Positive solutions for a Kirchhoff type problem with fast increasing weight and critical nonlinearity

Positive solutions for a Kirchhoff type problem with fast increasing weight and critical nonlinearity

Xiaotao Qian

and Wen Chao

1Department of Basic Teaching and Research, Yango University, Fuzhou 350015, P. R. China

2School of Management, Fujian University of Technology, Fuzhou 350108, P. R. China

Received 21 January 2019, appeared 17 April 2019 Communicated by Dimitri Mugnai

Abstract. In this paper, we study the following Kirchhoff type problem

−

a+b Z

R³K(x)|∇u|²dx

div(K(x)∇u) =λK(x)|x|^β|u|^q−2u+K(x)|u|⁴u, x ∈R³,

where K(x) = exp(|x|^α/4) with α ≥ 2, β = (α−2)(6−q)/4 and the parameters a,b,λ > 0. When 6−⁴_α < q < 6, we obtain a positive ground state solution for any λ>0. When 2 <q< 4, we obtain a positive solution forλ>0 small enough. In the proof we use variational methods.

Keywords: Kirchhoff type equation, critical nonlinearity, variational methods.

2010 Mathematics Subject Classification: 35B33, 35J75.

1 Introduction

In this paper, we consider the existence of positive solutions for the following Kirchhoff type problem

−

a+b Z

R³K(x)|∇u|²dx

div(K(x)∇u) =λK(x)|x|^β|u|^q−2u+K(x)|u|⁴u, x∈_R³, (1.1) where K(x) =exp(|x|^α/4)withα≥2,β= (α−2)(6−q)/4 and the parametersa,b,λ>0.

It is well known that Kirchhoff type problems are presented by Kirchhoff in [9] as an extension of the classical d’Alembert wave equation for free vibrations of elastic strings. When K(x)≡1, the general Kirchhoff type problem involving critical exponent

−

a+b Z

R³|∇u|²dx

∆u= f(x,u) +u⁵, x∈_R³, (1.2) has been studied by many researchers. Under different assumptions on f(x,u), some inter- esting studies for (1.2) can be found in [12–14,23,25]. There are also several existence results

BCorresponding author. Email: qianxiaotao1984@163.com

for (1.2) on a bounded domain Ω ⊂ _R³. For this case, we refer the interested readers to [4,10,11,16,22].

On the other hand, as pointed out in [3,8], one of the motivations for studying problem (1.1) due to the fact that, forα= q= 2,a =1,b=0 and λ=1/5, (1.1) arises naturally when one seeks self-similar solutions of the form

w(t,x) =t^−1/5u(xt^−1/2) to the evolution equation

w_t−_∆w= |w|⁴w on (0,∞)×_R³. For more detailed description, see [3,8].

In [5], Furtado et al. concerned the following equation

−div(K(x)∇u) =λK(x)|x|^γ|u|^q−2u+K(x)|u|^2∗−2u, x ∈_R^N, (1.3) where 2^∗ =2N/(N−2), N≥3,γ= (α−2)⁽₍²₂^∗_∗⁻₋^q₂⁾₎ and 2< q<2^∗. In that article, the authors obtained the existence of a positive solution for (1.3) by using Mountain Pass Theorem. In particular, when N = 3, they proved that there is a positive solution for large value of λ if 2<q≤6− ⁴

α, and no restriction onλif 6− ⁴

α <q<6.

Subsequently, Furtado et al. [6] studied the number of solutions for the following problem

−div K(x)∇u

=K(x)f(u) +λK(x)|u|^2∗−2u, x∈_R^N, (1.4) where f(u)is superlinear and subcritical. More precisely, for any given k ∈ N, the authors shown that there exists λ^∗ = λ^∗(k) > 0 such that (1.4) has at least k pairs of solutions for λ∈ 0,λ^∗(k). But they can not give any information about the sign of these solutions.

Recently, we investigated the following Kirchhoff type of problem with concave-convex nonlinearities and critical exponent (see [21])

−

a+e Z

R³K(x)|∇u|²dx

div(K(x)∇u) =λK(x)f(x)|u|^q−2u+K(x)|u|⁴u, x∈ _R³, where 1 < q < 2, and e > 0 is small enough. Under some conditions on f(x), we gave the existence of two positive solutions and obtained uniform lower estimates for extremal values for the problem. For more results of related problem, please see [2,7,17–20] and the references therein.

From these results above, we do not see any existence of positive solutions for problem (1.1) in the case of 2 < q < 6, the term R

K(x)|∇u|²dx and critical nonlinearity, hence it is natural to ask what the case would be. Our aim of this paper is to show how variational methods can be employed to establish some existence of positive solutions for the Kirchhoff type problem (1.1).

In order to state our main results, letHdenote the Hilbert space obtained as the completion ofC^∞₀ (_R³)with respect to the norm

kuk= _Z

R³K(x)|∇u|²dx 1/2

. Define the weighted Lebesgue spaces for eachq∈[_{2, 6}]

L^q_K(_R³) =

umeasurable inR³ : Z

R³K(x)|x|^β|u|^qdx<_∞

with the norm

kuk_q= _Z

R³K(x)|x|^β|u|^qdx 1/q

.

By [5, Proposition 2.1], we have that the embedding H,→ L^q_K(_R³)is continuous for 2≤q≤6, and compact for 2≤ q<6. This enables us to define for eachq∈[2, 6]

Sq=inf Z

R³K(x)|∇u|²dx : u∈ H, Z

R³K(x)|x|^β|u|^qdx=1

. (1.5)

In particular, when q = 6, we put S = S₆ for simplicity. It is worth mentioning that this constant is equal to the best constant of the embeddingD^1,2(_R³),→L⁶(_R³), see [2].

By the above embedding, it is easy to see the following functional associated to (1.1) I(u) = ^a

2kuk²+ ^b

4kuk⁴−^λ q

Z

R³K(x)|x|^β|u|^qdx− ¹ 6 Z

R³K(x)|u|⁶dx

is well defined on Hand I ∈ C¹(H,R). It is commonly known that there exists a one to one correspondence between the critical points of I and the weak solutions of (1.1). Here, we say u∈H is a weak solution of (1.1), if for anyφ∈ H, there holds

a+bkuk²

Z

R³K(x)∇u∇φdx−λ Z

R³K(x)|x|^β|u|^q−2uφdx−

Z

R³K(x)|u|⁴uφdx=_0.

Additionally, we say a nontrivial solution u ∈ H to (1.1) is a ground state solution, if I(u)≤ I(v)for any nontrivial solutionv ∈ Hto (1.1).

Our main results for (1.1) are the following theorems.

Theorem 1.1. Assume that a,b> 0, α≥ 2and6− ⁴_α < q <6. Then for any λ> 0, problem(1.1) has at least a positive ground state solution.

Theorem 1.2. Assume that a,b> 0, α≥ 2 and2 < q < 4. Then there existsλ∗ > 0 such that for anyλ∈ (0,λ∗), problem(1.1)has at least a positive solution.

Kirchhoff type problems are often treated as nonlocal in view of the presence of the term R K(x)|∇u|²dx which implies that equation (1.1) is no longer a pointwise identity. And so, the methods employed in [5] cannot be used here. For Theorem 1.1, motivated by [23] (see also [15]), we shall use Nehari Manifold method to prove the existence of a positive ground state solution for problem (1.1). For Theorem1.2, we cannot proceed as in proof of Theorem 1.1 since 2<q<4. We also remark that the method used in [5] by lettingλsufficiently large do not apply here, due to the appearance of the term R

K(x)|∇u|²dx. On the contrary, we overcome this difficulty by lettingλsmall enough, which is inspired by [12].

This paper is organized as follows. In the next section, we give some notations and pre- liminaries. Then we prove Theorem 1.1in Section 3, and Theorem1.2 in Section 4.

2 Notations and preliminaries

Throughout this paper, we writeR

uinstead ofR

R³u(x)dx. B_r(x)denotes a ball centered atx with radius r>_{0. Let}→denote strong convergence. Let* denote weak convergence. O(ε^t) denotes |O(ε^t)|/ε^t ≤ C as ε → 0, and o(ε^t) denotes |o(ε^t)|/ε^t → 0 as ε → 0. All limitations hold as n → _∞ unless otherwise stated. C and C_i denote various positive constants whose values may vary from line to line.

Lemma 2.1. Let a,b>0and2<q<6, then the functional I satisfies the mountain-pass geometry:

(i) There existρ,θ>0such that I(u)≥ θ>0for anykuk=ρ.

(ii) There exists e∈ H withkek>ρsuch that I(e)<0.

Proof. (i)By (1.5), we have that I(u) = ^a

2kuk²+ ^b

4kuk⁴−^λ q

Z

K(x)|x|^β|u|^q− ¹ 6

Z

K(x)|u|⁶

≥ ^a

2kuk²−^λ

qS⁻_q^q/2kuk^q− ¹

6S⁻³kuk⁶.

Therefore, since 2 < q < 6, it follows that there areρ,θ > 0 such that I(u) ≥ θ > 0 for any kuk=ρ.

(ii)Letu ∈H\ {0}. Thus, we have for 2<q< 6

t→+lim_∞I(tu) = _lim

t→+_∞

a

2t²kuk²+ ^b

4t⁴kuk⁴−^λ qt^q

Z

K(x)|x|^β|u|^q−¹ 6t⁶

Z

K(x)|u|⁶

=−_∞.

Thus, there existse:=tusuch thatkek>ρandI(e)<0.

3 Positive ground state solution for 6 −

α

< q < 6

In this section, we will employ Nehari method to prove the existence of a positive ground state solution of the considered problem for 6− ⁴

α <q< 6. And, suppose that the assumptions of Theorem1.1 hold throughout this section.

Define the Nehari manifold

Λ={u∈ H\ {0}: G(u) =0}, where

G(u) =hI⁰(u),ui=akuk²+bkuk⁴−λ Z

K(x)|x|^β|u|^q−

Z

K(x)|u|⁶. Let

c∗ = inf

u∈Λ

I(u) (3.1)

be the infimum of I on the Nehari manifold.

Lemma 3.1. For any u∈Λ, there areδ,σ>0such thatkuk ≥δandhG⁰(u),ui ≤ −σ.

Proof. For anyu∈Λ,

0=hI⁰(u),ui

=akuk²+bkuk⁴−λ Z

K(x)|x|^β|u|^q−

Z

K(x)|u|⁶

≥akuk²−λS⁻_q^q/2kuk^q−S⁻³kuk⁶.

From q > 6− ⁴

α and α ≥ 2, we have q > 4 and hence, there exists some δ > 0 such that kuk ≥δ. Furthermore,

hG⁰(u),ui=2akuk²+4bkuk⁴−qλ Z

K(x)|x|^β|u|^q−6 Z

K(x)|u|⁶

= (2a−qa)kuk²+ (4b−qb)kuk⁴−(6−q)

Z

K(x)|u|⁶

<(2a−qa)kuk²

<−(qa−2a)δ² <0.

Setσ = (qa−2a)δ², this finishes the proof.

Lemma 3.2. The functional I is coercive and bounded from below onΛ.

Proof. Foru∈Λ, it follows from Lemma3.1and 6− ⁴_α <q<6 that I(u) = ^a

2kuk²+ ^b

4kuk⁴− ^λ q

Z

K(x)|x|^β|u|^q−¹ 6

Z

K(x)|u|⁶

= a

2 − ^a q

kuk²+ b

4 −^b q

kuk⁴+ 1

q−¹ 6

Z

K(x)|u|⁶

≥ a

2 − ^a q

kuk²

≥ a

2 − ^a q

δ²>0.

Thus, the coercivity and lower boundedness of I hold. The proof of Lemma3.2 is completed.

Lemma 3.3. Given u∈ Λ, there existρ_u> 0and a continuous function gρu : Bρu(0)→_R⁺ defined for w∈ H, w∈ Bρu(0)such that

g_ρu(₀) =_1, g_ρu(w)(u−w)∈Λ and

hg_ρ⁰_u(0),φi=

(2a+4bkuk²)

Z

K(x)∇u∇φ−qλ Z

K(x)|x|^β|u|^q−2uφ−₆

Z

K(x)|u|⁴uφ akuk²+3bkuk⁴−λ(q−1)

Z

K(x)|x|^β|u|^q−5 Z

K(x)|u|⁶

.

Proof. Fixu∈Λand defineF:R⁺×H→_Ras below F(t,w) =atku−wk²+bt³ku−wk⁴−λt^q−1

Z

K(x)|x|^β|u−w|^q−t⁵ Z

K(x)|u−w|⁶. Sinceu∈_{Λ, we have}F(_{1, 0}) =0. Moreover, using Lemma3.1, we also have for 6− ⁴

α <q<₆ F_t(1, 0) =akuk²+3bkuk⁴−λ(q−1)

Z

K(x)|x|^β|u|^q−5 Z

K(x)|u|⁶

= a(2−q)kuk²+b(4−q)kuk⁴−(6−q)

Z

K(x)|u|⁶

< a(2−q)kuk²

≤ a(2−q)δ² <0.

Using the implicit function theorem forF at the point(1, 0), we can conclude that there exists ρ_u > 0 satisfying for w ∈ H, kwk < ρ_u, the equation F(t,w) = 0 has a unique continuous solutiont= gρu(w)>0 withgρu(0) =1. SinceF(gρu(w),w) =0 forw∈ H,kwk<ρ_u, we get

agρu(w)ku−wk²+bg³_ρu(w)ku−wk⁴−λg_ρ^q−_u¹(w)ku−wk^q_q−g⁵_ρu(w)ku−wk⁶₆

= ^akgρ_u(w)(u−w)k²+bkgρ_u(w)(u−w)k⁴−λkgρ_u(w)(u−w)k^q_q− kgρ_u(w)(u−w)k⁶₆ gρu(w)

=0, that is,

gρu(w)(u−w)∈Λ, for allw∈ H, kwk<ρ_u. Furthermore, we have for allφ∈ H, r>0

F(1, 0+rφ)−F(1, 0)

=aku−rφk²+bku−rφk⁴−λ Z

K(x)|x|^β|u−rφ|^q−

Z

K(x)|u−rφ|⁶

−akuk²−bkuk⁴+λ Z

K(x)|x|^β|u|^q+

Z

K(x)|u|⁶

= −a Z

K(x) 2r∇u∇φ−r²|∇φ|²

−b

"

2 Z

K(x)|∇u|²

Z

K(x) 2r∇u∇φ−r²|∇φ|²− _Z

K(x) 2r∇u∇φ−r²|∇φ|² 2#

−λ Z

K(x)|x|^β|u−rφ|^q− |u|^q−

Z

K(x)|u−rφ|⁶− |u|⁶ and consequently

hFw,φi|_t=1,w=0

= lim

r→0

F(1, 0+rφ)−F(1, 0) r

= −(2a+4bkuk²)

Z

K(x)∇u∇φ+qλ Z

K(x)|x|^β|u|^q−2uφ+6 Z

K(x)|u|⁴uφ.

Thus,

hg⁰_ρu(0),φi= −hF_w,φi F_t

t=1,w=0

=

(2a+4bkuk²)

Z

K(x)∇u∇φ−qλ Z

K(x)|x|^β|u|^q−2uφ−6 Z

K(x)|u|⁴uφ akuk²+3bkuk⁴−λ(q−1)

Z

K(x)|x|^β|u|^q−5 Z

K(x)|u|⁶

.

This completes the proof of Lemma3.3.

Lemma 3.4. For any u ∈ H\ {0}, there exists a unique t(u) > 0 satisfying t(u)u ∈ Λ and I(t(u)u) =max_t>0I(tu).

Proof. The proof is similar to [24, Lemma 4.1], and is omitted here.

Lemma 3.5. c∗ <c₁:= ^abS₄³ +^b3₂₄^S6 +^(b2S4+₂₄^4aS)3/2.

Proof. Let ϕ(x)∈ C₀^∞(_R³)be a cut-off function satisfying ϕ(x)≡1 in B_η(0), ϕ(x)≡0 outside B2η(0)and 0≤ ϕ≤1. Define

u_ε(x) =K^−1/2ϕ(x) 1

ε+|x|² 1/2

, and set

v_ε(x) = ^uε(x) ku_εk₆^. According to [2], we have that

kvεk² =

Z

K(x)|∇vε|²= S+O(ε^1/2) +O(ε^α/2) (3.2) and

kuεk⁶₆ =

Z

K(x)|uε|⁶ =ε^−3/2A0+O(1), withA0 =

Z 1

(1+|x|²)^{3. (3.3)} Then, we obtain the following estimate

kuεk₆^q = ε^−3/2A0+O(1)^q/6 = A^q/6₀ ε^−q/4+O

ε^−(q−6)/4

. (3.4)

In addition, we also have Z

K(x)|x|^β|u_ε|^q =

Z

B_2η(0)

K(x)|x|^βK(x)^−q/2ϕ^q(x) (ε+|x|²)^q/2

≥C Z

B2η(0)

|x|^βϕ^q(x) (ε+|x|²)^q/2

=C _Z

B2η(0)

|x|^β (ε+|x|²)^q/2 +

Z

B2η(0)

|x|^β(ϕ^q(x)−1) (ε+|x|²)^q/2

=C ε

β 2−^q₂+³₂ Z

B_2η/^√_ε(0)

|x|^β (1+|x|²)^q/2 +

Z

B2η(0)

|x|^β(ϕ^q(x)−1) (ε+|x|²)^q/2

!

=O ε

β 2−₂^q+³₂

+O(1),

whenever ₂^6α_+α < q < 6. This and (3.4) imply that for ₂^6α_+α < q < 6 and ε small enough, we have

Z

K(x)|x|^β|vε|^q=

R K(x)|x|^βu^q_ε kuεk^q₆

≥ ^O

ε

β 2−^q₂+³₂

+_O(₁) A^q/6₀ ε^−q/4+O ε^−(q−6)/4

=O ε

β 2−^q₄+³₂

+O ε^q/4

.

(3.5)

By Lemma3.4and the definition ofc∗, it is easy to see that Lemma3.5follows if we can show that

sup

t>0

I(tv_ε)< ^abS

3

4 + ^b

3S⁶

24 + (b²S⁴+4aS)^3/2

24 .

To this goal, let

g(t) =I(tv_ε) = ^a

2t²kv_εk²+ ^b

4t⁴kv_εk⁴− ^λ qt^q

Z

K(x)|x|^βv^q_ε − ^t6

6, t ≥₀ and

g₁(t) = ^a

2t²kv_εk²+ ^b

4t⁴kv_εk⁴− ^t

6

6, t ≥0.

Note thatkvεk₆ = 1. Thus, by Lemma3.4, we know that g(t) has a unique maximum point t_ε := t(v_ε) > 0. We claim that t_ε ≥ C₀ > 0 for some positive constant C₀ and any ε > 0.

Otherwise, there is some sequence ε_n → 0 satisfying t_εn → 0 and g(t_εn) = _supt>0I(tv_εn)_. Then, by Lemma2.1 and the continuity of I, we conclude that

0< θ≤c∗ ≤ lim

n→_∞I(t_εnv_εn) =0 which is a contradiction. Hence, the claim holds.

By (3.2), we also have that sup

t>0

g₁(t) = ^abkvεk⁶

4 + ^b

3kvεk¹²

24 + ^b

2kvεk⁸+4akvεk^23/2 24

= ^abS

3

4 +^b

3S⁶

24 + (b²S⁴+4aS)^3/2

24 +O(ε^1/2).

(3.6)

Obviously, we have 6−⁴_α > ₂^6α₊

α provided α ≥ 2. Furthermore, by using (3.5) and (3.6), we obtain for 6− ⁴

α <q<_{6 and}εsmall enough sup

t>0

I(tvε) = I(tεvε)

=g₁(t_ε)− ^λ qt^q_ε

Z

K(x)|x|^βv^q_ε

≤sup

t>0

g₁(t)−^λ qC₀^q

Z

K(x)|x|^βv^q_ε

≤ ^abS

3

4 +^b

3S⁶

24 + (b²S⁴+4aS)^3/2

24 +O(ε^1/2)−O ε

β 2−^q₄+³₂

+O ε

q 4

< ^abS

3

4 +^b

3S⁶

24 + (b²S⁴+4aS)^3/2

24 .

This completes the proof.

Lemma 3.6. Let{u_n} ⊂Λbe a(PS)_c∗ sequence for I with c∗ <c₁, where c₁ is given in Lemma3.5.

Then{un}has a convergent subsequence.

Proof. Let{un} ⊂Λbe a(PS)_c∗ sequence for I, namely

nlim→_∞I(u_n) =c∗ and lim

n→_∞I⁰(u_n) =0. (3.7) Firstly, we show thatku_nkis bounded. By (3.7), we have that for q>4

c∗+₁+o(₁)ku_nk ≥ I(u_n)−¹

4hI⁰(u_n)_,u_ni

= ^a

4ku_nk²− λ

q − ^λ 4

_Z

K(x)|x|^β|u_n|^q− 1

6−¹ 4

_Z

K(x)|u_n|⁶

≥ ^a 4ku_nk²

which implies that ku_nk is bounded. Up to a subsequence (still denoted by {u_n}), we may assume that

un *u∗ in H,

u_n →u∗ inL^r_K(_R³)_{, 2}≤r <_6, u_n →u∗ a.e. onR³

and

ku_nk²→ι². Sinceu_n∈Λ, it then follows from Lemma3.1that ι² >_0.

Secondly, we prove thatu∗ 6≡0. If not, we haveu∗ ≡0 and soR

K(x)|x|^β|u_n|^q= o(1). On the other hand, by (3.7) and the boundedness of{un}, we have

o(1) =hI⁰(u_n),u_ni= aku_nk²+bku_nk⁴−λ Z

K(x)|x|^β|u_n|^q−

Z

K(x)|u_n|⁶ and thus from (1.5),

akunk²+bkunk⁴ =

Z

K(x)|un|⁶+o(1)≤S⁻³kunk⁶+o(1). (3.8) Lettingn→_∞in (3.8), we have

ι² ≥ ^bS

3+√

b²S⁶+4aS³

2 .

Consequently,

c∗ = lim

n→_∞I(u_n)

= lim

n→_∞

ha

2kunk²+ ^b

4kunk⁴−^λ q

Z

K(x)|x|^β|un|^q−¹ 6

Z

K(x)|un|⁶ⁱ

= lim

n→_∞

ha

3ku_nk²+ ^b

12ku_nk⁴−λ6−q 6q

Z

K(x)|x|^β|u_n|^qi

= ^a 3ι²+ ^b

12ι⁴

≥c₁

in contradiction to the assumption c∗< _c1.

Finally, we claim that ku_nk² → ku∗k². Indeed, if to the contrary, it follows from Fatou Lemma that

ι² >ku∗k². (3.9)

Since I⁰(un)→0, we also have for allv ∈H o(1) = a+bkunk²

Z

K(x)∇un∇v−λ Z

K(x)|x|^β|un|^q−2unv−

Z

K(x)|un|⁴unv.

Then, passing to the limit as n→_{∞, we get} 0= a+bι²

Z

K(x)∇u∗∇v−λ Z

K(x)|x|^β|u∗|^q−2u∗v−

Z

K(x)|u∗|⁴u∗v.

Takingv=u∗in the above equation, we obtain 0= a+bι²

ku∗k²−λ Z

K(x)|x|^β|u∗|^q−

Z

K(x)|u∗|⁶_.

This together with (3.9) imply that hI⁰(u∗),u∗i< 0. By Lemma3.4, then it is easy to see that there existst₀ ∈(0, 1)such thathI⁰(t₀u∗),t₀u∗i=0. Therefore,

c∗ ≤sup

t>0

I(tu∗)

= I(t0u∗)

= I(t₀u∗)− ¹

4hI⁰(t₀u∗)_,t₀u∗i

= ^a

4t²₀ku∗k²+ 1

4 −¹ q

t^q₀

Z

K(x)|x|^β|u∗|^q+ ¹ 12t⁶₀

Z

K(x)|u∗|⁶

< ^a

4ku∗k²+ 1

4− ¹ q

_Z

K(x)|x|^β|u∗|^q+ ¹ 12

Z

K(x)|u∗|⁶

≤lim inf

n→_∞

a

4ku_nk²+ 1

4 −¹ q

_Z

K(x)|x|^β|u_n|^q+ ¹ 12

Z

K(x)|u_n|⁶

=lim inf

n→_∞

I(un)−¹

4hI⁰(un),uni

= c∗

a contradiction. Hence,ku_nk → ku∗k. This and the weak convergence of {u_n}in H implies thatun →u∗ in H, and Lemma3.6 is proved.

Lemma 3.7. For anyλ>0, there exists a sequence{un} ⊂Λsuch that:

un ≥0, I(un)→c∗ and I⁰(un)→0.

Proof. In view of Lemma3.2, we can apply Ekeland variational principle to construct a mini- mizing sequence{u_n} ⊂_Λsatisfying the following properties:

(i) I(un)→c∗,

(ii) I(z)≥ I(u_n)− ¹_nku_n−zkfor allz∈ _Λ.

Since I(|u|) = I(u), we can assume thatun ≥0 onR³. Let 0<ρ <ρn≡ ρun,gn≡ gun, where ρ_un and g_un are defined according to Lemma 3.3. Let vρ = ρu with kuk = 1. Fix n and let z_ρ = g_n(v_ρ)(u_n−v_ρ). Sincez_ρ∈ _Λ, by the property(ii), one gets

I(zρ)−I(un)≥ −¹

nkzρ−unk. It then follows from the definition of Fréchet derivative that

hI⁰(u_n),z_ρ−u_ni+o(kz_ρ−u_nk)≥ −¹

nkz_ρ−u_nk. Thus,

hI⁰(u_n),−v_ρ+ g_n(v_ρ)−1

(u_n−v_ρ)i ≥ −¹

nkz_ρ−u_nk+o(kz_ρ−u_nk)

which implies

−ρhI⁰(un),ui+ gn(vρ)−1

hI⁰(un),un−vρi ≥ −¹

nkzρ−unk+o(kzρ−unk). Therefore,

hI⁰(un),ui ≤ ¹ n

kz_ρ−u_nk

ρ + ^o(kz_ρ−u_nk)

ρ + ^gn(v_ρ)−₁

ρ hI⁰(un),un−vρi. (3.10) Fromkuk=1, Lemma3.3and the boundedness of{un}, it follows that

lim

ρ→0

|g_n(v_ρ)−1|

ρ = lim

ρ→0

|g_n(0+ρu)−g_n(0)|

ρ

=hg⁰_n(0),ui

≤ kg⁰_n(0)k

≤C₁. Note that

kz_ρ−u_nk=kg_n(v_ρ)(u_n−v_ρ)−(u_n−v_ρ)−v_ρk

=k g_n(v_ρ)−1

(u_n−v_ρ)−v_ρk

≤ |g_n(vρ)−1| · ku_n−vρk+kvρk

=|gn(vρ)−1|C₂+ρ.

Furthermore, for fixedn, since hI⁰(u_n)_,u_ni=_{0 and}(u_n−v_ρ)→u_n asρ→0, by lettingρ →0 in (3.10) we can deduce that

hI⁰(u_n),ui ≤ ^C n,

which shows that I⁰(u_n)→0. This finishes the proof of Lemma3.7.

With the previous preparations, we are now ready to prove Theorem1.1.

Proof of Theorem1.1. By Lemma3.7, we see that there exists a minimizing sequence{un} ⊂_Λ satisfying u_n ≥ 0, I(u_n)→ c∗ and I⁰(u_n)→ 0 for anyλ> 0. It then follows from Lemma3.6 thatu_n→u∗,I(u∗) =c∗andu∗ ≥0 is a weak solution of (1.1). By standard elliptic regularity argument and the strong maximum principle we have that u∗ >0. This and the definition of c∗imply thatu∗ is a positive ground state solution of (1.1) and the proof is complete.

4 Positive solution for 2 < q < 4

In this section, we apply Mountain Pass Theorem to obtain the existence of a positive solution for problem (1.1) when 2<q<4.

Define

c˜∗ := inf

γ∈_Γmax

t∈[0,1]I(γ(t)) (4.1)

where

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

Lemma 4.1. Assume2<q<4. Then there existsλ∗ >0satisfying for allλ∈(0,λ∗)

˜

c∗ <c₁−D₀λ

4 4−q

where c₁given in Lemma3.5and D₀ = ⁴⁻₄^{q 3q}_b

q 4−q(6−q)

6q S⁻_q^q/2₄₋⁴_q . Proof. We first recall that

uε(x) =K^−1/2ϕ(x) 1

ε+|x|² 1/2

. (4.2)

According to [5], we have that kuεk²=

Z

K(x)|∇uε|²=ε^−1/2A₁+O(1), with A₁=

Z |x|²

(1+|x|²)^{3. (4.3)} Let

h(t) = ^a

2t²kuεk²+ ^b

4t⁴kuεk⁴−¹ 6t⁶

Z

K(x)|uε|⁶, t≥0

By (3.3), (4.3) and the fact A₁A⁻₀^1/3 = S(see [1]), we have that forε>0 small enough sup

t>0

h(t) = ^abku_εk⁶ 4ku_εk⁶₆ + ^b

3ku_εk¹² 24kuεk¹²₆ + ^b

2ku_εk⁸+4aku_εk²ku_εk⁶₆^3/2 24kuεk¹²₆

= ^ab 4

A₁+O(ε^1/2)³ A₀+O(ε^3/2) + ^b

3

24

A₁+O(ε^1/2)⁶ (A₀+O(ε^3/2))² + ¹

24 h

b² A₁+O(ε^1/2)⁴+4a A₁+O(ε^1/2) A₀+O(ε^3/2)^i3/2 (A₀+O(ε^3/2))²

= ^abA

31A⁻₀¹

4 + ^b

3A⁶₁A⁻₀²

24 +

b²A⁴₁A⁻₀^4/3+4aA₁A₀^−1/33/2

24 +O(ε^1/2) +O(ε^3/2)

= ^abS

3

4 + ^b

3S⁶

24 +(b²S⁴+4aS)^3/2

24 +O(ε^1/2).

(4.4)

By the definition ofc₁ andD₀, we may choose smallλ₁such that for allλ∈(0,λ₁), c₁−D₀λ

4

4−q >0. (4.5)

Clearly, lim_t→0⁺h(t) =0 and hence, there exists t₁ >0 satisfying for allλ∈(_0,λ₁) sup

0<t≤t₁

I(tuε)< c₁−D₀λ

4

4−q. (4.6)

We consider the caset >t₁ next. Letε< η², then Z

K(x)|x|^β|u_ε|^q≥

Z

Bη(0)

K(x)|x|^βK(x)^−q/2ϕ^q(x) (ε+|x|²)^q/2

≥exp 1−^q

2

η^α/4 ^Z

Bη(0)

|x|^β (ε+|x|²)^q/2

≥exp 1−^q

2

η^α/4 1 (2η²)^q/2

Z

Bη(0)

|x|^β

=exp 1−^q

2

η^α/4 4π (2η²)^q/2

Z _η

0 r^2+βdr

= ^4πη

3+β−q

(3+β)2^q/2exp 1− ^q

2

η^α/4 .

(4.7)

By using (4.4) and (4.7), we have for allε=λ

8

4−q ∈(0,η)andt>t₁ I(tu_ε) =h(t)−λt^q

q Z

K(x)|x|^βu^q_ε

≤ _sup

t>0

h(t)−λt^q₁ q

Z

K(x)|x|^βu^q_ε

≤ ^abS

3

4 +^b

3S⁶

24 + (b²S⁴+4aS)^3/2

24 +O(ε^1/2)

−λt₁^q q

4πη^3+β−q

(3+β)2^q/2 exp 1− ^q

2

η^α/4

= c₁+O(λ

4

4−q)−λt^q₁ q

4πη^3+β−q

(3+β)2^q/2 exp 1− ^q

2

η^α/4 .

(4.8)

From q∈ (2, 4), we have ₄₋⁴_q > 2. Thus, there existsλ₂ > 0 sufficient small such that for all λ∈ (0,λ₂)

O(λ

4−4q)−λt^q₁ q

4πη^3+β−q

(3+β)2^q/2exp 1− ^q

2

η^α/4

<−D₀λ

4−4q. (4.9)

Let λ∗ = _min{λ₁,λ₂,η^(4−q)/8}andε =λ

8

4−q, then by (4.6), (4.8) and (4.9), we have that for all λ∈ (0,λ∗)

sup

t>0

I(tuε)<c₁−D₀λ

4 4−q. This completes the proof of Lemma4.1.

Lemma 4.2. Let{u˜n} ⊂H be a(PS)_csequence for I with c<c₁−D0λ

4

4−q, where c₁is defined as in Lemma3.5. Then {u˜_n}has a convergent subsequence.

Proof. Let{u˜n} ⊂Hbe a(PS)_c sequence for I, that is

nlim→_∞I(u˜_n) =c and lim

n→_∞I⁰(u˜_n) =_{0. (4.10)} Firstly, we show thatku˜nkis bounded. By (4.10), we have that

c+1+o(1)ku˜nk ≥I(u˜n)− ¹

6hI⁰(u˜n), ˜uni

= ^a

3ku˜_nk²+ ^b

12ku˜_nk⁴−λ6−q 6q

Z

K(x)|x|^β|u˜_n|^q

≥ ^a

3ku˜_nk²+ ^b

12ku˜_nk⁴−λ6−q

6q S⁻_q^q/2ku˜_nk^q

which means that ku˜_nkis bounded as 2 < q < 4. After passing to a subsequence, we may assume that

˜

u_n *u˜∗ in H,

˜

u_n →u˜∗ inL^r_K(_R³), 2≤r <6,

˜

u_n →u˜∗ a.e. onR³.

Writev_n=u˜_n−u˜∗and we claim thatkv_nk →0. Otherwise, up to a subsequence (still denoted by{vn}), we may supposekvnk → lwithl >0. From (4.10), we have thathI⁰(u˜n), ˜u∗i=o(1) and hence

0=aku˜∗k²+b(l²+ku˜∗k²)ku˜∗k²−λ Z

K(x)|x|^β|u˜∗|^q−

Z

K(x)|u˜∗|⁶. (4.11) On the other hand, byhI⁰(u˜_n), ˜u_ni= o(1), we can use Brézis–Lieb lemma to obtain

0=a(kvnk²+ku˜∗k²) +b(kvnk⁴+2kvnk²ku˜∗k²+ku˜∗k⁴)

−λ Z

K(x)|x|^β|u˜∗|^q−

Z

K(x)|v_n|⁶−

Z

K(x)|u˜∗|⁶+o(₁)_. ^(4.12) Combining (4.11) and (4.12), we obtain

o(1) =akvnk²+bkvnk⁴+bkvnk²ku˜∗k²−

Z

K(x)|vn|⁶ (4.13) and so, from (1.5) it follows that

akv_nk²+bkv_nk⁴+bkv_nk²ku˜∗k²=

Z

K(x)|v_n|⁶+o(1)≤S⁻³kv_nk⁶+o(1). Taking the limit asn→∞, we obtain that

l²≥ ^bS3+^pb²S⁶+4(a+bku∗k²)S³

2 ≥ ^bS3+√

b²S⁶+4aS³

2 . (4.14)

By (4.11) and Hölder’s inequality, we obtain I(u˜∗) = ^a

2ku˜∗k²+^b

4ku˜∗k⁴− ^λ q Z

K(x)|x|^β|u˜∗|^q−¹ 6

Z

K(x)|u˜∗|⁶

= ^a

3ku˜∗k²+ ^b

12ku˜∗k⁴−λ6−q 6q

Z

K(x)|x|^β|u˜∗|^q− ^b

6l²ku˜∗k²

≥ ^b

12ku˜∗k⁴−λ6−q

6q S⁻_q^q/2ku˜∗k^q− ^b

6l²ku˜∗k²_. For A:=λ⁶_6q^−qS_q^−q/2, define

ψ(t) = ^b

12t⁴−At^q.

By easy calculation, it follows thatψ(t)achieves its minimum value att_min = ^3pA_b ^1/(4−q)and

ψ(t_min) =− 3q

b ₄₋^q_q

4−q 4 A^4−4q. Thus, we have

I(u˜∗)≥ −⁴−q 4

3q b

₄₋^q_q (₆−q) 6q S⁻_q^q/2

₄₋⁴_q λ

4 4−q − ^b

6l²ku˜∗k²

= −D₀λ

4−4q − ^b

6l²ku˜∗k².

(4.15)