±∞
⊆
∂ n
13
Annali di Matematica (2012) 191:395–430 DOI 10.1007/s10231-011-0188-z
Neumann problems resonant at zero and infinity
LeszekGasin´ski·NikolaosS.Papageorgiou
Received: 8 December 2010 / Accepted: 20 January 2011 / Published online: 16 February 2011
© The Author(s) 2011. This article is published with open access at Springerlink.com
AbstractWeconsider a semilinear Neumann problem with a reaction which is resonant at both zero and . Using a combination of methods from critical point theory, together withtruncationtechniques,theuseofupper–lowersolutionsandoftheMorsetheory(critical
groups),weshowthattheproblemhasatleastfivenontrivialsmoothsolutions,fourofwhich haveconstantsign(twopositiveandtwonegative).
KeywordsResonance at zero and infinity·Critical point theory·Morse theory·
Truncation techniques·Regularity theory·Multiple solutions·Solutions of constant sign Mathematics Subject Classification (2000)35J20·35J60·58E05
1 Introduction
LetQ RNbe a bounded domain with aC2-boundary∂Q.In this paper, westudythefollowing semilinear Neumannproblem:
−1'.u(z)=f(z,u(z))inQ,
∂u=0 on∂Q. (1.1)
This research has been partially supported by the Ministry of Science and Higher Education of Poland under Grant no. N201 542438.
L.Gasin´ski(
B
)Institute of Computer Science, Jagiellonian University, ul. Łojasiewicza 6, 30-348 Kraków, Poland e-mail:Leszek.Gasinski@ii.uj.edu.pl
N. S. Papageorgiou
Department of Mathematics, National Technical University, Zografou Campus, 15780 Athens, Greece
e-mail:npapg@math.ntua.gr r
±∞
1−
→
∂ n
=
·
=
· ±∞ ·
= ·
{
39 L.Gasin´ski,N.S.Papageorgiou
Here,f (z,ζ)isameasurablefunctionwhichisC1intheζ-variable.Theaimofthisworkis to prove a multiplicity theorem when resonance occurs at both zero and. S u c h p r o b l e m s h a v e b e e n s t u d i e d e x t e n s i v e l y i n t h e c o n t e x t o f D i r i c h l e t
e q u a t i o n s . I n t h i s d i r e c t i o n , w e
m e n t i o n theworksofCostaandSilva[8],HiranoandNishimura[14],Landesmanetal.[20], Liang and Su [22], Liu [24], Li and Su [25], Li and Zou [26], Su andTang[34], Zou [37], andZouandLiu[38].ForthecorrespondingNeumannproblem,thebibliographyisnotthat rich. There have been some existence and multiplicity results for resonant semilinear Neu- mann problems.Wemention the works of Filippakis and Papageorgiou [10], Iannacci and Nkashama[15,16],Kuo[19],Li[21],LiandLi[23],Mawhin[27],Mawhinetal.[28],Qian [33], andTangandWu[35]. Iannacci and Nkashama [15] and Kuo [19] use variants of the well- knownLandesman–Lazercondition.IannacciandNkashama[16]useasigncondition, while Mawhin [27]
and Mawhin et al. [28] use a monotonicity condition on the functionζf(z,ζ).All the aforementioned works prove existence theorems, but do not address the question of multiplicity of the nontrivial solutions. Multiplicity results can be found in theworksofFilippakisandPapageorgiou[10],Li[21],LiandLi[23],Qian[33],andTangandWu[35].
InLi[21],LiandLi[23],andQian[33],theauthorsdealwithequationsofthe form
− 1'.u(z)+au(z)=f(z,u(z)) inQ,
∂u=0 on∂Q. (1.2)
In(1.2),thepresenceintheleft-handsideofthetermau(witha>0)facilitatestheanal-
ysisoftheequation,sinceinthiscasethedifferentialoperatorin(1.2)iscoercive.Whena0 , t h i s i s n o l o n g e r t r u e ( r e c a l l t h a t t h e P o i n c a r é i n e q u a l i t y f a i l s i n t h e S o b o l e v s p a c e1
H(Q)). Li [21] and Li and Li [23] produce an infinity of nodal (i.e., sign changing) solu- tions, by assuming an oscillatory behavior for the reactionf(z, ). Their approach uses critical point theory, Leray–Schauder degree on order intervals, and Morse theory. Qian [33] deals with equations which are superlinear at , using the so-called Jeanjean condi- tion. He produces a sequence of nodal solutions assuming a symmetry condition onf(z,) (namely thatf(z,)is odd). His arguments are based on the critical point theory. Finally, Filippakis and Papageorgiou [10] andTangandWu[35] assume thata0. Filippakis and Papageorgiou [10] permit the resonance at zero to be only with respect to the princi- pal eigenvalueλ00, impose a global sign condition onf(z,), and produce only three nontrivial smooth solutions.TangandWu[35] employ an anticoercivity condition on the potential
ζ
F(z,ζ)=
0 f(z,s)ds
and using the local linking theorem (see e.g., [12, p. 665]), they establish the existence of two nontrivial solutions.
In this paper, using a combination of variational methods based on the critical pointthe- ory,withMorsetheory(criticalgroups),weestablishtheexistenceofatleastfivenontrivial smooth solutions for problem (1.1), four of which have constant sign (two positive andtwo negative).
Inthenextsection,fortheconvenienceofthereader,werecallsomeofthemainmathe- maticaltoolsthatwewilluseintheanalysisofproblem(1.1).
r
∈
(· ·⊕
∈
Kϕc= u∈Kϕ:ϕ(u)=c.
ϕ
⊆
(
∈ −∞
2 Mathematicalbackground
LetXbe a Banach space and letX∗be its topological dual. By,we denote the duality brackets for the pair(X∗,X). LetϕC1(X). We say thatϕsatisfies the Cerami condition if the following is true:
“Every sequence{xn}n�1⊆X, such that
|ϕ(xn)|�M1and(1+xn)ϕ (xn)−→0 inX∗, for someM1>0, admits a strongly convergent subsequence.”
Using this compactness type condition, we can have the following minimax characteriza- tion of certain critical values of aC1-functional. The result is known in the literature asthe“mountain passtheorem”.
Theorem2.1I f XisaBanachspace,ϕ∈C1(X)satisfiestheCeramicondition,u0,u1∈Xaresuch thatu0−u1>r>0,
max{ϕ(u0), ϕ(u1)}<inf{ϕ(u):u−u0=r} =ηr, c=inf maxϕ (γ (t)) ,
wher e
γ∈0t1
= {γ∈C([0,1];X):γ (0)=u0, γ (1)=u1}, then c�ηrand c is a critical value ofϕ.
Throughoutthiswork,wewillusethefollowingnotation.Letϕ C1(X)andletc R.Weset ϕc=
{u∈X:ϕ(u)�c},Kϕ=f fu∈
X:ϕ( u)=0,
Let(Y1,Y2)be a topological pair andY1Y2X. For everyk�0, byHk(Y2,Y1), we denotethekthrelativesingularhomologygroupwithintegercoefficientsforthepair(Y1,Y2). Recall that for all integersk<0, we haveHk(Y2,Y1)=0. The critical groups ofϕat an isolated critical pointu∈Kcare definedby
Ck(ϕ,u)=Hkϕc∩U,ϕc∩U\{u}
∀k�0,whereUis a neighborhood ofx, suchthat Kϕ∩ϕc∩U= {u}
(see[7,29]).Theexcisionpropertyofsingularhomologytheoryimpliesthattheabovedefi- nitionofcriticalgroupsisindependentoftheparticularchoiceoftheneighborhoodU.
Supposethatϕ C1(X)satisfies the Cerami condition and infϕ(Kϕ)> .Choose c<infϕ(Kϕ). The critical groups ofϕat infinity are defined by
Ck(ϕ,∞)=Hk(X, ϕc)∀k�0
∈ ∈
∈ = ∈
∈
=r
n ∂
n
n n
2 2
n
n
f
f
(see[5]).Thedeformationtheorem(seee.g.,[12,p.636])impliesthattheabovedefinition isindependentofthechoiceoftheparticularlevelc<infϕ(Kϕ).IfKϕisfinite,thenweset
M(t,u)= rankCk(ϕ,u)tk∀t∈R,u∈Kϕ,
k�0
P(t,∞)= rankCk(ϕ,∞)tk∀t∈R.
k�0
Using these quantities, we have the Morse relation
M(t,u)=P(t,∞)+(1+t)Q(t), (2.1)
u∈Kϕ
where
Q(t)=
βktkk�0
is a formal series int∈Rwith nonnegative integer coefficients (see[7,29]). 2 LetX Hbe a Hilbert space,uH,Ua neighborhood ofuinH, andϕC(U).IfuKϕ, then the Morse index ofuis defined to be the supremum of the dimensions of the vector subspaces ofHon whichϕ(u)is negative definite.Wesay thatuKϕis nondegenerate, ifϕ(u)is invertible. Suppose thatuKϕis a nondegenerate critical point with Morse indexm.Then
where
Ck(ϕ,u)=δk,mZ∀k�0, δk,m 1 ifk=m,
0 ifk/=m.
In the analysis of problem (1.1), we will use the following two “natural” spaces:
C1(Q)=ru∈C1(Q):∂u
(z)=0 on∂Q (wheren(·)denotestheoutwardunitnormalon∂Q)and
H1(Q)=C1(Q)·, where·denotes the usual Sobolev norm ofH1(Q), i.e., u2=u2+ ∇u2∀u∈H1(Q).
The spaceC1(Q)is an ordered Banach space, with positive cone C+=u∈C1(Q):u(z)�0 for allz∈Q . This cone has a nonempty interior, given by
intC+=u∈C+:u(z)>0 for allz∈Q .
For a large class ofC1-functionals, we can identify theC1(Q)andH1(Q)local minimizers.
n n
More precisely, letg0:Q×R−→Rbe a Carathéodory function, such that
|g0(z,ζ)|�a0(z)+c0|ζ|r−1for almost allz∈Q,allζ∈R,
N−2
{
1 n
n Cn(Q)
n
0 0
n +∞
∈ +
/=
r
�
∂ n
�
�=� �
= f�
�
� ⊥
witha0∈L∞(Q)+,c0>0 and 1<r<
2∗=
r2 N ifN�3, +∞ ifN=1,2 (subcriticalgrowthforg0(z,·)).Weset
ζ
G0(z,ζ)=
0 g0(z,s)ds and consider theC1-functionalψ0:H1(Q)−→R, defined by
ψ0(u)= ∇u2−{G0(z,u(z))dz∀u∈H1(Q).
2 2 Q n
Proposition 2.2If u0∈H1(Q)is a local C1(Q)-minimizer ofψ0, i.e., there exists r0>0,
n n
such that
ψ0(u0)�ψ0(u0+h)∀h∈C1(Q),h1 �r0,
then u0∈C1(Q)and it is a local H1(Q)-minimizer ofψ0, i.e., there exists r1>0, such that
n n
ψ0(u0)�ψ0(u0+h)∀h∈H1(Q),h�r1.
Remark 2.3For the “Dirichlet” spaceH1(Q), this result was first proved by Brezisand Nirenberg [6] and was extended to the spacesW1,p(Q)(with 1<p<+∞) by Garcia Azoreroetal.[11](seealso[13]).Forthe“Neumann”spacesW1,p(Q)(1<p<) , the
resultcanbefoundinMotreanuetal.[30](forsmoothfunctionalsψ0)andinIannizzotto andPapageorgiou[17]
(fornonsmoothfunctionalsψ0).Asimplifiedproofoftheresultfor
more general operators than thep-Laplacian can be found in the recent work of Motreanu and Papageorgiou [31].
Next,werecallsomebasicfactsaboutthespectrumofthenegativeNeumannLaplacian. So, letmL∞(Q),m0 (a weight function), and consider the following weighted linear eigenvalueproblem:
−1'.u(z)=λm(z)u(z)inQ,
∂u=0 on∂Q. (2.2)
Evidently a necessary condition forλ∈Rto be an eigenvalue is thatλ�0.Moreover,λ0λ0(m)0 is an eigenvalue of (2.2) with corresponding eigenspaceR(the space of constant functions). Using the spectral theorem for compact operators, we can show that problem (2.2) has a sequenceλk(m)k�0of distinct eigenvalues, such thatλk(m)
−→+∞ask→+∞.Ifm≡1,wewrite�λ(k(1)=�λkforallk�0.
Foreveryintegerk�0,byE �λk(m), wedenotetheeigenspacecorrespondingtothe eigenvalue�λk(m).Theregularity(theory(seee.g.,[12])impliesthatE(
�λk(m)⊆ C1(Q).
Moreover,weknowthateachE�λk(m)hasthe“uniquecontinuationproperty”,namelyif u∈E(
�λk(m)v
anishesonasetofpositivemeasure,thenu(z)=0forallz∈Q.Weset
Hi=E(
�λk(m)and
Hi=Hi=
i
k�i+1
E(
�λk(m).
r∇u
λk(m)=max r
Qmu
2dz2 :u∈Hk,u/=0
2
�
(
�then
�
��
�
η�
∈ � ∈
/=�
2
f� �
=
N
Neumann problems resonant at zero and infinity
401
Wehavethefollowingvariationalcharacterizationfortheeigenvalues�λk(m):
0=�λ0(m)=min 22 :u∈H(Q),u /=0l
1 n
(2.3)
and fork�1, wehave
r
Qmudz
�
r∇u 2 l
r∇�u2 �
l
=minr
mu2dz:�u∈Hk,�u/=0. (2.4)
In(2.3),theminimumisattainedonE(
�λk(m)=
R,whilein(2.4)themaximumand minimumarerealizedonE�λk(m),k�1.
As a consequence of these variational characterizations and of the unique continuation property, we have the following useful facts (see e.g., [10]).
Proposition2.4Ifm,m∈L∞(Q)+\{0},m(z)�m�(z)foralmostallz∈Qandm/=m�, λk(m) < λk(m)∀k�0.
Proposition 2.5(a)If k�−1is aninteger,η∈L∞(Q)+,η(z)�λk+1for almostallz ∈Qandη/=�λk+1,thenthereexistsξ0>0,suchthat
∇�u2−{ u2d
z�ξ0�u2∀�u∈H�k.
Q
(b)If k�0is aninteger,η L∞(Q)+,η(z)�λkforalmostallz Qandη λk,thenthereexistsξ1>0,suchthat
∇u2−{ηu2dz�−ξ1u2∀u∈Hk.
Q
From the eigenvaluesλk(m)k�0only the firstoneλ0(m) 0 hasc o n s t a n t s i g n eigenfunction. All the other eigenvalues have nodal (i.e., sign changing) eigenfunctions.In
what followsu0denotes theL2-normalized, positive principal eigenfunction, i.e.,u0=11
(hereafter,by�| ·|NwedenotetheLebesguemeasureonRN).
�
|Q|2
Thenextresult,duetoLiangandSu[22](seealso[18]foranextensiontoBanachspaces),
ishelpfulincomputingcriticalgroups.ItisageneralizationofanearlierresultofPereraandSchechter[32].
2
Q
2
{ }∈[ ]
⊆
400 L.Gasin´ski,N.S.Papageorgiou
Proposition 2.6If H is a Hilbert space, ht t0,1C1(H)is a family of functionals, suchthat(ht)and∂thtare both locally Lipschitz, h0and h1satisfy the Cerami condition andthere exist a∈Randδ >0, such that
ht(u)�a⊆⇒(
(1+u)ht(u)∗�δfor all t∈ [0,1 ], (2.5) the
n Ck(h0,∞)=Ck(h1,∞)∀k�0.
p p
+|| ∈ ∈ ∈
+
ζ ζ→0 ζ
Remark 2.7Note that, if there existsR>0, such that inff
(1+u)ht(u)∗:t∈[0,1],u>R>0 and
inf{ht(u):t∈ [0,1],u�R}>−∞, then (2.5) holds.
In the sequel, we will use the notationr±=max{±r,0}for allr∈R. Also, by·we denote the norm of the Sobolev spaceH1(Q)and by|·|Nthe Lebesgue measure onRN. Finally by·p(1<p<∞), we denote the norm ofLp(Q)ofLp(Q;R)andp>1 is
the conjugate exponent ofp>1, i.e.,1+1=1.
3 The Ceramicondition
The hypotheses on the reaction termfare the following:
Hf:f:Q×R−→Ris a measurable function, such that for almost allz∈Q, we have f(z,0)=0,f(z,·)∈C1(R)and
(i) fζ(z,ζ)�a(z) cζr−2for
almostallz Q,allζ Rwitha L∞(Q),c>0and2�r�2∗;
(ii) thereexistintegeri�1,α∈(0,1),andη∞∈L∞(Q)withη∞(z)�0foralmostall z∈Q, η∞/=0, such that
lim
|ζ|→+∞
f(z,ζ)
ζ =�λi
uniformlyforalmostallz∈Qand,iff∞(z, ζ )=f(z, ζ )−�λiζ,then lim f∞( z ,ζ)
=0 and lim supf∞( z ,ζ)ζ �η∞(z)
|ζ|→+∞ |ζ|α |ζ|→+∞ |ζ|2α
uniformly for almost allz∈Q;
(iii) there exist integerm�1,m/=i,β>1, andη0∈L allz∈Q, η0/=0, such that
∞(Q)withη0(z)�0 for almost f ( z ,ζ)
f(z,0)=lim =�λm
uniformlyforalmostallz∈Qandiff0(z, ζ )=f(z, ζ )−�λmζ,then limf0( z ,ζ)
=0 and lim supf0( z ,ζ)ζ
�η0(z)
ζ→0 |ζ|β ζ→0 |ζ|2β
uniformly for almost allz∈Q;
(iv) thereexistnumbersa−<0<a+,suchthat
f(z,a+)�0�f(z,a−)for almost allz∈Q andf(·,a−)/=0,f(·,a+)/=0.
40 L.Gasin´ski,N.S.Papageorgiou
=r q−2
( ( �
α
∈
q−1,2q ,β∈ r,
2r−1 ,a−= −1, anda+=1.
n
1
ϕ (u)=A(u)−Nf(u)∀u∈H1(Q),
n
n
n
n
n n
�
n n
n
� �
Remark3.1H y p o t h e s i s Hf(ii)impliesthattheproblemisresonantatinfinity,whilehypoth-
esisHf(iii)implies that the problem is resonant at zero. So, we have a kind of “double resonance”.
Example 3.2The following functionfsatisfies hypothesesHf(for the sake of simplicity we drop thez-dependence):
f(ζ) λmζ−ξ|ζ|r−2ζ if|ζ|�1,
�λiζ−�ξ|ζ| ζ if|ζ|>1,
with1<q< 2<r<+∞,ξ>�λmand�ξ= ξ+�λi−�λm>0.Forthisexample,wetake
Letϕ:H1(Q)−→Rbe the energy functional for problem (1.1), defined by ϕ(u)= ∇u2−{F(z,u(z))dz∀u∈H1(Q).
2 2 Q n
We know thatϕ∈C2(
H1(Q). Moreover whereA∈L( n
H1(Q),H1(Q)∗is defined by
n n
(A(u),y⊕={(∇u,∇y)RNdz∀u,y∈H1(Q)
Q
andNf(y)(·)=f( ·,y(·))forally∈H1(Q).Also
(ϕ (u)y, v={(∇y,∇v)RNdz−{fζ(z,u(z))yvdz∀u,y,v∈H1(Q).
Q Q
Note thatϕ (u)∈L(
H1(Q),H1(Q)∗is a Fredholm operator.
Usingtheeigenvalueλi>0,wecanhavethefollowingorthogonaldirectsumdecompo- sitionoftheSobolevspaceH1(Q):
H1(Q)=Hi−1⊕E(�λi)⊕H�i+1, where
i−1
Hi−1= E(λk)and
k=0
Then for everyu∈H1(Q), we have
Hi+1=
k�i+
1
E(�λk).
withu∈Hi−1,u0 u=u+u0+�u,
∈E(�λi),�u∈H�i+1.Thisdecompositionis unique.
Proposition 3.3If{un}n�1⊆H1(Q)is a sequence, such that
n u n + u n 1
Neumann problems resonant at zero and
infinity 40
un−→+∞and�
un −→0in Hn(Q),
{
�
�
�
� f
2
0 n
0 n
n
un
the n
lim sup
n→+∞Q
f∞( z , u n) u n
dz<0.
un2α
ProofFromBartoloetal.
[3],weknowthatforagivenε>0,wecanfindm1(ε)>0smallenoughandm2(ε)>0largeenough,suchthat fz∈Q:u0(z)<m1u0N< ε∀u0∈E(�λi) (3.1)
an d
|{z∈Q:|�u(z)+u(z)|>m2�u+u}|N 2
α 1 1
<m1−αε1−α�ε∀u∈Hi+1,u∈Hi−1. (3.2) For everyn�1, we introduce the followingsets
D1n=z∈Q:u0(z)�m1u0,
D2n= {z∈Q: |u(z)+u(z)|�m2u+u}.
From (3.1) and (3.2), it follows that
|Q\D1n|N< ε,|Q\D2n|N<ε an
d
|D1n∩D2n|N�|Q1|N−|Q\D2n|N�|Q|N−2ε. (3.3) Choosingε∈(
0,1|Q|N, we see that
|D1n∩D2n|>0 and soD1n∩D2n/= ∅. Letz∈D1n∩D2n. Then
| u n ( z ) | 0
=|un(z)+�un(z)+un(z)|
�|u(z)|
−|� u n ( z ) + u n ( z ) |
un un
�m1u0
−m 2 � u n+u n un un n
un
Next letz∈D2n\D1n. Then
un . (3.4)
| u n ( z ) | 0
=|un(z)+�un(z)+un(z)|
�|u(z)|
−|� u n ( z ) + u n ( z ) | un
m1u0 un
m 2 � u n+u n
un
un un
un
< − . (3.5)
1
By virtue of hypothesesHf(i)and(ii), we can findc1=c1(ε) >0, such that f∞(z,ζ)ζ�(
η∞(z)+m2αε|ζ|2α+c1for almost allz∈Q,allζ∈R. (3.6)
{
{ I \
+ ||N
1
I u\ {
{ I \
+ ||N
{
{ I \
+ ||N
1
Iu\
2
I� \
I \
+ ||N
{ α
{ 2 �2 {
Then, using (3.6), wehave
D1n∩D2n
f∞( z , u n) u n
un2α dz
�
D1n∩D2n (η∞(z)+ε) | u n|2α un
dz c1
Q un2α
�m2α 0n 2α
un η∞(z)dz
D1n∩D2n 2α
I� u n+u n\ 2α{
−m2 un η∞(z)dz
D1n∩D2n
+ε
D1n∩D2n | u n|2α un
dz c1
Q (3.7)
un2α
(sinceη∞�0; seeHf(ii)). Also, hypothesesHf(i)and(ii), imply that
|f∞(z,ζ)ζ|�c2
(|ζ|2α+1for almost allz∈Q,allζ∈R, (3.8)
withc2>0. Hence, using (3.8), wehave
D2n\D1n
f∞( z , u n) u n
un2α dz
�c2
D2n\D1n
| u n|2α un
dz c2
Q un2α
�c2m2α 0 2αn
un
ε+c2m2α un+ u n2α
un |Q|N+ c 2
un2α |Q|N (3.9)
(see (3.5) and (3.3)). Moreover, we have f ∞ ( z , u n ) u n
d z c
unα | u n|2α
un
dz c2
un2α Q
Q\D2n Q\D2n
(see(3.8)).Notethat(u
n (·)2α∈L1(Q).So,byvirtueofHölderinequalityand(3.3),we hav
e
Q\D2n un
I \
{
I \
| un|
2α
un
dz=
Q
χQ\D2n | u n|2α un
Iun2\2α
1−α 2α
� un |Q\D2n|N �m1ε . (3.10) d
{
{ {
+ {
1
I u\ {
1n
2
I� \
1
{ I \
+ |Q|N
1
Iu\
2
I� \
2 α
{
n
un
un2α 1
� u n + n u 1 1 r
to finish theproof. un n
�c+1 Q
n f u
13
�therefore, finally we have f∞( z , u n) u n
un2α dz
Q
=
D1n∩D2n
+
Q\D2n
f∞( z , u n) u n
dzun2
α
D2n\D1n
f∞( z , u n) u n un2α dz
f∞( z , u n) u n
un2α dz
�m2α 0n 2α
un D∩D η∞(z)dz+m
2α un+ u n2α
un η∞1
+m2αε
D1n∩D2n
| u n|2α
un dz c1
un2α
+c2m2α 0 2αn
un
ε+c2m2α un+ u n2α
un |Q|N
c 2
+ |Q|N+mε (3.11)
un2 α 1
(see (3.7), (3.9) and (3.10)). Note that
χD1n∩D2n(z)−→χQ(z)=1 almost everywhere onQasε\.0 (see (3.3)).
So, if in (3.11) we pass to the limit asn→ +∞, wehave f∞( z , u n) u n
un2α dz
Iu0\2α{ I
u n+� u n\2α un
c 1 2α
I� u n+u n\2α
c 2
+un2α|Q|N+c2m2 so
un |Q|N+
un2α|Q|N, limsup{
n→+∞ f ∞ ( z , u n ) u n
dz�m2α⎛⎝
{ η∞(z)dz+ε(�c+1)⎞⎠
Q Q
for somec>0. (recallthat� −→0 inH(Q)). Let us chooseε<− η∞(z)dz ForR>0 andϑ∈(0,1), we introduce the set
Q
�m2α1 η∞(z)dz+m2α
2
η∞1 Q
13
C∞(R, ϑ, α)=u∈Hn 1(Q):u�R,u+u�ϑuα.
Proposition 3.4If hypotheses Hf(i)and(ii)hold, then there exist R>0,ϑ∈(0,1)and
�δ>0,suchthat
(ϕ(
u0),u0��δ∀u∈C∞(R,ϑ,α).
� n
unα
{
f∞(z,un)u
Q un2α
{ �
dz+c4
unα
un2α ∀n�1, (3.17)
I \
ProofWe argue indirectly. So, suppose that the proposition is not true. Then for anyϑ=
1 1
�δ=n,
n�1,wecanfindun∈Hn(Q),suchthat un�n,un+un�
From (3.12), we see that
unαand(
ϕ (un),u0< ∀n�1. (3.12)
an d
un−→+∞, �u n+u n −→0 (3.13)
(ϕ (un),u0= −{f∞(z,un)u0dz< (3.14)
n Q n n
02 02
(since∇un2=�λiun2foralln�1).From(3.14),itfollowsthat
lim inf
n�+∞
0ndz�0. (3.15)
un2α Q
On the other hand, by virtue of hypothesesHf(i)and(ii), for a givenε >0, we can find c3=c3(ε) >0, such that
|f∞(z,ζ)|�ε|ζ|α+c3for almost allz∈Q,allζ∈R. (3.16) Then, using (3.16), wehave
{ f ∞ ( z , u n)( � u n+u n ) dz
� (ε|un|α+cun32α)|un+un|dz {I| u n|\α|
� u + u |
Q un
� u n+u n
Q un
for somec4>0. Note that u n ( · ) α
un
2
∈Lα(Q)and I 2\
α 2
=2−α�2
2 N
=N−2ifN�3.
2
Hence,un+un∈L2−α(Q)and we can apply Hölder inequality and obtain
1 1
1
� ε
n n
Q
∗
dz�c5
unα
unα ∀n�1, (3.18)
un2α unα
� {I| u n|\α|
� u n+u n| � u n+u n
for somec5>0. Using (3.18) in (3.17), we have { f ∞ ( z , u n)( � u n+u n )
dz �c6� u n+u n ∀n�n0,
Q un
Q un
Q
{
−{
{
=
⎛⎝{
�
⎞⎠
ndz {
�
=
13
for somen0�1 andc6>0, so lim
n→+∞
(see (3.13)). Therefore
f ∞ ( z , u n )( u n+ u n )
d z 0 (3.19)
un2α
li m su p
n
→ +
∞
Q
f
∞
( z , u
n
) u
0
u
n 2 α
lim
su p
n
→ +
∞
Q
f∞( z , u n) u n
d z un2α
f∞( z , u n)( u n+ u n) un2α dz
� l i m s u p
n→+∞Q
f ∞ ( z , u n ) u n dz<0 (3.20)
un2α
(see(3.19)andProposition3.3).Comparing(3.15)and(3.26),w ereachacontradiction.Thisprovestheproposition.
nu
Using this proposition, we can now establish the Cerami condition for the energy func-tionalϕ.
13
{
2
dz�ε, 1+un n
{}
→ +∞
n
n n
Proposition 3.5If hypotheses Hf(i)and(ii)hold, thenϕsatisfies the Cerami condition.
ProofLet{un}n�1⊆H1(Q)be a
sequence, such
that{ϕ(un)}n�1⊆Ris bounded and
(1+un)ϕ (un)−→0 inH1(Q)∗.
(3.21)
Weshow that the
sequenceunn�1H1(Q)is
bounded.Weargue indirectly.
So, suppose that by passing to a suitable subsequence if necessary, wehaveun
.Notethat(3.21)implies (ϕ (un),h�ε
nh
∀h∈H1 (Q), (3.22)
withεn\ .0.In(3.22),wechooseh=
�un∈Hn(Q)andexploitingtheorth ogonalityofthe
component spaces, we have (ϕ(
un),�un
=∇�un 2−�λi�un2− {f∞(z,un)�un n
2 2
Q
so
ξ0�un�εn+ {
�εn+
Q
f∞(z,un)�un
(ε|un|α+c3
|�un| dz∀n�1 (3.23)
(see Proposition2.5(a) and (3.16)). Since
α 2
I2\ 2
∗
d
1
13
|un|
∈Lα(Q)a n d
α
=2−α�2,
∀n�1. (3.24)
�
ε +
unα un2 α
n
un2α+c7
{
n n n�
1 un
�
2
then from (3.23) and Hölder inequality, we have ξ0�un�εn+c7(
εun+1�
un∀ n�1, for somec7>0, so
I� u n\ 2
εn I�
u n � u n\
WeclaimthatthesequenceJ μn=�un
α
1 isbounded.Indeed,ifμn→+∞(atleast for a subsequence), then dividing (3.24) withμ2, we obtain
c7 1
ξ0�ε+ ε +c7 ,
n μn μn
withεn\.0.Passingtothelimitasn→+∞,weobtainξ0�0,acontradiction.Hence,
the sequence{μn}n�1is bounded and we may assume thatμn−→μ�0. Passing to the limit asn→ +∞in (3.24), we obtain
ξ0μ2�εc7μ so
ξ0μ�εc7.
Sinceε >0 was arbitrary, we letε\.0, to conclude thatμ=0. Therefore, un
u�nα −→0 inH1(Q). (3.25)
Next in (3.22), we chooseh= −un∈Hi−1. Then reasoning as above, we obtain
−( ϕ(
un),un
=−∇un2+�λiun2+{f∞(z,un)undz�εn,
2 2
Q
so using Proposition2.5(b), we have ξ1un2�εn+
Q |f∞(z,un)| |un|dz∀n�1.
Using also (3.16) and (3.24), we have Iun\2
εn
Iu
n u n \ +
andso
ξ unα �
un2α+c7
εunα un2 α ∀n�1
un
unα −→0 inH1(Q) (3.26)
(as before). LetR>0,ϑ∈(0,1)andδ >0 be as postulated in Proposition3.4. Then from (3.25), (3.26) and sinceun−→ +∞, we have thatun∈C∞(R, ϑ, α)for alln�n0and
unα
ξ0
n
so
so (
ϕ(
un),u0��δ∀n�n0 (3.27)
(ϕ (un),u0�εn, (3.28)
n �
n
un−→uinH1(Q) (3.29)
n
2 2
w
n
n
�
n n
n
� �
Hf(iii), we can have a result analogous to Proposition3.3.
(see Proposition3.4). If in (3.22), we chooseh=u0∈E(λi), then
n
withεn\.0. Comparing (3.27) and (3.28), we reach a contradiction. This proves that the sequence{un}n�1⊆H1(Q)is bounded. So, we may assume that
w
n
un−→ui n L2(Q). (3.30)
In (3.22), we chooseh=un−u∈H1(Q). We have
(A(un),un−u⊕−{f(z,un)(un−u)dz�εnu n−u ,
Q
so
lim
r→+∞(A(un),un−u⊕=0
1+un
(see (3.29)),so
∇un2 −→ ∇u2
(sinceA(un)−→A(u)inH1(Q)∗; see (3.29)). From the Kadec–Klee property of Hilbert spaces, wehave
so
∇un−→∇uinL2(Q;RN),
un−→uinH1(Q)
(see(3.29)).Therefore,ϕsatisfiestheCeramicondition. nu
4 Criticalgroups
Usingtheeigenvalueλm>0fromhypothesisHf(iii),wecanhavethefollowingorthogonal directsumdecompositionofH1(Q):
where
H1(Q)=Hm−1⊕E(�λm)⊕H�m+1,
m−1
Hm−1= E(λk)and
k=0
Then for everyu∈H1(Q), we have
Hm+1=
k�m+
1
E(�λk).
withu∈Hm−1,u0 u=u+u0+
�u,
∈E(�λm),�u∈H�m+1andthedecompositionisunique.Usinghypothesis
n
{
�
�
n n
�
�
2
0 n
u−
�
0
n
n
un
(
410 L.Gasin´ski,N.S.Papageorgiou
Proposition 4.1If{un}n�1⊆H1(Q)is a sequence, such that u n + u n 1
then
un−→0and�
un −→0in Hn(Q),
lim sup
n→+∞Q
f0( z , u n) u n
dz<0.
un2β
ProofI t isclearfromhypothesisHf(iii)thatwecantakeβ>1small,suchthat2∗>2β.AsintheproofofProp osition3.3,fromBartoloetal.[3],weknowthatforagivenε>0,
wecanfindm1(ε)>0smallenoughandm2(ε)>0largeenough,suchthat fz∈Q:u0(z)<m1u0N< ε∀u0∈
E(�λm) (4.1)
and
|{z∈Q:|�u(z)+u(z)|>m2�u+u}|N 2∗
2∗−2 β
2β
<m1 ε�ε∀u∈Hm+1,u∈ Hm−1. (4.2)
For everyn�1, we introduce the followingsets:
Q1n=f
z∈Q:u0(z)�m1u0,
Q2n= {z∈Q: |un(z)+un(z)|�m2un+un}.
From (4.1) and (4.2), we have
|Q\Q1n|N< ε,|Q\Q2n|N< ε (4.3) and
|Q1n∩Q1n|N�|Q1n|N− |Q\Q2n|N�|Q|N−2ε. (4.4) Choosingε∈(0,1|Q|N), we see that|Q1n∩Q2n|>0, henceQ1n∩Q2n/=0. Let
z∈Q1n∩Q2n. Then
| u n ( z ) | 0
=|un(z)+�un(z)+un(z)|
�|u(z)|
−|� u n ( z ) − u n ( z ) |
un un un un
�m1 0 n
un m2un+ u n
un .
(4.5) Next, letz∈Q2n\Q1n. Then
| u n ( z ) |
=|un(z)+�un(z)+un(z)|
�|u0(z)|
+| u � n ( z ) + u n ( z ) | un
u0 un
� u n+u n
un
un un
HypothesisHf(iii)implies that we can findδ=δ(ε)>0, such that
<m1 +m
2
. (4.6)
1 Neumann problems resonant at zero and infinity
411 f0(z,ζ)ζ�η0(z)+m2βε|ζ|2βfor almost allz∈Q,all|ζ|�δ. (4.7) On the other hand, by virtue of hypothesisHf(i), we have
f0(z,ζ)ζ�c8|ζ|μfor almost allz∈Q,all|ζ|> δ (4.8)
