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Annals of Mathematics

Plateau's Problem and Dirichlet's PrincipleAuthor(s): Richard CourantSource: The Annals of Mathematics, Second Series, Vol. 38, No. 3 (Jul., 1937), pp. 679-724Published by: Annals of MathematicsStable URL: http://www.jstor.org/stable/1968610 .

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ANNALS OF MATHEMATICSVol. 38,No. 3,July,1937

PLATEAU'S PROBLEM AND DIRICHLET'S PRINCIPLEBy RICHARDCOURANT

(Received December 15, 1936; revised May 11, 1937)CONTENTrS

1. Introduction.2. Preliminary Lemmas on the Dirichlet Integral.3. Plateau's Problem forone contour (k = 1).4. Plateau's Problem fork contours. Preparations.5. Solution of Plateau's Problem fork = 2.6. Plateau's Problem for arbitrary k.7. Solution of Plateau's Problem and the problem of least area based on conformalmapping.8. Plateau's Problem for one-sided minimal surfaces and for minimal surfaces ofhighergenus.

1. INTRODUCTIONThe constructionfa surface f east area bounded by a givencontourr in

the3-dimensional uclidean xl, x2,x3-spaces one of the classical problems nthe calculusofvariations.1 To formulateheproblem nalyticallywe supposethe surface nderconsiderationepresented yfunctions ,(u, v) oftwoparam-etersu, v orbya vectorg(u,v)withthex, as components)n a givendomainBoftheu, v-planewiththe boundaryC,;thesefunctionshallbe continuousntheclosed domainB + C, have piece-wise ontinuous econd derivatives n B andmap C on r. Then theproblemrequiresus to minimizeby one of thesead-missiblefunctionshe integral(1) A(S) = ff\EG - F2) dudv,wherewe havewiththe usual notations

E ?2 E /aXN2. G 2 ax(;2;F = -u X_

,A au avThis integralA(g) is invariantunder arbitrary ransformationsf the param-etersu, v and theirdomainB, which, f r is a simpleJordancurve,may bechosen s theunitcircleu2+ v2 1.

1 Cf. throughoutthis paper as a referencethe excellent reportby Rad6 "The problem ofPlateau," Ergebnisse der Mathematik, II, 2, Berlin, 1933.679


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680 RICHARD COURANTThe Euler conditions fthisvariationalproblem orm systemofnon-linearpartialdifferentialquations n B withadditionalnon-linearonditions,xpress-ingnot onlythat the required urface s bounded by r, but also that it hasmeancurvature eroor s a "minimal urface." Replacingtheproblem f eastareabythatof minimal urface oundedby r-this problemscalledPlateau'sproblem2-onehas to meetfirst he difficultyonsistingn thenon-linearityfthedifferentialquationsand secondthedifficultyonsistingnthenon-linearityof the additional conditions. Riemann and afterhim Weierstrass,H. A.Schwarz,Darbouxandothers avealready ntirely isposed f hefirst ifficulty.By takingadvantage ofthe freedomn the choice of the parameters ne canlinearizethe differentialquations of the least area problem. Suppose it ispossible to introducesometric arameters , v on thesurfaces consideredn

ourvariationalproblem, .e. parameters orwhich(2) E-G =O; F = O.or, n otherwords,parameterswhichcorrespond o a conformalmappingof thesurface on theu,v-plane. Thenwe have(3) A(S) = D(S) = DB(S)with(4) D(S) = (E + G)du d = _ + x') duAwhereD(r) is theclassicalDirichlet ntegral.In general, fu and v are notnecessarilysometric arameters,we have thatA S) f VEG duv, and herefore(5) A(S)


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 681This leads to the definition f a minimal urface s one forwhich, n a suitableparametric epresentation, e have the ineardifferentialquation(7) At= 0with he additionalnon-linear onditions(8) E-G=F=O.In otherwords, Plateau's problem requires us to finda "potential vector"X(u,v) in B + C whichgives a conformalmapping of B on a surfaceboundedby r.3If we consider otential unctions ,, s therealparts9tf,(u+ iv) = Tf,(w) ofanalyticfunctions,(w) of the complexvariable w = u + iv, t follows t oncethat(8') (E-G) -2iF = E [f,,(w)]2 p(w)is an analytic function fw = u + iv for any potential vectorX. The con-dition 2) that our potential urfaceS withAX= 0 be a minimal urface anbe expressedby(9) (w) = 0.

It may be remarked hat the boundary conditions epresent wo non-linearrelationsbetweenx1, X2, X3on C while the two additional relations 2) or (9),which have the appearance of two additional non-linearpartial differentialequations,amountto only one morerequirement f the type of a boundarycondition. For, sincesp(w) s an analyticfunctionwithany potentialvector ,this functionmustvanishidenticallyf its real partE - G has theboundaryvalues zeroand if naddition he maginary artvanishes t one point.On thisdefinitionf a minimal urfaceRiemann, Schwarz,and othershavebased thesolutionofPlateau's Problemformany nteresting articular ases-thereby eviatingfrom he calculusofvariations s the starting oint. MorerecentlyGarnierobtained a rathergeneral resultby pursuing he line of Rie-mann's method. But it was only by combining iemann's dea againwiththeviewpoint nd withthe methodof the calculus of variations that eventually,in 1932,T. Rado and J. Douglas independentlyucceeded n giving satisfacto-rilygeneral xistence roof.4

3This linearization of thenon-linear Euler equations corresponds exactly to that of theEuler equations of geodesic lines by choosing the arc length as parameter. Incidentallythe linearization of the differentialequations of a minimal surface is only a special caseof a much more general fact concerning quasi-linear partial differentialequations of thesecond order in two independent variables (cf. Courant-Hilbert, Methodendermathemat-gschenPhysik, vol. II, Chapter III, in press).4Quite a different pproach to the problem of least area fromthat of the calculus ofvariations was attempted by Lebesgue. More recently McShane has also pursued thisline, and has established the solution fork = 1 under remarkably general conditions forthe surfaces in competition. Cf. Rado's report.


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682 RICHARD COURANTThe merit of Douglas's works extends beyond thesolution of the originalproblem f Plateau. Apart from somewhat reater eneralitywithrespect oadmissibleboundaries and apart fromhis remark hat the method pplies to

anynumberm ofdimensions f thexi, - *, xspace he has attacked theprob-lemntself n a broaderfront. He envisagedthe moregeneral nd much moredifficultask of the constructionf a minimal urfacewhich s bounded by kgiven contours Jordan urves) P1, ... , ,rkand whichhas a prescribed opo-logical structure, .g. is required to be one-sidedor two-sided nd to have aprescribed enus. Douglas has so far publisheda solutionof the problemfortwo-sidedminimal urfaces fthegenuszerofork = 1 and k = 2 and also forone-sided urfaces f the type of a Moebius stripwithk = 1. Moreoverhe hasannounced hepublication fa solutionnthe general ase.5Rado achieved his goal by first pproximating o the lower bound of theintegralA S) by means of polyhedral urfaces and then by mapping thesesurfaces onformally n the unit circle. Douglas, on the contrary, ays muchemphasis on avoiding the use of conformalmappingand rather on includingRiemann's mapping theoremas a consequenceof the solution of Plateau'sproblem orm = 2; i.e. for two-dimensional-space.6The representation f the minimal urfaceby a potential vector s and theconsideration f the relationships 5) (5a) betweenA s) and D(S) makes itplausible that Xis the solution of the variationalproblem:To minimizethe

Dirichlet ntegral (S) under he condition hat s maps thecircleB ona surfaceS boundedby P.7 This problemwillbe called Problem .However,Douglas does not make such a variationalproblem f the classicalRiemann-Dirichletype hebasisofhis reasoning. Instead,from hebeginning,he substitutes n the Dirichlet ntegralfor potentialvectors solely and then5 Journal of Math. and Phys., vol. XV (1936) p. 55-64 and p. 106-123. The second ofthese papers gives more detailed information about the proof, which is based upon thetheoryofRiemann's multiply-periodic9-functionson a Riemann-surface.A complete reference o Douglas' previous papers is included in his article "The problemof Plateau," Bull. Amer. Math. Soc. (1933) p. 227-251.6 However, Douglas also applies the theoryofconformal mapping ofpolyhedral surfacesto show that his solution gives the east area.7 If we assume the possibility of a conformalmapping on the unit circle for all surfacesadmitted to competition in the original variational problem A (S) = min., then fromourinequalities (5) and (5a) it follows immediately that the lower limits of A(r) and D(T)must be identical. Therefore, the solution of the problem for the Dirichlet integral alsosolves the original problem forthe area and satisfies E - G = F = 0 in addition to ax = 0,because for the solution Twe have A(T) = D(T). This reasoning fork = 1 which plays animportant role in Rado's proof was later emphasized also by Douglas "The mappingtheorem of Koebe and the problem of Plateau," Journ. ofMath. and Phys. vol. X (1931)pp. 106-130. It is true that the initial assumption requires some discussion. But it canbe verified n a rather elementarymanner and with sufficient enerality,even in the highercases fork > 1.


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 683transformshe ntegralD(S) bymeans ofPoisson's formulanto his well-knownfunctional, hichcontainsonlyboundaryvalues ofXon C:(10) H( ) = A 2f f 4(C) X(o)]2 dado,whereX(z) are theboundaryvalues ofgon C as functions fthe angle?Y. Nowhestartswiththeproblem fminimizing (X) ifall thosevectorsXon C whichmap C ina continuousmonotonicwayon r are admitted o competition.This ingeniousdeparturefromclassical lines to a variational problemnotimplying erivativesmakes t easy to establish heexistence f a minimizing .Therebythe complicationsn the method are shifted o the task ofexcludingharmfulingularitiesrom hesolution nd of dentifyinghesolutionwith theboundary alues ofa potentialvector atisfying2).In the case of two contourswherethegeneralization ftheboundaryfunc-tional H(X) becomes less elementary, hese complications re moremarked;and thiscertainly s trueall the more n the case of morecontours r ofhighergenus, whereDouglas, accordingto his announcement,will make essentialuse of thetheory f Abelianfunctions n Riemann surfaces farbitrary enus,also consideringheirdependenceon the moduliof thesurface.To link Plateau's problem with these deep and beautifultheorieswill be,when presented n detail, an achievementof highest nterest. But it seemsworthwhile o avoid thecomplications risingfrom heexplicit xpression y aboundary-functional,nd ratherto start directlywith a DirichletProblem Ias above,where hevectorsX ncompetition eed not be potentialvectors.8The methodwhich thushave developed on theline ofDirichlet'sPrinciplenot only solvesPlateau's problem nd themost generalproblem ormulated yDouglas for ny numberofcontours nd any prescribed opologicaltype, n avery simpleway, but it also allows us to solve Plateau's problem n cases ap-parently otaccessible oDouglas' originalmethod,n whichparts of thebound-aries are free nprescribedmanifolds fany dimension essthanm. Moreover,otherproblemsn geometry nd hydrodynamics, ot as yetsolved, seemacces-sible to thismethod.The method onsists ssentially ftwoparts. In thefirst artthe variationalProblem I is solved. Accordingto the fundamentalDirichletPrinciple thesolutionmustbe a potentialvector. In the secondpart thesolution s shown,by variationalmethods, o satisfy he condition 9) characterizing minimal

8Rado (1.c.p. 86 ff)has already observed that in the case of one contour, by a reasoningbased on such a variational problem, an alternative of Douglas' proof of the relationsE - G = F = 0 can be given. Moreover, as also was noticed by Rado (l.c.p. 77), in exclud-ing the possibility of a degeneration from his solution, Douglas uses an argument equiva-lent to a reasoning which has been applied in some of my formerpapers on the Dirichletprinciple and conformalmapping. In the present paper this reasoning is of basic impor-tance for the construction of the solution itself. Cf. Lemmas 5 and 6.


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684 RICHARD COURANTsurface. The domainB of the parametersu, v for~(u, v) in all cases can besupposed s boundedby circles, ne-sidednessr higher enusbeingtakencareofby proper oordination f certainboundaries. (Also othertypesof domainsBare suitable,and even, in case of highergenus,sometimespreferable.) Thevariationalproblem s dealt with,on the basis ofelementaryemmas. In thecase k > 1 or forhighergenusetc. someadditionalconditions re required opreventdegeneration f the domainB.For thetask of thesecondpart twoessentiallydifferent ethodsare offered.On the one hand one can avoid any use of theoremson conformalmap-ping nwhichcase thecondition - G = F = 0 appearsas a "naturalboundarycondition,"risingby suitablevariationswithrespectto thedegreeoffreedomleft n the boundaryrepresentation,nd inhigher ases by additionalvariationwithrespectto the domainB. For m = 2 then, ndeed,generaltheorems nconformalmappingresult s special cases.On the otherhand forthe secondpart one can use a quitedifferent ethodbasedonsimple, ut general heoremsnconformalmapping f domainsntheu,vplane,bounded byJordan urves. By consenting o the use ofthose lightlylesselementaryools one can simplifyhe taskof the second partconsiderably.Rado's and Douglas'smethodsyield minimal urfacewhich t thesame timehas the least possiblearea under the prescribed onditions. The same is truefor the methodpresented n this paper. However thereexist in many casesminimal urfaceswithgivenboundarieswhichdo notfurnish he least possiblearea (e.g. the catenoid, f the endpointsof the generating urve are nearlyconjugate). Aswas first oticedbyMax Shiffman9uchcasesofrelativeminimaare also accessibleto themethodofthispaper.An outline fthemethodhas alreadybeenpublished."0 In thepresent aperthe problem ork contours nd genuszero s treated n fulldetailwith the firstmethod voidingconformalmapping n ??3, 4, 5, 6. The second variantbasedontheoremsn conformalmapping sgiven n ?7; how themethod ppliesto thecase ofhigher enusand to thecase of onesidedminimal urfacess discussed n?8. The proof f a mappingtheorem sed in ?7, the detailed discussion fthetopologicaldegenerationsfthedomainsB used in ?8, theamplificationorthecase of freeboundaries nd other pplications nd extensions f the method redeferredo subsequentpublication.

2. PRELIMINARY LEMMAS ON THE DIRICHLET INTEGRAL1. Boundaryvalue problemand minimum f the Dirichlet integral. Wesupposethat wecan solve theboundaryvalue problem fthepotential quation9 Cf. a subsequent publication and footnote24.10Proceedings of the Natl. Acad. of Sciences, vol. XXII (1936) pp. 367-372, 373-375.Douglas's notes indicating his method for the general case of a prescribed topologicalstructure (see footnote 5) were eitherpublished or submitted to the editors while mycom-munications were in proofs. Therefore it seems fitting o state expressly that mypublica-tions do not contest Douglas's claim to priority.


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 685Ap(u,v) 0 for domainB in the u, v-planeboundedbykcirclesCO, , Ckor,what s equivalent, hatweknowGreen'sfunction or ucha domain. Fork = 1 the solutionis given by the elementary oisson integral. We furtherconsider s known'11hefollowingLEMMA 1. Let the unction (u, v) range ver he etofallfunctionsontinuousin thedomainB and on its boundary, aving iecewise12continuous irst eriva-tivesn B, and assuming rescribedoundary alueson theboundary 1, , C>.Let theDirichletntegral

D(q) = f (q + q) du vadmit inite alues. Thentheminimalvalued = d(B) ofD(q) is attained or ndonly or the unction = p, which olves hecorrespondingoundary alue prob-lemofAp = 0.

LEMMA la. Let thedomainB have n additional ircularboundaryC*. Letthe unction (u, v) satisfyhe ame conditionss in Lemma1forB and Ci, **.Ck, but withno boundary aluesprescribedn C*. Then the minimal valuesdoofD(q) is attained orand only orthepotentialunction = po, whichhas theprescribed oundary alueson Cl , ** *, Ck and has a vanishing ormalderiva-tiveon C*. TWe ave D(p0) = do = do(B).Heredo= do(B) refers o theminimumwithrespect o the prescribed ound-ary values on C, , C2 ... Ckand, in addition, arbitraryvalues on C*.The functiono can be obtained s the solution f an ordinary oundaryvalueproblem or circulardomainconsisting fB and theimageofB by reflectionon C*, wherebyboundarypoints correspondingy reflection arrythe sameboundaryvalues.We add the remark:The statementsubsists if instead of functions , pvectors , X re substitutedn theDirichlet ntegral.The followingemma notused for olvingPlateau's Problem n the simplestcase k = 1 in ?3) states that in thevariationalproblemD(q) = min = d orD(q) = min = do he ower imitd ordo s notnoticeably ffected,f we restrictthe rangeof competition orq by imposing he condition hat q shall vanish (orbe constant) n a pointset contained n a sufficientlymall neighborhood fafixed point P of B.LEMMA 2. Ifd is the ower imit fd(B) ord0(B)as in Lemma1 or la and d, thesame ower imit nder he dditional onditionhat vanishes na prescribedoint11A simple proof can be found e.g. in Hurwitz-Courant, Funkltionentheorie, rd ed.Berlin, 1930, part 3 and in Courant-Hilbert, Mlethoden er mathematischen hysik, Bd. II,Chapter 7, ?1 (in press).12 A function is called piecewise-continuous n B, if it is continuous except for isolatedpoints and a finitenumberofarcs ofcurves withcontinuous tangents in B whereitmay havediscontinuities of the firstkind.


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686 RICHARD COURANTsetA insidea circle E of radius r = e aroundP, thenwe have im, o d, = d, andmore recisely ? d, < d + o(e), where (e) is a quantity ependingnlyon e andtending o zerowith . This relationholdsuniformlyor all boundary aluesforwhich q I < llI with ixedM.The same s true f B contains hepoint t infinity,, denotes circle f radius1/e around he rigin nd thepoint etA liesoutside . -

PROOF. We certainly ave dE ? d because the minimum roblemdefiningEoriginates rom hat defining bythe additionofnew restrictions,husnarrow-ing the range of competition. We have to show that for sufficientlymall e,functionsf can be foundforwhichtheadditionalconditions satisfied nd forwhichD(qE) < d + a-,where -= a(e) can be chosen arbitrarilymall. Indeed,ifp is the functionn the original roblem orwhichD(p) = d we obtain fromanotherfunction satisfying he additionalconditions y putting E(u,v) = prwhere n polar coordinates, being the distance from he pointP, we have"

T(r) = O for r er(r)= 1 for r >\/e

2 log-r for e r e.loge eWe have further

D(r) = - -. as E Olog eand

D(qE) < D(p) + D'(pT)where he symbolD' denotesthat the Dirichlet ntegral s to be extended verr < V/e. Considering he inequalities T i < 1; I p < M and usingthe in-equality

[D((p, 4t)]2 [ff (pt-U (Vp4IV)udv]2< [ff I pu.1. + I (V'v 1)dudv]2_ D (f) D~f

we findfromD'(pT) = JJ 2(p2 + p2)dudv + ffV p(T2Q+ TV)dud< d. Now D(^) can beexpressedn thefollowing ay by r, insteadofr,a as independent ariables,if r~(r,p) is substituted:

2D(j) = r + 3i ) rdrdJ=I t (Ur + (crip)2 + 2(1 + xY)2t2}drdg2;:i 9 ~~1 6XL= ffT gr + 2[2) rdrdlp+ ef' f>2Xr Xr , Xi - _ rdrdso

+ e2R.Here, thefirsterm sequal to 2d. The coefficient of e2remains oundedas iseasily seen by Schwarz's inequality. Thereforewe inferfrom he conditionD(^) > d,by passing o the imite -O 0,that the coefficientf e:

fl fw { r2X7r r + X4, - 2rdrd) ptendsto zerowithe. AgainusingSchwarz's inequality,for a boundarystripp < r ? 1 and realizing hat


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694 RICHARD COURANTwhere he imit suniform ithrespect oall functions , orwhich X , |X7 I Iare equallybounded. This equation, by theusual product ntegration f thecalculus of variationand on accountofthepotentialequation At = 0 is im-mediately ransformednto

2wr(14) lim| X r,i3)rZg,,di5 0.r-1 JONow, from hedefinitionf theanalyticfunctionf w = u + iv

(p(w) = (gu- ig)2 = (E - G) - 2iF = E [fl(w)I2by ntroducingogwas new ndependent ariablewe find

2 _W _____\2 (x ax_w2


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PLATEAU S PROBLEM AND DIRICHLET' S PRINCIPLE 6954. Theorems on continuousdependence. The following PPROXIMATIONTHEOREM rovedby Douglas and Rado2' is an immediate onsequenceofthepreceding theory:If theJordanontoursm convergeith -. 0 to a Jordanontour in the

strongense nd if rFe spans a minimal urface . whose rea d. is bounded yaquantityM not dependent n E, then here xists limitingminimalsurfacebounded y with n area not xceedinghe owerimit fdeFor, let Of be the minimizingectorwhichmaps the unit circleB conformallyon S, and satisfies suitablethreepoint condition hen by No. 2 the equi-continuityf thevectorsXes established. Hence a subsequenceof the Xecanbe chosenwhichconverges niformlyo a potentialvectorXmappingB on asurface bounded by r. In this passage to the limitthe relations - G =F = 0 arepreserved. The relation (t) < lim nfd,follows xactly s in No. 2.In generalwe can,withrespect o the lower imitsd, state only ower semi-continuity,nd not continuity,n their dependenceon the boundary. Butundersuitablerestrictionsoncerninghe convergence f rib)to r the lowerlimitd can be shown o be continuous y a simpledirect easoning.CONTINUITY THEOREM. Let

Xi, = Xi + {i (XI, **X.) (i ,**m)bea transformationf heX-spacento heX'-space ith

axktransforming into F' and theminimizing ector belongingo F into anothervector 'continuousn B + C, with iecewise ontinuous irst erivativesn B andmappingB on a surface ounded y F'. Let the ower imitsbelongingo P andr' respectivelye d and d'. Then wehaved' < d(1 + 6), with = (1 + me)2 -_1tending o zerowithE.

PROOF. On accountofSchwarz's nequalitywe have,rfr~~~~a ~ ataXA I a~iaxA .D(W - I) =ZE J+ Zdudv < me2D(Q).iJ~k aXk au /\k axk av/Hence (VD(X') - /D(X))2 < m2eD(S)

and therefore ecause of D(t) = d, D(X') _ d' we have d' < (1 + m E)2d.5. Riemann'smapping heorem. AsDouglashas observed, heresult fNo. 3contains, or hespecialcase m = 2, (xi = x, x2 = y) whenr is a Jordan

curve n the x, y-plane,Riemann'smappingtheorem: hereexists conformalmapping ftheunit circle fthew-plane n the nteriorfa Jordan urveF in thex, y-plane.21 For referenceee Rad6's report. Douglas alsoproved moregeneral heorem, heretheboundedness fd, s not required.


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696 RICHARD COURANTThe following emarkmay be added: In thecasem = 2 the ntegral

A = (Xuyt- xyYu)dudvrepresentsor ll admissible ectorsX n the Problem the fixed rea A includedin the Jordan curve r. Therefore nstead of attempting o minimizetheDirichlet Integral we could just as well start with the variational problemD(T) - A = min. orILj (xu- yv)2 + (xv + yu)2}dudv min.This is exactly he famousvariationalproblemwhichRiemann consideredn hisdoctor's hesis. Of course, heminimum alue here s zero, nd accordinglyhesolution atisfies he Cauchy-Riemann quations.Riemann'smapping heorem or he nterior fJordan urves s obtainedbythismethodmerely nthe basis of the knowledge f Poisson's ntegral. Moreover,our method stablishesdirectly: The conformal appingof the nterior of theunit circle n the nterior ofa Jordan urver implies continuousne-to-onemapping f theboundaries and r (See also No. 6). Indeed, that C is mappedcontinuouslyn r was shown bove as a consequence f emma5. Conversely,toeach point on r correspondsnlyone pointon C. This also followsfromlemma5. For, fthe nverse onformalmapping fG onB is given by thefunc-tionsu(x, y), v(x,y)wehaveJJuX+ u2 + v Yz)dxdy = 2r.Now we consider pointQ on r and small circular rcs C, with the radiusraround Q which ogetherwith an arc of r bound simply onnected ubdomainsG,of G havingQ as boundarypoint. On accountof emma 5 we can make thediameter fthe mageofCr nB arbitrarilymallfor ufficientlymallvalues ofr.Hence the diameter f the imageB, of GCmust tend to zero togetherwith r.Consequentlythe nested domainsB, define single limitingpoint P on Cwhich orrespondsoQ.22

6. The solution f Plateau's problemfurnishes one-to-onemapping f theboundaries. For thegeneral aseofPlateaus problemwithm > 2 theone-to-onecorrespondencefthe boundaries an be proved s follows:Sinceaccording o our constructionhemappingofC on r is continuous nd22Forthe readerwho s interestedn conformalmapping s such, t maybe statedthatthesamereasoning ubsistsforthemappingofthe unitcircleon a generaldomain notnecessarily oundedbya Jordan urve. Thenourcircular rcsC define esteddomainsG. in G whose imiting ointsform "prime nd,"and it is sucha prime nd to whichsinglepointP onC corresponds. Ourreasoning stablishes heone-to-oneorrespondencebetween heprime-endsfG and thepointsofC.


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PLATEAU S PROBLEM AND DIRICHLET 'S PRINCIPLE 697eo ipso monotonicwe only have to prove that a whole arc b on C never cor-responds o the samepoint, .g. s = 0, on r.If b corresponds othe pointX= Owe could,by an elementary ransformation,mapB conformallyn a halfunit circleB' so that b corresponds o the diameterPQ. The minimizing ector s transformednto a vectorX' in B' for which,becauseofthe nvariance fthe Dirichlet ntegral, B'(X') = d.Now we completeB' to the wholeunitcircleby affixinghe complementaryhalf ircleB" and defineA'= 0 n B". Thenfor he unitcircleB = B' + B" wehaveD(X') = d, andX s anadmissible ectornourvariationalproblem. SinceA' s certainly ot a regularpotential vector n the whole circleB we have an-other admissiblepotentialvector with the same boundary values and withD(a) < d (Lemma 1), whichcontradicts he minimum-characterf d. There-foreno sucharcb can occur, nd our statements proved.234. PLATEAU'S PROBLEM FOR k CONTOURS. PREPARATIONS

1. Heuristic considerations. If the boundary r of the requiredminimalsurface consists fkoriented ordan urvesrP, * *, rk nd S is tobe mappedon a domainB in the u, v-planewith the boundary C consisting f k corre-spondingly riented urvesC1, *-., Ck,then we have no longer ompletefree-dom n thechoiceofthis domain. For,twodomains, n particular wo circulardomains, re conformallyquivalent, fand only fcertain onformalnvariants,the moduli, coincide. Withoutbasing our theoryon the knowledgeof thisfactwe therefore ill n our variationalproblem eave sufficienteewayforthechoiceofour circular omainB.In thecase k > 1, wemustexpectthat the problem tselfrequires ertainconditions or ts solvability. For, ifthecurvesare too farapart or otherwisemisplaced, hecorrespondingeast area problemmayhave a solutionconsistingof several separatedsurfaces ach bounded by only a part of the systemofcurvesr. In thiscase for he ower imit of the area in consideration e havea = 6' + 6" where ' and 6" are thecorrespondingower imits or hesystemrPconsistingfonepartof the r, and for hecomplementaryystemrP".It is easily eenthatunder ll circumstanceshe nequality ? 6' + 6" holds.Indeed, a systemof two surfacesS' and S" boundedby F', rP" esp. alwayscan be modifiednto a surface withr as boundaryby joining S' and S" by atiny pipelikeconnectionwitharbitrarily mall area; therefore he lower imitcorrespondingo r cannot xceedthesum6' + 5".In the case a = 6' + 6" a minimizingequence i, 82, *.. ofsurfaces ounded

23 This proof does not make use of the character of our solution as a minimal surface.Taking advantage of the result E - G = F = 0 Douglas gives another proof in the fol-lowing way: Since the potential vector X s constant on b it can be analytically extended,by the principle ofreflection,beyond b, and therefore s regular on b. If s denotes the arclength n bwe have Xr X' = 0. But since , = 0 on b,theconsequence s Xr = 0 on b.This implies that on b all the firstderivatives of the potential vector Xvanish which leadsto the absurd consequence that X s constant throughout ts domain of existence.


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698 RICHARD COURANTby r and withareas tending o a mustbe expectedto have the tendency odegeneratentoat least two separateparts n thefollowingway: On Snthereexists closedcurveTnwith ts diameter ending o zero,so that T. separatesSn ntotwo parts, ne boundedby F', the otherby r".These heuristic onsiderationsorthe area do not apply immediately o ourDirichletntegral. We therefore illgive n No. 3 and in thefollowingectionpreciseformulationnd a proofof correspondingtatements orthe DirichletIntegral.

2. The variational roblem. We considern theu, v-plane circular omainB withboundaryC consistingf kcirclesC1, ** , Ck. Whether hisdomain senclosedn oneof thesecirclesC1or contains he nfiniteointof theplaneofthecomplexvariablew = u + iv doesnot matter. In thedomainB again ~(u, v)denotes vector, ontinuousnB + C, havingpiecewise ontinuous irst eriva-tives in B and mappingthe boundarycirclesmonotonicallyn the prescribedsenseon theprescribed ordan urves rF, * , rk in the m-dimensionalpace.The lower imitof the Dirichlet ntegralforall theseadmissiblevectorsg andfixedB is called d(B). The correspondingower imit, f not onlyA,but alsothecircular omainBA an be chosenarbitrarily, aybe calledd, the "absoluteminimum." (d is thereforehe ower imitof thevaluesd(B)). Now we againconsider

PROBLEM I. To find circular omainB and in it an admissible ector forwhichD(s) attains ts absoluteminimum . We expressly uppose as in thecase k = 1,that ourboundaryr allows finite aluesof the Dirichlet ntegral.If in our variationalproblemthe domainB is fixedwe shall occasionallydenote tbyProblem ' orbyProblem (B).It may again be emphasized hat because ofthe invarianceofthe DirichletIntegralunderconformalmappingwe can replaceB by any domainobtainedby a lineartransformationf the w-plane. In particularwe can chooseone ofthe circlesC, as the unit circleand establishon thisunitcirclea three-pointconditionust as fork = 1. Or we can requirethat two circlesCi and C2beconcentric,1being heunitcircle, nd that onefixed ointonC1be transformedintoa fixed ointon r1. Or wemay replacea domainB contained n a circleC1byanother ircular omainwhich ontains he nfiniteoint.We shall assumeB inside the unit circleC1. Then we consider subsetC' of the set of boundarycircles, nd among themC1. The correspondingcurves r1, - - - form subset rFof the boundaryr. The circlesC' definecirculardomainB' included n C1. The complementaryet C" ofboundarycirclesC, defines circular omainB" containinghepoint at infinity,nd thecorrespondingurves rF form subset F" oftheboundaryr so thatwe haveC' + C"I C, rF rF" F. Theoriginalomain istheproductfB' andB", that s the point et common othese wodomains.If in our variationalproblem or I(B) we replacethe domainB byB' orB"


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 699and r by F' or IF"we have to considerower imits (B'), d(B") for ixed omainsand the absoluteminimad' or d" whichdepend onlyon r, r r" respectively.

3. Lemmas concerning he lowerlimits ofD(T).LEMMA a. If r' and rI" are twocomplementaryets of boundary urvesF,i.e. r,+ I' = r and if d, d' and d" are the orrespondingbsoluteminima henwehaved < d' + d".PROOF. We consider a circular domain B' included in the unit circle C1so that there xists n admissible otentialvectorX'whichmapsB' on a surfacebounded by r' and whichfor a prescribed mall a satisfies BEX') < d' + 6.If 0 is an arbitrary oint n B', e.g. the origin,we cut outfrom ' a concentriccirculardisc Kc boundedby the small circle C* with the radius E around 0.For the remaining omainB' = B' - K, we consider the new variationalproblem btained by imposing he newboundary-condition = 0 on C*. ByLemma 2 we have for the correspondingower imit d'(B') the inequality

d'(B') < D(') + a(E) < d' + 6 + aC(E)

FIG. 2where (E) - 0 forE 0. Accordingly e have in B' an admissiblevector3'for which

DB;(3') < d' + a + a(E).In the same way we consider domain B" containing he pointat infinity;fromB" we cut out the exterior f a sufficientlyarge circlewith the radius1/E round0; in the remaining omainB" we have a vector " which vanishesat thecircumferencefthe argecircle,which atisfiesntheotherboundary ir-cles and in B" the conditions oradmission n our variationalproblem nd thecondition

D4'(a") ? d" + 6 + a(E).Now by a similarity ransformationith the factor 2 we contractB" into adomain ncluded nthecircleC* with the radius E around0 and again call thenew domain B"' and the correspondingector3". The value of theDirichlet


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700 RICHARD COURANTintegralDBE (3") is not changed by this transformation.For the domainB = B' + B" wehave certainly,f3denotes ', 3"respectively,

DB(J) < d' + d" + 23 + 2ar(E).If we impose on the vectors n the variationalproblemdefining (B) theadditional ondition = 0 on C* and thereforearrow he rangeofcompetition,weobtain lower imit (B, E)which ertainlysnot smaller han d(B). But ourcondition = 0 on C* obviously stablishes woindependent ariationalprob-lemsforB' andB" and g- is admissible. Therefore ehave

d < d(B) < d(B, e) _ DB(8) < d' + d" + 26 + 2o(e)and since6 as wellas E couldbe chosen rbitrarilymallwhiled doesnot dependon3and Ewehave d ? d' + d" as stated ntheLemma.The considerations f the precedingLemma are supplemented y anotherLemmapointingn the oppositedirectionnd useful aterfork > 2 in excludingthedegenerationf ourparameter-domain.We consider gain a finite irculardomainB = BE with the following roperty: he set of boundarycirclesCis divided into two complementaryets C' and C". One of these sets C"consists f circles nsidea circleC. oftheradiusr = e arounda fixed oint0 ofB. For all the other irclesC' thedistancer from shallremain bove a fixedquantityh. If Etends o zerowesaythatthedomainB, tends o eparation.

LEMMA b. If thedomainBE tends o separation hen orthe ower imitofthevaluesd(B,) the nequalitylim nfd(B,) _ d' + d"o

holds. Preciselyd(B,) > d' + d" -(e)

where (E) tends ozerouniformlyora fixedboundh andfora fixedboundM foron r. In theparticular asek = 2 themore recise imiting quationlim d(BE) = d' + d"C-0

is true.PROOF. Let B' be thefinite omain bounded by C'. Removing he nteriorK, ofthecircleC,,= C*with he radiusr = = Ve around0 from ' we obtainthe domainB' - K, = B.. Now we consider he lower imitdo(B,) of theDirichlet ntegralDB,(g) ifthevectorginB. maps C' onr , but does not haveprescribed oundaryvalueson C*. Lemma3 of ?2 statesthatdo(Bf) > d(B') - o'(rq) _ d' - (,q).whereo'(-) tends to zerowithi uniformlyora fixedboundh and fora fixedbound M for Ion r.


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 701In thesamewaywe obtain withcorrespondingotations

do(B$f')? d(B") - o"(-) ?> d" -a"(7)wherethe index -qrefers o a domainbounded by C" and the circleC, = C*and where heboundaryvalues on C* of thevectors n competitionre free.Now we return o our variationalproblem (Be) defining (B,). While intheproof f Lemma a we increased he lower imitby narrowinghe rangeofcompetition, e nowwiden the rangeof competition y easing the conditionsfor dmission, nd we thereforebtaina not arger owerbound in thefollowingway:We permit headmissible ectorsgto be discontinuoust thecircleC, = C*.In thisnewvariationalproblem he twodomainsB, and B;' are entirelyeparateand thereforehe new ower imit s equal to thesumofthecorrespondingower\?

P

R

FiG.3limitsdo(B,) and do(B$,'). We therefore btain, considering he inequalitiesabove,

d(B.) ? do(Bf) + d(B"') _ d' + d"X- '(i) -a(),which or ->0 anda(e) = '(i) + a"(I,) contains hestatement four emma.23aLEMMA C. If in a sequence fdomainsB. withe -> 0 twoboundary ircles

Cl, C2withradii above fixedboundcome loser hanE,then im, o d(B,) -> mthats the ower imits (B,) arenotbounded.PROOF. Withoutrestricting he generalitywe may suppose that C, is theunit circleand C2 approaches,not shrinking o zero, a fixedpoint R on C,.23a For k = 2 we knowfrom he considerations f Lemma a) that d(B.) : d(B') +d(B") + a(e) witha(e) tending o zero. Hence in thiscase wereally have the equalitysign imbo d(B) = d(B') + d(B") = d' + d".


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702 RICHARD COURANTThen forsufficientlymall a all circleswithradii r betweena fixed, mall,aand Va6 round R intersectC1 and C2 in two pointsP and Q. Since C1 ismappedon r1 nd C2 on r2we certainlyhave for very dmissible ectorXnB, thatI (P) -~(Q) I > a where is the positivedistancebetween rPand r2.But on accountof Lemma 5, ?2 there xistsa pair of pointsP, Q forwhich,withDB = M,

XP) - X(Q) < [i 109 a I]hence2

M, >-a I ogThis shows that also d(B,) > '(a2/r)* log a I, and since a can be chosenarbitrarilymall,our emma s proved.

5. SOLUTION OF PLATEAU'S PROBLEM FOR k = 21. Solutionof Problem . In the case of two contours k = 2) the solutionproceedsvery implybecauseB can be chosen s an annularring,C1 beingtheunitcircle, 2a concentricircle f the radiusq. Let d be the ower imitof theDirichlet ntegral n Problem of ?4,d' = d1, d" = d2the correspondingower

limitsforthe one-contourase withC1, r1respectively 2, r2 as boundaries.By Lemmaa we have d1+ d2 > d. If the equalitysignholds,thenthe lowerlimitis certainly pproximatedby a minimizing equence of surfaces end-ingto degeneration. We exclude hispossibilityyexpresslymakingn ProblemI the ssumption24for heboundaryr(16) d < di + d2.Now let i, , ***be a minimizing equence ofvectors;B1, B2 ... a corre-sponding equenceof annularrings. Lemma b of?4 showsthat the radii q. ofC2cannothave zero as lower imit, ecausefor uchsequencesofrings he owerlimitd(B,) mustbe not less thand1+ d2, whileaccording o our assumption(16), limDB,,(SP)< d1+ d2

24This condition can be interpretedas follows. We consider on a surface S bounded byrP and r2all closed curves topologically equivalent on S to these boundary curves. Let thelower limit oftheir diameters be a. Then, ifwe impose in Problem I the condition a = 0,the lower limitwill be di + d2 . Our inequality now requires that the additional conditiona = 0 increases the value of the lower limit d. This has been generalized by AM.hiffman(See also footnote9) in the followingway: Instead ofProblem I we consider a Problem Iaby requiringthat forour admissible surfaces we have a 2 a where a is any given number.If thenby imposing thenew condition a = a the lower limit is actually increased, it can beshown that Problem Ia has a solution which gives a minimal surface with a > a. Thissolution may be different rom that obtained by Problem I.The same principle can be applied to obtain other solutions of Plateau's problem if wereplace a by another suitable continuous functional of the surface g(u, v).


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 703On accountofLemma c in ?4 no subsequenceof theq, can tendto 1 becausethe Dirichlet ntegrals (^,) are bounded. Hence at leastfor subsequenceofthevthedomainsB, tend to a domainB whereB is an annularringbetween he

unitcircleCi and a circleC2having radiusqwith0 < q < 1.Theboundary alues fA,onthe oundaries i andC2 ofB, areequicontinuous.PROOF. If thiswerenot truee.g. on C, we could,on account of Lemma 6 of?2,find t least a subsequenceof the t, forwhich n arbitrarilymall fixed rcbon C1 containing fixedpointR wouldbe mapped by S, on an arc rP - evnearly coveringrPentirely; y,s a small part of rPtending o a single point,whichwe may assume,without oss ofgenerality,obe thepointX= 0. We alsomay assume that the minimizing equence X, consistsof potential vectors:A4Y= 0. Now we put X.= = + awhere and j are also potentialvectorsand where p= XonCi; a=0onC1

p=0onC2; a= XonC2.We haveD(x,) = D(s) = D(b) + D(3) + 2 f (jup + 3viv)dudv.

ApplyingGreen'sformuland observinghat = 0 on C1, 00 on C2,weobtain(17) D(x,) = D(b) + D(3) - 2 X rdswherehe ast ntegrals extendedver he ontour1 s denoteshe rc engthonC1,a/ardifferentiationith especto theradius .On thiscircle as/ar is equallybounded ya certain oundM. For thecomponentsf3vanish n C1 ndhence his ector an beanalyticallyxtendedas a potential ector, ytheprinciplefreflection,eyond 1 nto concentricring etweenheradii and 1/q. Since he omponentsf3areboundednthisring y131 A, where is thegreatest istance f r from he origin,hecomponentsf hederivativesnC1havetheboundA/(1 - q) = M. Forthevalues Jnbwe have ustas well heboundA,while nthecomplementaryarc C1 - b we havewith rbitrarilymallE and for ufficientlyargev theinequality 1X? e. Therefore,f1 s the ength fb,

2A2l A2 f 13ds < +2e 2r=a(v).1 a1-r1-Since1canbechosenrbitrarilymall s well s E, fv s arge nough,-(v) endstozero s v ncreases.Hencewehavefrom17)D ,) > D(t) + D(3) -a(v)or sinceD(t) > di,D(&) > d2,

D(&,) > di + d2 -a(V);


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704 RICHARD COURANTand forv - cc d > d, + d2, in contradictiono our assumption 16). Thisprovesthe equicontinuity f the boundaryvalues of If on C1, and similarlyon C2 -

On account of this factwe can choosea subsequenceof the Ifforwhichtheboundaryvalues convergeuniformly.Therefore he corresponding otentialvectors lso converge niformlyn B to a potentialvector whichmaps C on rand forwhich,with hesame reasonings in ?3, wehave D(S) < d and therefore,sincedisthe ower imit or dmissible ectors ndsince is admissible, (S) = d.Thus, Problem is solved by a potentialvector I, or by the surfaceSdefinedby I.2. The solution urnishes minimal urfaceS. Since C1, C2 are concentric

circleswe can proceed s in ?3 fork = 1. We perform variation fthe mini-mizing vector&(r,A) (in concentric olar coordinates)again replacing by3(r, ) =(r, sn),with sp= a + EX(r,A). Thus we findexactly as before,(18) lirn f (ri; 6)ri , odA - X(r2; ty)r2 d}= 0.

r2 - qAs before n ?3, No. 3 we concludefrom his equation: The analyticfunctionp(w) = E - G - 2iF is expressednB by p(w) = c/w2where is a real constant.For: we may choose X in the neighborhood f the two boundarycircles asidenticalwiththe normal derivativeof Green'sfunction ora fixed ingularpoint Q in B and the ring r2< r < ri. Then (18) expresses he fact thatthe maginaryart fW2y(w) vanishesnQ.25We have to showthat in the equationW2 (W) = c the constantc vanishes.Here our formerrgument f ?3 collapsesbecause the pointw = 0 does notbelongto B. But we obtain the desiredrelationc = 0 as the variationalcondition,not yet exploited,which states that we cannot reach a smallerlower imitd byvariation fthe domainB, i.e. ofthe radiusq. This variationwithrespect o the sizeof the circleC canbe performedythefollowingmethod.

To get ridofthe difficultyhich risesfrom ur lack ofknowledge bout thederivatives f Xat the boundaries,we replacethe domainB by anothercon-centric annular ringB' between the circlesCl: r = ri < 1 and C: r = = p > q;r, a again are concentric olar coordinates. We show first hat theminimumpropertyf oursolutionwithrespect o thedomainB implies similar ropertywithrespect to B'. The potentialvectorX(r, ) whichsolves ProblemI isanalytic on Cl and C' so that the values X(r1;A) = fi(ta) and X(p,A) = f2(6) areanalyticfunctions f the angle A~. The domainB consistsof the domainB'plusthe twoannularringsR1:r, < r < 1 and R2:q < r < p. Now weperform

25 If we do not want to make use of Green's functionfor a ring, but only of Poisson'skernel we may choose X in the neighborhood of one of the circles as this Poisson kernel andas zero otherwise; then our equation yields (cf. the general case discussed later in ?6,No. 2) that the imaginary part ofw2p(w)vanishes first n Cl and second on C2, thereforeidentically in B.


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 705variation withthe annular ringB' replacingthe innercircleC2 by anothercircleC*: r= a- tp. In the newringB* wedefine potential ector (r,9, a) =3(u, v,a) as thevectorwhichhas theboundary aluesfi(z9)on C' andf2&?)on C*.Obviouslywe have j(u, v, p) = (u, v). Now we state: The Dirichlet ntegralD*(^) = 2ff,2 + 2)dudv D(a-)(wherethesuperscript*ndicatesthatB* is the domainofintegration)s, as afunctionf a, a minimum or -= p; B* = B'. (In otherwords: Theminimumpropertybvious or he riginal ingB itself olds lso for oncentricings.)To prove our statementwe first ffix o our domain B* the originalringR,again; further e affixheannularringR*: tq < r < tp= a-to the nnercircleC* ofB*. R2 is transformednto R* justbya dilation. Now weconsidern theringdomainG = B* + R1 + R thevector which s identicalwithX n R1, to

FIG. 4S(u,v,a) inB* and whosevalue at everypointof R* is identicalwith he value ofX at the correspondingointofR2. The newdomainG and in it the vectorareadmissible or ompetitionnProblem . Therefore e have

Da(3) = DR1(j) + DR;(j) + D*(^) > d = DR1(X)+ DR2(X) + DB'(X);but according o ourdefinition,R8(a) = DR1(X) and DR;(a) = DR2(X). There-foretfollowsmmediatelyhatD*(^) > DB'(X), which s ourstatement.Now since our potentialvector3 depends analyticallyon the parameterathe same is truefor heDirichlet ntegralD*(^) = D(o-). Hencewe can differ-entiatewithrespect o theparameter -according o therulesoftheelementaryintegral alculus. Thuswe obtain theequation0 =- D(a) |= -f P(2 + F2)da + 2ff (3u o + &vivy)dudVBy usingGreen'sformula, utting s = pdA nd observing hat ja = 0 on Cl,we transformhis nto

0 = -f (X2+ X) ds -2 f Xrds with X = 3ol c-p.rp or=ffp


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706 RICHARD COURANTNow 2 + r2 + X.and according o thedefinitionfg(u,v, a)

=an ,o(p,A, cr)lamp - rrHence(19) f g _ &)ds = 0.By virtuefW29(W) = r2(&2 _ -2) - 2ir27r sourformula19) becomes

9Z w2p(w)ds = 0IWI-Pwhere J denotes the real part.Therefore ur variationof thedomainB results n thefollowing undamentalcondition:For every oncentricircle = p we have(20) 9 I w25p(w)ds= 0.

Substitutingn 20)c = w2p(w), here isreal,wefindmmediately= 0 or5p(w) 0. Thereforedefines minimalurface,nd Plateau's Problem ork = 2 is solved.3. Additionalemarks. Firstweobserve:Theminimumis the reaoftheminimalurface.Further: heminimumdependsn a loweremicontinuousay nthe ounda-ries r. Here again theboundariesr(e) whichapproach r in thestrong enseas E tends ozeroneednot onsistf imple urves. To prove he tatement emayassume hattheminima , belongingo theboundaries(e) areboundedinE. Weconsider subsequenceor such hatd. * dwherehebar ndicatesthe ower imitwith especto E. Wehavetoprove _ d.Now wehavetwo lternativeossibilities1) d, > d' + d' - X(e)with -Ofor E 02) There xists positive ixed anda subsequencefe tendingozero uchthatfor his ubsequence, < d' + d'e' a.In the irstasewehave mmediatelyor he owerimitshe elation > d' + d"sincefork = 1 already ' > d' d" > d" is known rom3,No. 2 andsinced' + d" > d on ccountfLemma ,the elation > d sestablished.In thesecond asewe consideror ursequence theminimizingotentialvectors f belonging o the variationalproblem for p(e) and the domainB.

withD(Xf) = de. Lemma (b) of?4 excludesdegenerationfB, because of theassumption) with ixed ositive . Henceusing lso Lemma of?2we canchooset east subsequenceforwhich ,tends o a limitingircularomain .Equicontinuityfthe f on theboundaryircles ollows xactly s beforenNo. 1 and thereforee haveat leasta subsequencefvectors e uniformly


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 707converging o a potentialvector r for whichd < d and which satisfies heconditions f the problem for the boundaryP and for the domain B. Ourtheorems thusproved.

At the same timewe recognize hat t defines minimal urface nd that wehaveproved he followingAPPROXIMATION THEOREM. Underthe assumption < d' + d - a a sub-sequencet least ftheminimal urfacesE forthe ontours(a) convergesoaminimal urfaceolving lateau's problemor he ontour, providedhat he reasofthe urfaces , remain ounded.26Exactlyas in ?3,No. 4 fork = 1ourstatement oncerningemicontinuityfdcanbe replacedby a Continuityheoremheformulationnd the proof f whichare dentical o that n ?3.4. Conformalmappingfordoubly connecteddomains. As an applicationofourtheory or hespecialcase ofm = 2 dimensions f the p-spacewe obtaintheresult:Every wofoldonnectedomainG in an x, y-plane ounded y woJordan urvesrF, r2 can be mappedconformallyn an annularringB. The conformalmap-pingof the pen domainsG and B establishes continuousne-to-oneorrespon-dencebetweenheboundaries.The mappingtheorem nd the continuity fthemappingofC on r follows

fromour generaltheoryform = 2 if we can verify priorithe inequalityd < d1+ d2wheredi and d2are theminimafor the correspondingminimumproblemsfor the singlecontours rF, r2 resp. Since for singlecontourstheproblemwas solvedand thusthe possibility fthemapping heoremwas provedwe know thatdi and d2 re theareas included n I, and P2 respectively. Ifweknew hatd also representsheareaofG wewouldhave immediately = d1 d2if, as we shall assume, P2 is included n IF; and all themored < d1+ d2.But the relationd = d1 - d2follows nlyafterwe have solved problem fork = 2 and thushaveprovedthe possibility fthe conformalmapping fB onG.Thereforet snowour taskto verifyhe nequality < di + d2directly.To thisendwe use a typical rgument fcontinuityased on the theorems fNo. 3. Suppose problem is solved for a domainG, thenthe equalityd =d- d2 and hencethe inequalityd < d1+ d2holds. Now let G' be a neigh-bouringdomain bounded by rP and IF and originating romG by a smalldeformation fthe x, y-plane s described n the continuityheorem f No 3.Then by thistheorem heinequalityd' < d' + d' willbe obtainedalso forthedomainG'; thisremark nables us, fwe startwith circular omainG,forwhichthe solvability fproblem and the conformalmapping s trivial, o pass withafinitenumberof steps to an arbitrarypolygonaldomain G. Precisely,wesuppose that for all domainsG in consideration he area di included in the

26 For k = 2 Douglas has shown that this theorem can be extended to cases ofunboundedareas. Cf. Rado's report.


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708 RICHARD COURANToutercontourrF is equally boundedby d1< M, whileall the inner ontoursr2 shall includean area d2 > p wherep is a fixedpositivenumber. If forGproblem is solved wehave certainly = di - d2.

Now we consider transformationf the x, y-plane nto the x', y' plane:xi = x + t(x, y), y' = y + 71(x, ), wherethefunctions, -q re continuous nG + r and thefirst erivatives fI, -q re allowedto be discontinous nlyalongstraightines nG. G', rI, d', dj, d2 refero the transformedomain. Withasmallpositive and 6 = (1 + 2E)2 _ 1we requireI Tz < 1E, I ty < 1E, 1 77z < Ei I ny < E,) 4 < E, 171< e

and we choose E so small that di < M, d > p, AM< 2p and I i-d'I < 'p.To thestraight ines ofdiscontinuitynG correspondnalytic ines n B by thesupposed conformalmapping and therefore ur continuity heorem can beapplied. It statesfor he ower imits nproblem forG'

d' < d(l + 2E)2 = d + d6 < d + MA < d + 2P,hence,since accordingto our assumption,d = di - d2 < di - p, we haved' < d - p + 2p < di < di + d'. ThereforeorG' the nequality ' < di + d2is proved;consequently roblem canbe solvedand d' is the area ofG'. Thusthe nequality ' < di + d2 mplies gainthemuch trongerelations

di = d - d2 < d + d - 2p.We apply thisresultby oining n arbitraryolygonal omainG = GN withcirculardomainGo by a sequence of domainsG1, G2, *-- with the followingproperties: ) G1, *** GN are bounded by polygons, ) for all thesedomainswehave the fixedbounds M, p described bove, c) Gris transformedntoG,+1bya transformationfthetypedescribed bove.26aSince forGoproblem issolvedbya domainB congruentoGO thesolvabilityfollowsuccessivelyorG,, G2, *-*, GN= G. By a passageto a limitwithpolygonaldomainsG we finally an solve the problem,on the basis of theapproximationheorem fNo. 3, for nydomainGboundedby arbitraryJordancurves. Therefore hepossibilityfthe conformalmappingof the nterior fBonthatofG is established nd at the ametime hecontinuousmapping fC on r.That also themappingof r on C is a continuous orrespondences seenexactlyas inthecase ofonecontourn ?3.Incidentally,we now canshow for rbitrarym nexactlythesame manner sinthecase k = 1 that: Our solution fPlateau's problemstablishes one-to-onecorrespondenceetweenhe oundaries and r.

26aThe possibility fconstructinguch a chain is obvious on thebasis ofelementarytopology. E.g. we nayassumethat twoconsecutive olygonal omainsdiffernly nonecorner. Thetransformationsfonepolygonntothenextone can be effectedy piecewiselineartransformations.


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 7096. PLATEAU'S PROBLEM FOR ARBITRARY k

In the case k > 2 wemay ease our taskby assuming hatPlateau's Problem salready olved for mallervalues of k and further hat the ower emicontinuityof the ower imitd in its dependence n the boundary r is already establishedforsmallervalues of k.Also thepossibility f conformalmappingof any domainB of the u, v-planeon a circular omain can be assumedprovided hatB is bounded by less thankJordan urves. This mapping mplies one-to-one orrespondencelso on theboundaries. Therefore t follows rom he initialremarksn ?2 that in varia-tionalproblems for essthankcontourstdoes not matterwhetherwe chooseBas a circulardomainor as any domainwith a correspondingumber fJordancontours.

In the Problem we may take C, as theunit circle, 2 as a concentric ircleinside of C, and the other circlesC, in the annular ringbetween C1 and C2.Again let Bn, g be a minimizingequenceof such circulardomainsand cor-responding otentialvectors. Again we makethe assumption(21) d < d' + d"forevery partitionof C in C' + C" and correspondinglyf r in r' + F"(cf.notations f ?4).Underthisconditionwe shall show: Theproblem has a solutionwhich ep-resents minimal urface ounded y r and conformallyquivalento a circulardomain B. Further: The theorems f ?5 concerning ower semicontinuityand continuity f the owerbound d withrespect o theboundary, heapproxi-mation heorem,hetheoremnthepossibility f conformal apping f any planedomainbounded yk Jordan urves n a circular omain, nd theone-to-oneor-respondencef he oundaries,ubsist lsofork > 2.

1. Solutionofthe variationalproblem.We state:a) The sequence ofdomainsB,-or a suitablesubsequence-convergesto acircular omainB.b) On each boundarycircleofB theXnre equicontinuous unctions f thearclength rtheangle.c) The potentialvectorsgn-or a subsequence-convergeuniformlyn B to apotential ectorXwhich olvesProblem .To provea) we have to excludethefollowing ossibilities:First: B tends oseparationnthe senseexplained n ?4, n which ase Lemmab) of?4wouldyieldd = limd(Bn) > d' + d" which s incompatiblewith 21).Second: Two boundarycirclesof B, X none of themshrinking o a point,become arbitrarilynear each other for certain increasing ndices n. Thiscontradicts emma c) of ?4.Third: Certainboundary circles of B,, shrinkto a point on one boundary


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710 RICHARD COURANTcircle .g. C1whoseradiusremains bove a positiveboundand which herefore-ifwechoosea suitable ubsequence-converges o a limitingircle.To excludethis last typeof irregularitys a somewhatmoredelicatetask.'We llustrate ur reasoning y thefigure for he typical asewhere he circleC3converges o a pointR on the unit circleC1. AroundR we again introducepolarcoordinates ,tA. On accountofLemma5 of ?2we see: For an arbitrarilysmallfixed -and sufficientlyargen-there existsa circular rc c ofradiusrwith6 < r < \/V round R joiningtwo pointsP, Q on C1 and including 3,on which heoscillation fthe vector = Sn iS essthan 27rM/Iog6 I]' = e(b),Mbeinga bound of the valuesD(gn). Without oss of generalitywe may assumethaton thearc cwe have Ign < e(b) for ll n.The arcc togetherwiththearc b': PRQ of C1 forms closedcurvec + b' andtogetherwith the complementaryrc b" ofC1 forms he closedcurve c + b".IfR is a pointofequicontinuity n C1 thentheoscillation four vectorgon thecurveT' = c + b' remains ess than a quantitywhichtendsto zerowith 6.IfR is a pointofnonequicontinuityhesameis truefor he curveT" = c + b"

( C (t(5)\g

FIG. 5on accountofLemma 6 ?2 at least afterwe choosea proper ubsequence .The imageofthearc b' by the representationn(U, v) is calledp3', hatofb" iscalled f3" o that we have f3'+ 13" = r,. One of thesearcs f3' r f3"has adiameter ending o zero with6,which lso is truefor hediameter fthe mage'y f c.c dividesB = Bn intothedomainsB' and B", withB' + B" = B and withconnectivities ' < k and k" < k so that our theory s establishedforthemaccording o our assumption. The boundariesof B' are mapped by gnon asystemof continuouscurves rI' and the boundariesofB" on another ystemof such curves r"5. If 6 = 8n ends to zero, rP'and rP" tend to a partitionof r into rI and rI" plus an isolated point g = 0 which howeveraccordingto Lemma 2 of ?2 will not affect he lowerlimitof the Dirichlet ntegral.Now in the minimum roblemdefining (B) = d(Bn) we case the conditionsfor dmission y allowinggto be discontinuousn c but only n sucha waythatthe boundaries fB' and B" remainmappedon P'a and rP". Then the lowerlimitcertainlywill becomesmalleror at most equal to the value d(B). The


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 711new minimum roblem efers bviously o two ndependent omainsB' and B"and the correspondingower imit s equal to the sum ofthe two ower imitsforB' and B" under the condition hat theirboundariesare mapped on P'a andIF"'. Thus in an obvious denotationwe have

d(B) ? d(B', rF') + d(B", F"6),hence all themore

d(B) > d(r') + d(Fr"b)ifon the right ide we introduce he absolute owerboundscorrespondingo theboundariesP'a and r"6. Now we letn tendto infinitynd at the same timeto zero; then the left ide of our inequality endsto d whileon accountof thesemicontinuityfd fork' < k and V" < k thelower imitofthe right ide cannot be lessthan d' + d". Therefore e obtain mmediately ? d' + d" whichcontradictsurcondition < d' + d". Hence thethirdpossibilityfdegenera-tionof ourminimizingequenceBn s excluded.Thereforehe sequenceofdomainsBn, or at least a subsequence,convergesto a circular omainB boundedby k circles. The equicontinuityfthebound-ary values of the vectors nnow follows xactly s in ?6 and also exactly n thesame way we obtain an admissible imiting otentialvectorT in B forwhichD(t) = d. This potential ector hereforeolves Problem defining surfaceinthe n-space.

2. The solution S is a minimal urface.To identifyhepotential urfaceS as a minimal urfacerequires omewhatdeeperreasoning ork > 2. We firststablish henaturalboundary onditionsonC, originatingrom he degree f freedomnthe boundary epresentationorthe curvesrI . For that we consider ne of theboundary ircles .g. C, (C1mayincludethe others). Let r, a again be concentric olar coordinates nd letX(r,A)be a function ithcontinuous erivativesnB + C, vanishing denticallyexcept in a neighboring ing adjacent to Ci. Then the natural boundarycondition orC1 s exactly s beforen ??4,5 uniformlynX:f2Xlim X(r,)rgrgodtd 0.r-1 JOOr ifweput as before

r~44 = X(w2P(w)) = p(u, v) = p(r,A),p(u, v) beinga potentialfunction,

f2r(22) lim J X(r, 2 p(r, 5) 3 =0.-1 JOIf Q is, as in ?3, a point n B we may again chooseX n thestrip djacent to C1as the Poisson kernelbelonging o the interior f the circlewiththe radius rand the pointQ. Then


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712 RICHARD COURANT2ro

)(p, t)p (p, ) dd = H~p(Q)= p(r,a)is a potentialfunction fu, v or r, a regularforr < p and equal to thefixedpotentialfunction on r = p. Now formula 22) states that H,(Q) tendstozero uniformlyor every closed subdomainas p tends to I. Fromthistherefollows yan elementaryeasoning:The potential unctionp has theboundaryvalueszero on C1.

PROOF. The difference(Q) - IHp(Q) = =p(Q) wp(r, a) is a potentialfunctionwhich has the boundaryvalues zero for r = p and existsin a ringl ? r < p withfixed which ies entirelyn B. On r = l the functionW, sboundedbecause p is regularnB, and forp -+ 1we have ll,(Q) -> 0. Now thepotential unction ,(Q) canbe analytically xtended yreflectioneyondr pinto hering < r = p2/l. Thereforen everynterior oncentricingRthederiv-atives of cp are equally boundedwith the same bound N. Now if Q in Bhas a distance essthan hto Cl and all themoreto Cp thenwe certainly ave forsufficientlymall h, Iwp(Q) < Nh. Therefore orp -* 1 because ofWp * pwe obtain,for Q fixed, p(Q) I ! Nh, whichproves hat p(Q) indeedapproacheszero uniformly henthedistanceh to Cl tendsto zero.Therefore is regularalsoonC1 and assumesthere hevalueszero.In thesamewaywe show: If a, is thecenter fthecircleC, ofradiusp,thenthe maginary artofthe functionw - a')2 so(W) vanishesonC, .(23) W(w- a,)2(c(w)] 0 on C,.These boundary onditionswhichfork = 2 and k = 1, a, = 0 amountto thefact that '(w2A(w)) = 0 throughout (not onlyat the boundaries)have herea morecomplex ppearance.Ourtaskis now to obtainthedesiredrelation '(w) = 0 inB from hesecon-ditions 23) by combining hemwith the furtheronditionswhichexpressthefact that variationsof the size and the positionof the boundarycirclesC,cannot mprove he lower imitd.

The variation ftheradiiof thecirclesC, yields nexactly hesamewayas in?6 theconditions(24) 9 f w -_a)2 o(w)dW 0where his ntegral s extendedovera circleC' concentric o C, and sufficientlynear to it.We shall showby a similar easoning hat the variation f the center , ofC,yieldsforC, the two additionalconditions(25) 9f (w- a) so(w)d = 0(26) $?f(w-a,,)sc(w)d = 0,


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PLATEAU S PROBLEM AND DIRICHLET 'S PRINCIPLE 713where again the integrals re extended over concentric irclesC' sufficientlynear toC,.To establish hesevariationalconditionswe perform s in ?6 a simple imul-taneousvariation f theminimal ector and theminimaldomainB. First wereplaceB by anotherdomainB' which s bounded by circlesC' concentricndsufficientlyloseto C, . Let R, denotetheringbetweenC and C'. The valuesof the minimizing otentialvectorXon C' are analytic functions,(6) of thepolaranglecorrespondingothis ircle.We nowreplaceB' byanother ircular omainB* depending n a parameterwith boundariesC* which originate rom he boundariesC' by parallel trans-formations.We furtherransferheboundary aluesf,t) from ' to C* bythecorresponding arallel transformation.With thesefixedboundaryvalues wesolvetheboundary alueproblem orB* and thusdefinen B* a potential ector

/~

FIG. 6a(u,v, e) whereby (u,v, o) =(u, v) is the originalminimizing ector. AffixingtheringR, to C* by a parallel transformation e obtain a domainBe = B* +E R*, R* denoting he ring R, after he parallel transformation. n B, wedefine vector which s equal to 3(u,v, E) n B* and which n the ring R* hasthesame values as S in thecorresponding ointsof the ringR,. Obviously hevectora and the domainB. are admissible n our Problem and hence becauseoftheminimum roperty fB and S wehave

DB4(3) = DB*(J) + Z DR:(a)= DB.(3) + > DR,( )> DB(Q) = DB'(X) + E DR,( ).


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714 RICHARD COURANTTherefore e obtainDB*(3) > DE (c). In otherwords: The minimum ropertyof Xwithrespectto the position of the centers ubsistsalso forsubdomainsboundedby circles oncentrico the boundary irclesC, .After hispreparationwe again can take advantageof the analyticcharacterof the boundary aluesofS onC'. We supposenowthat B* originates rom 'by shiftingnly one circleC' parallel to theu-axisby the distance , theothercirclesremaining ixed. Then a (u, v, e) has continuousderivatives f everyorderwithrespect o u, v,e27n B* and on its boundaryand from ur result tfollows hat

a DB*() =O fore = O.aAgain applying elementary ules for the differentiationf an integralwithrespect o a parameterwe get

2ff (auaue+ jV3V.)dudv (a + ^)dv = 0 fore = 0where hesecond ntegral s a line ntegral roundtheclosed curveC*. Green'stheorem n accountofAi = 0 now gives, f again r and t are polar coordinatesconcentrico C, and s = r4,

-2f3eards +f*(+ ) cos Mds=O fore = O.Now onC* for = 0we have that3r(u,v,o) = rand on accountofthe definition3f U, v,o) = U. But from r = 6osa + S, sin aweinfer-(u + cs) os 6 + 2u = (E - G) cosa + 2F sin

= (1/r) J[(w a,)X(w)I.Hence our equationbecomes(25) Jf.w-a,) p(w) 6= 0.In thesamewaywe find y performingvariation paralleltransformation)fC' parallelto the v-axisthat(26) (w - a)(w) d6 = 0.

The totality f ourvariationalconditions an be written n a conciseform.First 25) and (26) can be combinednthe form(27) J (w - a,) (w)dt6 0.

27 o verifyhisweonlyneedthePoissonkernel or single ircleC,


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PLATEAU S PROBLEM AND DIRICHLET IS PRINCIPLE 715Further, n C' we have w - a, = rei0 nd i(w - a,) d6 = dw. This last con-dition hereforean be writtenn the form(28) f


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716 RICHARD COURANTis a realperiodic unction ftheangle a forwhich, y (30) and (28),

2ro(31) f(6)fdY= 02v(32) f(f ) cos (a -Ao)dA = 0

holdsfor very o. But such a periodic unction (G) as indeed at least 4 zerosas described bove.3'Now let N ? 0 be the numberof zerosof p(w) in the interior fB. Thisnumbers expressed,f thecircleC, contains heother irclesC., by theformula2wriN= dlog o + .d logo(w)cl t~~v2 IV

FIG. 7wherenow, C' denotes path of ntegrationoincidentwithC, exceptfor ittlehalfcirclesircumventinghezeros ofyp n C, as infigure . But27riN d log [(w - al)2s0(w)]- d log (w_ a)2

+ E{f, d log [(w - av)20(W)] d log (w avx)2}or2riN = d log[(w - ai)2 (W)]- 47ri(33) Clk + I log [(w - )2(W)] + 47ri}.

31Theequation 31)shows hat here reat leasttwo uch ntervals )) > 0 andf(ty) 0separatedby2 zeros. If therewerenot at least four uch intervalswithalternatinglydifferentigns their umbermustbe even)with t least 4 zerosas endpoints,hat s, fweonlyhad twozeroswe could bychanging heorigin - 0 supposethesezeros n theform, and-,8. Then the function (cos iY-cos g) f(t) would have the same sign for all values oand the integralf cos t7 cos ,3)f(,Y)dtYould not vanish, in contradiction to (32).


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 717But each of the small halfeircles ircumventing zerowinds around this zeroin the negative ense as indicated n figure and leads from eal values of theintegrand w - a,)2'p(w) o real values and thereforeontributeso the ntegralmultiple f - ri while theotherpart of C' on which w - c)2 (w) is real doesnot contributenything o the imaginary art. Each of the ntegrals hereforecontributes t least - 4ri to the maginarypart; hencethe right ide of 33) is,exceptfor he factor , less than or equal to - 8Srand itwould follow hat

0 < 2grN - 8Sr,which s absurd. Consequentlywe have (p(w)= 0 identically nd our theoremis proved.

4. Additional emarks. a) Exactly in the same way as in the previousparagraphsby applyingourresultto the case m = 2 we obtain the followingtheoremnconformal apping:Everydomainof thex, y-planeboundedbyk Jordancurves an be mappedconformallyn a circulardomainof theu, v-plane.32The mapping mplies aone-to-oneorrespondencef theboundaries.b) Our solutionofPlateau's Problemfurnisheslso form > 2 a one-to-onecorrespondenceftheboundaries and r. The proofproceeds exactly on thesame lines as in ?3.c) The lower imitdependson theboundaryr in a lower emicontinuousay.Alsothisfactfollows xactly longthesamelines as in the case k = 2.d) The approximationheorem nd the continuityheorem f ??3, 5 remainvalid fork > 2.Thus the statementsn the introduction f ?6, assumed as true for k - 1contours,re established orkcontours nd hence theproof s completed.Finallywe notice: the owerimit is equaltothe reaoftheminimal urface .

7. SOLUTION OF PLATEAU'S PROBLEM AND THE PROBLEM OF LEAST AREA BASEDON CONFORMAL MAPPING1. Preliminary emarks. If insteadofbasingthe proofof Riemann'smap-ping theorem nd its generalizationsn thegeneral heory f Plateau's Problemwedo notdecline aking dvantageof fundamental actsconcerning onformalmapping of domains in a plane we can considerably implify he solutionofPlateau's Problem.33Thus we also can easilyestablishthefact: Oursolution fPlateau's Problemfurnishestthe ametime he urface f east rea bounded yr.

32 This theoremforarbitraryk has previously been proved in differentways by Koebeand by the author (cf. Hurwitz-Courant, Funktionentheorie,Berlin 1931). The proofcontained in the present paper seems to be essentially more elementary.33 For thesefacts cf.Hurwitz-Courant, Funktionentheorie, art III, Berlin, 1931,Courant,Crelle's Journ. Vol. 165 (1931) p. 247 ff. nd a paper to be published.


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718 RICHARD COURANTWe again start by solvingProblem as before. Then theuse of conformalmapping ervesto identifyhesolution s a minimal urface34The domainB can eitherbe chosenas a circulardomainas beforeor-in away more elementary rom the point of view of conformalmapping-as a"Schlitz" domain. A "Schlitz" domain s the whole u, v-plane ncluding hepointat infinity, ut cut along k straight egments, ll parallel to the u-axis.The two edgesofeachofthecuts form component , oftheboundaryC.

(C1[C2

'C3fC4FIG. 8The newmethodfor he solution f Plateau's Problem s based uponthetwofollowing heorems:

THEOREM (a). Everyk-fold onnected omain of the plane withk simpleJordan ontoursan be mappedconformallyn a domainB e.g. of the Schlitz"-typeB wherebyheboundariesre mapped n a one-to-oneontinuous ay.35THEOREM (f). Let a k-foldonnectedomainG in theu, v-plane onsist ftwoor morepartsG1, G2, ***whichmay be separatedor adjacentand whichareconnectedn the ollowing ay: Someanalytic rc oftheboundaryf partG,shallcorrespondy an analytic ransformationfw = u + iv toan analytic rc of theboundary f (another r the ame)G, and correspondingointsmaybeconsideredas identical. (Thus interior ointsof these rcs becomenterior oints of G.)Then thedomainG can bemapped onformallyna simpledomainB ofthe bovetype o that orrespondingoints nthe orrespondingarts ftheboundariesfthe

G, becomedentical nd that heboundary fG correspondsn a one-to-one aytothatofB. In this t is assumedthat ucha mapping s topologicallyossible.36In otherwords: Theorem$ says that a domain G connected n abstractoanbe realized n concretoeometricallyy conformalmappingof the composingdomains.An examplefor domainG as consideredn ourtheorem s theunit circle nthe u, v-plane, ut along the v-axisbetweenv = 0 and v = 2, the twoedges34Another method using conformalmapping was as mentioned before, pointed out byDouglas and Rad6 and in a somewhat differentway by McShane.35 The one-to-one correspondence of the boundaries has' to be interpretedin such a waythat the opposite points at a cut have to be counted as different.36 This theorem s also of a quite elementary character although apparently never statedexplicitly. I shall give a simple proof for (a) and (I) and for some generalizations onpiecewise Riemannian manifolds in a separate paper.


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 719beingcoordinated ythe transformation' = 2v2. TheoremA states: The unitcirclewiththis cut can be mapped on the unit circleG' with anothercut sothatafter hemapping orrespondingointsbecomecoincident.

v

UG G

FIG. 9Fromthe pointofview ofmethod tmaybe worthwhilementioninghat (a)and also (3) can be obtaineddirectly y nearlythe same reasoning s in??2-6onlyon the basis of the boundaryvalue problemofpotentialtheory nd thecorresponding inimum ropertyf theDirichlet ntegral. By thisremark hesubsequentproofof the least area property ould be made independentofpreliminarynowledgen conformalmapping.2. Solutionof Plateau's problem. SupposeProblem is solvedby thevectorX n the domainB. Ourtaskis to prove (w) = 0. We cutB byan arbitrary

straightegment through pointP; e.g. a segment = const.Thuswe obtaina new domainB' with he boundaryC' consistingfC and thetwo edgesof the cut L. BetweenthetwoedgesofL we establishan analyticone-to-one orrespondenceeavingthe endpoints fL intact. (In otherwordswe define domainG wherecorrespondingointson the two edges of L areidentified.)Now we replaceProblem by the following:PROBLEM II. Let 8 be a vectorn B' continuousn B' and its boundary ',

mappingC on r, having ontinuous oundary alues dentical t correspondingpoints f he wo dges f he utL andhaving iecewiseontinuousirst erivativesnB'. Thenwe askfortheminimum aluedoofDB' (8) obtained yadmittingllpossible orrespondencesetweenhe dges fL.In otherwords:werenouncen Problem continuityf alongL, butwe stillrequirethat Xmap the two edges ofL on the same curve n the n-space, hemappingsof the two edges being equivalentby an analytic transformation.Nowwe statethe followingTHEOREM: Theminimumo n Problem I is the ame as in Problem and is

attainedbytheoriginal olution ofProblem . This follows asilyby meansofour theoremsa) and (fi).We consider ny vector8 correspondingo a domainB' complyingwiththeconditions f Problem I for our domainB', and defining long the cut L aone-to-one nalyticcorrespondence.On accountof the theorems a) and (a)


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720 RICHARD COURANTwe can map B' conformallyn a new domainB* so that correspondingointsof the edges ofL becomeidentical. Let the vectorgpass by this conformalmapping ntothe transformedector3*. We certainly ave,by the nvarianceof the integralunderconformalmapping,DB*(3*) = DB'(3). By adding theimagepointsofL to B* we obtain a newdomainB and we are now surethat3* is continuous n B and that 3*satisfies he conditions f Problem forB.Hence

DB'(3) = DB*(3*) = DB(3*) > dwhich roves urtheoremince n theparticular ase ofProblem for heoriginaldomainB we have DB(9) = d.Problem I givesus the advantageofmorefreedom or hevariation hanwehad in Problem I. We exploitthe minimumproperty f the solution g ofProblem withrespect o Problem I in thefollowing ay: Assuming urcutLthrough as a segmentv = constantwe replace g(u, v) by 3(u, v) = &(u+EX(u,v), v) wheree is a smallparameter nd X(u,v) vanishes dentically very-where except n a small rectangleR adjacent to one edge of L; X is analytic,otherwiserbitrarynR, e.g. X = X(u)ju(v)whereX(u) = 0 at theendpoints fL and ji(v) = 0 at theupper dge ofR.

R? [LFIG. 10Nowwehave 3e(u, v,o) = Xgund thevariational quation

afaD(^) =0 fore= 0becomes, ince3andD depend nalytically n E, nd differentiationaybe per-formed nder he ntegral ignJJ3u ue + 3ovav)dudv0 fore = 0,or on accountofGreen'sformula

f 9xuvdu = 0.Because of thearbitrarinessf Xwe obtainon L, hence nP, gug, = F = 0.If insteadwe take L as a segment - v = const.we getexactly n thesameway (gu- g) u + g,) = E - G = 0wherebyur taskof dentifyingursolution s a minimal urfaces completed.The preceding easoning hows n general:If theconditionsoradmission n


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PLATEAU S PROBLEM AND DIRICHLET S PRINCIPLE 721Problem are lessened yadmittingiscontinuitiesf ofthe ype escribedbove,along finite umberf nalytic rcs, heminimumn thisnewProblemI remainsequalto d and is attained ythe olution ofProblem.

3. Our solutionfurnishes he least possible area. To provethe least areapropertyf our solutionwe recallthe followingefinitionfthearea of a surfaceS: If Hn is a sequenceofpolyhedralurfaceswhich pproachS, then the owestpossible imit fthe areasof all the polyhedras theareaofS.Now let us suppose first hat the boundary curves rF are polygons. Weconsider n arbitrary olyhedronH boundedby r withtriangular ides S,-whichis no restriction f generality-and the edges Al We subdivide thedomainB intoa setoftriangles , theedgesLI of which re straightinesorarcsofcircles o that the edgesLI and the triangles , correspond opologically oAl and S, .Now we pass from roblem with he ower imitd toProblem I whichdiffersfrom roblem inso far s theadmitted ectors areallowedto be discontinuousalongtheedgeLI mappingboth sidesofthe edgeLI on the same analyticcurveand establishingn analyticcorrespondenceetweenthe twoedges. Then weknowfromNo. 1 thatthe ower imit nProblem I againis d.Finallywe replaceProblem I by Problem II imposingon the vectorstheadditionalcondition hat they shall map LI on Al. For the lowerlimit d*of Problem II we certainly ave d* > d because the rangeof competitionnProblem II is narrowedwith respectto that in Problem I. But the lowerlimit nProblem II simplys equal to thesumof the correspondingower imitsreferringo themappingof the triangles , on triangularurfacesboundedbythe same linesA, as the plane triangle , and theminimums the area ofthetriangles ,. (The boundaryvalues in thismappingare analyticon accountof the principle f reflection.) Therefore * is exactlythe area of the poly-hedronH whiled is the area of the solutionof Problem . This proves ourstatement mmediately.

If P does notconsistof polygons t can be approximated y polygons, ndon accountof the definitionf the area S as the lower imit of the areas ofapproximating olyhedrawe obtain the desired generaltheoremmmediatelyfrom he preceding esult oncerninghe special case of polygonalboundaries.Withthisresultwe areable to interpreturpreviousnequalities < d' + d"etc. as inequalitiesreferringo lower imitsof areas,which ntuitivelysmoresatisfactoryhan theoriginaldefinition.8. PLATEAU'S PROBLEM FOR ONE-SIDED MINIMAL SURFACES AND FOR MINIMAL

SURFACES OF HIGHER GENUSIt mayhappen-as was emphasized yDouglas-that a smallerowerbounddfor heDirichlet ntegral s obtained f urfaces f a higher opological tructureboundedby r are admitted n ourvariationalproblems, .g. one-sided urfacesof the type of a Moebius stripor surfacesof highergenus topologicallynot


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722 RICHARD COURANTequivalent oa simpledomainB of he u,v-plane. In this?8we will howbrieflythatour methodcan disposealso of thesecases in fullgenerality.To this ndwehave to representurfaces with he boundaryr = r1+ rkby vectorsg(u, v) in a domain B of the u, v-planewith the prescribedtopological tructure. Such a domain B can be obtainedfrom simpledomainB* by establishing ne-to-one nalytic correspondencesetweencertainpartsof the boundariesof B* and by consideringorrespondingointsas identical,that s,attaching o them he samevaluesof thevectors . E.g. ifS and hence

CZ

KIG.1FIG. 11

_ ,,, ~ FI. 12

B shall be an orientable two sided)surface fgenuspwe mayconsider circulardomainB* containing he pointat infinity,oundedby k circlesC1, ***Ckwhich orrespondo r-and by p pairsK,, K' of circles. K, shall correspondtoK' by a linear ransformationfwwhich ransforms* into another irculardomainoutsideB*. Corresponding oints reidentifiednd thusB* becomesdomainofgenusp.Another uitablenormalform f a plane domainofgenusp withk boundarycurves s the "Schlitzdomain." (Cf. ?7). Such a Schlitz-domain onsistsofthe whole u, v-planecut by k straight egmentsC1, - , Ck parallel to the


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PLATEAU'S PROBLEM AND DIRICHLET S PRINCIPLE 723u-axis,further y p quadruples of such infinite uts as in figure12. Eachquadrupleof the nfiniteypeof cuts consists f two pairsof cutsparallelto theu-axisextending o u -> + X and lyingvertically bove the other,one pairseparating he otherpair. The edgesof the cuts correspond o each otherasindicatednfigure 2so thatthe correspondingointshave thesamecoordinate .Here also correspondingointsare identified. Also limiting ases, whencutsofdifferentairs partlycoincide mustbe considered.Non-orientableone-sided) urfaces an be topologically epresentedy planedomains n a similarway. E.g. we consider gain a circulardomain B* with aboundarycircleM which is transformednto itselfby an analytical (linear)transformationof the variablew- u - iv forwhichthe terated ransforma-tion T T is the identity nd which eaves no point on M invariant. Corre-spondingpointson M shall againbe identified; henB* becomesone sided. AdomainB boundedbyk circlesC1,by p pairsK,, K' ofcircles nd inadditionby qcirclesM1 ... Mq as described bove is, forq > 0, a one sided surfaceofthecharacteristic" = 2p + q. This characteristic umberN and thecharacterof orientabilityefine hetopological ypeof the surface.In a similarway one can consider chlitzdomainsB representingon-orien-table surfaces f givencharacteristic.A surface in ther-spaceboundedbythe fixed ontoursr can-as a limit fasequenceof such surfaces-degenerate nto a surfaceof smallercharacteristicor into two or more separate surfacesbounded by complementaryets rI,rat, .. of r respectivelynd havinga sumofcharacteristicsot exceedingN.If the prescribedopological tructures one sided,the degenerationmayyieldone or more ne sidedparts. Correspondinglye have to consider egenerationof the domainB in theu, v-plane. E.g. a sequenceof circulardomainsBnasabove maytendto separation xactly s in ?6, orone or moreofthe circlesM,mayshrink opointsorthesamemay happento a pairofcirclesK,, K'., etc.Similardegenerationsmay occurwith Schlitz domains. We shall call suchdegenerateurfaces ornon-degenerateurfaceswith lowercharacteristic iththe total boundaryr surfaces fa lower opological tructure.To investigate he possibility f a minimalsurfacewith the boundary rand the prescribedopological tructurewe consider gain thevariationalprob-lem for omainsB of one ofthe types definedabove, where the vectors ad-mittedmust have thesame values in equivalentboundarypoints.Exactly as in ?4 we findthe following esult. For a givenboundaryr wecannotmprovehe owerimit in Problem bypassingto a lower opologicalype.Now the following eneraltheoremholds:38If togivenr andgiven opologicaltructurehe owerimitd in Problem -orthe orrespondingrea of surface-isreally mallerhan or ower opologicalypes,

37 The characteristic and the character of orientability are the only topological invari-ants of our surfaces except forthe numberkof contours. For one sided surfaces it is alwayspossible to choose q eitheras 1 or as 2, or else to choose p as 0.38 See Douglas loc. cit. footnote 5.


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724 RICHARD COURANTthen roblem has a solution whichconstitutes minimal surface boundedby r having he prescribedopologicaltructurend moreoveriving he eastpos-siblearea spannedby r in comparisono ll surfaces fnothigher opologicalype.

To prove this theoremwe first ave to solve problem along the lines of ?6excluding egenerationfthe minimizingequenceB. of domains and securingequicontinuity f the minimizing ectorsby means of the supposed nequalities.Then we have to verify hat the solution s a minimal urface, itherby themethodof ?6 orby the methodof ?7, the latterof whichrequires slightgen-eralization f the mapping heorems f ?7. The details,not essentially ifferentfrom hose in the preceding ectionsof this paper, will be given in anotherpublication.In thesameway as forN = 0 thesolution an be recognized s the solutionofthe problemof least area.Concluding emarks. The methodsdeveloped n this paper permit pplica-tions ovariousothervariational roblems. At thisplace itmay onlybe statedthat a slightmodificationllowsus also to solve Plateau's Problem fthe bound-aries of the minimal urface re not entirely ixed, ut ifpartsofthe boundaryarcs are free o moveon prescribed urfaces f ess thanmdimensions. In thiscase, undersuitable assumptions, or a minimizing equence of domains andvectors he existence fa limiting omainB again can be established. On theboundaryC of the domain equicontinuityubsistson thosepartswhich corre-spondto fixed artsoftheboundary ,whilefor he otherpartsoftheboundaryC ofB a certainuniformityn theapproximationf thepointsX(u,v) to pointson the prescribed urfaces an be establishedwhereafterhe solutionproceedsin the same way as in this paper. The methodbecomes particularly imple fthenon fixed oundary urfacesre inear ndthereforeheprinciple freflectioncan be appliedto the constructionfthe solution.39NEW YORK UNIVERSITY.

39See a paper byE. Ritter oon to appear.

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