = 0, we know φ (t) has a unique solution. By (2.20), for n large enough, we have t0 = 1 + o(1), that is φ(t0 ) = φ(1 + o(1)). By (2.20), (2.23) and (f2 ), it is obvious that there exists a 0 < ε < 21 such that ∫ ∫ ∫ p |un | r r−1 f un dx < t0 ∥un ∥2 −(t0 +ε)p−1 dx−(t +ε) K(x)|un | dx 0 s |x| N N N R R R ∫ ∫ ∫ p p |un | |un | r p−1 ≤ t0(p−1) dx−t dx+t (r−1) K(x)|un | dx 0 s s 0 RN |x| RN RN |x| ∫ ∫ p |un | r p−2 r−1 − t0 K(x)|un | dx−(p−1)t0 ε s dxer RN |x| RN ∫ r r−2 − (r − 1)t0 ε K(x)|un | dx RN ∫ ∫ p |un | r ≤ (p − 2) dx + (r − 2) K(x)|un | dx s RN |x| RN ( ) ∫ ∫ p |un | r − ε (p − 1) K(x)|un | dx + o(1) s dx + (r − 1) RN RN |x| ∫ ∫ p |un | r ≤ (p−2) K(x)|un | dx− εγ + o(1). (2.24) s dx+(r−2) |x| N N R R for n large enough. By (2.24) and γ > 0, we have ∫ ∫ ∫ p ε |un | r f un dx < (p − 2) K(x)|un | dx − γ s dx + (r − 2) 2 RN RN RN |x| for n large enough, which contradicts (2.22). Thus, {|t′n (0)|} is bounded uniformly for any n. The proof is complete. □

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537

3. Proof of the main results Proof of Theorem 1. If hypothesis of Theorem 1 holds, then from Lemma 2.2 to Lemma 2.7 are correct. Therefore, for V satisfies (V1 ) or (V2 ), the process of proof for Theorem 1 is similar. Assume that (f2 ) holds. From Lemma 2.7, we have a minimizing sequence {un } ⊂ N satisfying lim I(un ) = c0 and lim ∥I ′ (un )∥H −1 = 0.

n→∞

n→∞

Let u0 ∈ H 1 be the weak limit of {un } in H 1 . According to (2.17), we get that ∫ f u0 > 0. RN

It follows from Lemma 2.7 that ⟨I ′ (u0 ), ω⟩ = 0, ∀ω ∈ H 1 , which means that u0 is a weak solution of (1.1) and u0 ∈ N . Thus ) ( )∫ ( )∫ ( 1 1 1 1 1 r 2 − ∥u0 ∥ − − K(x)|u0 | dx − 1 − f u0 dx c0 ≤ I(u0 ) = 2 p r p p RN RN ≤ lim I(un ) = c0 . n→∞

Moreover, we can deduce that un → u0 strongly in H 1 , that is c0 = I(u0 ) = inf I(u). u∈N

Therefore, u0 is the ground state solution for I. On the other hand, we have u0 ∈ N + . Assume u0 ∈ N − , by Lemma 2.4, there exists a unique t+ := t+ (u0 ) such that t+ u0 ∈ N − . So t+ = 1. From Lemma 2.4, we get t− := t− (u0 ) < 1 such that t− u0 ∈ N + . Since dI(t− u0 ) d2 I(t− u0 ) = 0, > 0, dt dt2 there exists t+ ≥ t > t− such that I(tu0 ) > I(t− u0 ). By Lemma 2.4, one has I(t− u0 ) < I(tu0 ) ≤ I(t+ u0 ) = I(u0 ), which is contradictory with the fact that u0 is the least energy solution. Hence, u0 ∈ N + . Now, we prove ∫ that u0 is a local minimum for I. From Lemma 2.5, for any u ∈ H 1 satisfying RN f u > 0, there exists a unique t− (u) ∈ (0, t0 (u)) such that t− (u)u ∈ N + and I(su) ≥ I(t− u), ∀s ∈ (0, t0 (u)). For every u0 ∈ N + , we have t− (u0 ) = 1 < t0 (u). Taking ε small enough such that the following inequality (3.1) holds for any ∥ω∥ < ε: 1 < t0 (u0 − ω).

(3.1)

By Lemma 2.4, for any ω : ∥ω∥ < ε, we get t(ω)(u0 − ω) ∈ N . Since t(ω) → 1 as ∥ω∥ → 0. We can assume the inequality t(ω) < t0 (u0 − ω) holds for every ∥ω∥ < ε, ω ∈ H 1 . Thus, t(ω)(u0 − ω) ∈ N + and for all s : 0 < s < t0 (u0 − ω). We deduce that I(s(u0 − ω)) ≥ I(t(ω)(u0 − ω)) ≥ I(u0 ).

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From (3.1), we can take s = 1 and we have I(u0 − ω) ≥ I(u0 ), for ω ∈ H 1 and ∥ω∥ < ε.

(3.2)

Therefore, u0 is a local minimum for I. To prove results when f satisfies (f1 ), we shall use an approximation argument. Since f satisfies (f1 ), then fε = (1 − ε)f satisfies (f2 ), ∀ε ∈ (0, 1). Define ) ∫ ( ∫ p u2 1 1 |u| 2 2 dx Iε (u) = |∇u| − µ 2 + Vk (x)u dx − 2 RN p RN |x|s |x| ∫ ∫ 1 r − K(x)|u| dx − (1 − ε) f udx. r RN RN Let uε ∈ Nε+ , where Nε+

1

:= {u ∈ H :

⟨Iε′ (u), u⟩

∫

2

= 0, ∥u∥ −(p−1) RN

p

|u| s dx−(r−1) |x|

∫

r

K(x)|u| dx > 0}, RN

which satisfies Iε (uε ) = inf I(u) := cε u∈Nε

and ⟨Iε′ (uε ), ω⟩ = 0, ∀ω ∈ H 1 .

(3.3)

We can easily get that ∥uε ∥ ≤ C3 , for 0 < ε < 1 and C3 is a suitable constant. Take u ∈ N + , necessarily ∫ f udx > 0 and we have ∫ RN (1 − ε) f u > 0, 0 < ε < 1. RN

According to Lemma 2.5 with f = fε , we get 0 < t− ε < t0 + with t− ε u ∈ Nε . By the fact that t0 > 1 and Lemma 2.5, we can deduce that

Iε (t− ε u) ≤ Iε (u). Moreover, cε ≤ Iε (t− ε u) ≤ Iε (u) ≤ I(u) + ε∥f ∥H −1 ∥u∥ ≤ I(u) + εC4 ,

(3.4)

where C4 is a positive constant. From (2.14) with f = fε , combining with the inequality (3.4), we have −

(r − 1)2 (r − 1)2 ∥f ∥2H −1 ≤ − ∥fε ∥2H −1 ≤ cε ≤ c0 + εCε . 2r(r − 2) 2r(r − 2)

As n → ∞, taking εn → 0 such that u0 ∈ H 1 . Therefore, we get cεn → c ≤ c0 and uεn ⇀ u0 weakly in H 1 as n → ∞. It follows from (3.3) that ⟨I ′ (u0 ), ω⟩ = 0, ∀ω ∈ H 1 . Moreover, I(u0 ) ≤ c0 and u0 ∈ N , which imply I(u0 ) = c0 . Thus uεn → u0 strongly in H 1 . The proof of this theorem is complete. □ Lemma 3.1. Suppose that k = 0, N ≥ 7, 0 ≤ µ < µ − 4, 0 ≤ s < 2, (V4 ), (K), (f2 ) hold and f (x) ̸≡ 0, then I satisfies the (PS)c condition for any c, c < c0 +

N −s 2−s Sµ,s (RN ) 2−s . 2(N − s)

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539

Proof . Assume {un } ⊂ E is a (PS)c sequence for I at the level c, that is N −s 2−s Sµ,s (RN ) 2−s . 2(N − s) ′ ∥I (un )∥ → 0 as n → ∞.

I(un ) → c < c0 +

Moreover, we can easily get that ∫ ∫ ∫ p 1 1 |un | 1 r 2 dx − ∥un ∥E − K(x)|un | dx − f un dx = c + o(1). 2 p RN |x|s r RN RN ∫ ∫ ∫ p |un | r K(x)|un | dx − f un dx = ⟨I ′ (un ), un ⟩. ∥un ∥2E − s dx − |x| N N R RN R

(3.5) (3.6)

(3.7) (3.8)

According to (3.7), (3.8) and the H¨ older inequality, one has ) ( )∫ ( )∫ ( p 1 1 |un | 1 1 1 1 − ∥un ∥2E + − dx− 1 − f un dx = c+o(1)− ⟨I ′ (un ), un ⟩. 2 r r p RN |x|s r RN r Moreover, (

) ( ) 1 1 1 2 − ∥un ∥E − 1 − ∥f ∥E −1 ∥un ∥E ≤ C. 2 r r Therefore, there exists a positive constant C such that ∥un ∥2E ≤ C.

(3.9)

Moreover, combining the boundedness of {un } with Lemma 2.4, by usual arguments we may assume that, up to a subsequence if necessary, there exists u ∈ E such that ⎧ un ⇀ u weakly in E as n → ∞, ⎪ ⎪ ⎨ un → u strongly in Lploc (RN ) for all p ∈ [2, 2∗ ) as n → ∞, (3.10) un → u strongly in Lr (RN , K(x)) as n → ∞, ⎪ ⎪ ⎩ N un (x) → u(x) a.e. in R as n → ∞. Since |

un

s |x| p

p−1

|

p

is bounded in Lp (RN )−1 = L p−1 (RN ), we also get |un | |x|

p−2

un

s(p−1) p

p−2

⇀

|u| |x|

u

s(p−1) p

p

weakly in L p−1 (RN ).

From (3.5) and (3.6), we deduce that ∫ ∫ ∫ p |u| r ∥u∥2E − dx − K(x)|u| dx − f udx = 0, s RN |x| RN RN which shows that u ∈ N and I(u) ≥ c0 .

(3.11)

On the other hand, ∫ ∫ p |un | 1 r K(x)|un | dx − f un dx = c + o(1). s dx − r RN N |x| RN ∫ ∫ ∫R p |un | r ∥un ∥2E − K(x)|un | dx − f un dx = o(1). s dx − |x| N N R R RN Let vn := un − u, according to the Brezis–Leib Lemma, we get ∫ ∫ ∫ p p p |un | |vn | |u| dx = dx + s s s dx + o(1). RN |x| RN |x| RN |x| 1 1 ∥un ∥2E − 2 p

∫

(3.12) (3.13)

(3.14)

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

540

∫

∫

r

∫

r

K(x)|un | dx =

r

K(x)|vn | dx +

RN

K(x)|u| dx + o(1).

RN

(3.15)

RN

∥un ∥2E = ∥vn ∥2E + ∥u∥2E + o(1).

(3.16)

N

Since V ∈ C(RN ) ∩ L 2 (RN ), then for any ε > 0, there possesses M > 0 large enough such that (∫

)2

N

N 2

|V (x)| dx

< ε.

(3.17)

{x:|V (x)|≥M }

By (3.10), (3.17), H¨ older inequality and boundedness of {vn }, as n → +∞, one has ∫ ∫ ∫ V (x)vn2 dx = V (x)vn2 dx + V (x)vn2 dx BR (0)∩{x:|V (x)≤M }

BR (0)

∫

vn2 dx

≤M

BR (0)∩{x:|V (x)≥M }

) 2 (∫ N |V (x)| dx

(∫

N 2

+ {x:|V (x)|≥M }

BR (0)

2∗

)

2 2∗

|vn | dx

BR (0)

ε ε + 2 2 ≤ ε. ≤

(3.18)

N

It follows from V ∈ L 2 (RN ) that there exists R > 0 for any ε > 0 such that (∫

N 2

)

|V (x)| dx

1 N 2

< ε.

(3.19)

RN \BR (0)

By H¨ older inequality, (3.9), (3.10), (3.18), (3.19) and boundedness of {vn }, as n → +∞, we get ∫ ∫ ∫ V (x)vn2 dx = V (x)vn2 dx + V (x)vn2 dx RN

RN \BR (0)

BR (0)

∫ ≤

(x)vn2 dx

V

≤ ε+ε

N 2

+ RN \BR (0)

BR (0)

(∫

) 2 (∫ N |V (x)| dx

(∫

2∗

)

2 2∗

|vn | dx

RN \BR (0)

) 2∗ 2 |vn | dx 2∗

RN

≤ ε.

(3.20)

Therefore, by the (3.20), we have ∫ lim

n→∞

RN

V (x)vn2 dx = 0.

From (3.12)–(3.16), (3.21) and Lemma 2.3, we get that ) ∫ ( ∫ 2 p 1 1 |vn | |vn | 2 I(u) + |∇vn | − µ 2 dx − dx = c + o(1) 2 RN p RN |x|s |x| ) ∫ ( ∫ 2 p |vn | |vn | 2 |∇vn | − µ 2 dx − s dx = o(1). |x| RN RN |x| Without loss of generality as n → ∞, we may assume that ) ∫ ( ∫ 2 p |vn | |vn | 2 |∇vn | − µ 2 dx → ζ, s dx → ζ. |x| RN RN |x|

(3.21)

(3.22) (3.23)

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541

According to (2.2) and (3.23), we obtain 2

ζ ≥ Sµ,s (RN )ζ p . If ζ > 0, then we can get that N −s

p

ζ ≥ Sµ,s (RN ) p−2 = Sµ,s (RN ) 2−s . Combining (3.5) and (3.22), we deduce that I(u) = c −

N −s 2−s 2−s ζ ≤c− Sµ,s (RN ) 2−s < c0 , 2(N − s) 2(N − s)

which contradicts (3.11). Hence, ζ = 0 and un → u strongly in Hµ as n → ∞. Moreover, by (3.21), we have un → u strongly in E. Therefore, the lemma is complete. □ Remark 3.2. According to the result in [11], we have if u ∈ E is a solution of the problem (1.1), then u ∈ Lσloc (RN ) for any σ ∈ (0, √N ). µ−β

Lemma 3.3. Assume k = 0, N ≥ 7, 0 ≤ µ < µ − 4, 0 ≤ s < 2, (V4 ), (K), (f2 ) hold and f (x) ̸≡ 0, then there exists v ∈ E, v ̸= 0 such that sup I(u0 + tv) < c0 + t≥0

N −s 2−s Sµ,s (RN ) 2−s . 2(N − s)

(3.24)

ε defined in (2.3). Denote Proof . Considering the extremal functions Uµ,s √

( Cε =

2

2εβ (N − s) √ µ

µ ) 2−s

, Uε =

ε Uµ,s . Cε

Let φ(x) ∈ C0∞ (RN ), 0 ≤ φ(x) ≤ 1 and φ(x) satisfies: { φ(x) ≡ 1 for |x| ≤ R, φ(x) ≡ 0 for |x| ≥ 2R. Define uε (x) = φ(x)U ε (x), V ε (x) = ( ∫

uε (x)

) 2∗1(s) .

∗

|uε |2 (s) dx |x|s RN

According to Remark 3.2 and Lemma 3.4 in [3], we get the following estimates 2∗ (s)

∫ RN

(

∫

2

ε 2

|∇V | − µ

RN

|V ε |

2

dx = 1.

(3.25)

)

N −2

dx = Sµ,s (RN ) + O(ε 2−s ).

|x| ⎧ ( √ ) µα ⎪ ⎪ ⎪ O ε 2−s , ⎪ ⎪ ⎪ ⎪ ) ⎨ ( √µα ε α |V | dx = O ε 2−s | ln ε| , ⎪ ⎪ ⎪ ( √µ(N −α√µ) ) ⎪ ⎪ ⎪ ⎪ , ⎩O ε (2−s)β

RN

∫

|V ε | s |x|

1 ≤ α < √N

µ+β

α = √N

µ+β

√N µ+β

(3.26)

,

,

< α < 2∗ .

(3.27)

542

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

∫ RN

⎧ ( N −2 ) ⎪ ⎪ ⎪O ε 2−s , ⎪ ⎪ ⎪ ⎨ (

0 ≤ β < 1,

) N −2 2 |V ε | dx = O ε 2−s | ln ε| , ⎪ ⎪ ( N −2 ) ⎪ ⎪ ⎪ ⎪ ⎩O ε (2−s)β , ∫ RN

∫ RN

∫ RN

∫

β = 1,

(3.28)

β > 1.

( N −2 ) r−1 |u0 | V ε dx = O ε (2−s)β . s |x| ( N −2 ) r−1 u0 |V ε | dx = O ε (2−s)β . s |x| ( N −2 ) r−1 ε |u0 | V dx = O ε (2−s)β . ( N −2 ) ε r−1 u0 |V | dx = O ε (2−s)β .

(3.29) (3.30) (3.31) (3.32)

RN

Define Φ := {ρ ≥ 0 : I(u0 + ρV ε ) ≥ c∗ }, where c∗ < 0 is a constant small enough. We can easily get that ρ is bounded on Φ. By the definition of I, we can deduce that there exists a constant ρε > 0 such that I(u0 + ρε V ε ) = sup I(u0 + ρV ε ) = sup I(u0 + ρV ε ). ρ>0

ρ∈Φ

By [20], we get that the following elementary inequality (3.33) holds: q

q

q

q−1

|a + b| ≥ |a| + |b| − C(|a| Therefore, we have ∫

|u0 + ρε V ε | dx ≥ s |x|

∫ RN

), ∀a, b ∈ R, C = C(q) > 0.

∫ p p |u0 | |ρε V ε | s dx + s dx |x| RN |x| RN (∫ ) ∫ p−1 p−1 |u0 | |V ε | |u0 | |V ε | −C dx + dx . s s |x| |x| RN RN ∫ ∫ r r r |u0 + ρε V ε | ≥ |u0 | dx + |ρε V ε | dx N R(∫ RN ) ∫ r−1 ε ε r−1 −C |u0 | |V |dx + |u0 | |V | dx . p

RN

q−1

|b| + |a| |b|

∫

RN

(3.34)

(3.35)

RN

Since N ≥ 7 and β > 2. It is easy obtain that √N

µ+β

∫

(3.33)

< r < 2∗ . By (3.27) and (K), we get

( √µ(N −r√µ) ) r K(x)|V ε | dx = O ε (2−s)β .

(3.36)

RN N

Since u0 ∈ N and u0 is a solution of problem (1.1), combining (3.25)–(3.35) and V (x) ∈ C(V ) ∩ L 2 (R we can deduce that I(u0 + ρε V ε ) ∫ p 1 1 |u0 + ρε V ε | = ∥u0 + ρε V ε ∥2E − dx s 2 p RN |x|

N

)

,

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

− ≤ − − ≤

∫ r K(x)|u0 + ρε V ε | dx − f (x)(u0 + ρε V ε )dx N N R R ( ) 2 ∫ 2 ∫ ε 2 |V | |ρε | |ρε | 2 2 ε |∇V | − µ V (x)|V ε | dx I(u0 ) + dx + 2 2 2 N N |x| R R (∫ ) ∫ r∫ |ρε | r r−1 r−1 K(x)|V ε | dx+C K(x)|V ε | |u0 |dx+ K(x)|V ε | |u0 | dx r RN N RN (∫ R ) ∫ p−1 p−1 p ∫ ε ε p |u0 | |V | |V | |u0 | |V ε | |ρε | dx + dx s dx + C s s p |x| |x| RN RN |x| RN ( √µ(N −r√µ)) ( N −2 ) ( N −2) N −s 2−s N 2−s (2−s)β 2−s c0 + −O ε (2−s)β . Sµ,s (R ) +O ε +O ε 2(N − s) 1 r

543

∫

By the computation with N ≥ 7 and β > 2, we can easily get that √ √ N −2 µ(N − r µ) > . (2 − s) (2 − s)β √ √ N −2 µ(N − r µ) > . (2 − s)β (2 − s)β

(3.37)

(3.38) (3.39)

It follows from (3.36)–(3.39), we obtain that I(u0 + ρε V ε ) < c0 +

N −s 2−s Sµ,s (RN ) 2−s . 2(N − s)

for ε small enough. Hence, we take v0 = V ε , the proof of the lemma is complete.

□

Proof of Theorem 2. According to Remark 1.5, the proof of the first solution is similar with Theorem 1. Therefore, we only show the proof of the second solution. By the definition of I and Theorem 1, we can get that (1) There exists δ0 > 0 small and r > 0 such that I(u0 + ω) − I(u0 ) ≥ δ0 holds for all ω: ω ∈ E with ∥ω∥ = r. (2) There exists v ∗ = ρ∗ v0 such that I(u0 + v ∗ ) − I(u0 ) < 0 holds for ρ∗ large enough. Define c = inf sup I(u), Γ ∈P u∈Γ

where P denotes the set of all continuous path joining u0 and v ∗ . By the Mountain Pass Theorem without (PS) condition in [1], we can get a sequence {un } ⊂ E satisfying I(un ) → c and I ′ (un ) → 0 as n → ∞. Moreover, c0 < c ≤ sup I(u0 + tv ∗ ) ≤ sup I(u0 + ρv0 ) < c0 + t∈[0,1]

ρ≥0

N −s 2−s Sµ,s (RN ) 2−s . 2(N − s)

By Lemma 3.1 and passing to a subsequence, we get that un → u strongly in E, which implies that u is a solution of problem (1.1) and u ̸= u0 . □

544

C. Wang, Y. Shang / Nonlinear Analysis: Real World Applications 46 (2019) 525–544

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