# Existence of Solutions for a Fractional Laplacian Equation with Critical Nonlinearity

Authors: Zifei Shen; Fashun Gao
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the fractional Laplacian equation(−Δ)su+λA(x)u=μu+|u|2*(s)-2u, x∈ℝN, here N>2s, s∈(0,1), 2*(s)=2N/(N-2s) is the critical exponent, and A(x)≥0 is a real potential function. Employing the variational method
we prove the existence of nontrivial solutions for μ small and λ large.


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## Body

## 1. Introduction

We consider the nonlinear Schrödinger equation:(1)iℏ∂ψ∂t=-ℏ2Δψ+A(x)ψ-|ψ|p-2ψ,x∈ℝN,
whereℏis the Planck constant. When looking for stationary waves of the formψ(t,x)=e-iμ(ℏt)φ(x)withμ∈ℝN, one is led to considering the elliptic equation inℝN; namely, replacingℏbyε, one sees thatφmust satisfy
(2)-ε2Δφ+A(x)φ=ε2μφ+|φ|p-2φ.
Settingu(x):=ε-2/(p-2)φ(x)andλ=ε-2, this equation is transformed into
(3)-Δu+λA(x)u=μu+|u|2*-2u,x∈ℝN.
Problem (3) has been widely studied in the literature (see, for instance, [1, 2] and references therein), where2*=2N/(N-2)is the critical exponentN≥4,andA(x)≥0is a potential well.The study of existence and concentration of the semiclassical states of Schrödinger equation goes back to the pioneer work [3] by Floer and Weinstein. Ever since then, equations of (3) type with subcritical nonlinearities (p<2*=2N/(N-2)forN≥3) have been studied by many authors. For critical nonlinearity (p=2*forN≥4), Clapp and Ding [1, 2] established the existence and multiplicity of positive solutions and minimal nodal solutions which localize near the potential well forμsmall andλlarge.The fractional Schrödinger equation is a fundamental equation of fractional quantum mechanics. It was discovered by Nick Laskin as a result of extending the Feynman path integral, from the Brownian-like to Lévy-like quantum mechanical paths. The term fractional Schrödinger equation was coined by Nick Laskin.Recently, a great attention has been devoted to the fractional and nonlocal operators of elliptic type, both for their interesting theoretical structure and in view of concrete applications in many fields such as combustion and dislocations in mechanical systems. This type of operator seems to have a prevalent role in physical situations and has been studied by many authors [4–9] and references therein. In [5], Di Nezza et al. deal with the fractional Sobolev spaceWs,pand analyze their role in the trace theory. They prove continuous and compact embeddings, investigating the problem of the extension domains and other regularity results. In [8], Felmer et al. proved the existence of positive solutions of nonlinear Schrödinger equation involving the fractional Laplacian inℝN. They further analyzed regularity, decay, and symmetry properties of these solutions. Servadei and Valdinoci [9] studied the existence of nontrivial solutions for equations driven by a nonlocal integrodifferential operatorLKwith homogeneous Dirichlet boundary conditions. They give more general and more precise results about the eigenvalues of a linear operator.The aim of this paper is to study the fractional Laplacian equation:(4)(-Δ)su+λA(x)u=μu+|u|2*(s)-2uinℝN,
whereN>2s,λ>0,μ∈ℝ,  s∈(0,1),andHs(ℝN)is the usual fractional Sobolev space, and2*(s)=2N/(N-2s)is the corresponding critical exponent. SupposeA(x)satisfies the following assumptions.(A1)A∈C(ℝN,ℝ),A≥0,Ω:=intA-1(0) is a nonempty bounded set with smooth boundary, andΩ-=A-1(0).(A2)There existsM0>0such that
(5)L{x∈ℝN:A(x)≤M0}<∞,whereLdenotes the Lebesgue measure inℝN.The fractional Laplace operator(-Δ)sin (4) can be defined as
(6)-(-Δ)su(x)=12∫ℝN‍u(x+y)+u(x-y)-2u(x)|y|N+2sdy,x∈ℝN.We say that a functionu∈Hs(ℝN)solves (4) in the weak sense if
(7)∫ℝ2N(u(x)-u(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λ∫ℝN‍A(x)u(x)φ(x)dx=μ∫ℝN‍u(x)φ(x)dx+∫ℝN‍|u(x)|2*(s)-2u(x)φ(x)dx,11111111111111111∀φ∈Hs(ℝN).Define the energy functional by(8)Iλ(u)=12∫ℝ2N‍|u(x)-u(y)|2|x-y|N+2sdxdy+12λ∫ℝN‍A(x)|u(x)|2dx-12μ∫ℝN‍|u(x)|2dx-12*(s)∫ℝN‍|u(x)|2*(s)dx.
Then we know the critical points ofIλare exactly the weak solutions of (7). In this sense we will prove the existence of the critical points of the functionalIλ. Fréchet derivative ofIλis
(9)〈Iλ′(u),φ〉=∫ℝ2N‍(u(x)-u(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λ∫ℝN‍A(x)u(x)φ(x)dx-μ∫ℝN‍u(x)φ(x)dx-∫ℝN‍|u(x)|2*(s)-2u(x)φ(x)dx11111111111111∀φ∈Hs(ℝN).Concerning the Schrödinger equation:(10)-Δu+λA(x)u=μu+|u|2*-2uinℝN,u>0inℝN,u∈H1(ℝN).
Clapp and Ding [1] proved the following.(a)
Assume(A1)and(A2)hold andN≥4. Then, for every0<μ<μ11(Ω), there existsλ(μ)>0such that (4) has a least solutionuλfor eachλ≥λ(μ), whereμ11(Ω)is the first eigenvalue of-ΔonΩwith boundary conditionu=0.(b)
Assume(A1)and(A2)hold andN≥4. Then, there exist0<μ*<μ11(Ω)and for each0<μ≤μ*there exist two numbersΛ(μ)>0and0<c(μ)<(1/N)SN/2such that ifλ≥Λ(μ), then (4) has at leastcat(Ω)(the number of solutions is bounded from below by a topological invariant) solutions with energyIλ,μ≤c(μ).(c)
Every sequence of solutions(un)of (10) such that0<μ<μ11(Ω),λn→∞andIλn,μ(un)→c<(1/N)SN/2asn→∞concentrates at a solution of
(11)-Δu=μu+|u|2*-2uinΩ,u>0inΩ,u=0on∂Ω,
whereSis the best Sobolev constant.Our aim is to show that (a) and (c) can be extended to problem (4). In this paper, we have the following results.Theorem 1.
Assume(A1)and(A2)holdN>2sands∈(0,1). Then, for every0<μ<μ1(Ω), there existsλ(μ)>0such that (4) has at least a solutionufor eachλ≥λ(μ), whereμ1(Ω)is the first eigenvalue of(-Δ)sonΩwith boundary conditionu=0. There is a great deal of work onμ1(Ω); see for example [9]. We have
(12)μ1(Ω)=infu∈H0s(Ω)∖{0}∫ℝ2N(|u(x)-u(y)|2/|x-y|N+2s)dxdy∫Ω‍|u(x)|2dx.Theorem 2.
Every sequence of solutions(un)of (4) such that0<μ<μ1(Ω),λn→∞,andIλ(un)→c<(s/N)SsN/2sasn→∞concentrates at a solution of
(13)(-Δ)su=μu+|u|2*(s)-2uinΩ,u=0onℝN∖Ω,
whereΩis defined as in(A1).HereSsis defined as
(14)Ss:=infu∈E∖{0}∫ℝ2N‍(|u(x)-u(y)|2/|x-y|N+2s)dxdy|u|2*(s)2,
whereEis anL2(ℝN)space with potential and will be defined in Section 2.There is a great deal of work on (13); see, for example, [4, 6, 7] and the references therein. Among them Servadei and Valdinoci [4, 6, 7] studied the problem
(15)LKu+λu+|u|2*(s)-2u+f(x,u)=0inΩ,u=0inℝN∖Ω,
whereΩis an open bounded set with Lipschitz boundary inℝN,  N>2s,s∈(0,1),λ>0is a real parameter.LKis defined as follows:
(16)LKu(x)=12∫ℝN‍(u(x+y)+u(x-y)-u(x))K(y)dxdy,1111111111111111111111111111111111111111111x∈ℝN.
HereK:ℝN∖{0}→(0,+∞)is a function such that
(17)m(x)·K∈L1(ℝN),wherem(x)=min{|x|2,1};
there existsθ>0such thatK(x)≥θ|x|-(N+2s)andK(x)=K(-x)for anyx∈ℝN∖{0}. They proved that problem (15) admits a nontrivial solution for anyλ>0. They also studied the casef(x,u)≡0andK(x)=|x|-(N+2s), respectively.Clapp and Ding [1] proved the existence of minimizing sequence for energy function of (10) on Nehari manifold and assumed that it is a Palais Smale sequence by Ekeland’s variational principle. Since Palais Smale conditions hold, this finished the proof of (a). For (c), they analyzed the problem directly. We will show that their method can be extended to the case0<s<1.This paper is organized as follows. In Section2, we give some preliminary results. In Section 3, we finish the proof of Theorem 1. In Section 4, we finish the proof of Theorem 2.


## 2. Preliminary Results

Throughout this paper we write|·|qfor theLqnorm forq∈[1,∞]. We always assume that(A1)-(A2)hold,N>2s,λ>0,μ∈ℝ,ands∈(0,1).μ1(Ω)is the first eigenvalue of(-Δ)sonΩ.Ωis a nonempty bounded set with smooth boundary.We consider the fractional Sobolev space:(18)Hs(ℝN)={u∈L2(ℝN)∣∫ℝ2N‍|u(x)-u(y)|2|x-y|N+2sdxdy+∫ℝN‍|u(x)|2dx<+∞|u(x)-u(y)|2|x-y|N+2s}
with norm
(19)∥u∥Hs=(∫ℝ2N‍|u(x)-u(y)|2|x-y|N+2sdxdy+∫ℝN‍|u(x)|2dx)1/2.
And let
(20)E={u∈Hs(ℝN)∣∫ℝN‍A(x)u2dx<+∞}
be the Hilbert space equipped with norm
(21)∥u∥E=(∥u∥Hs2+∫ℝN‍A(x)u2dx)1/2.
Ifλ>0, then it is equivalent to the norms
(22)∥u∥λ=(∥u∥Hs2+λ∫ℝN‍A(x)u2dx)1/2.
ThusEis continuously embedded inHs(ℝN).Remark 3.
We know the embeddingHs(ℝN)↪Lν(ℝN)is continuous; see [5] or [8]. So the embeddingE↪Lν(ℝN)is also continuous for anyν∈[2,2*(s)].Thanks to Remark3, we can define the constantSsas in formula (14) and get thatSs>0.Lemma 4.
Letun∈Ebe such thatλn→∞and∥un∥λn2<C. Then, there is au∈H0s(Ω)such that, up to a subsequence,un→uinL2(ℝN).Proof.
Ifun→ustrongly inL2(ℝN), we proveu∈H0s(Ω). SetFm={x:|x|≤m,A(x)≥1/m},andm∈ℕ. Fornlarge enough thatλn≥1, thanks toλn→∞. So∥un∥E2≤∥un∥λn2<C, we get
(23)∫Fm|un|2dx≤m∫Fm‍A(x)|un|2dx≤mCλn⟶0111111111111111111111111111111asn⟶∞
for everym. This implies thatu(x)=0for a.e.x∈ℝN∖Ω. Hence, since∂Ωis smooth,u∈H0s(Ω).
We will show thatun→ustrongly inL2(ℝN). LetF={x∈ℝN:A(x)≤M0}withM0as in(A2), and letFc=ℝN∖F. Then
(24)∫Fcun2dx≤1λnM0∫FcλnA(x)un2dx≤CλnM0⟶0
asn→∞. SettingBRc=ℝN∖BR, whereBR={x∈ℝN:|x|≤R}, and choosingr∈(1,N/(N-2s)),and   r′=r/(r-1), we have
(25)∫BRc∩F(un-u)2dx≤|un-u|2r2L(BRc∩F)1/r′≤C1∥un-u∥E2L(BRc∩F)1/r′⟶0
asR→∞, thanks to(A2). Sinceun→uinLloc2(ℝN),
(26)∫BR(un-u)2dx⟶0
asn→∞. Byu∈H0s(Ω),
(27)∫ℝN(un-u)2dx=∫Fcun2dx+∫F(un-u)2dx≤∫Fc‍un2dx+∫BRc∩F‍(un-u)2dx+∫BR‍(un-u)2dx⟶0
asn→∞. Thusun→ustrongly inL2(ℝN).We denoteAλ:=(-Δ)s+λA(x)and by〈·,·〉theL2-inner product and write
(28)〈Aλu,v〉=∫ℝ2N‍(u(x)-u(y))(v(x)-v(y))|x-y|N+2sdxdy+λ∫ℝN‍A(x)uvdx
foru,v∈E. Setaλ:=infσp(Aλ), the infimum of the point spectrum ofAλ. Observe that
(29)0≤aλ=inf{〈Aλu,u〉:u∈E,|u|2=1}
and thataλis nondecreasing inλ.Lemma 5.
For each0<μ<μ1(Ω), there existsλ(μ)>0such thataλ≥(μ+μ1(Ω))/2forλ≥λ(μ). Consequently,
(30)cμ∥u∥λ2≤〈(Aλ-μ)u,u〉
for allu∈E,  λ≥λ(μ), wherecμ>0is a constant.Proof.
Assume, by contradiction, that there exists a sequenceλn→∞such thataλn<(μ+μ1(Ω))/2for allnandaλn→cλ≤(μ+μ1(Ω))/2. Letun∈Ebe such that|un|2=1and〈(Aλn-aλn)un,un〉→0. Then
(31)∥un∥λn2=∫ℝ2N‍|un(x)-un(y)|2|x-y|N+2sdxdy+∫ℝN‍|un(x)|2dx+λn∫ℝN‍A(x)|un(x)|2dx=〈(Aλn-aλn)un,un〉+(1+aλn)|un|22≤2(1+μ1(Ω))
for allnlarge. By Lemma 4 there is au∈H0s(Ω)such that, up to a subsequence,un→uinL2(ℝN), and thus|u|2=1. Using Fatou’s theorem, we know
(32)∫ℝ2N|u(x)-u(y)|2|x-y|N+2sdxdy-cλ∫Ω|u(x)|2dx≤limn→∞inf(∫ℝ2N|un(x)-un(y)|2|x-y|N+2sdxdy111111111111-aλn∫ℝN‍|un(x)|2dx|un(x)-un(y)|2|x-y|N+2s)≤limn→∞inf〈(Aλn-aλn)un,un〉=0.
Consequently,
(33)∫ℝ2N|u(x)-u(y)|2|x-y|N+2sdxdy≤cλ<(μ+μ1(Ω))2<μ1(Ω).
Sinceμ1(Ω)is the first eigenvalue of(-Δ)sonΩwith boundary conditionu=0, we haveμ1(Ω)≤∫ℝ2N(|u(x)-u(y)|2/|x-y|N+2s)dxdy. This is a contradiction.In the following, enlargingλ(μ)if necessary, we assumeλ(μ)≥μ/M0; thus
(34)λM0-μ≥0∀λ≥λ(μ).


## 3. The Proof of Theorem1

In this section we will finish the proof of Theorem1.The critical points ofIλlie on the Nehari manifold
(35)M={u∈E∖{0}:〈Iλ′(u),u〉=0}.
Since0<μ<μ1(Ω)and2<2*(s), the functiont∈ℝ+→Iλ(tu)has a unique maximum pointt(u)>0andt(u)u∈M. Define
(36)c1:=infMIλ,
and we observe that
(37)c1=infu∈E,u≠0maxt≥0Iλ(tu).
From Lemma 5, the constantc1is positive. On the other hand, we define
(38)c=infγ∈Γmaxt∈[0,1]Iλ(γ(t)),
where
(39)Γ:={γ∈C([0,1],E):γ(0)=0,Iλ(γ(1))<0}.Proposition 6.
Considerc=c1.Proof.
Proposition is proved, for instance, in [8, see Section2].Mis radially diffeomorphicV={v∈E:|v|2*(s)=1}. Foru∈M, the functionalIλis
(40)Iλ(u)=sN〈(Aλ-μ)u,u〉=sN|u|2*(s)2*(s).
So,
(41)c1:=infu∈MIλ(u)=infv∈VsN〈(Aλ-μ)v,v〉N/2s.We consider the functional(42)IΩ(u)=12∫ℝ2N‍|u(x)-u(y)|2|x-y|N+2sdxdy-12μ∫Ω‍|u(x)|2dx-12*(s)∫Ω‍|u(x)|2*(s)dx=12〈(A0-μ)u,u〉-12*(s)|u|2*(s)2*(s)
onH0s(Ω). Its Nehari manifold
(43)MΩ={u∈H0s(Ω)∖{0}:〈IΩ′(u),u〉=0}
is radially diffeomorphicVΩ={v∈H0s(Ω):|v|2*(s)=1}. Set
(44)c(Ω):=infu∈MΩIΩ(u)=infv∈VΩsN〈(A0-μ)v,v〉N/2s.Proposition 7.
If0<μ<μ1(Ω)andλ≥λ(μ),then
(45)sN(cμSs)N/2s≤c<c(Ω)<sNSsN/2s,
whereSsis defined in formula (14) andcμis given in Lemma 5.Proof.
By Lemma5,cμ∥v∥E2≤cμ∥v∥λ2≤〈(Aλ-μ)v,v〉for allv∈E. Taking infima overv∈Vgives the first inequality. SinceVΩ⊂Vand〈Aλv,v〉=〈A0v,v〉forv∈VΩ, it follows thatc≤c(Ω). By [6, see Section7] and [10, see Section8], we knowc(Ω)<(s/N)SsN/2sandc(Ω)is achieved at someu0. Thusc<c(Ω),because othercwould be also achieved atu0which vanishes outsideΩ, contradicting the maximum principle.
Hence, Proposition7 is proved.By definition ofc1and Proposition 6, there exists a minimizing sequence forIλonM,and we note{uj}. By Ekeland’s variational principle, we may assume that it is a Palais Smale sequence. So we have
(46)Iλ(uj)⟶c,(47)sup{|〈Iλ′(uj),φ〉|:φ∈E,∥φ∥λ=1}⟶0
asj→+∞.Proposition 8.
I
λhas at least one critical point with critical valuecfor each0<μ<μ1(Ω)andλ≥λ(μ).Proof.
We proceed by steps.
Step 1. The sequence{uj}is bounded inE.
Proof. For anyj∈ℕby (46) and (47) it easily follows that there existsC1>0such that
(48)|Iλ(uj)|≤C1,|〈Iλ′(uj),uj∥uj∥λ〉|≤C1.
As a consequence of (48) we have
(49)Iλ(uj)-12*(s)〈Iλ′(uj),uj〉=sN〈(Aλ-μ)uj,uj〉≤C1(1+∥uj∥λ).
By (49) and the definition ofIλwe have
(50)∥uj∥λ2≤C2(1+∥uj∥λ).
Thus{uj}is bounded inE.
Step
2. Problem (7) admits a solutionu∞∈E.
Proof. By Step1andEis a reflexive space, up to a subsequence, still denoted byuj, there existsu∞∈Esuch thatuj→u∞weakly inE; that is,
(51)∫ℝ2N(uj(x)-uj(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λ∫ℝNA(x)uj(x)φ(x)dx⟶∫ℝ2N(u∞(x)-u∞(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λ∫ℝN‍A(x)u∞(x)φ(x)dx
asj→+∞. Since Step1and Remark 3, we have thatujis bounded inL2*(s)(ℝN). SinceL2*(s)(ℝN)is a reflexive space, up to a subsequence
(52)uj⟶u∞weaklyinL2*(s)(ℝN)
asj→+∞. While by Lemma 4, up to a subsequence,
(53)uj(x)⟶u∞(x)inL2(ℝN),(54)uj⟶u∞a.e.inℝN
asj→+∞. By (52) and the fact that|uj|2*(s)-2ujis bounded inL2*(s)/(2*(s)-1)(ℝN),we have
(55)|uj|2*(s)-2uj⟶|u∞|2*(s)-2u∞weaklyinL2*(s)/(2*(s)-1)(ℝN)
asj→+∞.
Since (47) holds true, for anyφ∈E(56)0⟵〈Iλ′(uj),φ〉=∫ℝ2N‍(uj(x)-uj(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λ∫ℝN‍A(x)uj(x)φ(x)dx-μ∫ℝN‍uj(x)φ(x)dx-∫ℝN‍|uj(x)|2*(s)-2uj(x)φ(x)dx.
Passing to the limit in this expression asj→+∞and taking into account (51), (53), and (55), we get
(57)∫ℝ2N‍(u∞(x)-u∞(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λ∫ℝN‍A(x)u∞(x)φ(x)dx-μ∫ℝN‍u∞(x)φ(x)dx-∫ℝN‍|u∞(x)|2*(s)-2u∞(x)φ(x)dx=0
for anyφ∈E; that is,u∞is a solution of problem (7).
Step
3. The following equality holds true:
(58)Iλ(u∞)=sN∫ℝN‍|u∞(x)|2*(s)dx≥0.
Proof. By Step2, takingφ=u∞∈Eas a test function in (7), we have
(59)∫ℝ2N|u∞(x)-u∞(y)|2|x-y|N+2sdxdy+λ∫ℝNA(x)|u∞(x)|2dx=μ∫ℝN|u∞(x)|2dx+‍∫ℝN‍|u∞(x)|2*(s)dx.
So we get
(60)Iλ(u∞)=sN∫ℝN‍|u∞(x)|2*(s)dx≥0.
Hence, Step3is proved.
Now, we conclude the proof of Proposition8.
We writevj:=uj-u∞, and thenvj→0weakly inE. Moreover, since (54) holds true, by the Brézis-Lieb Lemma, we get
(61)∫ℝ2N‍|uj(x)-uj(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|uj(x)|2dx=∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|vj(x)|2dx+∫ℝ2N‍|u∞(x)-u∞(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|u∞(x)|2dx+∘(1),∫ℝN‍|uj(x)|2dx=∫ℝN‍|vj(x)|2dx111111111111111+∫ℝN‍|u∞(x)|2dx+∘(1),∫ℝN‍|uj(x)|2*(s)dx=∫ℝN‍|vj(x)|2*(s)dx11111111111111111+∫ℝN‍|u∞(x)|2*(s)dx+∘(1),11111111111111111111111111111asj⟶+∞.
Then,
(62)c⟵Iλ(uj)=12∫ℝ2N‍|uj(x)-uj(y)|2|x-y|N+2sdxdy+12λ∫ℝN‍A(x)|uj(x)|2dx-12μ∫ℝN‍|uj(x)|2dx-12*(s)∫ℝN‍|uj(x)|2*(s)dx=12∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy+12λ∫ℝN‍A(x)|vj(x)|2dx+12∫ℝ2N‍|u∞(x)-u∞(y)|2|x-y|N+2sdxdy+12∫ℝN‍A(x)|u∞(x)|2dx-12μ∫ℝN‍|u∞(x)|2dx-12μ∫ℝN‍|vj(x)|2dx-12*(s)∫ℝN‍|vj(x)|2*(s)dx-12*(s)∫ℝN‍|u∞(x)|2*(s)dx+∘(1)=Iλ(u∞)+12∫ℝ2N|vj(x)-vj(y)|2|x-y|N+2sdxdy+12∫ℝN‍A(x)|vj(x)|2dx-12μ∫ℝN‍|vj(x)|2dx-12*(s)∫ℝN‍|vj(x)|2*(s)dx+∘(1),(63)〈Iλ′(uj),uj〉=∫ℝ2N‍|uj(x)-uj(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|uj(x)|2dx-μ∫ℝN‍|uj(x)|2dx-∫ℝN‍|uj(x)|2*(s)dx=∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|vj(x)|2dx+∫ℝ2N‍|u∞(x)-u∞(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|u∞(x)|2dx-μ∫ℝN‍|u∞(x)|2dx-μ∫ℝN‍|vj(x)|2dx-∫ℝN‍|vj(x)|2*(s)dx-∫ℝN‍|u∞(x)|2*(s)dx+∘(1)=〈Iλ′(u∞),u∞〉+∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|vj(x)|2dx-μ∫ℝN|vj(x)|2dx-∫ℝN‍|vj(x)|2*(s)dx+∘(1).
By〈Iλ′(u∞),u∞〉=0and〈Iλ′(vj),vj〉→0, we get
(64)∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy+λ∫ℝN‍A(x)|vj(x)|2dx-μ∫ℝN‍|vj(x)|2dx⟶b,∫ℝN|vj(x)|2*(s)dx⟶b.
As in the proof of Lemma 4 one shows that
(65)∫F|vj(x)|2dx⟶0
asj→∞, whereF={x∈ℝN:A(x)≤M0}. LetFc=ℝN∖F. Then, by (34),
(66)Ss∥vj∥L2*(s)(ℝN)2≤∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy≤∫ℝ2N‍|vj(x)-vj(y)|2|x-y|N+2sdxdy+∫Fc‍(λA(x)-μ)|vj(x)|2dx≤〈(Aλ-μ)vj,vj〉+μ∫F‍|vj(x)|2dx=〈(Aλ-μ)vj,vj〉+∘(1).
Passing to the limit yieldsb≥Ssb2/2*(s). Eitherb=0orb≥SsN/2s. Ifb=0, the proof is complete. Assumingb≥SsN/2s, we obtain from Step3, (45), and (62) that
(67)sNSsN/2s≤(12-12*(s))b=c<sNSsN/2s,
which is a contradiction. Thusb=0, and
(68)∥uj-u∞∥λ⟶0
asj→+∞. This ends the proof of Proposition 8.We have finished the proof of Theorem1 by Proposition 8.


## 4. The proof of Theorem2

Proof of Theorem2.
Let(un)be a sequence of solutions of (4) such that0<μ<μ1(Ω),λn→∞,andNIλn(un)=〈(Aλn-μ)un,un〉→Nc<sSsN/2s. Then, by Lemma 4, there is au∈H0s(Ω)such that, up to a subsequence,un→uinE. Byunthat is a solution of (4), we have
(69)∫ℝ2N‍(un(x)-un(y))(φ(x)-φ(y))|x-y|N+2sdxdy+λn∫ℝN‍A(x)un(x)φ(x)dx-μ∫ℝN‍un(x)φ(x)dx=∫ℝN‍|un(x)|2*(s)-2un(x)φ(x)dx
for anyφ∈E. Ifφ∈H0s(Ω),thenλn∫ℝN‍A(x)un(x)φ(x)dx=0for alln, so lettingn→∞we obtain
(70)∫ℝ2N‍(u(x)-u(y))(φ(x)-φ(y))|x-y|N+2sdxdy-μ∫ℝN‍u(x)φ(x)dx=∫ℝN‍|u(x)|2*(s)-2u(x)φ(x)dx
for anyφ∈H0s(Ω). So,uis a solution of (13). We writevn:=un-u. Then,vn→0inL2(ℝN).
SinceA(x)=0forx∈Ω, we get
(71)〈(Aλn-μ)un,un〉=〈(A0-μ)u,u〉+〈(Aλn-μ)vn,vn〉.
Byvn→0inEand the Brézis-Lieb Lemma, we have
(72)∫ℝN|un(x)|2*(s)dx=∫ℝN‍|u(x)|2*(s)dx+∫ℝN‍|vn(x)|2*(s)dx+∘(1).
So, we can get
(73)〈(Aλn-μ)vn,vn〉-∫ℝN‍|vn(x)|2*(s)dx=∘(1).
We claim that∫ℝN‍|vn(x)|2*(s)dx→0. Assume∫ℝN‍|vn(x)|2*(s)dx→b>0. Then,
(74)Ss(∫ℝN‍|vn(x)|2*(s)dx)2/2*(s)≤∫ℝ2N‍|vn(x)-vn(y)|2|x-y|N+2sdxdy≤〈(Aλn-μ)vn,vn〉=∫ℝN‍|vn(x)|2*(s)dx+∘(1),
thanks to (73). It follows that
(75)Ss≤(∫ℝN‍|vn(x)|2*(s)dx)(2*(s)-2)/2*(s)+∘(1)≤(∫ℝN‍|un(x)|2*(s)dx)(2*(s)-2)/2*(s)+∘(1),SsN/2s≤limn→∞∫ℝN‍|un(x)|2*(s)dx=limn→∞〈(Aλn-μ)un,un〉=Nc<sSsN/2s.
This is a contradiction. Thus∫ℝN‍|vn(x)|2*(s)dx→0and〈(Aλn-μ)vn,vn〉→0, by (73). Hence, by (71)
(76)limn→∞〈(Aλn-μ)un,un〉=〈(A0-μ)u,u〉.
Sinceun=vninℝN∖ΩandA(x)=0forx∈Ω,
(77)∫ℝN‍A(x)|un(x)|2dx≤∫ℝN‍λnA(x)|un(x)|2dx=∫ℝN‍λnA(x)|vn(x)|2dx≤〈(Aλn-μ)vn,vn〉.
Therefore,∫ℝNA(x)|un(x)|2dx→0and (76) implies thatun→uinE.



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*Source: 638425-2013-12-05.xml*