Entire solutions of the degenerateMonge-Ampe`re equation with a finite number of singularities. 6 1 Jose´ A.Ga´lveza,BarbaraNellib1 0 2 n a a Departamento de Geometr´ıa y Topolog´ıa, Facultad de Ciencias, Universidad de Granada, E-18071 Granada, J Spain;e-mail: jagalvez@ugr.es 7 b Dipartimento Ingegneria e Scienze dell’Informazione e Matematica, Universita´ di L’Aquila, via Vetoio - Loc. ] G Coppito,E-67010L’Aquila,Italy;e-mail: nelli@univaq.it D . Abstract h t WedeterminetheglobalbehaviorofeveryC2-solutiontothetwo-dimensionaldegenerateMonge-Ampe`re a m equation,uxxuyy−u2xy =0,overthefinitelypuncturedplane.Withthis,weclassifyeverysolutionintheonce ortwicepuncturedplane. Moreover,whenwehavemorethantwosingularities,ifthesolutionuisnotlinearin [ ahalf-strip,weobtainthatthesingularitiesareplacedattheverticesofaconvexpolyhedronP andthegraphof 1 uismadebypiecesofconesoutsideofP whicharesuitablygluedalongthesidesofthepolyhedron. Finally, v ifwelookforanalyticsolutions,thenthereisatmostonesingularityandthegraphofuiseitheracylinder(no 4 singularity)oracone(onesingularity). 7 4 MathematicsSubjectClassification: 35K10,53C21,53A05. 1 0 . 1 1 Introduction. 0 6 1 AcelebratedresultprovedbyA.V.Pogorelov[16],andindependentlybyP.HartmanandL.Nirenberg[8],states : thatalltheglobalsolutionstothedegenerateMonge-Ampe`reequation v i X (1) u u −u2 =0, xx yy xy r a 1TheauthorswerepartiallysupportedbyINdAM-GNSAGA,PRIN-2010NNBZ78-009,MICINN-FEDERGrantNo.MTM2013-43970-P, andJuntadeAndaluc´ıaGrantNo.FQM325. 1 whereu:R2 −→RisafunctionofclassC2,aregivenby u(x,y)=α(x)+c y, 0 uptoarotationinthe(x,y)-plane,thatis,thegraphofuisacylinder(seealso[18,19]). ThisdegenerateMonge-Ampe`reequationhasbeenextensivelystudiedfromananalyticpointofviewandalso fromageometricpointofviewsincethegraphofeverysolutiondeterminesaflatsurfaceintheEuclidean3-space. Local properties of the solutions of (1) have been analyzed in many papers (see, for instance, [20, 21] and referencestherein). Ourobjectiveistostudytheglobalbehaviorofthesolutionsof(1)forthelargestnonsimply- connecteddomains,thatis,inthefinitelypuncturedplane. Observethat,whentheMonge-Ampe`reequationisellipticorhyperbolic,alargeamountofworkintheunder- standing of these solutions in the punctured plane has been achieved from a local and global point of view (see, amongothers,[1,2,3,4,5,6,7,9,10,11,13,14,17]). The paper is organized as follows. After a first section of preliminaries, in Section 3 we establish the global behaviorofeveryC2solutionu:R2\{p ,...,p }−→RtotheMonge-Ampe`reequation(1)withisolatedsingu- 1 n laritiesatthepointsp .Weshowthatforeverysingularpointp thereexistsatleastasectorS ⊆R2\{p ,...,p } i i i 1 n suchthatthegraphofuoverS isapieceofacone. Moreover,thereexistsatmostamaximalstripSsuchthatthe i graphofuoverS isacylinder. Asaconsequenceoftheseresults,weobtainthatifuisananalyticsolutionthen thereisatmostonesingularity,andthegraphofuiseitheracylinderifthereisnosingularityoraconeifthereis onesingularity. InSection4weclassifyallsolutionsto(1)withoneortwosingularities. Inparticular,ifuisasolutionwith onesingularityatp then,eitherthegraphofuisaconeorthereexistsalinercontainingtop suchthatthegraph 0 0 ofuisacylinderoveronehalf-planedeterminedbyrandaconeovertheotherhalf-plane. Wealsodescribeevery solutioninthetwicepuncturedplane. Althoughthebehaviorofasolutionuto(1)withmorethantwosingularitiescanbecomplicated, duetothe existenceofsomeregionsinthe(x,y)−planewhereuisalinearfunction,weclassifyinSection5everysolution whichdoesnotadmitalargeregionwhereuislinear,thatis,everysolutionsuchthatthedomainwhereuislinear does not contain a half-strip of R2. In such case, we show that the singular points are the vertices of a convex compact polyhedron C ⊆ R2 with non empty interior, the solution u must be linear over C and the graph of u is made by some cones over R2\C which are suitably glued along the segments of the boundary of the planar polyhedronu(C). 2 Preliminaries. Let Ω ⊆ R2 be a domain, and u : Ω −→ R be a function of class C2 satisfying the degenerate Monge-Ampe`re equation u u −u2 = 0, in Ω. As we mentioned in the introduction, it is well known that every solution xx yy xy u(x,y)tothedegenerateMonge-Ampe`reequationinthewholeplaneR2isgivenbyu(x,y)=α(x)+c y,upto 0 arotationinthe(x,y)-plane(see[8,12,16]),thatis,itsgraphisacylinder. Thus, it is natural to study the solutions to the degenerate Monge-Ampe`re equation in the possible largest domains. In other words, we consider solutions to (1) in the non simply-connected domains R2\{p ,...,p }. 1 n 2 Here,weassumeuhasanisolatedsingularityatanyp ,thatis,ucannotbeC2-extendedtop . i i Forthestudyofthesesolutions,wewillneedthefollowingresultwhichisaconsequenceof[8,Lemma2]. Lemma1. Letu(x,y)beaC2 solutiontothedegenerateMonge-Ampe`reequation(1)and(x ,y ) ∈ Ωapoint 0 0 where the Hessian matrix of u does not vanish. Then, there exists a line r in the (x,y)−plane such that the connectedcomponentr ofr∩Ωcontaining(x ,y )satisfies: 0 0 0 1. r ismadeofpointswithnonvanishingHessianmatrix, 0 2. thegradientofuisconstantonr . 0 Moreover,thereexistsanopensetU ⊆Ωcontainingr suchthatifapoint(x ,y )∈U hasthesamegradientas 0 1 1 apointofr then(x ,y )∈r . 0 1 1 0 Fromnowon,givenasolutionuto(1)inΩ,wewilldenotebyN theopensetgivenbythepointsinΩwhose Hessianmatrixdoesnotvanish,andbyU =Ω\N. NotethatthegraphΣofasolutionuto(1)corresponds,fromageometricpointofview,toaflatsurfaceinR3, andthepoints Σ :={(p,u(p)): p∈N}, Σ :={(p,u(p)): p∈U}, N U correspond,respectively,tothenonumbilicalpointsandtotheumbilicalpointsofthegraphΣ. Givenapointp∈N wewilldenotebyr(p)thepieceofliner determinedbyLemma1. Moreover,sincethe 0 gradientofuisconstantonr(p),theimageofr(p)givenby R(p):={(q,u(q))∈R3 : q ∈r(p)} isapieceoflineinR3. Thus,thepreviouslemmaassertsthatΣ ismadeofpiecesofpairwisedisjointlinesofR3. Inaddition,every N connected component of Σ with an interior point is contained in a plane, because its Hessian matrix vanishes U identically. AfirstconsequenceofthepreviouslemmaisthateachsolutionofthedegenerateMonge-Ampe`reequationcan becontinuouslyextendedtothesingularitybecausethenormofitsgradientisboundedaroundthesingularity. Lemma2. LetΩ ⊆ R2 beadomain,p ∈ ΩandubeaC2 solutionof(1)inΩ\{p }. Thenthemodulusofthe 0 0 gradientofuisboundedinaneighborhoodofp . Inparticular,ucanbecontinuouslyextendedtop . 0 0 Proof. Considerε>0suchthatthecloseddiskD (ε)centeredatp withradiusεiscontainedinΩ. Let p0 0 m=max{(cid:107)grad (u)(cid:107): (cid:107)p−p (cid:107)=ε}, p 0 wheregrad (u)denotesthegradientofuatapointp. p Letusseethatthemodulusofthegradientofuislessthanorequaltomforeverypointp∈D (ε)\{p }. p0 0 Let p ∈ N, then the piece of line r(p) has at least a point q in the boundary of the disk D (ε). So, from p0 Lemma1onehasthat(cid:107)grad (u)(cid:107)=(cid:107)grad (u)(cid:107)≤m. Hence,thepreviousinequalityhappensintheclosureof p q N. 3 On the other hand, if p is a point in the interior of U, we can consider the connected component U of U to 0 whichpbelongs. TheHessianmatrixofuvanishesidenticallyinU ,so,thegradientofuisconstantintheclosure 0 of U . Moreover, since the closure of U intersects the closure of N, we obtain that (cid:107)grad (u)(cid:107) ≤ m, as we 0 0 p wantedtoshow. 3 Global behaviour of the graphs. Now, consider a C2 solution u : R2\{p ,...,p } −→ R to the degenerate Monge-Ampe`re equation (1) with 1 n singularitiesatthepointsp . Letp∈N,thenfromthepreviousconsiderations,weknowthatr(p)mustbe: i 1. aline,or 2. ahalf-linewithendpointatasingularpointp ,or i 3. asegmentwithendpointsattwodifferentsingularpointsp ,p . i j Asaconsequenceweobtainthefollowingresult. Proposition1. Letu:R2\{p ,...,p }−→RbeaC2 solutionof(1)withisolatedsingularitiesatthepointsp . 1 n i Assumethereexistsapointp ∈ N suchthatr(p)isalineandletS(p) ⊆ R2\{p ,...,p }bethemaximalopen 1 n stripcontainingr(p). ThenuisgiveninS(p)by (2) u(xe +ye )=α(x)+v y, y ∈R 1 2 0 foraC2functionα(x),aconstantv ∈R,andanorthonormalbasis{e ,e }ofR2wheree isparalleltor(p). 0 1 2 2 Moreover, 1) ifq ∈N satisfiesthatr(q)isalinethenr(q)⊆S(p), 2) ifrisalinecontainedinU thenr ⊆S(p). Proof. Letq beapointintheopenstripS(p),withq ∈ N. SinceS(p)hasnosingularpointthenr(q)intersects r(p)orr(q)isparalleltor(p). FromLemma1,intheformercaser(q)=r(p)andinthelattercaser(q)mustbe alinecontainedinS(p). Now,letr beaparallellinetor(p)containedinS(p). Then,fromthepreviousdiscussionwehaveobtained 0 thatr ⊆ N orr ⊆ U. Asaconsequence,ifr hasapointqintheinteriorofU,int(U),thenr iscontainedin 0 0 0 0 int(U). Hence,ifr = r(q) ⊆ N thentheimageofr isthelineR(q),andifr ⊆ int(U)thentheimageofr isa 0 0 0 0 line,becausetheimageofeachconnectedcomponentofint(U)isapieceofaplane. Thus,theimageofr must 0 alsobealineifr ⊆∂N =∂(int(U)). 0 Therefore,uisgiveninS(p)by u(xe +ye )=α(x)+v(x)y, y ∈R 1 2 4 forcertainfunctionsα(x),v(x),and{e ,e }orthonormalbasisofR2,withe paralleltor(p). Inaddition,from 1 2 2 (1),v(x)mustbeaconstantv ∈R. 0 Let us prove now the case 1), that is, assume q ∈ N such that r(q) is a line and see that r(q) ⊆ S(p). If r(q)(cid:54)⊆S(p),thenr(q)isparalleltor(p)becauseotherwiser(q)intersectsr(p),andsor(p)=r(q). DenotebyS theuniqueconnectedcomponentofR2\S(p)containingr(q). ObservethatS isaclosedhalf- 0 0 planeofR2,paralleltor(p),sothatitsboundary∂S isalinecontainingatleastasingularpointp ,duetothe 0 i0 maximalityofS(p). Letε>0suchthattheopendiskD (ε)centeredatp withradiusεdoesnotcontainanothersingularpoint, pi0 i0 andD (ε)∩r(q)=∅. Moreover,sincethenumberofsingularpointsisfinite,wecanchooseεsmallenoughso pi0 thatifrisanylineparalleltor(p)intersectingD (ε)thenr =∂S orrhasnosingularpoint. ThesetD+ (ε) = D (ε)∩int(S )mustcpoin0tainsomepointa0 ∈ N. Otherwise,theimageofD+ would beapieceofapip0lane,andpfir0om(2)oneh0asthatp isasingularpointif,andonlyif,everypointinD pi∩0 ∂S is i0 pi0 0 singular. Thisisacontradictionwhichclaimstheexistenceofthepreviouspointa. SinceN isanopensetandthesetofsingularitiesisfinite,wecanassumethatr(a)isnotasegmentjoining twosingularpoints. So, r(a)isahalf-lineoralinewhichdoesnotintersectsr(p)orr(q), thatis, r(a)mustbe paralleltor(p). Then,fromthechoiceofε,onehasthatr(a)isalsoalineparalleltor(p). Finally,ifwechoosethemaximalstripS(a)⊆R2\{p ,...,p }containingtheliner(a),weobtainfromthe 1 n choiceofεthat∂S(p)∩∂S(a)isthelineparalleltor(p)containingp . Butthisisnotpossible,becauseinthis i0 caseonehasfrom(2)thatp isasingularpointifandonlyifeverypointin∂S(p)∩∂S(a)issingular. i0 Therefore,thereisnoq ∈N suchthatr(q)isalinewhichisnotcontainedinS(p). Inthecase2),theliner ⊆U mustbeparalleltor(p)becauseotherwiser∩r(p)(cid:54)=∅,contradictingLemma1. So,thiscasecanbeexactlyprovenasthepreviouscasereplacingr(q)byr. ThepreviousresultassertsthatthegraphofuoverS(p)isacylinderinR3foliatedbylines,allofthemparallel toR(p). Observethatthiscylinderhasnonumbilicalpoints,andprobablyumbilicalpointsaswell. Inparticular, ifthereisnosingularpointthenS(p)=R2anduisgloballydeterminedby(2). Definition1. WesaythatuisacylindricalfunctioninastripS ifuisgivenby(2)inS. Moreover,wesaythatu isaconicalfunctioninanopensector Sθ2(p)={p+ρ(cosθ,sinθ): ρ>0, θ <θ <θ }, 0<θ −θ <2π, θ1 1 2 2 1 orinR2\{p},ifuisgivenby (3) u(p+ρ(cosθ,sinθ))=u +ρα(θ) 0 inoneofthepreviousdomains,foracertainC2functionα(θ). Now,letusprovesometechnicallemmaswhichwillbenecessaryinourstudy. Lemma3. LetΩbeadomain,p ∈ Ω,S ⊆ Ωbeanopensectorwithvertexatp ,andu : Ω\{p } −→ Rbea 0 0 0 C2 solutionof(1)withisolatedsingularityatp . Ifforeveryp ∈ S∩N onehasthatr(p)isahalf-linewithend 0 pointatp thenuisaconicalfunctioninS. 0 5 Proof. Ifp ∈ S ∩N thenthegraphofr(p)isthehalf-lineR(p),andifr ⊆ int(U)∩S isahalf-linewithend 0 pointatp thentheimageofr isahalf-lineinR3,becauseeachconnectedcomponentofthegraphofint(U)is 0 0 apieceofaplane. Asaconsequence, theimageofahalf-liner withendpointatp isalsoahalf-lineofR3 if 0 0 r ⊆∂N =∂(int(U)). 0 Consequently,u(p +ρθ)=u (θ)+ρα(θ),withρ>0. And,fromLemma2,u (θ)mustbeconstantsince 0 0 0 uiscontinuousatp . 0 Lemma4. Letu:R2\{p ,...,p }−→RbeaC2 solutionof(1)withisolatedsingularitiesatthepointsp . Let 1 n i S ,S ⊆R2\{p ,...,p }betwodisjointopensetssuchthat 1 2 1 n 1) uisacylindricalfunctioninthestripS anduisaconicalfunctioninthesectorS ;or 1 2 2) uisaconicalfunctioninthesectorS andalsointhesectorS ,bothwithdifferentvertices. 1 2 If∂S ∩∂S containsasegment,theneverynonsingularpointin∂S ∩∂S iscontainedinU. 1 2 1 2 Proof. If S is a strip and e is a normal unit vector to a connected component of ∂S then, from (2), one has that Hess(u)(e,e)isconstantforallpointinthiscomponent. Ontheotherhand,ifS isasectorwithvertexatp andthehalf-linep +ρθ ,withρ>0,iscontainedin∂S 0 0 0 then,from(3),onehasthatHess(u)(e,e)= α(cid:48)(cid:48)(θ0)+α(θ0),whereeisanormalunitvectortothehalf-line. ρ Inthecase1)or2),itisclearthatα(cid:48)(cid:48)(θ )+α(θ )vanishesateverypointin∂S ∩∂S . Thisisequivalentto 0 0 1 2 showthattheHessianmatrixofuvanishesateverynonsingularpointin∂S ∩∂S . 1 2 Proposition2. Letu:R2\{p ,...,p }−→RbeaC2 solutionof(1)withisolatedsingularitiesatthepointsp , 1 n i andp∈N. Thenr(p)cannotbeasegment. Moreover,foranysingularpointp , i (4) N ={p∈N : r(p)isahalf-linewithendpointp } pi i isanopenset. Proof. Let p ∈ N such that r(p) is either a segment or a half-line. Up to a translation and a rotation, we can assumepistheoriginandasingularendpointofr(p)is(x ,0)withx <0. 1 1 SinceN isanopenset,weconsiderε>0suchthattheopendiskD(ε)centeredattheoriginwithradiusεis containedinN. Moreover, usingthatthesetofsingularpointsisfinite, εcanbechosensmallenoughinsucha waythereisnosingularpointp =(x ,y )with0<|y |<ε. k k k k Let q = (x ,y ) be a sequence in D(ε)+ = {(x,y) ∈ D(ε) : y > 0} going to p = (0,0). Observe that n n n thenumberofsegmentsoftheformr(q)containedinN canonlybefinitebecausethenumberofsingularitiesis finite. Thus,replacingq byapointclosetoit,wecanassumer(q )isahalf-lineoraline,foreverypointq of n n n thesequence. Sincer(q )cannotintersectr(p),theanglebetweenr(q )andr(p)tendstozero. n n Ifthereexistsasubsequenceq suchthatr(q )arelines,thenallofthemmustbeparalleland,fromPropo- m m sition1,r(q)isahorizontallineforallq =(x,y)∈D(ε)+. 6 Otherwise,sincethenumberofsingularitiesisfinite,thereexistsasubsequenceq suchthatr(q )arehalf- m m lines with the same singular end point p . Thus, as the angle between r(q ) and r(p) tends to zero, p is i1 m i1 necessarily placed on the line y = 0. In particular, taking ε small enough, we can assume that the sector made ofallhalf-linescontainingapointq ∈ D(ε)+ andwithendpointatp doesnotcontainanothersingularpoint. i1 Hence,eachhalf-lineofthissectorwithendpointp isoftheformr(q)forq ∈D(ε)+. i1 Analogously, we can follow the same reasoning for D(ε)− = {(x,y) ∈ D(ε) : y < 0}. Thus, r(q) is a horizontallineforall q ∈ D(ε)−, orr(q)isahalf-lineforallq ∈ D(ε)− withthesamesingularendpointp , i2 placedattheliney =0. Now,letusdenotebyS+ thestrip{(x,y) ∈ R2 : 0 < y < ε}ifr(q)isahorizontallineforallq ∈ D(ε)+, orthesectorcontainingD(ε)+ withvertexatp ifr(q)isahalf-lineforallq ∈ D(ε)+. WealsodenoteS− the i1 correspondingsetforthepointsinD(ε)−. From Proposition 1, S+ and S− cannot be two strips because the point (x ,0) is singular. Moreover, from 1 Lemma 4, S+ and S− are two sectors with the same vertex p = p because otherwise p = (0,0) would be i1 i2 containedinU. Thus, fromLemma3, ifr isthehalf-linecontainingp = (0,0)withendpointp = p thena i1 i2 pointinrisasingularpointifandonlyifeverypointinrissingular. Hence,p =p =(x ,0)andr(p)cannot i1 i2 1 beasegment. Thisprovesthatifp∈N suchthatr(p)isahalf-linethenthereexistsε>0suchthatD (ε)⊆N andforall p q ∈D (ε)thesetr(q)isahalf-linewiththesameendpoint. Thatis,N isanopenset. p pi AsaconsequenceofPropositions1and2,wegetthefollowingresult. Corollary1. Letu:R2\{p ,...,p }−→RbeaC2solutionof(1)withisolatedsingularitiesatthepointsp . If 1 n i V isaconnectedcomponentofN theneitherV ismadeofparallellinesorV ismadeofhalf-lineswithcommon endpointatsomep . i Corollary2. Letu : R2\{p ,...,p } −→ RbeaC2 solutionof(1)withisolatedsingularitiesatthepointsp . 1 n i ThenthesetN ,givenby(4),isanonemptyunionofopensectorswithvertexatp ,foreverysingularityp . In pi i i particular,uisaconicalfunctionineachconnectedcomponentofN . pi Proof. FromProposition2andLemma3,weonlyneedtoshowthatN isnonemptyforeachsingularpointp . pi i So,fixasingularpointp andassumeN isempty. i pi IfthereexistedapuncturedneighborhoodU ofp containedinU thentheimageofU wouldlieinaplaneand i p wouldnotbeasingularpoint. Thus,theremustexistasequenceq ∈N goingtop . Moreover,asubsequence i n i q mustsatisfythatr(q )isalineforallm(andsothelinesareparallel)orr(q )isahalf-lineforallmwith m m m acommonsingularendpointp (cid:54)= p . Uptoatranslationandarotation,wecanassumep istheoriginandthe j i i limitlineofthelinescontainingr(q )isy =0. Inparticular,inthelattercasep isplacedintheliney =0. m j Observe that q cannot be placed in the segment joining the singular points p and p , that is, in any case, m i j every point q is in the open upper half-space R2 = {(x,y) ∈ R2 : y > 0} or in the open lower half-space m + R2 ={(x,y)∈R2 :y <0}. − Passingtoasubsequence,ifnecessary,wecanassumeq iscontainedinR2 forallm,orq iscontainedin m + m R2 forallm. Supposeq ∈ R2 foreverym. Ifr(q )isalineforeachmthenwedenotebyS+ themaximal − m + m 7 stripdeterminedbyProposition1, whichsatisfiesthattheliney = 0isinitsclosure. Ifr(q )isahalf-linefor m allmthenformbigenoughtheopensectordeterminedbyr(q )andy = 0doesnotcontainanysingularpoint, m andwecallS+tothissector. InthefirstcasethefunctionuwouldbecylindricalinS+(Proposition1),andinthe secondcaseuwouldbeconicalinS+(Lemma3)withvertexatp (cid:54)=p . j i Ontheotherhand,theremustexistasequenceofpointsinN ∩R2 goingtop . Otherwisetherewouldexist − i ε > 0suchthattheopenhalf-diskD(ε)∩R2 iscontainedinU. Thus,theimageofD(ε)∩R2 liesinaplane. − − But, since u is cylindrical or conical in S+ then the origin p is a singular point if and only if every point of a i neighborhoodofp intheliney =0issingular,whichgivesusacontradiction. Thus,theprevioussequenceexists i andwecandefineS−inananalogousway. So,uwillbecylindricalorconicalinS+ andinS−. But,inanycase,thisisagainacontradictionbecausep i wouldbeasingularpointifandonlyifeverypointofaneighborhoodofp intheliney = 0issingular. Hence i N (cid:54)=∅. pi Asaconsequenceweobtainafirstglobalclassificationresult. Theorem1. Letu:R2\{p ,...,p }−→Rbeananalyticsolutionof(1)withisolatedsingularitiesatthepoints 1 n p . Then,eitherthereisnosingularpointanduisacylindricalfunctioninR2,orthereisauniquesingularpoint i anduisaconicalfunctioninR2\{p }. 1 Proof. Theresultiswellknownifthereisnosingularpoint[8,16]. Otherwise,uptoatranslation,wecanassume that the origin is a singular point. So, for polar coordinates (ρ,θ), the function u(ρ,θ) is analytic and, from the previousCorollary,u vanishesidenticallyinthenonemptyopensetN .Therefore,u mustvanishglobally, ρρ (0,0) ρρ thatis,uisaconicalfunctioninR2\{(0,0)}. 4 Entire solutions with one or two singularities. We devote this Section to the characterization of the entire solutions to the degenerate Monge-Ampe`re equation u u −u2 = 0,whereuisaC2 functioninR2 minusoneortwopoints. Theseresultswillbeaconsequence xx yy xy oftheglobalresultsstudiedinthepreviousSection. Wefirstprovethatthereonlyexisttwodifferentkindsofsolutionsinthepuncturedplane. Theorem2. Letu:R2\{(0,0)}−→RbeaC2 solutionof(1)withanisolatedsingularityattheorigin. Then,u isgivenbyoneofthefollowingcases(Figure1): 1. uisaconicalfunctioninR2\{(0,0)},or 2. thereexistsalinercontainingtheoriginsuchthatuisacylindricalfunctioninonehalf-planedetermined byr,anduisaconicalfunctionintheotherhalf-plane. Proof. FromCorollary2,N cannotbeempty. Thus,ifthereisnopointp∈N suchthatr(p)isalinethen,from Lemma3,uisaconicalsolutioninR2\{(0,0)}. 8 Ontheotherhand, ifthereisapointp ∈ N suchthatr(p)isalinethen, fromProposition1, thereexistsan open half-plane S(p), containing r(p), such that u is a cylindrical function in S(p). Moreover, the boundary of S(p)inR2isalinercontainingthesingularpoint(0,0).IfS−istheotheropenhalf-planedeterminedbyr,again Proposition1givesthatr(q)mustbeahalf-lineforeverypointq ∈ N ∩S−. Hence, uisaconicalfunctionin S−,byLemma3. Figure1: solutionswithonesingularity. Nowwepresentsomefamiliesofexamplesofsolutionstotheequation(1)inthetwicepuncturedplane. Example4.1. ConsideranopenstripS inR2sothatitsboundaryisgivenbytwolinesr ,r . Choosetwopoints 1 2 p ∈ r , and p ∈ r . Then, a solution u to the equation (1) can be obtained in R2\{p ,p } such that u is a 1 1 2 2 1 2 cylindricalfunctioninS,anduisaconicalfunctionineachhalf-planeofR2\S withvertexatp (Figure2). i Example4.2. LetS beanopenhalf-planeinR2 withboundarygivenbyaliner,andp ,p ∈ r. Leth ,h be 1 2 1 2 twoparallelhalf-linescontainedinR2\S withrespectiveendpointsp ,p . Asolutionuto(1)canbedefinedin 1 2 R2\{p ,p }suchthatuisacylindricalfunctioninS,uisalinearfunctioninthehalf-stripdeterminedbyh ,h 1 2 1 2 andr,anduisaconicalfunctionintheremainingtwodomainswithverticesatp ,p ,respectively(Figure3). 1 2 Example4.3. AssumeS isanopenhalf-planeinR2 withboundarygivenbyaliner, p ∈ r, andp ∈ R2\S. 1 2 Let h ,h be two half-lines in R2\S with end point at p such that at most one of them is contained in r, and 1 2 1 p belongs to the closed sector determined by h and h . Let h be the half-line parallel to h with end point 2 1 2 i i at p . Then, a solution u to (1) can be defined in R2\{p ,p } such that u is a cylindrical function in S, u is a 2 1 2 linearfunctionintheunboundeddomaindeterminedbyh ,h ,h andh ,anduisaconicalfunctionintherestof 1 2 1 2 domains(Figure4). 9 S p 1 p 2 r r 1 2 Figure2: theconfigurationofr(p)linesforexample4.1. S p 1 h 1 p h 2 2 r Figure3: theconfigurationofr(p)linesforexample4.2. Example 4.4. Consider p ,p ∈ R2, and S ,S two disjoint open sectors with vertex at p ,p , respectively. A 1 2 1 2 1 2 solutionuto(1)canbeobtainedinR2\{p ,p }suchthatuisaconicalfunctioninS andS ,anduislinearin 1 2 1 2 theopensetdeterminedbyR2\(S ∪S )(ifnotempty). SeeFigure5. 1 2 Ourobjectiveistoshowthateverysolutionto(1)inthetwicepuncturedplaneisgivenbyoneoftheprevious examples. Forthat,wewillneedsomepropertiesassociatedwiththesetsN definedin(4). pi Letu : R2\{p ,...,p } −→ RbeaC2 solutionof(1)withisolatedsingularitiesatthepointsp . Wedefine 1 n i 10

Entire solutions of the degenerateMonge-Ampere equation with a finite number of singularities

Similar Entire solutions of the degenerateMonge-Ampere equation with a finite number of singularities