A class of fourth-order elliptic equations with concave and convex nonlinearities in R N
Zijian Wu and Haibo Chen
BSchool 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|q−2u+|u|p−2u, 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= r−rq for somer∈ (2, 2∗). Associated with (1.1), we consider theC1-functional If, for each u∈ H2(RN),
If(u) = 1
2kuk2−1 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
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=λuq−1+up−1, in Ω,
u>0, in Ω,
u∈ H10(Ω),
(1.2)
where Ω is a bounded domain in RN with 1 < q < 2 < p < 2∗ (2∗ = N2N−2 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)uq−1+gµ(x)up−1, 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|q−2u+b(x)|u|p−2u, 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
2p−−q2 S
p(2−q) 2(p−2)
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.
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 H−2(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 p−2
. (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,
which implies that
kuk ≤ (p−q)|f|rq (p−2)S
q
r2
!2−1q
. (2.3)
Combining (2.2) and (2.3) we deduce that
|f|rq ≥
p−2 p−q
2−q p−q
2p−−q2 S
p(2−q) 2(p−2)
p S
q
r2 =σ, which is a contradiction. This completes the proof.
Lemma 2.2. If|f|rq ∈ (0,σ), then the setNf−is 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 p−2
>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 kuk2− p−q pq
Z
RN f(x)|u|qdx
≥ p−2
2p kuk2− p−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.
For eachu∈ H2(RN)\{0}, we write
tmax:= (2−q)kuk2 (p−q)R
RN|u|pdx
!p−12
>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+, t−u ∈ Nf− and
If(t+u) = inf
tmax≥t≥0If(tu), If(t−u) = sup
t≥tmax
If(tu). (ii) t−is 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) =t2−qkuk2−tp−q 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
!2p−−q2
kuk2− (2−q)kuk2 (p−q)R
RN|u|pdx
!pp−−q2 Z
RN|u|pdx
=kukq kukp R
RN|u|pdx
!2p−−q2
2−q p−q
2p−−q2 p−2 p−q
≥ kukq
kukp S−
p
p2kukp
2−q p−2
2−q p−q
2p−−q2 p−2 p−q
=kukq
(2−q)S
p
p2
p−q
2−q p−2
p−2 p−q
>|f|rqS−
q
r 2kukq
≥
Z
RN f(x)|u|qdx>0.
(2.4)
Hence, there are uniquet+andt−such 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=tq−1
s(t)−
Z
RN f(x)|u|qdx
and
hψ0f(tu),tui=tq+1s0(t) fortu∈ Nf.
We have t+u ∈ Nf+, t−u ∈ Nf−, and If(t−u) ≥ 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
tmax≥t≥0If(tu), If(t−u) = sup
t≥tmax
If(tu).
(ii)By the uniqueness oft−and the external property oft−, we have thatt−is 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+∪ Nf−and define α+f = inf
u∈Nf+If(u), α−f = inf
u∈Nf−If(u). Lemma 2.6. For|f|rq ∈(0,σ), we haveαf ≤α+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,αf ≤α+f <0.
3 Proof of Theorem 1.1
First, we will use the idea of Ni and Takagi [12] to get the following lemmas.
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|q−2uvdx−pR
RN|u|p−2uvdx
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−tq−1 Z
RN f(x)|u−ω|qdx−tp−1 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|q−2uvdx−p Z
RN|u|p−2uvdx 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 p−2
+on(1). (3.3)
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
!2−1q
+on(1). (3.4)
Combining (3.3) and (3.4) asn→+∞, we deduce that
|f|rq ≥
p−2 p−q
2−q p−q
2p−−q2 S
p(2−q) 2(p−2)
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|q−2uvdx−pR
RN|u|p−2uvdx
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 H−2(RN);
(ii) there exists a minimizing sequence{un} ∈ Nf−such that If(un) =α−f +on(1),
I0f(un) =on(1) in H−2(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)kH−2(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)(un−ω)∈ Nf, ∀ω ∈Ben(0).
We choose 0 < ρ< en. Let u∈ H2(RN)\{0}andωρ = kρuuk. Sincegn(ωρ)(un−ωρ)∈ Nf, we deduce from (3.8) that
1
n[|gn(ωρ)−1|kunk+ρgn(ωρ)]
≥ 1
nkgn(ωρ)(un−ωρ)−unk
≥ If(un)−If(gn(ωρ)(un−ωρ))
= 1
2kunk2− 1 q
Z
RN f(x)|un|qdx− 1 p
Z
RN|un|pdx−1
2 gn(ωρ)2kun−ωρk2 +1
q gn(ωρ)q
Z
RN f(x)|un−ωρ|qdx+ 1
p gn(ωρ)p
Z
RN|un−ωρ|pdx
= − gn(ωρ)2−1
2 kun−ωρk2− 1
2(kun−ωρk2− kunk2) + gn(ωρ)q−1
q
Z
RN f(x)|un−ωρ|qdx +1
q Z
RN f(x)|un−ωρ|qdx−
Z
RN f(x)|un|qdx
+ gn(ωρ)p−1 p
Z
RN|un−ωρ|pdx+ 1 p
Z
RN|un−ωρ|pdx−
Z
RN|un|pdx
.
(3.9)
Note that
lim
ρ→0+
gn(ωρ)−1
ρ = lim
ρ→0+
gn(0+ρkuuk)−gn(0)
ρ =
(gn)0(0), u kuk
.
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|q−2un
− u kuk
dx +
(gn)0(0), u kuk
Z
RN|un|pdx+
Z
RN|un|p−2un
− u kuk
dx
= −
(gn)0(0), u
kuk kunk2−
Z
RN f(x)|un|qdx−
Z
RN|un|pdx
− 1 kuk
Z
RN|un|p−2unudx + 1
kuk
Z
RN(∆un∆u+∇un∇u+unu)dx− 1 kuk
Z
RN f(x)|un|q−2unudx
= −
(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)kH−2(RN)= sup
u∈H2(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 H−2(RN). (3.10)
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+f ≥ αf, 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
n→∞If(un) =αf,
which implies thatIf(u+) =αf and limn→∞kunk2= 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 t−u+ ∈ Nf−. In particular, we have t+ < t− = 1 and so If(t+u+)< If(t−u+) = 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
2kuk2− 1 p
Z
RN|u|pdx.
From [17,21], we know thatS0 is achieved atu0∈ H2(RN). Moreover, S0 = I0(u0) =sup
t≥0
I0(tu0). Then, we have the following lemma.
Lemma 3.5. If|f|rq ∈ (0,σ), thenα−f <αf +S0.
Proof. From Lemma2.5(iii), Nf−disconnects 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++t1u0 ∈ Λ2. 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 = t20− t
p 0
R
RN|u++tu0|pdx ku++tu0kp . Thus
t0 ≤
"
ku+/t+u0k R
RN|u+/t+u0|p1/p
#p/(p−2)
→ 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,
that is,u++t1u0∈ Λ2. So there existsk ∈(0, 1)such thatu++kt1u0 ∈ Nf−. Furthermore, we have
α−f ≤ If(u++kt1u0)
= 1
2ku++kt1u0k2− 1 q
Z
RN f(x)|u++kt1u0|qdx− 1 p
Z
RN|u++kt1u0|pdx
< If(u+) + 1
2kkt1u0k2− 1 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 u−in Nf−and it satisfies
(i) If(u−) =α−f ;
(ii) u−is a solution of equation(1.1).
Proof. From Lemma3.3, let{un}be a(PS)α−
f sequence for If on N−f , i.e.,
If(un) =α−f +on(1), I0f(un) =on(1) in H−2(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, u−is 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→u−strongly 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
2kunk2− 1 q Z
RN f(x)|un|qdx− 1 p
Z
RN|un|pdx
= If(u−) + 1
2kvnk2− 1 q
Z
RN f(x)|vn|qdx− 1 p
Z
RN|vn|pdx+on(1)
= If(u−) + 1
2kvnk2− 1 p
Z
RN|vn|pdx+on(1),
(3.17)