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Ground state solutions for a quasilinear Kirchhoff type equation

Hongliang Liu

1

, Haibo Chen

B2

and Qizhen Xiao

1

1School of Mathematics and Physics, University of South China, Hengyang, Hunan 421001, PR China

2School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083, PR China

Received 15 January 2016, appeared 15 September 2016 Communicated by Dimitri Mugnai

Abstract. We study the ground state solutions of the following quasilinear Kirchhoff type equation

1+b Z

R3|∇u|2dx

∆u+V(x)u−[(u2)]u=|u|10u+µ|u|p−1u, xR3,

whereb0 andµis a positive parameter. Under some suitable conditions onV(x), we obtain the existence of ground state solutions of the above equation with 1<p<11.

Keywords: Kirchhoff type equations, ground state solution, quasilinear, variational methods.

2010 Mathematics Subject Classification: 35J20, 35J60.

1 Introduction and main results

Consider the following Kirchhoff type equation

1+b Z

R3|∇u|2dx

∆u+V(x)u−[(u2)]u=|u|10u+µ|u|p1u, x∈R3, (1.1) where b≥ 0, 1< p <11, µ> 0 is a parameter and the potentialV(x)satisfies the following condition:

(V) V∈C(R3,R)satisfies infV(x) =V0>0 and for each M>0, meas{x∈R3 : V(x)≤M}<

+, where meas denotes the Lebesgue measure inR3.

Problem (1.1) arises in an interesting physical context. In fact, if V(x) = 0 and replacing R3by a bounded domainΩ⊂R3in (1.1), problem (1.1) without the term[(u2)]ureduces to the following Dirichlet problem of Kirchhoff type

(− 1+bR

R3|∇u|2dx

∆u= f(u), x ∈Ω,

u=0, x ∈∂Ω, (1.2)

BCorresponding author. Email: math_chb@csu.edu.cn

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which is related to the stationary analogue of the equation

ρ∂2u

∂t2P0 h + E

2L Z L

0

∂u

∂x

2

dx

!

2u

∂x2 =0 (1.3)

presented by Kirchhoff in [7], where ρ,ρ0,h, EandLare constants. Problem (1.3) extends the classical D’Alembert’s wave equation, by considering the effects of the changes in the length of the strings during the vibrations. After Lions [10] proposed an abstract framework to the problem (1.2), many papers devoted to the existence of multiple nontrivial solutions, infinitely many solutions and ground solution to the semiliner (without the term [(u2)]u) Kirchhoff type problems by applying the modern variational methods. See for instance, Liu and He [15], Wu [23], Sun and Wu [21], Chen and Li [3], Li and Ye [8], He and Zou [6], Zhang et al. [26], Zhang and Zhang [27], Liu and Guo [12–14], Liu and Chen [4,11] and the references therein.

On the other hand, many papers concerned with the following quasilinear Schrödinger equation

∆u+V(x)u−[(u2)]u=h(u), x ∈RN. (1.4) Such equations arise in various branches of mathematical physics and have been extensively studied in recent years. For example, the problem (1.4) was transformed to be a semilinear one by a change of variables and the existence of positive solutions of problem (1.4) in [18] was obtained on an Orlicz space by using the mountain pass theorem. The same method was also used in [5], but the usual Sobolev space H1(RN)framework was used as the working space.

Liu, Wang and Wang [19] obtained the existence of both one sign and nodal ground state type solutions of problem (1.4) by the Nehari method. In [17], the authors presented an approach to study problem (1.4) and proved that the solutions of problem (1.4) can be obtained as limits of 4-Laplacian perturbations.

However, to the best of our knowledge, very few papers deal with problem (1.1) in the literature. More precisely, Liang and Shi [9] studied the problem (1.1) and obtained infinitely many solutions which tend to zero via a concentration-compactness principle and the mini- max methods. In [24], the authors got the infinitely many small energy solutions of problem (1.1) by applying Clark’s theorem.

Motivated by the reasons above, the aim of this paper is to show the existence of ground state solutions of problem (1.1). Different from the semilinear problems, the feature of the quasilinear problem is the appearance of the term [(u2)]u. This makes the problem more challenging and interesting because in general there is no suitable space in which the energy functional enjoys both smoothness and compactness. Therefore, the variational methods can not be applied directly. As we shall see in the present paper, problem (1.1) can be viewed as an elliptic equation coupled with a non-local term. The competing effect of the non-local term with the critical nonlinearity and the lack of compactness of the embedding ofH1(R3)into the space Lp(R3), prevent us from using the variational methods in a standard way. Following the idea of [5,18], we transform the problem to a semilinear one by a change of variables.

Note that the problem (1.1) becomes problem (1.4) when b= 0. It is worth pointing out that although the idea was used to solve the problem (1.4) above, the adaptation to the procedure to our problem is not trivial at all since the appearance of non-local term. To obtain the ground state solution of problem (1.1), however, some more delicate estimates are needed in the present paper.

Before stating our main results we need to introduce some notations and definitions.

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Notation 1.1. Throughout this paper, we denote byk · kr theLr-norm, 1≤r≤∞, and we use the notation→(*)to denote strong (weak) convergence. Also, if we take a subsequence of a sequence{un}we shall denote it again{un}. We useo(1)to denote any quantity which tends to zero whenn → ∞. C andCi express distinct positive constants which may vary from line to line.

Definition 1.2. A nontrivial solution of problem (1.1) is called a ground state solution if its energy is minimal among the energy of all nontrivial solutions.

Now, we give our main results.

Theorem 1.3. Suppose that condition (V)holds. Then problem(1.1)has a ground state solution for allµ>0when9< p<11.

Theorem 1.4. Suppose that condition(V)holds. Then there existsµ >0such that problem(1.1)has a ground state solution for allµ∈(µ,+)when1< p≤9.

Remark 1.5. It should be mentioned that the authors in [16] have proved the problem (1.1) with b= 0 has no nontrivial solution ifx· ∇V(x)≥0 and p≥11. This is the reason why we just consider the problem for 1< p<11.

Remark 1.6. Compared to the previous results (see e.g. [9]), the main novelty in this paper is that we are able to obtain the existence of the ground state solution of problem (1.1) with 1 < p < 11. Moreover, since we consider the critical case, our main results are also different from [24] in which the authors studied the nontrivial solutions.

The remainder of this paper is organized as follows. In Section 2, we present some pre- liminaries while the proofs of our main results is given in Section 3.

2 Preliminaries

Define the Hilbert space X=

u∈ H1(R3): Z

R3V(x)v2dx<

with the inner product

hu,vi=

Z

R3[∇u∇v+V(x)uv]dx and the norm

kuk=hu,ui12.

It is well known that the embedding X ,→ Ls(R3) for 2 ≤ s < 6 is compact under the con- dition (V). Problem (1.1) is the Euler–Lagrange equation associated with the natural energy functional

I(v) = 1 2

Z

R3(1+2v2)|∇v|2dx+1 2

Z

R3V(x)v2dx+ b 4

Z

R3|∇v|2dx 2

1 12

Z

R3|v|12dx− µ p+1

Z

R3|v|p+1dx. (2.1)

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Unfortunately,I in general is not well defined onXbecause∇(v2)is not always in L1(R3). To overcome this difficulty, based on the strategy developed in [5], we introduce aC-function f defined by

f0(t) = p 1

1+2|f(t)|2, fort ∈[0,+) and

f(−t) =−f(t), fort∈(−∞, 0].

Some properties of the function f are necessary in our arguments which we list below. The corresponding proofs can be found in [5,18,25]. We omit them here.

Lemma 2.1. The function f enjoys the following properties:

(1) f is uniquely defined and invertible C-function;

(2) 0< f0(t)≤1,∀t∈R;

(3) |f(t)| ≤ |t|,∀t∈R;

(4) f2(t)≤√

2|t|,∀t∈R;

(5) f(tt) is decreasing for t>0;

(6) There exists a positive constant C such that

|f(t)| ≥

(C|t|, |t| ≤1, C|t|12, |t| ≥1.

By making the change of variablesv = f(u), the functional I can be rewritten as J(u):= I(f(u)) = 1

2 Z

R3|∇u|2dx+ 1 2

Z

R3V(x)f2(u)dx+ b 4

Z

R3(f0(u)|∇u|)2dx 2

1 12

Z

R3|f(u)|12dx− µ p+1

Z

R3|f(u)|p+1dx, (2.2)

which is well defined onX. Moreover, by Lemma 2.1, standard arguments (see e.g. Proposi- tion 1.12 in [22]) show that J ∈C1(X,R)and

hJ0(u),ϕi=

Z

R3∇u∇ϕdx+

Z

R3V(x)f(u)f0(u)ϕdx

Z

R3|f(u)|10f(u)f0(u)ϕdx

µ Z

R3|f(u)|p1f(u)f0(u)ϕdx+b Z

R3

|∇u|2 1+2f2(u)dx

× Z

R3

∇u∇ϕ(1+2f2(u))−2|∇u|2f(u)f0(u)ϕ [1+2f2(u)]2 dx

, (2.3)

for allu, ϕ∈ X. As in [9], if uis a nontrivial critical point of J, thenuis nontrivial solution of problem

∆u+V(x)f(u)f0(u)−b Z

R3|f0(u)|2|∇u|2dx· f0(u)f00(u)|∇u|2+|f0(u)|2∆u =g(x,u), (2.4)

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where

g(x,u) = f0(u)hµ|f(u)|p1f(u) +|f(u)|10f(u)i. Let

B(ρ) =

u∈ X: Z

R3[|∇u|2+V(x)f2(u)]dx≤ρ2

and

S(ρ) =∂B(ρ) =

u∈ X: Z

R3[|∇u|2+V(x)f2(u)]dx=ρ2

.

The following two lemmas show that the functionalJhas a mountain pass geometric structure.

Lemma 2.2. There existρ,α>0such that J(u)≥αfor all u∈ S(ρ).

Proof. Since 1< p<11, for anyε>0, there exists a constantC(ε)>0 such that

|t|p+1ε|t|2+C(ε)|t|12, ∀t ∈R. (2.5) By (2.5), condition (V), Lemma 2.1(4) and the Sobolev inequality, for u ∈ S(ρ), it deduces that

J(u)≥ 1 2

Z

R3|∇u|2dx+ 1 2

Z

R3V(x)f2(u)dx− 1 12

Z

R3|f(u)|12dx− µ p+1

Z

R3|f(u)|p+1dx

1 4

Z

R3|∇u|2dx+ 1 4

Z

R3V(x)f2(u)dx+1 4

Z

R3V0f2(u)dx

µε p+1

Z

R3|f(u)|2dx− 1

12 + µC(ε) p+1

Z

R3|f(u)|12dx

1 4

Z

R3[|∇u|2+V(x)f2(u)]dx− 1

12 +µC(ε) p+1

Z

R3|f(u)|12dx

1 4

Z

R3[|∇u|2+V(x)f2(u)]dx− 1

12 +µC(ε) p+1

8

Z

R3|u|6dx

1 4

Z

RN[|∇u|2+V(x)f2(u)]dx−C Z

R3|∇u|2dx 3

1

4ρ2−Cρ6,

forε>0 small. Choose ρ>0 with 14ρ2−Cρ6= 18ρ2:=α>0. Then J(u)≥αfor allu∈ S(ρ). The proof is completed.

Lemma 2.3. There exists a u∈ X such that J(u)<0.

Proof. Choosing w ∈ X∩L12(R3) with 0 < |w| ≤ 1, it follows from Lemma 2.1(5) that f(tw)≥ f(t)wfort >0. Hence fort≥1, by Lemma2.1(3)and(6), we have

J(tw) = t

2

2 Z

R3|∇w|2dx+ 1 2

Z

R3V(x)f2(tw)dx+ b 4

Z

R3(f0(tw)|∇tw|)2dx 2

1 12

Z

R3|f(tw)|12dx− µ p+1

Z

RN|f(tw)|p+1dx

t

2

2 Z

R3|∇w|2dx+ t

2

2 Z

R3V(x)w2dx+bt

4

4 Z

R3|∇w|2dx 2

t

6C 12

Z

R3|w|6dx→ −∞, ast→+∞,

which implies that there exists a larget>0 such that J(tw)<0. We complete the proof.

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Define the mountain pass levelcof the functional J as c= inf

γΓmax

r∈[0,1]J(γ(t)), (2.6)

whereΓ={γ∈C([0, 1],X):γ(0) =0,J(γ(1))<0}. Let Φ(u) =

Z

R3

|∇u|2+V(x)f2(u)dx. (2.7) It follows from Lemma 2.2 that J(u) ≥ 0 for u ∈ B(ρ). This implies that Φ(γ(1)) > ρ for all γΓ. Hence there exists a tγ ∈ (0, 1) such that Φ(γ(tγ)) = ρ for every γΓ. By the definition ofc, we havec≥α>0, where αis given in Lemma2.2.

3 Proof of main results

Now, we are in the position to verify the main results. To this end, a further estimate of the mountain pass level value c is necessary. We recall that the best constant S for the Sobolev embeddingD1,2(RN),→ L2(RN)is given by

S= inf

vD1,2(RN),kvk2=1

k∇vk22. (3.1)

Consider the functionwε defined by

wε = [N(N−2)]N82

[ε+|x|2]N42 ,ε>0. (3.2) Let 0 < R < 1 and uε = φwε, where φ is a smooth cut-off function satisfying φ(x) = 1 for

|x| ≤Randφ(x) =0 for|x| ≥2R. For anyε>0, it is known that

(w2ε) =w

2(N+2) N2

ε

and the infimum in (3.1) is achieved by the function w2ε. Moreover, followed by [2], a direct computation yields that

Z

RN|∇(u2ε)|2dx=SN2 +O ε

N2 2

, (3.3)

Z

RNu22ε dx=SN2 +O ε

N 2

, (3.4)

Z

RN|∇uε|2dx≤O ε

N2

2 |lnε|, (3.5)

Z

RNu2εdx=O ε

N2 4

, (3.6)

and

Z

RNuqεdx=O ε

N

218q(N2)

, ∀2 <q<22. (3.7) Taking uε as a test function, we have the following estimates for the level value c given in (2.6).

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Lemma 3.1.

(i) If9< p<11,then c< 16S32 for anyµ>0.

(ii) If1< p≤9,then there exists a constantµ >0such that c< 16S32 for anyµ>µ.

Proof. LetIbe the functional defined in (2.1). ThenI(uε)is well defined sinceuε ∈ X∩L(R3). Firstly, we consider the case of 9 < p < 11. From I(u) and uε, we can define tε > 0 satisfying

I(tεuε) =sup

t0

I(tuε).

Here, we claim that there exist positive constants t1, t2 and ε0 such that t1 ≤ tε ≤ t2 for all ε∈(0,ε0). Indeed, by (3.3), (3.4), (3.5) and (3.6), there exists a smallε2>0 such that

0≤ I(tεuε) =sup

t>0

I(tuε)

= 1 2

Z

R3(1+2t2u2ε)|∇tuε|2dx+1 2

Z

R3V(x)t2u2εdx+b 4

Z

R3|∇tuε|2dx 2

1 12

Z

R3|tuε|12dx− µ p+1

Z

R3|tuε|p+1dx

= t

2

2 Z

R3|∇uε|2dx+ t

2

2 Z

R3V(x)u2εdx+ t

4

4 Z

R3|∇(u2ε)|2dx +bt

4

4 Z

R3|∇uε|2dx 2

t12 12

Z

R3|uε|12dx− µtp

+1

p+1 Z

R3|uε|p+1dx

t

2

2 Z

R3|∇uε|2dx+ t

2

2 Z

R3V(x)u2εdx+ t

4

4 Z

R3|∇(u2ε)|2dx +bt

4

4 Z

R3|∇uε|2dx 2

t

12

12 Z

R3|uε|12dx

≤ t2+ t

4

2(S32 +1) + bt

4

4 − t

12

12S32, (3.8)

which means that t2+ t24(S32 +1) + bt44t1212S32. Thus, there exists t2 > 0 small such that tε ≤ t2 < 1 for all ε ∈ (0,ε2). It follows from (V), (3.3), (3.4) and (3.7) that there exists ε1∈(0,ε2)such that

I(tuε) = t

2

2 Z

R3|∇uε|2dx+ t

2

2 Z

R3V(x)u2εdx+t

4

4 Z

R3|∇(u2ε)|2dx + bt

4

4 Z

R3|∇uε|2dx 2

t

12

12 Z

R3|uε|12dx− µt

p+1

p+1 Z

R3|uε|p+1dx

t

4

4 Z

R3|∇(u2ε)|2dx− t

12

12 Z

R3|uε|12dx− µt

p+1

p+1 Z

R3|uε|p+1dx

t

4

4S32µ

p+1tp+1ε

12p

8t

12

12S32. Let k := max0t1 14t4121t12

S32. Thenk > 0. We can find a small ε0 < ε1 with 11µε

12p

8k2

for all ε∈(0,ε0). Therefore, I(tεuε)≥ max

0t1

1

4S32t4µ p+1ε

12p

8 tp+11 12t12S32

k

2. (3.9)

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Combining (3.9) with (3.8) yields that 0< k

2 ≤ I(tεuε)≤t2ε + 1

2(S32 +1) + b 4

t4ε1

12S32t12ε ,

which implies that there exists at1>0 such thattε ≥t1for all ε∈ (0,ε0). Hence our claim is true.

For anyε∈(0,ε0), applying (3.3)–(3.7) again, we have I(tεuε) = t

2 ε

2 Z

R3|∇uε|2dx+ t

2 ε

2 Z

R3V(x)u2εdx+t

4 ε

4 Z

R3|∇(u2ε)|2dx + bt

4 ε

4 Z

R3|∇uε|2dx 2

t

12 ε

12 Z

R3|uε|12dx− µt

p+1 ε

p+1 Z

R3|uε|p+1dx

t

22

2 Z

R3|∇uε|2dx+ t

22

2 Z

R3V(x)u2εdx+t

4 ε

4 Z

R3|∇(u2ε)|2dx + bt

4 ε

4 Z

R3|∇uε|2dx 2

t

12 ε

12 Z

R3|uε|12dx− µt

p+1 1

p+1 Z

R3|uε|p+1dx

t

22

2O ε

1

4|lnε|+t

22

2|V|O ε

1 4

+ t

4 ε

4 h

S32 +O ε

3 2i

+ bt

4 ε

4 h

O ε

14|lnε|i2t

12 ε

12 h

S32 +O ε

32i

−CO ε

12p

8

1

6S32 +Ch O ε

1

4|lnε|−O ε

12p

8 i

1 6S32.

Hence we can find a small ¯ε>0 such that sup

t0

J(f1(tuε¯)) =sup

t0

I(tu¯ε) = I(tε¯uε¯)≤ 1 6S32.

Moreover, we conclude from (3.8) that J(f1(tuε¯)) = I(tuε¯)→ − ast → +, which shows that there exists a ¯t > 0 such that J(f1(tu¯ ε¯)) <0. Let ¯γ(t) = f1(ttu¯ ε¯). Then ¯γΓand for anyµ>0,

c≤ max

0t1J(γ¯(t))< 1 6S32.

Secondly, we consider the other case 1< p≤ 9. Now, we rewrite the functionalIas Iµ. Let u0 ∈C0 (RN)withu0 6=0 and definetµ >0 such thatIµ(tµu0) =supt0Iµ(tu0). We claim that tµ →0 asµ→+. Indeed, if the assertion does not hold, then there exists a constantt0> 0 and a sequence{µn}such thatµn →+andtµn ≥ t0 for alln. Without loss of generality, we assume thatµn ≥ 1 for alln. Let tn = tµn and I1 = Iµ|µ=1. Then 0≤ Iµn(tnu0)≤ I1(tnu0)for alln, which implies thattnis bounded from above. Moreover, we have

Iµn(tnu0) = t

2n

2 Z

R3|∇u0|2dx+ t

2n

2 Z

RNV(x)u20dx+ t

4n

4 Z

RN|∇(u20)|2dx + bt

4n

4 Z

R3|∇u0|2dx 2

t

12n

12 Z

R3|u0|12dx−µnt

p+1 n

p+1 Z

R3|u0|p+1dx

(9)

t

2n

2 Z

R3|∇u0|2dx+ t

2n

2 Z

R3V(x)u20dx+t

4n

4 Z

R3|∇(u20)|2dx +bt

4n

4 Z

R3|∇u0|2dx 2

µnt

p+1 n

p+1 Z

R3|u0|p+1dx

≤ C− µnt

p+1 n

p+1 Z

R3|u0|p+1dx→ −∞, asn→∞,

which contradicts Iµn(tnu0)≥0. Hence the claim holds. Sincetµ →0 asµ→+and Iµ(tµu0)≤ t

2 µ

2 Z

R3|∇u0|2dx+ t

2 µ

2 Z

R3V(x)u20dx+t

4 µ

4 Z

R3|∇(u20)|2dx,

we have Iµ(tµu0)→ 0 asµ→ + and hence there exists aµ > 0 such that supt0Iµ(tu0)<

1

6S32 for allµ>µ. This implies thatc< 16S32 for allµ> µ. The proof is completed.

Recall that, for any c ∈ R, we say {un} is a (C)c sequence of J if J(un) → c and (1+kunk)J0(un)→0 asn →∞. In order to obtain the existence of ground state solutions, we need to study some behaviors of a (C)c sequence of J carefully.

Lemma 3.2. Let c∈ Rand{un} ⊂ X be a(C)c sequence of J. Then{Φ(un)}is bounded, where Φ is defined in(2.7). In particular,{un}is bounded in H1(R3).

Proof. Letwn= f(un)

f0(un). Then Lemma2.1(3)and(4)imply that

∇wn =

1+ 2f

2(un) 1+2f2(un)

∇un. Hence,hJ0(un),wni →0 asn→∞.

Define two real functionsg(t) =|t|10t+µ|t|p1tandG(t) =Rt

0g(s)ds. Then there exists a constantλ∈(4, 12)such that

|limt|→0

tg(t)−λG(t)

t2 =0 and lim

|t|→

tg(t)−λG(t)

tλ = +. Therefore, there existsr >0 such that

tg(t)−λG(t)≥0, ∀ |t| ≥r. (3.10) Moreover, for anyε>0, there exists a positive constant C(ε)such that

|tg(t)−λG(t)| ≤ε|t|2+C(ε)|t|12, ∀t ∈R. (3.11) Then it can be deduced from (3.10) that

c+o(1) = J(un)− 1

λhJ0(un),wni

= 1 2 Z

R3|∇un|2dx− 1 λ

Z

R3

1+ 2f

2(un) 1+2f2(un)

|∇un|2dx+ 1

2− 1 λ

Z

R3V(x)f2(un)dx +

b 4− b

λ Z

R3(f0(un)|∇un|)2dx 2

+

Z

|f(un)|>r

1

λg(f(un))f(un)−G(f(un))

dx

+

Z

|f(un)|≤r

1

λg(f(un))f(un)−G(f(un))

dx

(10)

≥ 1

2− 2 λ

Z

R3|∇un|2dx+2 1

4− 1

Z

R3V(x)f2(un)dx +

Z

|f(un)|≤r

1

λg(f(un))f(un)−G(f(un))

dx. (3.12)

From (3.11), there exists a constant M>V0such that

1

λtg(t)−G(t)

≤ 1

4− 1

M|t|2, ∀ |t| ≤r, (3.13) whereV0 is the number given in the condition (V). Let A= {x ∈ R3 : V(x)≤ M}. Then it follows from condition(V)that meas(A)<∞. By (3.13) and condition(V), we have

1 4− 1

Z

R3V(x)f2(un)dx+

Z

|f(un)|≤r

1

λg(f(un))f(un)−G(f(un))

dx

≥ 1

4− 1

Z

{|f(un)|≤r}(V(x)−M)f2(un)dx

≥ 1

4− 1

Z

{|f(un)|≤r,V(x)≤M}

(V(x)−M)r2dx

≥ 1

4− 1

·meas

A ∩ {x∈RN :|f(un)| ≤r}·(V0−M)r2

≥ 1

4− 1

·meas(A)·(V0−M)r2. This combining with (3.12) implies that

1 2 − 1

λ Z

R3|∇un|2dx+ 1

4 − 1

Z

R3V(x)f2(un)dx

≤ 1

4 − 1

·meas(A)·(M−V0)r2+c+o(1).

which means that Z

RN

|∇un|2+V(x)f2(x)dx<+∞. (3.14) In particular, by Lemma2.1(6)and (3.14), we have

Z

R3|un|2dx=

Z

{|un|≤1}|un|2dx+

Z

{|un|>1}|un|2dx

≤C1 Z

R3V(x)f2(un)dx+

Z

R3|un|6dx

≤C1 Z

R3V(x)f2(un)dx+C2 Z

R3|∇un|2dx 3

<+∞,

which together with (3.14) implies that {un}is bounded in H1(RN). The proof is completed.

Lemma 3.3. Let {un} ⊂ X be a (C)c sequence of J. If c < 16S32, then there exist R,ξ > 0 and a sequence{yn} ⊂R3such that

lim sup

n Z

BR(yn) f2(un)dx≥ξ.

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Proof. Arguing by contradiction, we suppose that the conclusion is not true, i.e., lim sup

n Z

BR(yn)f2(un)dx=0.

Then Lion’s concentration compactness principle (Lemma 1.21 in [22]) implies that

f(un)→0, in Ls(R3)for alls ∈(2, 6). (3.15) By Lemma2.1(4), Lemma3.2and the interpolation, we have

Z

R3|f(un)|s→0, ∀s∈(2, 12). (3.16) In view of Lemma3.2, passing to a subsequence, we may assume that

Z

R3

1+ 2f

2(un) 1+2f2(un)

|∇un|2dx+

Z

R3V(x)f2(un)dx+b Z

R3(f0(un)|∇un|)2dx 2

→B (3.17)

and Z

R3|f(un)|12dx→ D. (3.18)

It follows from the definition of Sthat S

Z

R3|f(un)|12dx 13

Z

R3|∇f2(un)|2dx=

Z

R3

4f2(un)

1+2f2(un)|∇un|2dx

Z

R3

1+ 2f

2(un) 1+2f2(un)

|∇un|2dx

+

Z

R3V(x)f2(un)dx+b Z

R3(f0(un)|∇un|)2dx 2

,

which combining with (3.17) and (3.18) yields that SD13 ≤ B. In addition, from (3.16), (3.17) and (3.18) we have

0= lim

nhJ0(un),wni= B−D, wherewn= f(un)

f0(un). Hence,B= D≥S32. Moreover, we deduce that c= lim

nJ(un)

= lim

n

1 2

Z

R3|∇un|2dx+ 1 2

Z

R3V(x)f2(un)dx+b 4

Z

R3(f0(un)|∇un|)2dx 2!

+ lim

n

1 12

Z

R3|f(un)|12dx− µ p+1

Z

RN|f(un)|p+1dx

≥ lim

n

1 4

Z

R3(1+ 2f

2(un)

1+2f2(un))|∇un|2dx+ 1 4

Z

R3V(x)u2ndx

+ lim

n

b 4

Z

R3(f0(un)|∇un|)2dx 2

1 12

Z

R3|f(un)|12dx

!

= 1

4 − 1 12

D

1 6S32,

which contradicts c< 16S32. The proof is completed.

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In what follows, we shall give the proof of Theorem1.3and Theorem1.4. Since the proofs of them are similar, we just give the details of Theorem1.3.

Proof of Theorem1.3. Letcbe the mountain pass level given in (2.6). From Lemma2.2, Lemma 2.3 and the mountain pass theorem (see e.g. Theorem 3 in [20]), the functional J has a (C)c sequence {un} ⊂ X. In view of Lemma 3.2, we may assume that un * u in H1(R3) and f(un)* f(u)inX, which implies that un→u inLsloc(R3)for 2< s<6 and f(un)→ f(u)in Lsloc(R3)for 2<s <12. HencehJ0(un),ϕi → hJ0(u),ϕi=0 for any ϕ∈ C0(R3), that is,uis a weak solution of (2.4). Moreover, since the embeddingX,→L2(R3)is compact for 2≤s <6, we get f(un)→ f(u)in L2(R3)for 2 ≤s < 6. We conclude from Lemma3.1(i)thatc < 16S32 forp=10, µ>0. By Lemma3.3, there exists a constantξ >0 such that

Z

R3 f2(u) = lim

n Z

R3 f2(un)≥ξ,

which shows thatu is a nontrivial solution of problem (2.4). Hence u = f(v)is a nontrivial solution of problem (1.1). Finally, letting d = inf{J(u) : u ∈ X,u 6= 0,J0(u) = 0}, we know thatdis achieved by the lower semi-continuity. The proof is completed.

Acknowledgements

We would like to thank the reviewers and the handling editor for their comments and sugges- tions, which led to a great improvement in the presentation of this work.

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