亚洲免费av电影一区二区三区,日韩爱爱视频,51精品视频一区二区三区,91视频爱爱,日韩欧美在线播放视频,中文字幕少妇AV,亚洲电影中文字幕,久久久久亚洲av成人网址,久久综合视频网站,国产在线不卡免费播放

        ?

        Solutions to theσk-Loewner-Nirenberg Problem on Annuli are Locally Lipschitz and Not Differentiable

        2021-08-31 00:38:00YanYanLiandLucNguyen
        Journal of Mathematical Study 2021年2期

        YanYan Li and Luc Nguyen

        1 Department of Mathematics,Rutgers University,Hill Center,Busch Campus,110 Frelinghuysen Road,Piscataway,NJ08854,USA;

        2 Mathematical Instituteand St Edmund Hall,University of Oxford,Andrew Wiles Building,Woodstock Road,Oxford OX2 6GG,UK.

        Abstract.We show for k≥2 that the locally Lipschitz viscosity solution to theσk-Loew ner-Nirenberg problem on a given annulus{a<|x|

        Key w ords:σk-Loew ner-Nirenberg problem,σk-Yamabe problem,viscosity solution,regularity,conformal invariance.

        1 Introduction

        Let?be a smooth bounded domain in Rn,n≥3.For a positiveC2functionudef ined on an open subset of Rn,letAudenote its conformal Hessian,namely

        For 1≤k≤n,letσk:Rn→R denotek-th elementary symmetric function

        and letΓkdenote the coneΓk={λ=(λ1,...,λn):σ1(λ)>0,···,σk(λ)>0}.

        In[7,Theorem 1.1],it was shown that theσk-Loewner-Nirenberg problem

        Eq.(1.2)is a fully nonlinear elliptic equation of the kind considered by Caffarelli,Nirenberg and Spruck[3].We recall the following def inition of viscosity solutions which follows Li[20,Def initions 1.1 and 1.1’](see also[19])where viscosity solutions were f irst considered in the study of nonlinear Yamabe problems.

        Let

        Def inition 1.1.Let??Rn bean open set and1≤k≤n.Wesay that an upper semi-continuous(alower semi-continuous)function u:?→(0,∞)isasub-solution(super-solution)to(1.2)in the viscosity sense,if for any x0∈?,?∈C2(?)satisfying(u??)(x0)=0and u??≤0(u??≥0)near x0,thereholds

        Wesay that a positivefunction u∈C0(?)satisf ies(1.2)in theviscosity senseif it is both a sub-and asuper-solution to(1.2)in theviscosity sense.

        Eq.(1.2)satisf ies the following comparison principle,which is a consequence of the principle of propagation of touching points[23,Theorem 3.2]:Ifvandware viscosity sub-solution and super-solution of(1.2)and ifv≤wnear??,thenv≤win?;see[7,Proposition 2.2].The above mentioned uniqueness result for(1.2)-(1.3)is a consequence of this comparison principle and the boundary estimate(1.4).

        In the rest of this introduction,we assume that?is an annulus{a<|x|

        Our f irst result improves on the above non-existence ofC2solutions to(1.2)-(1.3).

        We observe the following result,which is essentially due to Gursky and Viaclovsky[12].We provide in the appendix the detail for the piece which is not directly available from[12].

        Theorem 1.3.Supposethat n≥3,2≤k≤n,and(Mn,g)is a compact Riemannian manifold.If λ(?Ag)∈Γk on M,then(1.7)hasa Lipschitz viscosity solution.

        Here viscosity solution is def ined analogously as in Def inition 1.1.

        Wesay that a positive function u∈C0(M)satisf ies(1.7)in theviscosity senseif it is both a sub-and asuper-solution to(1.7)in theviscosity sense.

        In both contexts,it is an interesting open problem to understand relevant conditions on?,or on(M,g),which would ensure that(1.2)-(1.3),or(1.7)respectively,admits a smooth solution.We make the following conjecture.

        Conjecture 1.1.Suppose thatn≥3,2≤k≤n,and??Rnis a bounded smooth domain.Then the locally Lipschitz viscosity solution to(1.2)-(1.3)is smooth near??.

        Some further questions are in order.

        Question 1.1.Suppose thatn≥3,2≤k≤n,and??Rnis a bounded smooth domain.If(1.2)-(1.3)has a smooth sub-solution,must(1.2)-(1.3)have a smooth solution?

        Question 1.2.Suppose thatn≥3,2≤k≤n,and??Rnis a smooth strictly convex(non-empty)domain.Is the locally Lipschitz viscosity solution to(1.2)-(1.3)smooth?

        If?is a ball,then thesolution to(1.2)-(1.3)issmooth and correspondsto the Poincar′e metric.

        Question 1.4.Suppose thatn≥3,2≤k≤n,and(Mn,g)is a Riemannian manifold such thatλ(?Ag)∈ΓkonM.Does(1.7)have a unique Lipschitz viscosity solution?

        It is clear that(1.7)has at most oneC2solution by the maximum principle.In fact,if(1.7)has aC2solution,then that solution is also the unique continuous viscosity solution in view of the strong maximum principle[2,Theorem 3.1].Equivalently,if(1.7)has two viscosity solutions,then it has noC2solution.

        Question 1.5.Suppose thatn≥3 and 2≤k≤n.Does there exist a Riemannian manifold(Mn,g)such thatλ(?Ag)∈ΓkonMand(1.7)has a Lipschitz viscosity solution which is notC2?

        Finally,we discussthe case where(1.3)is replaced by f inite constant boundary conditions

        Theorem 1.4.Supposethat n≥3and2≤k≤n.Let?={a<|x|

        (i)u is smooth in each of{a≤|x|

        (iii)and thef irst radial derivative?ru jumpsacross{|x|=m}:

        Remark 1.1.It is clear from Theorem 1.4(in Cases 1–3)that ifuis aC1and radially symmetric solution to(1.2)in theviscosity sensein someopen annulus?thenu∈C∞(?).

        Remark 1.2.In Case 4,theexact value ofmis

        wherepis the solution to

        The following question is related to Question 1.

        For comparison,we recall here a result of Bo Guan[8]on the Dirichletσk-Yamabe problem in the so-called positive case which states that the existence of a smooth subsolution implies the existenceof a smooth solution.

        We concludethe introduction with one more question.

        Question 1.7.Letn≥3,2≤k≤nandm/=n?1.Does there exist a smooth domain??Rnsuch that the locally Lipschitz solution to(1.2)-(1.3)isC2away from a setΣwhich has Hausdorff dimensionm?

        In Section 2,we prove all the results above except Theorem 1.3,whose proof is done in the appendix.Theorem 1.1 is proved f irst in Subsection 2.1.We then prove a lemma on the existence and uniqueness a non-standard boundary value problem for the ODE related to(1.2)in Subsection 2.3 and use it to prove Theorem 1.2 in Subsection 2.4 and Theorem 1.4 in Subsection 2.5.

        2 Proofs

        2.1 Proof of Theorem 1.1

        We will usethe following lemma.

        Mij(δij?mimj)≥0.

        the conclusion follows.

        Proof.Fix somep∈Σn?1and letνbe a unit vector atf(p)normal to the image of a small neighborhood ofp,.Recall that

        This means

        Using(2.1)yields the conclusion.

        Integrating overΣ,we thus have thatHΣ≡0 and?u≡const onΣ.In particular,f:Σn?1→?is a minimal immersion with respect to?g.This is impossible as there is no smooth minimal immersion in Rnwith codimension one.

        2.2 Preliminary ODEanalysis

        By the uniqueness result in[7,23],the solutionsuin Theorems 1.2 and 1.4 are radially symmetric,u(x)=u(r)wherer=|x|.

        As in[5,32],wework on a round cylinder instead of Rn.Namely,let

        where here and below′denotes differentiation with respect tot.

        Note that,fork≥2,at points whereuis twice differentiable,λ(?Au)∈Γkif and only ifσk(λ(?Au))>0 and|ξ′|>1.Indeed,ifσk(λ(?Au))>0 and|ξ′|>1,then(2.2)impliesσi(λ(?Au))>0for 1≤i≤kand soλ(?Au)∈Γk.Conversely,ifλ(?Au)∈Γkfor somek≥2,then

        Using(2.2),we seethat the f irst two inequalities imply|ξ′|>1.

        We are thus led to study the differential equation

        under the constraint that|ξ′|>1.

        It is well known(see[5,32])that(2.3)has a f irst integral,namely

        A plot of the contours ofHfork=2,n=7 is provided in Figure 1.See[5]for a more complete catalog.

        Before moving on with the proofs of our results,we note the following statement.

        Remark 2.1.As a consequence of Theorem 1.4,we have in fact thatH(ξ,ξ′)is(locally)constant along viscosity solutions.

        Figure 1:The contours of H for k=2,n=7.Each radially symmetric viscosity solution to(1.2)lies on a single contour of H but avoid the shaded region,i.e.the dotted parts of the contours of H are excluded.Every smooth solution stays on one side of the shaded region.Every non-smooth solution jumps(on one contour)from the part below the shaded region to the part above the shaded region at a single non-differentiable point.

        HenceH(ξ,ξ′)is also constant in{?a

        2.3 A lemma

        Proof.We use ideas from[5].

        By(2.6)we thus have

        In this proof,wewill only need to consider thecasethat(?1)k H(p,q)<0.Then by[5](Theorem 1,Case II.2 for evenkand Theorem 2,Case II.2 for oddk),we have thatTp,qis also f inite(corresponding tor?being f inite in the notation of[5])and

        Before moving on to the next stage,we note that,in view of(2.6),

        In particular,then length ofIp,qdependsonly onn,kand thevalueofH(p,q),rather thanpandqthemselves.

        2.4 Proof of Theorem 1.2

        solves(1.2)-(1.3)in{a

        (i)limt→±Tξ(t)=?∞,

        (ii)ξ′(0?)=1,ξ′(0+)=?1,

        (iii)and|ξ′|>1 in(?T,0)∪(0,T).Indeed,letξT:[0,T)→R be the solution obtained in Lemma 2.3,and def ine

        there is noC2function?such that?≥unearx0and?(x0)=u(x0).Therefore(a)holds.

        Suppose now that?is aC2function such that?≤unearx0and?(x0)=u(x0).Asuis radial,this implies that

        LetOdenote the diagonal matrix with diagonal entries 1,?1,...,?1.Note that,in block form,

        Thus,by(2.18),

        Also,?(x0)=ˉ?(x0)and,in view of(2.17),??(x0)=?ˉ?(x0).Hence

        2.5 Proof of Theorem 1.4

        (i)Suppose thatT

        (ii)SupposethatT=T(a,b,c1,c2).We show that Case 3 holds.

        (iii)Suppose thatT>T(a,b,c1,c2).We show that Case 4 holds.

        In this case,we selectp≥pa(≥pb)such that

        Suchpexists as the right hand side tends toT(a,b,c1,c2)whenp→paand diverges to∞asp→∞.Recall the solutionξpdef ined in the proof of Lemma 2.3.Let

        Then 2T±

        We then let

        Wecan then proceed as in theproof of Theorem 1.2 to show thatξis thedesired solution.

        AAppendix:Proof of Theorem 1.3

        By[12,Theorem 1.4],we have for all suff iciently smallτ>0 that the problem

        has a unique smooth solutionuτ.Furthermore,by[12,Propositions 3.2 and 4.1],the family{uτ}is bounded inC1(M)asτ→0.(C2bounds foruτwere also proved in[12],but these bounds are unbounded asτ→0.)Hence,along some sequenceτi→0,uτiconverges uniformly to someu∈C0,1(M).To conclude,we show thatuis a viscosity solution to(1.7).

        For notational convenience,we renameuτiasui.Fix someˉx∈M.

        Step 1:We show thatuis a sub-solution to(1.7)atˉx.More precisely,we show that for every?∈C2(M)such that?≥uonMand?(ˉx)=u(ˉx)there holds that

        Note that

        and so

        Recalling(A.1),we hence have

        Sinceδis arbitrary,this proves(A.2)after sendingδ→0.

        Step 2:We show thatuis a super-solution to(1.7)atˉx,i.e.if?∈C2(M)is such that?≤uonMand?(ˉx)=u(ˉx),then

        By(A.1),we hence have

        Acknowledgment

        The authors would like to thank Matt Gursky and Zheng-Chao Han for stimulating discussions.Theauthorsare grateful to therefereesfor their very careful reading and useful comments.YanYan Liis partially supported by NSFGrant No.DMS-1501004.

        亚洲先锋影院一区二区| 波多野结衣爽到高潮大喷| 末发育娇小性色xxxx| 538在线啪在线观看| 美腿丝袜av在线播放| 亚洲熟妇av一区二区在线观看| 国内免费高清在线观看| 婷婷成人基地| 亚洲国产精品无码久久九九大片健 | 精品国产a毛片久久久av| 狠狠色噜噜狠狠狠8888米奇| 中文字幕无码日韩专区免费| 伊人久久一区二区三区无码| 久久老熟女乱色一区二区| 人妻少妇偷人精品免费看| 中文字字幕在线精品乱码| 加勒比精品久久一区二区三区| 国产av一区麻豆精品久久| 无码熟妇人妻av在线网站| 先锋影音最新色资源站| 窄裙美女教师在线观看视频| 青青操视频手机在线免费观看| 色哟哟最新在线观看入口| 欧美熟妇精品一区二区三区| 亚洲AV永久无码精品表情包| 久久精品久99精品免费| 性裸交a片一区二区三区| 草莓视频一区二区精品| av手机天堂在线观看| av区无码字幕中文色| 色婷婷久久一区二区三区麻豆| 99久久久69精品一区二区三区| 一区视频免费观看播放| 欧美熟妇性xxx交潮喷| 欧美成人专区| 人妻av中文字幕精品久久| 亚洲三区在线观看内射后入| 色窝窝免费播放视频在线| 亚洲一区二区免费日韩| 青青草高中生在线视频| 人妻丰满熟妇av无码区hd|