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A class of fourth-order elliptic equations with concave and convex nonlinearities in R N

Zijian Wu and Haibo Chen

B

School of Mathematics and Statistics, Central South University, Changsha, Hunan 410083, P. R. China Received 15 October 2020, appeared 15 September 2021

Communicated by Roberto Livrea

Abstract.In this article, we study the multiplicity of solutions for a class of fourth-order elliptic equations with concave and convex nonlinearities inRN. Under the appropriate assumption, we prove that there are at least two solutions for the equation by Nehari manifold and Ekeland variational principle, one of which is the ground state solution.

Keywords:fourth-order elliptic equation, multiple solutions, Nehari manifold, Ekeland variational principle.

2020 Mathematics Subject Classification: 35J35, 35J60.

1 Introduction and main results

In this article, we consider the multiplicity results of solutions of the following fourth-order elliptic equation:

(∆2u−∆u+u= f(x)|u|q2u+|u|p2u, inRN,

u∈ H2(RN), (1.1)

where N > 4, 1 < q < 2 < p < 2(2 = 2N/(N−4)), the weight function f satisfies the following condition:

(F) f ≥0, f ∈ Lrq(RN)∩L(RN)whererq= rrq for somer∈ (2, 2). Associated with (1.1), we consider theC1-functional If, for each u∈ H2(RN),

If(u) = 1

2kuk21 q

Z

RN f(x)|u|qdx− 1 p

Z

RN|u|pdx, where kuk= R

RN |∆u|2+|∇u|2+u2 dx1/2

is the norm in H2(RN). It is well known that the solutions of (1.1) are the critical points of the energy functionalIf [14].

BCorresponding author. Email: math_chb@163.com

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In reality, elliptic equations with concave ang convex nonlinearities in bounded domains have been the focus of a great deal of research in recent years. Ambrosetti et al. [1], for example, considered the following equation:





u=λuq1+up1, in Ω,

u>0, in Ω,

u∈ H10(),

(1.2)

where Ω is a bounded domain in RN with 1 < q < 2 < p < 2 (2 = N2N2 if N ≥ 3; 2 =

∞ifN = 1, 2)and λ> 0. They found that there is λ0 > 0 such that (1.2) admits at least two positive solutions for λ∈ (0,λ0), has a positive solution for λ= λ0 and no positive solution exists forλ> λ0. Actually, many scholars have also obtained the same results in the unit ball BN(0; 1), see [2,6,10,13].

Furthermore, it is also an important subject to deal with elliptic equation with concave- convex nonlinearities when a bounded domain Ω is replaced by RN. Wu [18] studied the concave-convex elliptic problem:





u+u= fλ(x)uq1+gµ(x)up1, inRN,

u>0, inRN,

u∈ H1(RN),

(1.3)

where 1<q<2< p<2 (2 =2N/(N−2)if N≥3, 2 =if N=1, 2), fλ= λf++ f(f±=±max{0,±f} 6=0)

is sign-changing, gµ = a+µband the parameters λ,µ > 0. When the functions f+, f, a, b satisfy appropriate hypotheses, author obtained the multiplicity of positive solutions for the problem (1.3). Hsu and Lin [9] dealt with the existence and multiplicity of positive solutions for the following semilinear elliptic equation:





∆u+u=λa(x)|u|q2u+b(x)|u|p2u, inRN,

u>0, inRN,

u∈ H1(RN),

(1.4)

wherea, bare measurable functions and meet the right conditions. They obtained the result of multiple solutions of the equation (1.4).

Inspired by the existing literature [5,8,9,11,15,18–20], the main aim of this article is to study (1.1) involving concave-convex nonlinearities on the whole space RN. As far as we know, there are few articles dealing with this type of fourth-order elliptic equation (1.1) involving concave-convex nonlinearities. Using arguments similar to those used in [16], we will prove the existence of two nontrivial solutions by using Ekeland variational principle [7].

Let

σ=

p−2 p−q

2−q p−q

2pq2 S

p(2q) 2(p2)

p S

q

r2 >0,

whereSpandSr are the best Sobolev constant. Now, we state the main result.

Theorem 1.1. Assume that(F)holds. If|f|rq ∈ (0,σ), then(1.1)has at least two nontrivial solutions, one of which is the ground state solution.

This paper is organized as follows. In Section 2, we give some notations and preliminaries.

In Section 3, we are concerned with the proof of Theorem1.1.

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2 Notations and preliminaries

We shall throughout use the Sobolev space H2(RN) with standard norm. The dual space of H2(RN) will be denoted by H2(RN). h·,·i denotes the usual scalar product in H2(RN). Lr(RN) is the usual Lebesgue space whose norms we denote by |u|r = R

RN|u|rdx1/r

for 1 ≤ p < ∞. Moreover, we denote by Sr the best Sobolev constant for the embedding of H2(RN)in Lr(RN).

Now, we consider the Nehari minimization problem:

αf =inf{If(u)|u∈ Nf}, whereNf ={u∈ H2(RN)\{0}|hI0f(u),ui=0}. Define

ψf(u) =hI0f(u),ui=kuk2

Z

RN f(x)|u|qdx−

Z

RN|u|pdx.

Then foru∈ Nf,

hψ0f(u),ui=hψ0f(u),ui − hI0f(u),ui

=kuk2−(q−1)

Z

RN f(x)|u|qdx−(p−1)

Z

RN|u|pdx.

Similarly to the skill used in Tarantello [16], we splitNf into three parts:

Nf+={u∈ Nf | hψ0f(u),ui>0}, N0f ={u∈ Nf | hψ0f(u),ui=0}, Nf={u∈ Nf | hψ0f(u),ui<0} and note that ifu ∈ Nf, that is,hI0f(u),ui=0, then

hψ0f(u),ui= (2−p)kuk2−(q−p)

Z

RN f(x)|u|qdx

= (2−q)kuk2−(p−q)

Z

RN|u|pdx.

(2.1)

Then, we have the following results.

Lemma 2.1. If|f|rq ∈(0,σ), then the submanifoldN0 =.

Proof. Suppose the contrary. ThenNf0 6= ∅, i.e., there existu ∈ Nf such that hψ0f(u),ui= 0.

Then foru∈ N0by (2.1) and Sobolev inequality, we have (2−q)kuk2 = (p−q)

Z

RN|u|pdx≤ (p−q)S

p

p 2kukp, and so

kuk ≥

(2−q)S

p

p2

p−q

1 p2

. (2.2)

Similarly, using (2.1), Sobolev and Hölder inequalities, we have (p−2)kuk2 = (p−q)

Z

RN f(x)|u|qdx≤ (p−q)|f|rqS

q

r 2kukq,

(4)

which implies that

kuk ≤ (p−q)|f|rq (p−2)S

q

r2

!21q

. (2.3)

Combining (2.2) and (2.3) we deduce that

|f|rq

p−2 p−q

2−q p−q

2pq2 S

p(2q) 2(p2)

p S

q

r2 =σ, which is a contradiction. This completes the proof.

Lemma 2.2. If|f|rq ∈ (0,σ), then the setNfis closed in H2(RN).

Proof. Let {un} ⊂ Nf such thatun → u in H2(RN). In the following we show u ∈ Nf. In fact, byhI0f(un),uni=0 and

hI0f(un),uni − hI0f(u),ui=hI0f(un)−I0f(u),ui+hI0f(un),un−ui →0 asn→∞, we havehI0f(u),ui=0. Sou∈ Nf. For anyu ∈ Nf, that is,hψ0f(u),ui<0, from (2.1) we have

(2−q)kuk2<(p−q)

Z

RN|u|pdx ≤(p−q)S

p

p 2kukp, and so

kuk>

(2−q)S

p

p2

p−q

1 p2

>0.

Hence Nf is bounded away from 0. Obviously, by (2.1), it follows that hψ0f(un),uni → hψ0f(u),ui as n → +∞. From hψ0f(un),uni < 0, we have hψ0f(u),ui ≤ 0. By Lemma 2.1, for|f|rq ∈ (0,σ), Nf0 =∅, then hψ0f(u),ui<0. Thus we deduce u∈ Nf. This completes the proof.

Lemma 2.3. The energy functional If is coercive and bounded below onNf. Proof. Foru ∈ Nf, then, by Sobolev and Hölder inequalities,

If(u) = If(u)− 1

phI0f(u),ui

= p−2

2p kuk2p−q pq

Z

RN f(x)|u|qdx

p−2

2p kuk2p−q

pq |f|rqS

q

r 2kukq. This completes the proof.

The following lemma shows that the minimizers onNf are “usually” critical points forIf. The details of the proof can be referred to Brown and Zhang [4].

Lemma 2.4. Suppose thatu is a local minimizer for Ib f on Nf. Then, if ub∈ N/ f0,u is a critical pointb of If.

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For eachu∈ H2(RN)\{0}, we write

tmax:= (2−q)kuk2 (p−q)R

RN|u|pdx

!p12

>0.

Then, we have the following lemma.

Lemma 2.5. For each u∈ H2(RN)\{0}and|f|rq ∈ (0,σ), we have

(i) there exist unique 0 < t+ := t+(u) < tmax < t := t(u)such that t+u ∈ Nf+, tu ∈ Nf and

If(t+u) = inf

tmaxt0If(tu), If(tu) = sup

ttmax

If(tu). (ii) tis a continuous function for nonzero u.

(iii) Nf=nu∈ H2(RN)\{0}|k1

ukt

u kuk

=1o . Proof. (i)Fixu∈ H2(RN)\{0}. Let

s(t) =t2qkuk2−tpq Z

RN|u|pdx fort≥0.

We have s(0) = 0, s(t) → − as t → ∞, s(t)is concave and achieves its maximum at tmax. Moreover, for |f|rq ∈(0,σ),

s(tmax) = (2−q)kuk2 (p−q)R

RN|u|pdx

!2pq2

kuk2− (2−q)kuk2 (p−q)R

RN|u|pdx

!ppq2 Z

RN|u|pdx

=kukq kukp R

RN|u|pdx

!2pq2

2−q p−q

2pq2 p−2 p−q

≥ kukq

kukp S

p

p2kukp

2q p2

2−q p−q

2pq2 p−2 p−q

=kukq

(2−q)S

p

p2

p−q

2q p2

p−2 p−q

>|f|rqS

q

r 2kukq

Z

RN f(x)|u|qdx>0.

(2.4)

Hence, there are uniquet+andtsuch that 0< t+ <tmax<t, s(t+) =

Z

RN f(x)|u|qdx =s(t) and

s0(t+)>0>s0(t). Note that

hI0f(tu),tui=tq1

s(t)−

Z

RN f(x)|u|qdx

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and

hψ0f(tu),tui=tq+1s0(t) fortu∈ Nf.

We have t+u ∈ Nf+, tu ∈ Nf, and If(tu) ≥ If(tu) ≥ If(t+u) for each t ∈ [t+,t] and If(t+u)≥ If(tu)for each t∈[0,t+]. Thus,

If(t+u) = inf

tmaxt0If(tu), If(tu) = sup

ttmax

If(tu).

(ii)By the uniqueness oftand the external property oft, we have thattis a continuous function ofu6=0.

(iii)Foru∈ Nf, letv= kuuk. By part (i), there is a uniquet(v)>0 such thatt(v)v∈ Nf, that ist(kuuk)kuuk ∈ Nf. Sinceu∈ Nf, we havet(kuuk)kuuk =1, which implies

Nf

u∈ H2(RN)\{0}| 1 kukt

u kuk

=1

. Conversely, letu∈ H2(RN)\{0}such that ku1kt

u kuk

=1. Then t

u kuk

u

kuk ∈ Nf. Thus, Nf=

u∈ H2(RN)\{0}

1 kukt

u kuk

=1

. This completes the proof.

By Lemma2.1, for|f|rq ∈ (0,σ)we writeNf =Nf+∪ Nfand define α+f = inf

u∈Nf+If(u), αf = inf

u∈NfIf(u). Lemma 2.6. For|f|rq ∈(0,σ), we haveαfα+f <0.

Proof. Letu∈ Nf+. By (2.1) we have Z

RN|u|pdx< 2−q p−qkuk2, and so

If(u) = 1

2− 1 q

kuk2+ 1

q− 1 p

Z

RN|u|pdx

<

1 2 −1

q

+ 1

q− 1 p

2−q p−q

kuk2

=−(p−2)(2−q)

2pq kuk2 <0.

Therefore,αfα+f <0.

3 Proof of Theorem 1.1

First, we will use the idea of Ni and Takagi [12] to get the following lemmas.

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Lemma 3.1. If|f|rq ∈ (0,σ), then for every u ∈ Nf, there existe > 0and a differentiable function g:Be(0)⊂ H2(RN)→R+:= (0,+)such that

g(0) =1, g(ω)(u−ω)∈ Nf, ∀ω∈ Be(0) and

hg0(0),vi= 2(u,v)−qR

RN f(x)|u|q2uvdx−pR

RN|u|p2uvdx

hψ0f(u),ui (3.1)

for all v∈ H2(RN). Moreover, if0<C1≤ kuk ≤C2, then there exists C>0such that

|hg0(0),vi| ≤Ckvk. (3.2) Proof. We defineF:R×H2(RN)→Rby

F(t,ω) =tku−ωk2−tq1 Z

RN f(x)|u−ω|qdx−tp1 Z

RN|u−ω|pdx,

it is easy to see F is differentiable. Since F(1, 0) =0 and Ft(1, 0) = hψ0f(u),ui 6= 0, we apply the implicit function theorem at point (1, 0) to get the existence of e > 0 and differentiable functiong: Be(0)→R+ such thatg(0) =1 andF(g(ω),ω) =0 for∀ω∈ Be(0). Thus,

g(ω)(u−ω)∈ Nf, ∀ω ∈Be(0).

Also by the differentiability of the implicit function theorem, for allv∈ H2(RN), we know that

hg0(0),vi=−hFω(1, 0),vi Ft(1, 0) . Note that

−hFω(1, 0),vi=2(u,v)−q Z

RN f(x)|u|q2uvdx−p Z

RN|u|p2uvdx andFt(1, 0) =hψ0f(u),ui. So (3.1) holds.

Moreover, by (3.1), 0<C1≤ kuk ≤C2 and Hölder’s inequality, we have

|hg0(0),vi| ≤ Cekvk hψ0f(u),ui

for some Ce > 0. To prove (3.2), therefore, we only need to show that |hψ0f(u),ui| > d for some d > 0. We argue by contradiction. Assume that there exists a sequence {un} ∈ Nf, C1 ≤ kunk ≤ C2, we havehψ0f(un),uni = on(1). Then by (2.1) and Sobolev’s inequality, we have

(2−q)kunk2= (p−q)

Z

RN|un|pdx+on(1)

≤ (p−q)S

p

p 2kunkp+on(1), and so

kunk ≥

(2−q)S

p

p2

p−q

1 p2

+on(1). (3.3)

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Similarly, using (2.1) and Hölder and Sobolev inequalities, we have (p−2)kunk2 = (p−q)

Z

RN f(x)|un|qdx+on(1)

≤(p−q)|f|rqS

q

r 2kunkq+on(1), which implies that

kunk ≤ (p−q)|f|rq (p−2)S

q

r2

!21q

+on(1). (3.4)

Combining (3.3) and (3.4) asn→+, we deduce that

|f|rq

p−2 p−q

2−q p−q

2pq2 S

p(2q) 2(p2)

p S

q

r2 =σ,

which is a contradiction. Thus if 0<C1≤ kuk ≤C2, there existsC>0 such that

|hg0(0),vi| ≤Ckvk. This completes the proof.

Lemma 3.2. If |f|rq ∈ (0,σ)∈ (0,σ), then for every u ∈ Nf, there exist e> 0and a differentiable function g: Be(0)⊂ H2(RN)→R+ such that

g(0) =1, g(ω)(u−ω)∈ Nf, ∀ω∈ Be(0) and

h(g)0(0),vi= 2(u,v)−qR

RN f(x)|u|q2uvdx−pR

RN|u|p2uvdx

hψ0f(u),ui (3.5)

for all v∈ H2(RN). Moreover, if0<C1≤ kuk ≤C2, then there exists C>0such that

|h(g)0(0),vi| ≤Ckvk. (3.6) Proof. Similar to the argument in Lemma 3.2, there exist e > 0 and a differentiable function g: Be(0)→R+such thatg(0) =1 andg(ω)(u−ω)∈ Nf for allω ∈ Be(0). By u∈ Nf, we have

hψ0f(u),ui=kuk2−(q−1)

Z

RN f(x)|u|qdx−(p−1)

Z

RN|u|pdx<0.

Since g(ω)(u−ω)is continuous with respect to ω, when e is small enough, we know for ω∈ Be(0)

kg(ω)(u−ω)k2−(q−1)

Z

RN f(x)|g(ω)(u−ω)|qdx−(p−1)

Z

RN|g(ω)(u−ω)|pdx<0.

Thus, g(ω)(u−ω) ∈ Nf, ∀ω ∈ Be(0). Moreover, the proof details of (3.5) and (3.6) are similar to Lemma3.1.

Lemma 3.3. If|f|rq ∈ (0,σ), then

(i) there exists a minimizing sequence{un} ∈ Nf such that If(un) =αf +on(1),

I0f(un) =on(1) in H2(RN);

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(ii) there exists a minimizing sequence{un} ∈ Nfsuch that If(un) =αf +on(1),

I0f(un) =on(1) in H2(RN).

Proof. (i)By Lemma2.3and the Ekeland variational principle onNf, there exists a minimizing sequence {un} ⊂ Nf such that

αf ≤ If(un)<αf + 1

n (3.7)

and

If(un)≤ If(v) + 1

nkv−unk for each v∈ Nf. (3.8) And we can show that there exists C1,C2 > 0 such that 0 < C1 ≤ kunk ≤ C2. Indeed, if not, that is, un → 0 in H2(RN), then If(un) would converge to zero, which contradict with If(un) → αf < 0. Moreover, by Lemma 2.3 we know that If(u) is coercive on Nf, {un} is bounded in Nf.

Now, we show that

kI0f(un)kH2(RN) →0 asn→∞.

Applying Lemma 3.1 with un to obtain the functions gn(ω) : Ben(0)→ R+ for some en > 0, such that

gn(0) =1, gn(ω)(unω)∈ Nf, ∀ω ∈Ben(0).

We choose 0 < ρ< en. Let u∈ H2(RN)\{0}andωρ = kρuuk. Sincegn(ωρ)(unωρ)∈ Nf, we deduce from (3.8) that

1

n[|gn(ωρ)−1|kunk+ρgn(ωρ)]

1

nkgn(ωρ)(unωρ)−unk

≥ If(un)−If(gn(ωρ)(unωρ))

= 1

2kunk21 q

Z

RN f(x)|un|qdx− 1 p

Z

RN|un|pdx−1

2 gn(ωρ)2kunωρk2 +1

q gn(ωρ)q

Z

RN f(x)|unωρ|qdx+ 1

p gn(ωρ)p

Z

RN|unωρ|pdx

= − gn(ωρ)2−1

2 kunωρk21

2(kunωρk2− kunk2) + gn(ωρ)q−1

q

Z

RN f(x)|unωρ|qdx +1

q Z

RN f(x)|unωρ|qdx−

Z

RN f(x)|un|qdx

+ gn(ωρ)p−1 p

Z

RN|unωρ|pdx+ 1 p

Z

RN|unωρ|pdx−

Z

RN|un|pdx

.

(3.9)

Note that

lim

ρ0+

gn(ωρ)−1

ρ = lim

ρ0+

gn(0+ρkuuk)−gn(0)

ρ =

(gn)0(0), u kuk

.

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If we divide the ends of (3.9) byρand letρ→0+, we have 1

n

(gn)0(0), u kuk

kunk+1

≥ −

(gn)0(0), u kuk

kunk2

Z

RN∆un

u kuk

+∇un

u kuk

+un

u kuk

dx +

(gn)0(0), u kuk

Z

RN f(x)|un|qdx+

Z

RN f(x)|un|q2un

u kuk

dx +

(gn)0(0), u kuk

Z

RN|un|pdx+

Z

RN|un|p2un

u kuk

dx

= −

(gn)0(0), u

kuk kunk2

Z

RN f(x)|un|qdx−

Z

RN|un|pdx

1 kuk

Z

RN|un|p2unudx + 1

kuk

Z

RN(∆un∆u+∇un∇u+unu)dx− 1 kuk

Z

RN f(x)|un|q2unudx

= −

(gn)0(0), u kuk

hI0f(un),uni+ 1

kukhI0f(un),ui

= 1 kuk

D

I0f(un),uE , that is,

1 n

|h(gn)0(0),ui|kunk+kuk≥ hI0f(un),ui. By the boundedness ofkunkand Lemma3.2, there exists ˆC>0 such that

Cˆ n ≥

I0f(un), u kuk

. Therefore, we have

kI0f(un)kH2(RN)= sup

uH2(RN)\{0}

hI0f(un),ui kuk ≤ Cˆ

n, that is,I0f(un) =on(1)as n→+∞. This completes the proof of (i).

(ii)Similarly, by using Lemma3.2, we can prove (ii). We will omit the details here.

Now, we establish the existence of minimum for If on Nf+.

Theorem 3.4. Assume that(F)holds. If|f|rq ∈ (0,σ), then the functional If has a minimizer u+in Nf+and it satisfies

(i) If(u+) =αf =α+f ;

(ii) u+is a solution of equation(1.1).

Proof. From Lemma3.3, let{un}be a(PS)αf sequence for If on Nf, i.e.,

If(un) =αf +on(1), I0f(un) =on(1) in H2(RN). (3.10)

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Then it follows from Lemma2.3that{un}is bounded inH2(RN). Hence, up to a subsequence, there existsu+∈ H2(RN)such that





un*u+ inH2(RN);

un→u+ inLsloc(RN) (2≤s<2); un(x)→u+(x) a.e. inRN.

(3.11)

By(F), Hölder inequality and (3.11), we can infer that Z

RN f(x)|un|qdx=

Z

RN f(x)|u+|qdx+on(1) asn→∞. (3.12) In fact, for any e>0, there exists Msufficiently large such that

Z

|x|>M

|f(x)|rqdx 1

rq < e.

And from{un} ⊂ Nf inH2(RN)is bounded, we obtain that R

RN|un−u+|rdxqr

is bounded.

Therefore, we have Z

RN|f(x)(|un|q− |u+|q)|dx≤

Z

RN f(x)|un−u+|qdx

=

Z

|x|≤M f(x)|un−u+|qdx+

Z

|x|>M f(x)|un−u+|qdx

Z

|x|≤M

|f(x)|rqdx rq1 Z

|x|≤M

|un−u+|rdx qr

+ Z

|x|>M

|f(x)|rqdx rq1 Z

|x|>M

|un−u+|rdx qr

→ 0 asn→∞.

First, we can claim that u+ is a nontrivial solution of (1.1). Indeed, by (3.10) and (3.11), it is easy to see that u+ is a solution of (1.1). Next we show that u+ is nontrivial. Fromun ∈ Nf, we have that

If(un) = 1

2− 1 p

kunk2− 1

q− 1 p

Z

RN f(x)|un|qdx. (3.13) Letn→in (3.13), we can get

αf ≥ −p−q pq

Z

RN f(x)|u+|qdx.

In view of Lemma 2.6, we have 0 > α+fαf, which implies R

RN f(x)|u+|qdx > 0. Thus, u+ is a nontrivial solution of (1.1). Now we prove that un → u+ strongly in H2(RN) and If(u+) =α. In fact, byun,u∈ Nf, (3.12) and weak lower semicontinuity of norm, we have

αf ≤ If(u+) = 1

2 − 1 p

ku+k2− 1

q− 1 p

Z

RN f(x)|u+|qdx

≤ lim

n

1 2− 1

p

kunk2− 1

q− 1 p

Z

RN f(x)|un|qdx

= lim

nIf(un) =αf,

(12)

which implies thatIf(u+) =αf and limnkunk2= ku+k2. Noting thatun *u+in H2(RN), so un → u+ strongly in H2(RN). Furthermore, we have u+ ∈ Nf+. On the contrary, if u+ ∈ Nf, then by Lemma 2.5 (i), there are unique t+ and t such that t+u+ ∈ Nf+ and tu+ ∈ Nf. In particular, we have t+ < t = 1 and so If(t+u+)< If(tu+) = If(u+) = αf, which is a contradiction. By Lemma2.4 we may assume that u+ is a solution of (1.1). This completes the proof.

In order to obtain the existence of the second local minimum, we consider the following minimization problem:

S0=inf{I0(u)|u∈ H2(RN)\{0},I00(u) =0}, where

I0(u) = 1

2kuk21 p

Z

RN|u|pdx.

From [17,21], we know thatS0 is achieved atu0∈ H2(RN). Moreover, S0 = I0(u0) =sup

t0

I0(tu0). Then, we have the following lemma.

Lemma 3.5. If|f|rq ∈ (0,σ), thenαf <αf +S0.

Proof. From Lemma2.5(iii), Nfdisconnects H2(RN)\{0}in exactly two components:

Λ1=

u

1 kukt

u kuk

>1

, Λ2=

u

1 kukt

u kuk

<1

,

and Nf+Λ1. Moreover, there exists t1 such that u++t1u0Λ2. Indeed, denote t0 = t((u++tu0)/ku++tu0k). Since

t

u++tu0 ku++tu0k

u++tu0 ku++tu0k

∈ Nf, we have

0≤ t

q 0

R

RN f(x)|u++tu0|qdx

ku++tu0kq = t20t

p 0

R

RN|u++tu0|pdx ku++tu0kp . Thus

t0

"

ku+/t+u0k R

RN|u+/t+u0|p1/p

#p/(p2)

→ ku0k ast →∞.

Therefore, there existst2 >0 such thatt0 < lku0k, for somel > 1 andt ≥ t2. Set t1 > t2+l, then

t

u++t1u0 ku++t1u0k

2

<l2ku0k2

≤ ku+k2+t21ku0k2+2t1 Z

RN(u+∆u0+∇u+∇u0+u+u0)dx

=ku++t1u0k2,

(13)

that is,u++t1u0Λ2. So there existsk ∈(0, 1)such thatu++kt1u0 ∈ Nf. Furthermore, we have

αf ≤ If(u++kt1u0)

= 1

2ku++kt1u0k21 q

Z

RN f(x)|u++kt1u0|qdx− 1 p

Z

RN|u++kt1u0|pdx

< If(u+) + 1

2kkt1u0k21 p

Z

RN|kt1u0|pdx

= If(u+) +I0(kt1u0)

αf +I0(u0)

=αf +S0. This completes the proof.

Next, we establish the existence of minimum forIf onNf.

Theorem 3.6. Assume that(F)holds. If|f|rq ∈(0,σ), then the functional If has a minimizer uin Nfand it satisfies

(i) If(u) =αf ;

(ii) uis a solution of equation(1.1).

Proof. From Lemma3.3, let{un}be a(PS)α

f sequence for If on Nf , i.e.,

If(un) =αf +on(1), I0f(un) =on(1) in H2(RN). (3.14) From Lemma 2.3 we have {un} is bounded in H2(RN). Hence, up to a subsequence, there exists u∈ H2(RN)such that





un*u inH2(RN);

un→u inLsloc(RN) (2≤s<2); un(x)→u(x) a.e. inRN.

(3.15)

From (3.14) and (3.15), we havehI0f(u),vi = 0, ∀v ∈ H2(RN), that is, uis a weak solution of (1.1) andu∈ Nf. Letvn=un−u. Then





vn*0 in H2(RN);

vn→0 in Lsloc(RN) (2≤ s<2); vn(x)→0 a.e. inRN.

(3.16)

Now we prove thatun→ustrongly in H2(RN), that is,vn →0 strongly inH2(RN). Arguing by contradiction, we assume that there is c > 0 such that kvnk ≥ c > 0. By the Brézis–Lieb theorem [3],

If(un) = 1

2kunk21 q Z

RN f(x)|un|qdx− 1 p

Z

RN|un|pdx

= If(u) + 1

2kvnk21 q

Z

RN f(x)|vn|qdx− 1 p

Z

RN|vn|pdx+on(1)

= If(u) + 1

2kvnk21 p

Z

RN|vn|pdx+on(1),

(3.17)

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