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

1.1 MB·

by  Jose' Antonio Galvez

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: [email protected]
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b Dipartimento Ingegneria e Scienze dell’Informazione e Matematica, Universita´ di L’Aquila, via Vetoio - Loc.
]
G Coppito,E-67010L’Aquila,Italy;e-mail: [email protected]
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