Ground states for a class of asymptotically periodic Schrödinger–Poisson systems with critical growth

Ground states for a class of asymptotically periodic Schrödinger–Poisson systems with critical growth

Da-Bin Wang

, Hua-Fei Xie and Wen Guan

Department of Applied Mathematics, Lanzhou University of Technology, Lanzhou, Gansu, 730050, People’s Republic of China

Received 7 August 2017, appeared 2 January 2018 Communicated by Dimitri Mugnai

Abstract. The purpose of this paper is to study the existence of ground state solution for the Schrödinger–Poisson systems:

(−∆u+V(x)u+K(x)φu=Q(x)|u|⁴u+f(x,u), x∈R³,

−∆φ=K(x)u², x∈R³, whereV(x)_,K(x)_,Q(x)_and f(x,u)are asymptotically periodic functions inx.

Keywords: Schrödinger–Poisson systems, ground state solution, variational methods.

2010 Mathematics Subject Classification: 35J20, 35J60, 35J65.

1 Introduction

For past decades, much attention has been paid to the nonlinear Schrödinger–Poisson system (i¯h^∂_∂t^Ψ =−_2m^h¯2 _∆Ψ+_U(_x)_Ψ+φ(_x)_Ψ− |_Ψ|^q−1_Ψ, x∈_R³, t ∈_R

−_∆φ=|_Ψ|², x∈_R^{3 (1.1)}

where ¯h is the Planck constant. Equation (1.1) derived from quantum mechanics. For this equation, the existence of stationary wave solutions is often sought, that is, the following form of solution

Ψ(x,t) =e^itu(x), x∈_R³, t∈_R.

Therefore, the existence of the standing wave solution of the equation (1.1) is equivalent to finding the solution of the following system (m= ¹_{2, ¯h}=_{1 andV}(_x) =_U(_x) +₁₎

(−_∆u+V(x)u+φu=|u|^q−1u, x∈_R³,

−_∆φ=u², x∈_R³. (1.2)

BCorresponding author. Email:wangdb96@163.com

To the best of our knowledge, the first result on Schrödinger–Poisson system was obtained in [5]. Thereafter, using the variational method, there is a series of work to discuss the ex- istence, non existence, radially symmetric solutions, non-radially symmetric solutions and ground state to Schrödinger–Poisson system (1.2) and similar problems [1,3–5,8–17,20,28,32, 34,37–39,42,44–47].

As far as we know, in [4], Azzollini and Pomponio firstly obtained the ground state solu- tion to the Schrödinger–Poisson system (1.2). They obtained that system (1.2) has a ground state solution whenV is a positive constant and 2 < q < 5, or V is non-constant, possibly unbounded below and 3 < q < 5. Since it’s great physical interests, many scholars pay at- tention to study ground state solutions to the Schrödinger–Poisson system (1.2) and similar problems [1,8,11,12,14,15,20,37,38,45,46].

In [1], Alves, Souto and Soares studied Schrödinger–Poisson system (−_∆u+V(x)u+φu= f(u), x∈_R³,

−_∆φ= u², x∈_R³, (1.3)

where f ∈C(_R⁺,R)andVis bounded, local Hölder continuous and satisfies:

(1) V(x)≥ α>0,x∈_R³,

(2) V(x) =V(x+y), ∀x∈_R³, ∀y∈_Z³, (3) lim_|x|→∞|V(x)−V₀(x)|=0,

(4) V(x)≤V₀(x), ∀x ∈_R³, and there existsΩ⊂_R³withm(_Ω)>0 such that V(x)<V₀(x), ∀x∈_Ω,

where V₀ satisfies (2). Alves et al. studied the ground state solutions to system (1.3) in case the periodic condition under(1)–(2)and in case the asymptotically periodic condition under (₁)_,(₃)_and(₄)respectively.

In [45], Zhang, Xu and Zhang considered existence of positive ground state solution for the following non-autonomous Schrödinger–Poisson system

(−_∆u+V(x)u+K(x)φu= f(x,u), x∈_R³,

−_∆φ=K(x)u², x∈ _R³. (1.4)

In some weaken asymptotically periodic sense compare with that of in [1], they obtained the positive ground state solution to system (1.4) whenV,Kand f are all asymptotically periodic inx.

More recently, Zhang, Xu, Zhang and Du [46] completed the results obtained in [45] to Schrödinger–Poisson system with critical growth

(−_∆u+V(x)u+K(x)φu=Q(x)|u|⁴u+ f(x,u), x∈_R³,

−_∆φ=K(x)u², x∈_R³. (1.5)

In [46], V,K,Q satisfy: V,K,Q ∈ L^∞(_R³), inf_R3V > 0, inf_R3K > 0, inf_R3Q > 0 and V−V_p,K−K_p,Q−Q_p ∈ F, where V_p, K_p and Q_p are 1-periodic in x_i, 1 ≤ i ≤ 3, and F ={g∈ L^∞(_R³)_:∀ε>_{0, the set} {x∈_R³_:|g(x)| ≥ε}has finite Lebesgue measure}_.

On the other hand, whenK = 0 the Schrödinger–Poisson system (1.4) becomes the stan- dard Schrödinger equation (replaceR³withR^N)

−_∆u+V(x)u= f(x,u), x ∈_R^N. (1.6) The Schrödinger equation (1.6) has been widely investigated by many authors in the last decades, see [2,6,19,24,25,29–31,40,41,43] and reference therein. Especially, in [19,24,25, 29,40,41], they studied the nontrivial solution or ground state solution for problem (1.6) with subcritical growth or critical growth in which V,f satisfy the asymptotically periodic condition. Other context about asymptotically periodic condition, we refer the reader to [18, 21,35,36] and reference therein.

Motivated by above results, in this paper, we will study ground state solutions to sys- tem (1.5) under reformative condition about asymptotically periodic case of V,K,Qand f at infinity.

To state our main results, we assume that:

(V) there existV_p :R³→_R, 1-periodic in x_i, 1≤i≤3, such that V0 := inf

x∈_R³Vp >0, 0≤V(x)≤Vp(x)∈ L^∞(_R³) and V(x)−Vp(x)∈ A0, where

A₀ :={k(x): for anyε>0, m{x ∈B₁(y):|k(x)| ≥ε} →0 as|y| →_∞}; (K) there existK_p :R³→R, 1-periodic in x_i, 1≤ i≤3, such that

K0:= _inf

x∈_R³Vp>_{0, 0}<_K(_x)≤ Kp(_x)∈ L^∞(_R³) _{and K}(_x)−Kp(_x)∈ A0; (Q) there existQ_p ∈C(_R³_,R), 1-periodic in x_i, 1≤ i≤3, and pointx₀∈_R³_{such that}

0< Qp(x)≤Q(x)∈C(_R³,R), Q(x)−Qp(x)∈ A₀ and

Q(x) =|Q|_∞+O(|x−x0|), asx→ x0; and f ∈C(_R³×_R⁺,R)satisfies

(f₁) _lims→0+ f(x,s)

s =0 uniformly for x∈_R³_, (f₂) lim_s→+_∞ ^f(x,s)

s⁵ =0 uniformly forx∈ _R³, (f₃) s → ^f(x,s)

s³ is nondecreasing on (0,+_∞),

(f₄) there exists an open bounded setΩ⊂_R³, containingx₀ given by(Q), satisfies

s→+lim_∞

F(x,s)

s⁴ = +_∞ uniformly forx∈ _Ω, (f₅) there exists f_p ∈C(_R³×_R⁺_,R⁺), 1-periodic in x_i, 1≤i≤3, such that

(i) f(x,s)≥ f_p(x,s)for all(x,s)∈_R³×_R⁺and f(x,s)− f_p(x,s)∈ A, where A:={h(x,s): for anyε>0, m{x ∈B₁(y):|h(x,s)| ≥ε} →0

as|y| →_∞uniformly for|s|bounded}, (ii) s→ ^fp(x,s)

s³ is nondecreasing on (_0,+_∞)_.

The next theorem is the main result of the present paper.

Theorem 1.1. Suppose that conditions (V), (K), (Q)and (f₁)–(f5) are satisfied. Then the system (1.5)has a ground state solution.

Remark 1.2.

(i) Functional sets A₀ in V,Q,K and A in (f₅) were introduced by [24,25] in which Liu, Liao and Tang studied positive ground state solution to Schrödinger equation (1.6) with subcritical growth or critical growth.

(ii) Since F ⊂ A₀, our assumptions on V,Q and K are weaker than [46]. Furthermore, V(x)≥0 in our paper but in [46] they assumedV(x)>_0.

(iii) In [46], to obtained the ground state to system (1.5), they firstly consider the periodic system

(−_∆u+V_p(x)u+K_p(x)φu=Q_p(x)|u|⁴u+ f_p(x,u)_, x∈_R³,

−_∆φ=K_p(x)u², x∈_R³. (1.7)

Then a solution of system (1.5) was obtained by applying inequality between the energy of periodic system (1.7) and that of system (1.5). In this paper, we do not use methods of [46]

and prove Theorem1.1directly.

2 The variational framework and preliminaries

To fix some notations, the letterC andC_i will be repeatedly used to denote various positive constants whose exact values are irrelevant. B_R(z)denotes the ball centered atzwith radiusR.

We denote the standard norm of L^p by|u|_p = (R

R³|u|^pdx)^1p _and|u|_∞ = _{ess supx∈R3}|u|_{. Since} we are looking for a nonnegative solution, we may assume that f(x,s) = fp(x,s) = 0 for all (x,s)∈(_R³,R⁻).

The Sobolev space H¹(_R³)endowed with the norm kuk²_H :=

Z

R³(|∇u|²+u²)dx.

The space D^1,2(_R³)endowed with the standard norm kuk²_D1,2 :=

Z

R³|∇u|²dx.

Let E := {u ∈ L⁶(_R³) : |∇u| ∈ L²(_R³) and R

R³V(x)u²dx < _∞} be the Sobolev space endowed with the norm

kuk²:=

Z

R³(|∇u|²+V(x)u²)dx.

Lemma 2.1 ([24]). Suppose(V)holds. Then there exists two positive constants C₁ and C₂ such that C₁kuk²_H ≤ kuk ≤C2kuk²_H for all u∈E. Moreover, E,→ L^p(_R³)for any p ∈[2, 6]is continuous.

The system (1.5) can be transformed into a Schrödinger equation with a nonlocal term. In fact, for allu∈ E(thenu∈ H¹(_R³)), considering the linear functionalLudefined in D^1,2(_R³) by

Lu(v) =

Z

R³K(x)u²vdx.

By the Hölder inequality, we have

|L_u(v)| ≤ |K|_∞|u|²₁₂

5

|v|₆ ≤C|u|²₁₂

5

kvk_D1,2. (2.1)

Therefor, the Lax–Milgram theorem implies that there exists a uniqueφ_u∈ D^1,2(_R³)such that

Z

R³∇φ_u· ∇vdx= (φ_u,v)_D1,2 =L_u(v) =

Z

R³K(x)u²vdx for any v∈D^1,2(_R³). Namely,φ_u is the unique solution of−_∆φ= K(x)u². Moreover,φ_ucan be expressed as

φ_u=

Z

R³

K(y)u²(y)

|x−y| ^dy.

Substitutingφuinto the systems (1.5), we obtain

−_∆u+V(x)u+K(x)φ_uu= Q(x)|u|⁴u+ f(x,u), x∈_R³. (2.2) By (2.1), we get

kφ_uk_D1,2 =kL_uk ≤C|u|²₁₂

5

≤Ckuk². Then, we have

|

Z

R³K(x)φ_uu²dx| ≤ |K(x)|_∞|φ_u|₆|u|²₁₂

5

(2.3)

≤ C|K(x)|_∞kφ_uk_D1,2|u|²₁₂

5

≤ C|u|⁴₁₂

5

≤ C₀kuk⁴_.

So the energy functional I :E→_Rcorresponding to Eq. (2.2) is defined by I(u) = ¹

2 Z

R³(|∇u|²+V(x)u²)dx+ ¹ 4

Z

R³K(x)φuu²dx−¹ 6

Z

R³Q(x)(u⁺)⁶dx−

Z

R³ F(x,u)dx, where F(x,s) =Rs

0 f(x,t)dt.

Moreover, under our conditions, I belongs toC¹, so the Fréchet derivative ofI is hI⁰(u),vi=

Z

R³(∇u· ∇v+V(x)uv)dx+

Z

R³K(x)φ_uuvdx−

Z

R³Q(x)(u⁺)⁵vdx−

Z

R³ f(x,u)vdx and(u,φ)∈E×D^1,2(_R³)is a solution of system (1.5) if and only ifu∈ Eis a critical point of I andφ=φ_u.

For allu∈E, let ˜φ_u∈ D^1,2(_R³)is unique solution of the following equation

−_∆φ=K_p(x)u².

Moreover,φe_u can be expressed as

φe_u =

Z

R³

K_p(y)u²(y)

|x−y| ^dy.

Let

I_p(u) = ¹ 2

Z

R³(|∇u|²+V_p(x)u²)dx+ ¹ 4

Z

R³K_p(x)φe_uu²dx

−¹ 6

Z

R³Qp(x)(u⁺)⁶dx−

Z

R³Fp(x,u)dx, whereF_p(x,s) =Rs

0 f_p(x,t)dt. ThenI_p is the energy functional corresponding to the following equation

−_∆u+V_p(x)u+K_p(x)φe_uu= Q_p(x)|u|⁴u+ f_p(x,u), x∈_R³. (2.4) It is easy to see that(u,φ)∈ E×D^1,2(_R³)is a solution of periodic system (1.7) if and only ifu∈Eis a critical point of Ipandφ=φeu.

Lemma 2.2. Suppose(K)holds. Then, Z

R³Kp(x)φe_u(·+z)u²(·+z)dx=

Z

R³Kp(x)φeuu²dx, ∀z∈_Z³, u∈ E.

Lemma 2.3. Suppose that(f₁),(f3)and(f5)hold. Then (i) ¹₄f(x,s)s ≥F(x,s)≥0for all(x,s)∈_R³×_R, (ii) ¹₄fp(x,s)s≥ Fp(x,s)≥0for all(x,s)∈_R³×_R.

Proof. The proof is similar to that of in [27], so we omitted here.

Lemma 2.4. I⁰ is weakly sequentially continuous. Namely if un * u in E, I⁰(un) * I⁰(u) in E⁻¹(_R³).

Proof. The proof is similar to that of Lemma 2.3 in [45,46], so we omitted here.

Lemma 2.5([24]). Suppose that (f₁), (f₂)and (i) of (f₅)hold. Assume that{u_n}is bounded in E and un →0in L_loc^s (_R³), for any s∈ [2, 6). Then up to a subsequence, one has

Z

R³(F(x,u_n)−F_p(x,u_n))dx =o(1). (2.5) Lemma 2.6([24,25]). Suppose that (V),(Q), (f₁), (f₂)and(i)of(f₅)hold. Assume that {un}is bounded in E and|z_n| →∞. Then up to a subsequence, one has

Z

R³(V_p(x)−V(x))u_nϕ(· −z_n)dx= o(1), (2.6) Z

R³(f(x,u_n)− f_p(x,u_n))ϕ(· −z_n)dx= o(₁)_{, (2.7)} and

Z

R³(Q(x)−Q_p(x))(u⁺_n)⁵ϕ(· −z_n)dx= o(1), (2.8) whereϕ∈C^∞₀ (_R³).

Lemma 2.7. Suppose that (K), (f₁) and (f₂) hold. Assume that u_n * 0 in E. Then up to a subsequence, one has

Z

R³(K(x)φ_unu_nϕ(· −z_n)−K_p(x)φe_unu_nϕ(· −z_n))dx= o(1), (2.9) where|z_n| →_∞andϕ∈ C₀^∞(_R³).

Proof. Seth(x):= K(x)−K_p(x). By(K), we haveh(x)∈ A₀. Then for anyε> 0, there exists Rε >0 such that

m{x∈ B₁(y)_:|h(x)| ≥ε}<_{ε, for any} |y| ≥R_ε.

We cover R³ by balls B₁(y_i), i ∈ N. In such a way that each point of R³ is contained in at most N+1 balls. Without any loss of generality, we suppose that |y_i| < R_ε, i = 1, 2, . . . ,n_ε and|y_i| ≥R_ε, i=n_ε+_1,n_ε+_2,n_ε+_{3, . . . ,}+_{∞. Then,}

Z

R³(K(x)φununϕ(· −zn)−Kp(x)φeununϕ(· −zn))dx

=

Z

R³

Z

R³

Kp(y)un(y)ϕ(y−zn)

|x−y| ^dyh(x)u²_n(x)dx +

Z

R³

Z

R³

Kp(_y)_u²_n(_y)

|x−y| ^dyh(x)un(x)ϕ(x−zn)dx +

Z

R³

Z

R³

h(y)u²_n(y)

|x−y| ^dyh(x)un(x)ϕ(x−zn)dx

=:E₁+E₂+E₃. Like the argument of [45], we define

H(x):=

Z

R³

K_p(y)u_n(y)ϕ(y−z_n)

|x−y| ^dy

=

Z

{y:|x−y|≤1}

K_p(y)u_n(y)ϕ(y−z_n)

|x−y| ^dy+

Z

{y:|x−y|>1}

K_p(y)u_n(y)ϕ(y−z_n)

|x−y| ^dy.

By the Hölder inequality and the Sobolev embeddings, we have

|H(x)| ≤ |Kp|_∞|un|₃|ϕ|_{6 Z}

{y:|x−y|≤1}

1

|x−y|^{2dy 1}₂

+|Kp|_∞|un|₂|ϕ|_{4 Z}

{y:|x−y|>1}

1

|x−y|^{4dy 1}₄

≤C _Z

{z:|z|≤1}

1

|z|^{2dz 1}₂

+C _Z

{z:|z|>1}

1

|z|^{4dz 1}₄

. So, sup_x∈R3|H(x)|<∞. Then, we obtain

E₁ =

Z

R³H(x)h(x)u²_n(x)dx

≤

Z

{x:|h(x)|≥ε}|H(x)h(x)u²_n(x)|dx+

Z

{x:|h(x)|<ε}|H(x)h(x)u²_n(x)|dx

=:Q₁+Q₂,

Q₁=

Z

{x:|h(x)|≥ε}

|H(x)h(x)u²_n(x)|dx

≤

Z

{x:|h(x)|≥ε,|x|>Rε+1}|H(x)h(x)u²_n(x)|dx+

Z

{x:|h(x)|≥ε,|x|≤Rε+1}|H(x)h(x)u²_n(x)|dx

≤

∑

Z

{x∈B1(yi):|h(x)|≥ε,|x|>Rε+1}|H(x)h(x)u²_n(x)|dx+2 sup

x∈_R³

|H(x)||Kp|_∞

Z

BRε+1

|un(x)|²dx

=:Q₁₁+Q₁₂,

Q₁₁=

∑

Z

{x∈B1(yi):|h(x)|≥ε,|x|>Rε+1}|H(x)h(x)u²_n(x)|dx

≤2 sup

x∈_R³

|H(x)||Kp|_∞

∑

Z

{x∈B1(yi):|h(x)|≥ε,|x|>Rε+1}|u²_n(x)|dx

≤ C

∑

(m{x ∈B₁(y)_:|h(x)| ≥ε})²³ Z

{x∈B1(yi):|h(x)|≥ε,|x|>Rε+1}|u⁶_n(x)|dx ¹₃

≤ C₁ε

2 3

∑

Z

{x∈B₁(y_i):|h(x)|≥ε,|x|>Rε+1}

(|∇u_n|²+u²_n)dx

≤ C₁(N+₁)ε

2 3

Z

R³(|∇u_n|²+u²_n)dx

≤ C₂ε

2 3.

Letε→0, we obtainQ₁₁ →0. By the conditionu_n *0, one hasu_n →0 inL²_loc(_R³). Therefore Q₁₂→0. SoQ₁ →0.

Q₂=

Z

{x:|h(x)|<ε}|H(x)h(x)u²_n(x)|dx

≤ εsup

x∈_R³

|H(x)|

Z

R³|u²_n(x)|dx

≤Cε.

Letε→0, we haveQ2 →0. Then, we getE₁→0. In the same way, we can prove E2→0 and E₃→0.

Let F={u∈ E: u⁺6=0}, define

N :={u∈ E\ {0}:hI⁰(u),ui=0}={u∈ F:hI⁰(u),ui=0}. ThenN is a Nehari type associate to I, and set c:=inf_u∈N I.

Lemma 2.8. Suppose that(V),(K),(Q)and(f₁)–(f₃)hold. For any u∈ F, there is a unique t_u >0 such that t_uu∈ N. Moreover, the maximum of I(tu)for t≥0is achieved.

Proof. Fix u ∈ F, define g(t) := I(tu), t > 0. Using (f₁), (f₂), and (f₃), we can prove that g(₀) =_0,g(t)>_{0 for}tsmall andg(t)<_{0 for}tlarge.

In fact, by(f₁)and(f₂),∀δ>0 there exists aC_δ >0 such that

|f(x,s)| ≤δ|s|+C_δ|s|⁵, |F(x,s)| ≤ ^δ

2|s|²+ ^Cδ

6 |s|⁶ for any(x,s)∈ (_R³,R).

So, we get that g(t) = ^t

2

2kuk²+ ^t

4

4 Z

R³K(x)φuu²dx− ^t

6

6 Z

R³Q(x)(u⁺)⁶dx−

Z

R³F(x,tu)dx

≥ ^t

2

2kuk²− ^δt

2

2 Z

R³|u|²dx− ^Cδt

6

6 Z

R³|u|⁶dx−Ct⁶ Z

R³|u|⁶dx

≥ ^t

2

2kuk²−Cδt²kuk²−CC_δt⁶kuk⁶. Hence,g(t)>0 fortsmall.

On the other hand, letΘ={x ∈_R³ :u(x)>0}, we have that g(t) = ^t

2

2kuk²+ ^t

4

4 Z

R³K(x)φ_uu²dx− ^t

6

6 Z

R³Q(x)(u⁺)⁶dx−

Z

R³F(x,tu)dx

≤ ^t

2

2kuk²+ ^t

4

4 Z

R³K(x)φuu²dx− ^t

6

6 Z

ΘQ(x)(u⁺)⁶dx.

Hence, it is easy to see thatg(t)→ −_∞ast→+_∞.

Therefore, there exists a t_u such that I(t_uu) = _maxt>0I(tu) _and t_uu ∈ N. Suppose that there existt₁> t₂ >0 such thatt₁u, t₂u∈ N. Then, we have that

1

t²₁kuk²+

Z

R³K(x)φuu²dx= t²₁ Z

ΘQ(x)(u⁺)⁶dx+

Z

Θ

f(x,t₁u)u t³₁ dx, 1

t²₂kuk²+

Z

R³K(x)φ_uu²dx= t²₂ Z

ΘQ(x)(u⁺)⁶dx+

Z

Θ

f(x,t₂u)u t³₂ dx.

Therefore, one has that 1

t²₁ − ¹ t²₂

kuk² = (t²₁−t²₂)

Z

ΘQ(x)(u⁺)⁶dx+

Z

Θ

f(x,t₁u)

(t₁u)³ − ^f(x,t₂u) (t₂u)³

u⁴dx, which is absurd according to (f₃)_andt₁>t₂>_0.

Remark 2.9. As in [31,43], we have c= _inf

u∈N I(u) = _inf

u∈Fmax

t>0 I(tu) = _inf

γ(t)∈_Γmax

t∈[0,1]I(γ(t))>₀ where

Γ:={γ∈ C([0, 1],E):γ(0) =0, I(γ(1))<0}.

Lemma 2.10. Suppose that(V),(K),(Q)and(f₁)–(f₃)hold. Then there exists a bounded sequence {un} ∈E such that

I(u_n)→c and kI⁰(u_n)k_E−1 →_0.

Proof. From the proof of Lemma2.8, it is easy to see that I satisfies the mountain pass geom- etry. By [33], there exists an{u_n}such that I(u_n)→ cand(₁+ku_nk)kI⁰(u_n)k_E−1 →0, so we havehI⁰(un),uni= o(1). By(f₃), we can obtain

1

4f(x,s)s≥ F(x,s) for any (x,s)∈(_R³,R).

Then, we have that c= _I(_un)− ¹

4hI⁰(_un)_,uni

= ¹

4ku_nk²+ ¹ 12

Z

R³Q(x)(u⁺_n)⁶dx+

Z

R³(¹

4f(x,u_n)u_n−F(x,u_n))dx

≥ ¹ 4kunk².

Therefor,{u_n}is bounded and the proof is finished.

The proof of next lemma similar to that of [24,26]. For easy reading, we give the proof.

Lemma 2.11. Suppose that(V),(K),(Q)and(f₁)–(f₃)hold. If u∈ N and I(u) =c, u is a solution of Eq.(2.2).

Proof. Suppose by contradictionuis not a solution. Then there exists ϕ∈ Esuch that hI⁰(u),ϕi<−1.

Chooseε∈ (0, 1)small enough such that for all|t−1| ≤1 and|σ| ≤ε, hI⁰(tu+σ ϕ),ϕi ≤ −¹

2.

We define a smooth cut-off functionζ(t)∈ [0, 1], which satisfies ζ(t) = 1 for|t−1| ≤ ₂^ε and ζ(t) = 0 for |t−1| ≥ ε. For t > 0 we introduce a curve γ(t) = tu for |t−1| ≥ ε and γ(t) = tu+εζ(t)ϕfor |t−1| < ε. Obviously, γ(t) is a continuous curve and when ε small enough, kγ(t)k > 0 for |t−1| < ε. Next we prove I(γ(t)) < c, for t > 0. If |t−1| ≥ ε, I(γ(t)) = I(tu) < I(u) = c. If |t−1| < ε, we define A : σ 7→ I(tu+σζ(t)ϕ). Obviously, A∈ C¹. By the mean value therm, there existsσ ∈(0,ε)such that

I(tu+εζ(t)ϕ) = I(tu) +hI⁰(tu+σζ(t)ϕ),εζ(t)ϕi ≤I(tu)− ^ε

2ζ(t)<c.

Defineν(u):= hI⁰(u),ui, thenν(γ(1−ε)) =ν((1−ε)u)>0 andν(γ(1+ε)) =ν((1+ε)u)<

0. By the continuity oft →ν(γ(t)), there existst⁰ ∈(1−ε, 1+ε)such thatν(γ(t⁰)) =0. Thus γ(t⁰)∈ N andI(γ(t⁰))<c, which is a contradiction.

Define

N_p ={u∈ F:hI⁰_p(_u)_,ui=₀} and cp= _inf

u∈N_pIp(_u)_. In fact,c_p =inf_u∈Fmax_t>0I_p(tu).

Remark 2.12. For anyu ∈ F, by Lemma2.8, there exists t_u > 0 such thatt_uu ∈ N and then I(t_uu)≥ c. Using V(x)≤ V_p(x),Q(x)≥ Q_p(x)andF(x,s)≥ F_p(x,s), we have c≤ I(t_uu)≤ Ip(tuu)≤maxt>₀Ip(tu). Then we obtainc≤cp.

3 Estimates

In this section, we will estimate the least energyc, and the method comes from the celebrated paper [7].

Let

S= _inf

u∈D^1,2(_R³)\{0}

|∇u|²₂

|u|²₆ ^.

In fact,Sis the best constant for the Sobolev embeddingD^1,2(_R³),→L⁶(_R³).

Without loss of generality, we assume that x0 = 0. Forε > 0, the function wε : R³ → _R defined by

wε(x) = ³

1 4ε

1 4

(ε+|x|²)¹²

is a family of functions on which S is attained. Let ϕ ∈ C₀^∞(_R³,[0, 1]) be a cut-off function satisfying ϕ = _{1, for} x ∈ B^ρ

2 and ϕ = _{0, for} x ∈ _R³\B_ρ, where B_ρ ⊂ Ω. Define the test function by

vε = ^uε (R

R³Q(x)u⁶_εdx)^16, whereuε = ϕwε. Then one has

Z

R³|∇vε|²dx≤ |Q|⁻_∞¹³S+_O(ε

1

2)_{, as}ε→0⁺, (3.1)

Z

R³|v_ε|²dx=O(ε

1

2), asε→0⁺, (3.2)

Z

R³|uε|⁶dx =K₁+O(ε

3

2), asε→0⁺, whereK₁is some positive constant, (3.3) Z

R³ Q(x)v⁶_εdx=1, (3.4)

Z

R³|vε|¹²⁵ dx=O(ε

3

5), asε→0⁺. (3.5)

Lemma 3.1. Suppose(V),(K),(Q)and(f₁)–(f₄)are satisfied. Then c< ¹₃|Q|⁻_∞¹²S³². Proof. Fort >0, define

g(t):= I(tv_ε)

= ^t

2

2 Z

R³|∇v_ε|²dx+ ^t

2

2 Z

R³V(x)v²_εdx+^t

4

4 Z

R³K(x)φ_vεv²_εdx

−^t

6

6 Z

R³ Q(x)v⁶_εdx−

Z

R³F(x,tv_ε)dx.

By Lemma 2.8, there exists a unique t_ε > 0 such that g(t_ε) =_maxt>0g(t)_andg⁰(t_ε) =_{0. We} claim that there exists C₁, C₂ such that C₁ ≤ tε ≤ C₂ for ε small enough. Indeed, if tε → 0 as ε → 0, one has g(t_ε) → 0, which is a contradiction. If t_ε → +_∞ as ε → +_{∞, one has} g(_tε)→ −∞, which is a contradiction. Thus the claim holds. For s>_{0, define}

ψ(s):= ^s

2

2 Z

R³|∇vε|²dx− ^s

6

6.

Then there existss_ε := R

R³|∇v_ε|²dx¹₄

such that

ψ(sε) =max

s>0 ψ(s) = ¹ 3

Z

R³|∇vε|²dx ³₂

.

By (3.1) and the inequality (a+b)^α ≤a^α+α(a+b)^α−1b,a>_0,b>_0,α≥1, we have ψ(sε)≤ ¹

3|Q|⁻_∞¹²S³² +O(ε

12). (3.6)

We claim

lim

ε→0⁺

R

R³F(x,tεvε)dx

O(ε¹²) = +_∞. (3.7)

By (3.3), forε small enough, one has|uε|₆≤2K₁and then for |x|< ε

1 2 < ^ρ₂, tεvε ≥ ^C1

2|Q|_∞K₁uε = ^C1

2|Q|_∞K₁wε = ^C1 2|Q|_∞K₁

3¹⁴ε¹⁴

(ε+|x|²)¹² ≥ Cε⁻¹⁴.

It follows from (f₄) that for any R > 0, there exists A_R > 0 such that for all (x,s) ∈ _Ω× [A_R,+_∞),

F(x,s)≥Rs⁴. Thus forεsmall enough, one has

Z

{x:|x|<ε

12}F(x,tεvε)dx≥CR Z

{x:|x|<ε

12}ε⁻¹dx=CRε¹².

Combining withF(x,s)≥ 0 and the arbitrariness of R, we can obtain the claim. By (2.3) and (3.5), we get

Z

R³K(x)φ_vεv²_εdx

≤C₀|v_ε|⁴₁₂

5

≤C₂ε.

Hence forεsmall enough, by (3.2), (3.6) and (3.7), we have c≤ max

t>0 I(tvε)

= ^t

2 ε

2 Z

R³|∇v_ε|²dx+ ^t

2 ε

2 Z

R³V(x)v²_εdx + ^t

4 ε

4 Z

R³K(x)φ_vεv²_εdx−^t

6 ε

6 Z

R³ Q(x)v⁶_εdx−

Z

R³F(x,t_εv_ε)dx

≤ ¹

3|Q|⁻_∞¹²S³² +O(ε) +O(ε

1 2)−

Z

R³F(x,tεvε)dx

≤ ¹

3|Q|⁻_∞¹²S³² +O(ε

1 2)−

Z

R³F(x,t_εv_ε)dx

< ¹

3|Q|⁻_∞¹²S³².

4 The proof of main result

The proof of Theorem1.1. From Lemma2.10, there exists a bounded sequence{u_n} ∈Esatisfy- ing I(u_n)→ c andkI⁰(u_n)k_E−1 → 0. Then there existsu ∈ E such that, up to a subsequence,

u_n * uin E, u_n → uin L²_loc(_R³)andu_n(x) →u(x)a.e. inR³. By Lemma2.4, for anyv ∈ E, we have

0=hI⁰(u_n),vi+o(1) =hI⁰(u),vi, that isuis a solution of Eq. (2.2). Since

0=hI⁰(u),u⁻i=ku⁻k+

Z

R³K(x)φu|u⁻|²dx≥ ku⁻k, thenu ≥_0.

We next distinguish the following two case to prove Eq. (2.2) has a nonnegative ground state solution.

Case 1. Suppose that u6=_{0. Then} I(u)≥c. By the Fatou lemma, we obtain c=lim inf

n→_∞

I(un)− ¹

4hI⁰(un),uni

=lim inf

n→_∞

1

4ku_nk²+ ¹ 12

Z

R³Q(x)(u⁺_n)⁶dx+

Z

R³

1

4f(x,u_n)u_n−F(x,u_n)

dx

≥ ¹

4kuk²+ ¹ 12

Z

R³Q(x)(u⁺)⁶dx+

Z

R³

1

4f(x,u)u−F(x,u)

dx

= I(u)− ¹

4hI⁰(u),ui

= I(u)_.

Therefore, I(u) =candI⁰(u) =0.

Case 2. Suppose that u=_{0. Define}

β:=_{lim sup}

n→_∞

sup

z∈_R³ Z

B₁(z)u²_ndx.

Ifβ=0, by using the Lions lemma [22,23], we haveu_n→_{0 in}L^q(_R³)_{for all}q∈(_{2, 6})_{. By the} condition of (f₁)and(f₂),∀δ > 0 there exists aC_δ > 0 such that f(x,u)u ≤ δ(|u|²+|u|⁶) + C_δ|u|^α andF(x,u)≤ ^δ₂|u|²+^δ₆|u|⁶+C_δ|u|^α for any(x,s)∈_R³×_Randα∈(2, 6). So

Z

R³ f(x,u_n)u_ndx→0, Z

R³F(x,u_n)dx→0.

Then

c= ¹

2ku_nk²+ ¹ 4

Z

R³K(x)φ_unu²_ndx−¹ 6

Z

R³Q(x)(u⁺_n)⁶dx+o_n(1), (4.1) kunk²+

Z

R³K(x)φunu²_ndx=

Z

R³Q(x)(u⁺_n)⁶dx+on(1). (4.2) By (4.2), we have

ku_nk² ≤ |Q|_∞|u_n|⁶₆+o_n(₁)≤ |Q|_∞S⁻³ku_nk⁶+o_n(₁)_{, (4.3)} which deduces that (i)ku_nk →_{0 or}(ii)ku_nk ≥ |Q|⁻_∞¹⁴S³⁴ +o_n(₁)_.

If (i) holds, by (2.3), one has R

R³K(x)φunu²_ndx → 0. It follows from (4.1) and (4.2) that c=0, which is a contradiction withc>0.

If(ii)holds, by (4.2) we have Z

R³Q(x)(u⁺_n)⁶dx≥ |Q|⁻_∞¹²S³² +on(1). (4.4)