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林芳华韩青椭圆方程
INTRODUCTIONInFall1992,thesecondauthorgaveacoursecalled”intermediateP.D.E”attheCourantInstitute.ThepurposeofthatcoursewastopresentsomebasicmethodsforobtainingvariousAPrioriestimatesforthesecondorderpartialdi?erentialequationsofelliptictypewiththeparticularemphasisonthemaximalprinciples,Harnackinequalitiesandtheirapplications.Theequationsonedealswitharealwayslinearthough,obviously,theyapplyalsotononlinearproblems.ForstudentswithsomeknowledgeofrealvariablesandSobolevfunctions,theyshouldbeabletofollowthecoursewithoutmuchdi?cul-ties.Thelecturenoteswasthentakenbythe?rstauthor.In1995attheuniversityofNotre-Dame,the?rstauthorgaveasimilarcourse.Theoriginalnoteswasthenmuchcompleted,anditresultedinthepresentform.Wehavenointentiontogiveacompleteaccountoftherelatedtheory.Ourgoalissimplythatthenotesmayserveasbridgebe-tweenelementarybookofF.John[J]whichstudiesequationsofothertypetoo,andsome-whatadvancedbookofD.GilbargandN.Trudinger[GT]whichgivesrelativelycompleteaccountofthetheoryofellipticequationsofsecondorder.WealsohopeournotescanserveasabridgebetweentherecentelementarybookofN.Krylov[K]onclassicaltheoryofellipticequationsdevelopedbeforeoraround1960’s,andthebookbyCa?arelliandXivier[CX]whichstudiesfullynonlinearellipticequations,thetheoryobtainedin1980’s.CHAPTER1HARMONICFUNCTIONSGUIDEThe?rstchapterisratherelementary,butitcontainsseveralimportantideasofthewholesubject.Thusitshouldbecoveredthroughly.Whiledoingthesections1.1-1.2,theclassicalbookofT.Rado[R]onsubharmonicfunctionsmaybeaverygoodreference.Alsowhenonereadssection1.3,somestatementsconcerningHopfmaximalprincipleinsec-tion2.1canbeselectedasexercises.Theinteriorgradientestimatesofsection2.3followsfromthesameargumentsasintheproofoftheProposition3.2ofsection1.3.Inthischapterwewillusevariousmethodstostudyharmonicfunctions.Theseincludemeanvalueproperties,fundamentalsolutions,maximumprinciplesandenergymethod.Foursectionsinthischapterarerelativelyindependentofeachother.§1.MeanValuePropertyWebeginthissectionwiththede?nitionofmeanvalueproperties.Weassumethat?isaconnecteddomaininRn.1De?nition.Foru∈C(?)wede?ne(i)usatis?esthe?rstmeanvaluepropertyif1u(x)=ωnrn?1??u(y)dSy?Br(x)foranyBr(x)??;(ii)usatis?esthesecondmeanvaluepropertyifnu(x)=ωnrn??u(y)dyBr(x)foranyBr(x)??whereωndenotesthesurfaceareaoftheunitsphereinRn.Remark.Thesetwode?nitionsareequivalent.Infactifwewrite(i)asu(x)rn?11=ωn??u(y)dSy,?Br(x)wemayintegratetoget(ii).Ifwewrite(ii)asnu(x)rn=ωnwemaydi?erentiatetoget(i).Remark.Wemaywritethemeanvaluepropertiesinthefollowingequivalentways:(i)usatis?esthe?rstmeanvaluepropertyif1u(x)=ωn??u(x+rw)dSw|w|=1??u(y)dy,Br(x)foranyBr(x)??;(ii)usatis?esthesecondmeanvaluepropertyifnu(x)=ωn??u(x+rz)dz|z|≤1foranyBr(x)??.Nowweprovethemaximumprincipleforthefunctionssatisfyingmeanvalueprop-erties.2ˉsatis?esthemeanvaluepropertyin?,thenuassumesProposition1.1.Ifu∈C(?)itsmaximumandminimumonlyon??unlessuisconstant.????Proof.Weonlyproveforthemaximum.SetΣ=x∈?;u(x)=M≡maxu??.ˉ?ItisobviousthatΣisrelativelyclosed.NextweshowthatΣisopen.Foranyx0∈Σ,ˉr(x0)??forsomer&0.BythemeanvaluepropertywehavetakeB????nnM=u(x0)=u(y)dy≤Mdy=M.ωnrωnrBr(x0)Br(x0)Thisimpliesu=MinBr(x0).HenceΣisbothclosedandopenin?.ThereforeeitherΣ=φorΣ=?.De?nition.Afunctionu∈C2(?)isharmonicif??u=0in?.Theorem1.2.Letu∈C2(?)beharmonicin?.Thenusatis?esthemeanvaluepropertyin?.Proof.TakeanyballBr(x)??.Forρ∈(0,r),weapplythedivergencetheoreminBρ(x)andget???????u?u??u(y)dy=dS=ρn?1(x+ρw)dSwBρ(x)?Bρ?ν|w|=1?ρ??(?)?u(x+ρw)dSw.=ρn?1?ρ|w|=1Henceforharmonicfunctionuwehaveforanyρ∈(0,r)???u(x+ρw)dSw=0.?ρ|w|=1Integratingfrom0torweobtain????u(x+rw)dSw=|w|=1|w|=1u(x)dSw=u(x)ωn??u(y)dSy.?Br(x)or1u(x)=ωn??1u(x+rw)dSw=ωnrn?1|w|=1Remark.Forafunctionusatisfyingmeanvalueproperty,uisnotrequiredtobesmooth.HoweveraharmonicfunctionisrequiredtobeC2.Weprovethesetwoareequivalent.3Theorem1.3.Ifu∈C(?)hasmeanvaluepropertyin?,thenuissmoothandharmonicin?.??∞Proof.Choose?∈C0(B1(0))withB1(0)?=1and?(x)=ψ(|x|),i.e.??ωn01rn?1ψ(r)dr=1.??z??Wede?ne?ε(z)=1?forε&0.Nowforanyx∈?considerε&dist(x,??).Thenwehave????????y??1dyu(y)?ε(y?x)dy=u(x+y)?ε(y)dy=nu(x+y)?εε?|y|&ε??=u(x+εy)?(y)dy??==0|y|&11n?1??dr?B1(0)ru(x+εrw)?(rw)dSw??u(x+εrw)dSw|w|=1??01ψ(r)rn?1dr??01=u(x)ωnψ(r)rn?1dr=u(x)whereinthelastequalityweusedthemeanvalueproperty.Hencewegetu(x)=(?ε?u)(x)foranyx∈?ε={y∈?;d(y,??)&ε}.Thereforeuissmooth.Moreover,byformula(?)intheproofofTheorem1.2andthemeanvaluepropertywehave?????????n?1n?1??u=ru(x+rw)dSw=rωnu(x)=0foranyBr(x)??.?r?rBr(x)|w|=1Thisimplies?u=0in?.Remark.BycombiningTheorems1.1-1.3,weconcludethatharmonicfunctionsaresmoothandsatisfythemeanvalueproperty.Henceharmonicfunctionssatisfythemaximumprinciple,aconsequenceofwhichistheuniquenessofsolutiontothefollowingDirichletprobleminaboundeddomain?u=fin?u=?on??4forf∈C(?)and?∈C(??).Ingeneraluniquenessdoesnotholdforunboundeddomain.ConsiderthefollowingDirichletproblemintheunboundeddomain??u=0in?u=0on??.Firstconsiderthecase?={x∈Rn;|x|&1}.Forn=2,u(x)=log|x|isasolution.Noteu→∞asr→∞.Forn≥3,u(x)=|x|2?n?1isasolution.Noteu→?1asr→∞.Henceuisbounded.Next,considertheupperhalfspace?={x∈Rn;xn&0}.Thenu(x)=xnisanontrivialsolution,whichisunbounded.Inthefollowingwediscussthegradientestimates.ˉR)isharmonicinBR=BR(x0).ThenthereholdsLemma1.4.Supposeu∈C(B|Du(x0)|≤nmax|u|.ˉRRBˉR).Sinceuissmooth,then?(Dxu)=0,Proof.Forsimplicityweassumeu∈C1(Bii.e.,DxiuisalsoharmonicinBR.HenceDxiusatis?esthemeanvalueproperty.Bythedivergencetheoremwehave????nnDxu(y)dy=u(y)νidSyDxiu(x0)=ωnRnBR(x0)iωnRn?BR(x0)whichimplies|Dxiu(x0)|≤nnn?1max|u|?ωR≤max|u|.nˉRωnRn?BRRBˉR)isanonnegativeharmonicfunctioninBR=BR(x0).Lemma1.5.Supposeu∈C(BThenthereholdsn|Du(x0)|≤u(x0).RProof.Asbeforebythedivergencetheoremandthenonnegativenessofuwehave??nn|Dxiu(x0)|≤u(y)dS=u(x0)yωnRn?BR(x0)Rwhereinthelastequalityweusedthemeanvalueproperty.5Corollary1.6.AharmonicfunctioninRnboundedfromaboveorbelowisconstant.Proof.SupposeuisaharmonicfunctioninRn.Wewillprovethatuisaconstantifu≥0.Infactforanyx∈RnweapplyProposition1.5touinBR(x)andthenletR→∞.WeconcludethatDu(x)=0foranyx∈Rn.ˉR)isharmonicinBR=BR(x0).ThenthereholdsProposition1.7.Supposeu∈C(Bforanymulti-indexαwith|α|=mnmem?1m!|Du(x0)|≤max|u|.ˉRRmBαProof.Weprovebyinduction.Itistrueform=1byLemma1.4.Assumeitholdsform.Considerm+1.For0&θ&1,de?ner=(1?θ)R∈(0,R).WeapplyLemma1.4touinBrandgetn|Dm+1u(x0)|≤max|Dmu|.ˉrrBBytheinductionassumptionwehavenm?em?1?m!max|Du|≤max|u|.ˉrˉR(R?r)mBBmHenceweobtain|DTakeθ=m+1nnmem?1m!nm+1em?1m!u(x0)|≤?max|u|=max|u|.ˉRr(R?r)BRRθ(1?θ)Bm.Thisimplies1=θm(1?θ)??11+m??m(m+1)&e(m+1).Hencetheresultisestablishedforanysinglederivative.Foranymulti-indexα=(α1,???,αn)wehaveα1!???αn!≤(|α|)!.Theorem1.8.Harmonicfunctionisanalytic.Proof.Supposeuisaharmonicfunctionin?.For?xedx∈?,takeB2R(x)??andh∈Rnwith|h|≤R.WehavebyTaylorexpansion??????i??m?1??1??u(x+h)=u(x)+h1+???+hnu(x)+Rm(h)i!?x?x1ni=16where??????m??1??Rm(h)=h1+???+hnu(x1+θh1,...,xn+θhn)m!?x1?xnforsomeθ∈(0,1).Notex+h∈BR(x)for|h|&R.HencebyProposition1.7weobtain????mmm?121nem!|h|ne|h|m?nm?max|u|≤|Rm(h)|≤max|u|.ˉ2Rˉ2Rm!RmRBBThenforanyhwith|h|n2e&R/2thereholdsRm(h)→0asm→∞.NextweprovetheHarnackinequality.Theorem1.9.Supposeuisharmonicin?.ThenforanycompactsubsetKof?thereexistsapositiveconstantC=(?,K)suchthatifu≥0in?,then1u(y)≤u(x)≤Cu(y)Cforanyx,y∈K.Proof.Bymeanvalueproperty,wecanproveifB4R(x0)??,then1u(y)≤u(x)≤cu(y)cforanyx,y∈BR(x0)wherecisapositiveconstantdependingonlyonn.NowforthegivencompactsubsetK,takex1,...,xN∈Ksuchthat{BR(xi)}coversKwith4R&dist(K,??).ThenwecanchooseC=cN.We?nishthissectionbyprovingaresult,originallyduetoWeyl.Supposeuisharmonicin?.Thenwehavebyintegratingbyparts??2u??=0forany?∈C0(?).?Theconverseisalsotrue.Theorem1.10.Supposeu∈C(?)satis?es??2(1)u??=0forany?∈C0(?).?Thenuisharmonicin?.7Proof.WeclaimforanyBr(x)??thereholds????(2)ru(y)dSy=n?Br(x)Br(x)u(y)dy.Thenwehaveddr1u(y)dSyωnrn?1?Br(x)??????nd1=u(y)dyωndrrnBr(x)????????nn1=?n+1u(y)dy+nu(y)dSy=0.ωnrrBr(x)?Br(x)Thisimplies1ωnrn?1??u(y)dSy=const.?Br(x)??????Thisconstantisu(x)ifweletr→0.Hencewehave??1u(x)=u(y)dSyforanyBr(x)??.ωnrn?1?Br(x)Nextweprove(2)forn≥3.Forsimplicityweassumethatx=0.Set??(|y|2?r2)n|y|≤r?(y,r)=0|y|&r????andthen?k(y,r)=(|y|2?r2)n?k2(n?k+1)|y|2+n(|y|2?r2)for|y|≤rand2k=2,3,...,n.Directcalculationshows?(?,r)∈C0(?)and??2n?2(y,r)|y|≤r.?y?(y,r)=0|y|&rByassumption(1)wehave??u(y)?2(y,r)dy=0.Br(0)Nowweproveifforsomek=2,???,n?1,??(3)u(y)?k(y,r)dy=0Br(0)8then(4)??u(y)?k+1(y,r)dy=0.Br(0)Infactwedi?erentiate(3)withrespecttorandget??????ku(y)?k(y,r)dy+u(y)(y,r)dy=0.?r?Br(0)Br(0)For2≤k&n,?k(y,r)=0for|y|=r.Thenwehave????ku(y)(y,r)dy=0.?rBr(0)kDirectcalculationyields??(y,r)=(?2r)(n?k+1)?k+1(y,r).Hencewehave(4).Thereforebytakingk=n?1in(4)weconclude??????22u(y)(n+2)|y|?nrdy=0.Br(0)Di?erentiatingwithrespecttoragainweget(2).§2.FundamentalSolutionsWebeginthissectionbyseekingaharmonicfunctionu,i.e.,?u=0,inRnwhichdependsonlyonr=|x?a|forsome?xeda∈Rn.Wesetv(r)=u(x).Thisimpliesv????+andhencev(r)=??n?1??v=0rc1+c2logr,n=2c3+c4r2?n,n≥3whereciareconstantsfori=1,2,3,4.Weareinterestedinafunctionwithsingularitysuchthat???udS=1foranyr&0.?r?BrHencewesetforany?xeda∈Rn1log|a?x|forn=22π1Γ(a,x)=|a?x|2?nforn≥3.ωn(2?n)9Γ(a,x)=Tosummarizewehavethatfor?xeda∈Rn,Γ(a,x)isharmonicatx=a,i.e.,?xΓ(a,x)=0foranyx=aandhasasingularityatx=a.Moreoveritsatis?es???Br(a)?Γ(a,x)dSx=1?nxforanyr&0.NowweprovetheGreen’sidentity.ˉ∩C2(?).Theorem2.1.Suppose?isaboundeddomaininRnandthatu∈C1(?)Thenforanya∈?thereholds??u(a)=?????Γ(a,x)?u(x)dx??????u?ΓΓ(a,x)(x)?u(x)(a,x)dSx.?nx?nxRemark.(i)Foranya∈?,Γ(a,?)isintegrablein?althoughithasasingularity.ˉtheexpressionintherightsidegiveszero.(ii)Fora∈/?,???Γ(iii)Bylettingu=1wehavex(a,x)dSx=1foranya∈?.??Proof.WeapplyGreen’sformulatouandΓ(a,?)inthedomain?\Br(a)forsmallr&0andget??(Γ?u?u?Γ)dx=?\Br(a)??????Γ?u?Γ?u?n?n??dSx????Br(a)??Γ?Γ?u?u?n?n??dSx.Note?Γ=0in?\Br(a).Thenwehave??Γ?udx=???????Γ?u?Γ?u?n?n??dSx?lim??r→0?Br(a)??Γ?Γ?u?u?n?n??dSx.Forn≥3,wegetbyde?nitionofΓ????????????????1?u2?n????=??ΓdSr???n????(2?n)ωn?Br(a)???u??dS???n???Br(a)≤rsup|Du|→0asr→0n?2?Br(a)10and???Br(a)?Γ1udS=?nωnrn?1??udS→u(a)asr→0.?Br(a)Forn=2,wegetthesameconclusionsimilarly.Remark.WemayemploythelocalversionoftheGreen’sidentitytogetgradientesti-ˉ1)isharmonicinB1.Formateswithoutusingmeanvalueproperty.Supposeu∈C(B∞any?xed0&r&R&1chooseacut-o?function?∈C0(BR)suchthat?=1inBrand0≤?≤1.ApplytheGreen’sformulatouand?Γ(a,?)inB1\Bρ(a)fora∈Brandρsmallenough.WeproceedasintheproofofTheorem2.1andweobtain??????u(a)=?u(x)?x?(x)Γ(a,x)dxforanya∈Br(0).r&|x|&RHenceonemayprove(withoutusingmeanvalueproperty)????sup|u|≤cB1|u|pB11??andsup|Du|≤cmax|u|B1B1wherecisaconstantdependingonlyonn.NowwebegintodiscusstheGreen’sfunctions.Suppose?isboundeddomaininRn.ˉ∩C2(?).WehavebyTheorem2.1foranyx∈?Letu∈C1(?)??u(x)=???Γ(x,y)?u(y)dy????????u?ΓΓ(x,y)(y)?u(y)(x,y)dSy.?ny?nyIfusolvesthefollowingDirichletboundaryvalueproblem??(?)?u=fu=?in?on??ˉand?∈C(??),thenucanbeexpressedintermsoffand?,withforsomef∈C(?)oneunknownterm.WewanttoeliminatethistermbyadjustingΓ.Forany?xedx∈?,considerγ(x,y)=Γ(x,y)+Φ(x,y)11ˉwith?yΦ(x,y)=0in?.ThenTheorem2.1canbeexpressedforsomeΦ(x,?)∈C2(?)asfollowsforanyx∈???u(x)=???γ(x,y)?u(y)dy????????u?γγ(x,y)(y)?u(y)(x,y)dSy?ny?nysincetheextraΦ(x,?)isharmonic.NowbychoosingΦappropriately,weareledtotheimportantconceptofGreen’sfunction.ˉ∩C2(?)suchthatForeach?xedx∈?chooseΦ(x,?)∈C1(?)???yΦ(x,y)=0Φ(x,y)=?Γ(x,y)fory∈?fory∈??.Denotetheresultingγ(x,y)byG(x,y),whichiscalledGreen’sfunction.IfsuchGexists,thensolutionutotheDirichletproblem(?)canbeexpressedas??u(x)=???G(x,y)f(y)dy+???(y)?G(x,y)dSy.?nyˉforeach?xedNotethatGreen’sfunctionG(x,y)isde?nedasafunctionofy∈?x∈?.NowwediscusssomepropertiesofGasfunctionofxandy.FirstobservationisthattheGreen’sfunctionisunique.Thisisprovedbythemaximumprinciplesincethedi?erenceoftwoGreen’sfunctionsareharmonicin?withzeroboundaryvalue.Infact,wehavemore.Proposition2.2.Green’sfunctionG(x,y)issymmetricin?×?,i.e.,G(x,y)=G(y,x)forx=y∈?.Proof.Pickx1,x2∈?withx1=x2.Chooser&0smallsuchthatBr(x1)∩Br(x2)=φ.SetG1(y)=G(x1,y)andG2(y)=G(x2,y).WeapplyGreen’sformulain?\Br(x1)∪Br(x2)andget?????G2?G1(G1?G2?G2?G1)=G1?G2dS?n?n?\Br(x1)∪Br(x2)???????????????G1?G1?G2?G2?G2?G2?G1dS?G1dS.?n?n?n?n?Br(x1)?Br(x2)????SinceGiisharmonicfory=xi,i=1,2,andvanisheson??wehave???Br(x1)???G1?G2G1?G2?n?n??dS+??12?Br(x2)???G2?G1?G2G1?n?n??dS=0.Notetheleftsidehasthesamelimitastheleftsideinthefollowingasr→0?????????????G2?Γ?Γ?G1Γ?G2dS+G1?ΓdS=0.?n?n?n?n?Br(x1)?Br(x2)Whilewehave???G2ΓdS→0,?n?Br(x1)??Γ?Br(x2)?G1dS→0?nasr→0and???ΓG2dS→G2(x1),?n?Br(x1)??G1?Br(x2)?ΓdS→G1(x2)?nasr→0.ThisimpliesG2(x1)?G1(x2)=0,orG(x2,x1)=G(x1,x2).Proposition2.3.Thereholdsforx,y∈?withx=y0&G(x,y)&Γ(x,y)forn≥310&G(x,y)&Γ(x,y)?logdiam(?)2πforn=2.Proof.Fixx∈?andwriteG(y)=G(x,y).Sincelimy→xG(y)=?∞thenthereexistsanr&0suchthatG(y)&0inBr(x).NotethatGisharmonicin?\Br(x)withG=0on??andG&0on?Br(x).MaximumprincipleimpliesG(y)&0in?\Br(x)forsuchr&0.Next,considertheotherpartoftheinequality.Recallthede?nitionoftheGreen’sfunctionG(x,y)=Γ(x,y)+Φ(x,y)where?Φ=0Φ=?ΓForn≥3,wehaveΓ(x,y)=1|x?y|2?n&0fory∈??(2?n)ωnin?on??.whichimpliesΦ(x,?)&0on??.Bythemaximumprinciple,wehaveΦ&0in?.Forn=2wehave11Γ(x,y)=log|x?y|≤logdiam(?)fory∈??.2π2πHencethemaximumprincipleimpliesΦ&?1logdiam(?)in?.WemaycalculateGreen’sfunctionsforsomespecialdomains.13Proposition2.4.TheGreen’sfunctionfortheballBR(0)isgivenby(i)forn≥3G(x,y)=(ii)forn=2G(x,y)=12π??????????R|x|????log|x?y|?log??x?y??.|x|R??1(2?n)ωn??????2?n????|x|??R2?n??|x?y|???x?y??;|x|R??ˉRwithXthemultipleofxandProof.Fixx=0with|x|&R.ConsiderX∈Rn\B2|X|?|x|=R2,i.e.,X=Rx.InotherwordsXandxarere?exiveofeachotherwithrespecttothesphere?BR.Notethemapx?→Xisconformal,i.e.,preservesangles.If|y|=R,wehavebysimilarityoftriangles|x|R|y?x|==R|X||y?X|whichimplies(1)??????|x|??R|x|??|y?x|=|y?X|=??y?x??RR|x|??foranyy∈?BR.Therefore,inordertohavezeroboundaryvalue,wetakeforn≥3????????n?21R11G(x,y)=?.(2?n)ωn|x?y|n?2|x||y?X|n?2Thecasen=2issimilar.Next,wecalculatethenormalderivativeofGreen’sfunctiononthesphere.Corollary2.5.SupposeGistheGreen’sfunctioninBR(0).ThenthereholdsR2?|x|2?G(x,y)=?nωnR|x?y|nforanyx∈BRandy∈?BR.Proof.Wejustconsiderthecasen≥3.RecallwithX=R2x/|x|2????????n?21R2?nG(x,y)=|x?y|2?n?|y?X|forx∈BR,y∈?BR.(2?n)ωn|x|14Hencewehaveforsuchxandy1DyiG(x,y)=?ωn??xi?yi?|x?y|n??R|x|??n?2Xi?yi?|X?y|n??yiR2?|x|2=ωnR2|x?y|nfor|y|=Rby(1)intheproofofProposition2.4.Weobtainwithni=nyi???G1R2?|x|2(x,y)=niDyiG(x,y)=?.n?nwR|x?y|ni=1DenotebyK(x,y)thefunctioninCorollary2.5forx∈?,y∈??.ItiscalledPoissonkernelandhasthefollowingproperties:(i)K(x,y)issmoothforx=y.(ii)K(x,y)&0for|x|&R??(iii)|y|=RK(x,y)dSy=1forany|x|&R.ThefollowingresultgivestheexistenceofharmonicfunctionsinballswithprescribedDirichletboundaryvalue.Theorem2.6(PoissonIntegralFormula).For?∈C(?BR(0)),thefunctionude?nedby????K(x,y)?(y)dSy|x|&R?BR(0)u(x)=?(x)|x|=Rˉ∩C∞(?)andsatis?esu∈C(?)???u=0u=?Fortheproof,seeF.JohnP107-P108.Remark.InPoissonintegralformula,bylettingx=0,wehave1u(0)=ωnRn?1whichisthemeanvalueproperty.15???(y)dSy?BR(0)in?on??.Lemma2.7(Harnack’sInequality).SupposeuisharmonicinBR(x0)andu≥0.Thenthereholds??????n?2??n?2RRR?rR+ru(x0)≤u(x)≤u(x0)R+rR+rR?rR?rwherer=|x?x0|&R.ˉR).NotethatuisgivenbyPoissonIntegralProof.Wemayassumex0=0andu∈C(BFormula??1R2?|x|2u(x)=u(y)dSy.ωnR?BR|y?x|nSinceR?|x|≤|y?x|≤R+|x|for|y|=R,wehave1R?|x|?ωnRR+|x|??n?2??1u(y)dSy≤u(x)R+|x|?BR????n?2??1R+|x|1≤?u(y)dSy.ωnRR?|x|R?|x|?BR??u(y)dSy.?BR??Meanvaluepropertyimplies1u(0)=ωnRn?1This?nishestheproof.Corollary2.8.IfharmonicfunctionuinRnisboundedaboveorbelow,thenu≡const.Proof.Weassumeu≥0inRn.Takeanypointx∈RnandapplyLemma2.7toanyballBR(0)withR&|x|.Weobtain??RR+|x|??n?2R?|x|u(0)≤u(x)≤R+|x|??RR?|x|??n?2R+|x|u(0)R?|x|whichimpliesu(x)=u(0)bylettingR→+∞.Nextweprovearesultconcerningtheremovablesingularity.Theorem2.9.SupposeuisharmonicinBR\{0}andsatis?es??u(x)=o(log|x|),n=2o(|x|2?n),n≥316as|x|→0.Thenucanbede?nedat0sothatitisC2andharmonicinBR.Proof.Assumeuiscontinuousin0&|x|≤R.Letvsolve???v=0inBRv=uon?BR.Wewillproveu=vinBR\{0}.Setw=v?uinBR\{0}andMr=max|w|.Weproveforn≥3.Itisobviousthatrn?2|w(x)|≤Mr?n?2on?Br.|x|Notewand1?BrareharmonicinBR\Br.Hencemaximumprincipleimpliesrn?2|w(x)|≤Mr?n?2foranyx∈BR\Br|x|whereMr=max|v?u|≤max|v|+max|u|≤M+max|u|withM=max|u|.Hencewehaveforeach?xedx=0?Br?Br?Br?Br?BRrn?21n?2|w(x)|≤M+rmax|u|→0asr→0,?Br|x|n?2|x|n?2thatisw=0inBR\{0}.§3.MaximumPrinciplesInthissectionwewillusethemaximumprincipletoderivetheinteriorgradientestimateandtheHarnackinequality.ˉ1)isasubharmonicfunctioninB1,i.e.,Theorem3.1.Supposeu∈C2(B1)∩C(B??u≥0.Thenthereholdssupu≤supu.B1?B1Proof.Forε&0weconsideruε(x)=u(x)+ε|x|2inB1.Thensimplecalculationyields??uε=??u+2nε≥2nε&0.17Itiseasytosee,bycontradictionargument,thatuεcannothaveaninteriormaximum,inparticular,supuε≤supuε.B1?B1Thereforewehavesupu≤supuε≤supu+ε.B1B1?B1We?nishtheproofbylettingε→0.Remark.TheresultstillholdsifB1isreplacedbyanyboundeddomain.Nextresultistheinteriorgradientestimateforharmonicfunctions.ThemethodisduetoBernsteinbackin1910.Proposition3.2.SupposeuisharmonicinB1.Thenthereholdssup|Du|≤csup|u|B?B1wherec=c(n)isapositiveconstant.Inparticularforanyα∈[0,1]thereholds|u(x)?u(y)|≤c|x?y|αsup|u|?B1foranyx,y∈B1wherec=c(n,α)isapositiveconstant.Proof.Directcalculationshowsthat??(|Du|2)=2n??i,j=1(Diju)2+2n??i=1DiuDi(??u)=2n??i,j=1(Diju)2whereweused??u=0inB1.Hence|Du|2isasubharmonicfunction.Togetinterior1estimatesweneedacut-o?function.Forany?∈C0(B1)wehave??(?|Du|2)=(???)|Du|2+4n??i,j=1Di?DjuDiju+2?n??i,j=1(Diju)2.1Bytaking?=η2forsomeη∈C0(B1)withη≡1inB1/2weobtainbyH¨olderinequality??(η|Du|)=2η??η|Du|+2|Dη||Du|+8η??≥??1822222n??i,j=1DiηDjuDiju+2η2n??i,j=1(Diju)22η??η?6|Dη|2|Du|2≥?C|Du|2whereCisapositiveconstantdependingonlyonη.Note??(u2)=2|Du|2+2u??u=2|Du|2sinceuisharmonic.Bytakingαlargeenoughweget??(η2|Du|2+αu2)≥0.WemayapplyTheorem3.1(themaximumprinciple)togettheresult.NextwederivetheHarnackinequality.Lemma3.3.SupposeuisanonnegativeharmonicfunctioninB1.Thenthereholdssup|Dlogu|≤CB1whereC=C(n)isapositiveconstant.Proof.Wemayassumeu&0inB1.Setv=logu.Thendirectcalculationshows??v=?|Dv|2.Weneedinteriorgradientestimateonv.Setw=|Dv|2.Thenweget??w+2n??i=1DivDiw=2n??i,j=1(Dijv)2.Asbeforeweneedacut-o?function.Firstnote(1)n??i,j=1(Dijv)≥2n??i1|Dv|4w22(Diiv)≥(??v)==.nnn21Takeanonnegativefunction?∈C0(B1).WeobtainbyH¨olderinequalityn??i=1??(?w)+2=2?≥?n??DivDi(?w)n??i,j=1(Dijv)+42Di?DjvDijv+2w??n??i=1Di?Div+(???)w2i,j=1n??i,j=1(Dijv)2?2|D?||Dv|3?|???|+C19|D?|???|Dv|21if?ischosensuchthat|D?|2/?isboundedinB1.Choose?=η4forsomeη∈C0(B1).Henceforsuch?xedηweobtainby(1)n??i=1??(η4w)+2DivDi(η4w)1≥η4|Dv|4?Cη3|Dη||Dv|3?4η2(η??η+C|Dη|2)|Dv|2n1≥η4|Dv|4?Cη3|Dv|3?Cη2|Dv|2nwhereCisapositiveconstantdependingonlyonnandη.HencewegetbyH¨olderinequalityn??1??(η4w)+2DivDi(η4w)≥η4w2?Cni=1whereCisapositiveconstantdependingonlyonnandη.Supposeη4wattainsitsmaximumatx0∈B1.ThenD(η4w)=0and??(η4w)≤0atx0.Hencethereholdsη4w2(x0)≤C(n,η).Ifw(x0)≥1,thenη4w(x0)≤C(n).Otherwiseη4w(x0)≤w(x0)≤η4(x0).Inbothcasesweconcludeη4w≤C(n,η)inB1.Corollary3.4.SupposeuisanonnegativeharmonicfunctioninB1.Thenthereholdsu(x1)≤Cu(x2)foranyx1,x2∈B1whereCisapositiveconstantdependingonlyonn.bysimpleintegrationweProof.Wemayassumeu&0inB1.Foranyx1,x2∈B1obtainwithLemma3.3u(x1)log≤|x1?x2|u(x2)??01|Dlogu(tx2+(1?t)x1)|dt≤C|x1?x2|.NextweproveaquantitativeHopfLemma.20ˉ1)isaharmonicfunctioninB1=B1(0).IfProposition3.5.Supposeu∈C(Bˉ1andsomex0∈?B1,thenthereholdsu(x)&u(x0)foranyx∈B?????u(x0)≥Cu(x0)?u(0)?nwhereCisapositiveconstantdependingonlyonn.Proof.ConsiderapositivefunctionvinB1de?nedbyv(x)=e?α|x|?e?α.Itiseasytosee12ifα≥2n+1.Henceforsuch?xedαthefunctionvissubharmonicintheregionA=B1\B1/2.Nowde?neforε&0??v(x)=e?α|x|(?2αn+4α2|x|2)&0forany|x|≥22hε(x)=u(x)?u(x0)+εv(x).Thisisalsoasubharmonicfunction,i.e.,??hε≥0inA.Obviouslyhε≤0on?B1andhε(x0)=0.Sinceu(x)&u(x0)for|x|=1/2wemaytakeε&0smallsuchthathε(x)&0for|x|=1/2.ThereforebyTheorem3.1hεassumesatthepointx0itsmaximuminA.Thisimplies?hε(x0)≥0?nor?u?v(x0)≥?ε(x0)=2αεe?α&0.?n?nNotesofarweonlyusedthesubharmonicityofu.Weestimateεasfollows.Setw(x)=u(x0)?u(x)&0inB1.ObviouslywisaharmonicfunctioninB1.ByCorollary3.4(Harnackinequality)thereholdsBinfw≥c(n)w(0)or????maxu≤u(x0)?c(n)u(x0)?u(0).B1Hencewemaytake????ε=δc(n)u(x0)?u(0)forδsmall,dependingonn.This?nishestheproof.To?nishthissectionweproveaglobalH¨oldercontinuityresult.21ˉ1)isaharmonicfunctioninB1withu=?on?B1.IfLemma3.6.Supposeu∈C(Bˉ1).Moreoverthereholds?∈Cα(?B1)forsomeα∈(0,1),thenu∈Cα/2(B??u??Cαˉ(B1)≤C?????Cα(?B1)whereCisapositiveconstantdependingonlyonnandα.Proof.Firstthemaximumprincipleimpliesthatinf?B1?≤u≤sup?B1?inB1.Nextweclaimforanyx0∈?B1thereholds(1)supx∈B1α|?(x)??(x0)||u(x)?u(x0)|≤2sup.α|x?x0||x?x0|x∈?B1Lemma3.6followseasilyfrom(1).Foranyx,y∈B1,setdx=dist(x,?B1)anddy=dist(y,?B1).Supposedy≤dx.Takex0,y0∈?B1suchthat|x?x0|=dxandˉd/2(x)?Bd(x)?B1.|y?y0|=dy.Assume?rstthat|x?y|≤dx/2.Theny∈BxxWeapplyTheorem3.2(scaledversion)tou?u(x0)inBdx(x)andgetby(1)α|u(x)?u(y)|∞≤C|u?u(x0)|L(Bdx(x))≤Cdx?????Cα(?B1).dx|x?y|αHenceweobtain|u(x)?u(y)|≤C|x?y|?????Cα(?B1).Assumenowthatdy≤dx≤2|x?y|.Thenby(1)againwehave|u(x)?u(y)|≤|u(x)?u(x0)|+|u(x0)?u(y0)|+|u(y0)?u(y)|+|x0?y0|+dy)?????Cα(?B1)≤C(dxαααα≤C|x?y|?????Cα(?B1)since|x0?y0|≤dx+|x?y|+dy≤5|x?y|.Inordertoprove(1)weassumeB1=B1((1,0,???,0)),x0=0and?(0)=0.De?neK=supx∈?B1|?(x)|/|x|α.Note|x|2=2x1forx∈?B1.Thereforeforx∈?B1thereholdsααα?(x)≤K|x|≤2Kx1.De?nev(x)=2α/2Kx1α/2αinB1.Thenwehaveαααα?2??v(x)=2K?(?1)x1&0inB1.2222Theorem3.1impliesu(x)≤v(x)=2Kx1≤2K|x|foranyx∈B1.Considering?usimilarlyweget|u(x)|≤2K|x|foranyx∈B1.Thisproves(1).αααααα§4.EnergyMethodInthissectionwediscussharmonicfunctionsbyusingenergymethod.Ingeneralweassumethroughoutthissectionthataij∈C(B1)satis?esλ|ξ|2≤aij(x)ξiξj≤Λ|ξ|2foranyx∈B1andξ∈RnforsomepositiveconstantsλandΛ.Weconsiderthefunctionu∈C1(B1)satisfying??1aijDiuDj?=0forany?∈C0(B1).B1Itiseasytocheckbyintegrationbypartsthatharmonicfunctionssatisfyaboveequationforaij=δij.Lemma4.1(CacciopolliInequality).Supposeu∈C1(B1)satis?es??1aijDiuDj?=0forany?∈C0(B1).B11Thenforanyfunctionη∈C0(B1),wehave????η2|Du|2≤CB1B1|Dη|2u2whereCisapositiveconstantdependingonlyonλandΛ.1Proof.Foranyη∈C0(B1)set?=η2u.Thenwehave????λη2|Du|2≤Λη|u||Dη||Du|.B1B1WeobtaintheresultbyH¨olderinequality.23Corollary4.2.LetubeinLemma4.1.Thenforany0≤r&R≤1thereholds????C2|Du|2≤u(R?r)2BRBrwhereCisapositiveconstantdependingonlyonλandΛ.Proof.Takeηsuchthatη=1onBr,η=0outsideBRand|Dη|≤2(R?r)?1.Corollary4.3.LetubeinLemma4.1.Thenforany0&R≤1therehold????u2≤θu2BRBRand??|Du|2≤θB??|Du|2BRwhereθ=θ(n,λ,Λ)∈(0,1).1Proof.Takeη∈C0(BR)withη=1onBR/2and|Dη|≤2R?1.ThenLemma4.1yields??C|D(ηu)|2≤C|Dη|2u2≤2RBRBR????u2BR\BRbynotingDη=0inBR/2.HencebyPoincar′einequalityweget??(ηu)2≤c(n)R2BRBR??|D(ηu)|2.Thereforeweobtain??u2≤CBR??u2,BR\BRwhichimplies(C+1)??u2≤CB??u2.BRForthesecondinequality,observethatLemma4.1holdsforu?aforarbitraryconstanta.Thenasbeforewehave??????Cη2|Du|2≤C|Dη|2(u?a)2≤2(u?a)2.RBR\BRBRBR24Poincar′einequalityimplieswitha=|BR\BR|?1??BR\BRu|Du|2.??(u?a)2≤c(n)R2BR\BR??BR\BRHenceweobtain??|Du|2≤CB??|Du|2BR\Binparticular(C+1)??|Du|2≤CB??|Du|2.BRRemark.Corollary4.3implies,inparticular,thataharmonicfunctioninRnwith?niteL2-normisidenticallyzeroandthataharmonicfunctioninRnwith?niteDirichletintegralisconstant.Remark.ByiteratingtheresultinCorollary4.3,wehavethefollowingestimates.LetubeinLemma4.1.Thenforany0&ρ&r≤1therehold????ρu2u2≤C()μrBρBrand??ρ|Du|2≤C()μrBρ??|Du|2Brforsomepositiveconstantμ=μ(n,λ,Λ).Lateronwewillprovethatwecantakeμ∈(n?2,n)inthesecondinequality.Forharmonicfunctionswehavebetterresults.Lemma4.4.Suppose{aij}isaconstantpositivede?nitematrixwithλ|ξ|2≤aijξiξj≤Λ|ξ|2foranyξ∈Rnforsomeconstants0&λ≤Λ.Supposeu∈C1(B1)satis?es??1aijDiuDj?=0forany?∈C0(B1).(1)B1Thenforany0&ρ≤r,thereholds????ρ??n??|u|2(2)|u|2≤crBrBρ25and??(3)Bρ|u?uρ|2≤c??ρ??n+2??rBr|u?ur|2wherec=c(λ,Λ)isapositiveconstantandurdenotestheaverageofuinBr.????Proof.Bydilation,considerr=1.Werestrictourconsiderationtotherangeρ∈0,1,1since(2)and(3)aretrivialforρ∈(,1].Claim.2|u|2L∞(B)+|Du|L∞(B)≤c(λ,Λ)??|u|2.B1Thereforeforρ∈0,??1????Bρn|u|2≤ρn|u|2L∞(B1)≤cρ??|u|2B1and??|u?uρ|2≤Bρ??Bρn+2|u?u(0)|2≤ρn+2|Du|2≤cρ∞L(B1)??|u|2.B1Ifuisasolutionof(1),soisu?u1.Withureplacedbyu?u1intheaboveinequality,thereholds????|u?u1|2.|u?uρ|2≤cρn+2BρB1ProofofClaim.Method1.Byrotation,wemayassume{aij}isadiagonalmatrix.Hence(1)becomesn??λiDiiu=0i=1with??0&λ≤λi≤Λfori=1,???n.Itiseasytoseethereexistsanr0=r0(λ,Λ)∈??10,1suchthatforanyx0∈Btherectangle????|xi?x0i|x;&r0iiscontainedinB1.Changethecoordinatexixi?→yi=i26andsetv(y)=u(x).Thenvisharmonicin{y;estimatestoyield??ni=12λiyi&1}.Intheball{y;|y?y0|&r0}usetheinterior??|v(y0)|2+|Dv(y0)|2≤c(λ,Λ)Br0(y0)??v2≤c(λ,Λ){??ni=1v2.2&1}λiyiTransformbacktoutoget??|u(x0)|2+|Du(x0)|2≤c(λ,Λ)|x|&1u2.Method2.Ifuisasolutionto(1),soareanyderivativesofu.ByapplyingCorollary4.2toderivativesofuweconcludethatforanypositiveintegerk??u??Hk(B)≤c(k,λ,Λ)??u??L2(B1).1)iscontinuouslyIfwe?xavalueofksu?cientlylargewithrespectton,Hk(Bˉ1)andthereforeembeddedintoC1(B|u|L∞(B)+|Du|L∞(B)≤c(λ,Λ)??u??L2(B1).This?nishestheproof.27CHAPTER2MAXIMUMPRINCIPLESGUIDEMoststatementsinsection2.1arerathersimple.OneprobabillyneedstogooverTheorm1.8andProposition1.9.Section2.2isoftenthestartingpointoftheAPrioriestimates.Section2.4canbeomitedinthe?rstreadingaswewilllookatitagaininsection5.1.Themovingplanemethodexplainedinsection2.5hasmanyrecentappli-cations.Wechoiceaverysimpleexampletoillustaresuchmethod.TheresultgoesbacktoGidas-Ni-Nirenberg,buttheproofcontainssomerecentobservationsinthepaper[BNV].TheclassicalpaperofGilbarg-Serrin[GS]maybeaverygoodaddingforthischapter.ItmaybealsoagoodideatoassumetheHanackInequalityofKrylov-Safanovinsection5.2toaskstudentstoimprovesomeoftheresultsinthepaper[GS].Inthischapterwewilldiscussmaximumprinciplesandtheirapplications.Twokindsofmaximumprincipleswillbediscussed.OneisduetoHopfandtheothertoAlexandro?.TheformergivestheestimatesofsolutionsintermsoftheL∞-normofthenonhomogenoustermswhilethelattergivestheestimatesintermsoftheLn-norm.Applicationsincludevariousaprioriestimatesandmovingplanemethod.§1.StrongMaximumPrincipleSuppose?isaboundedandconnecteddomaininRn.ConsidertheoperatorLin?Lu≡aij(x)Diju+bi(x)Diu+c(x)uˉWealwaysassumethataij,biandcarecontinuousandhenceforu∈C2(?)∩C(?).ˉandthatLisuniformlyellipticin?inthefollowingsenseboundedin?aij(x)ξiξj≥λ|ξ|2foranyx∈?andanyξ∈Rnforsomepositiveconstantλ.ˉsatis?esLu&0in?withc(x)≤0in?.IfLemma1.1.Supposeu∈C2(?)∩C(?)ˉ,thenucannotattainthismaximumin?.uhasanonnegativemaximumin?ˉinx0∈?.ThenDiu(x0)=0Proof.Supposeuattainsitsnonnegativemaximumof?andthematrixB=(Dij(x0))issemi-negativede?nite.ByellipticityconditionthematrixA=(aij(x0))ispositivede?nite.HencethematrixABissemi-negativede?nitewithanonpositivetrace,i.e.,aij(x0)Diju(x0)≤0.ThisimpliesLu(x0)≤0,whichisacontradiction.Remark.Ifc(x)≡0,thentherequirementfornonnegativenesscanberemoved.Thisremarkalsoholdsforsomeresultsintherestofthissection.TypesetbyAMS-TEX27ˉsatis?esTheorem1.2(WeakMaximumPrinciple).Supposeu∈C2(?)∩C(?)ˉ.Lu≥0in?withc(x)≤0in?.Thenuattainson??itsnonnegativemaximumin?Proof.Foranyε&0,considerw(x)=u(x)+εeαx1withαtobedetermined.ThenwehaveLw=Lu+εeαx1(a11α2+b1α+c).Sinceb1andcareboundedanda11(x)≥λ&0foranyx∈?,bychoosingα&0largeenoughwegeta11(x)α2+b1(x)α+c(x)&0foranyx∈?.ThisimpliesLw&0in?.ByLemma1.1,wattainsitsnonnegativemaximumonlyon??,i.e.,supw≤supw+.???Thenweobtainsupu≤supw≤supw+≤supu++εsupeαx1.??????x∈??We?nishtheproofbylettingε→0.ˉtotheAsanapplicationwehavetheuniquenessofsolutionu∈C2(?)∩C(?)followingDirichletboundaryvalueproblemforf∈C(?)and?∈C(??)Lu=fin?u=?on??ifc(x)≤0in?.Remark.Theboundednessofdomain?isessential,sinceitguaranteestheexistenceˉTheuniquenessdoesnotholdifthedomainisofmaximumandminimumofuin?.unbounded.SomeexamplesaregiveninSection1inChapter1.Equallyimportantisthenonpositivenessofthecoe?cientc.Example.Set?={(x,y)∈R2;0&x&π,0&y&π}.Thenu=sinxsinyisanontrivialsolutionfortheproblem??u+2u=0in?u=0on??.28Theorem1.3(HopfLemma).LetBbeanopenballinRnwithx0∈?B.Supposeu∈C2(B)∩C(B∪{x0})satis?esLu≥0inBwithc(x)≤0inB.Assumeinadditionthatu(x)&u(x0)foranyx∈Bandu(x0)≥0.Thenforeachoutwarddirectionνatx0withν?n(x0)&0thereholds1??liminf[u(x0)?u(x0?tν)]&0.t→0+tRemark.Ifinadditionu∈C1(B∪{x0}),thenwehave?u(x0)&0.?νProof.WemayassumethatBhasthecenterattheoriginwithradiusr.Weassumeˉ)andu(x)&u(x0)foranyx∈Bˉ\{x0}(sincewecanconstructfurtherthatu∈C(BatangentballB1toBatx0andB1?B).Considerv(x)=u(x)+εh(x)forsomenonnegativefunctionh.Wewillchooseε&0appropriatelysuchthatvattainsitsnonnegativemaximumonlyatx0.Denote?α|x|2?αr21(x0).De?neh(x)=eΣ=B∩B?ewithαtobedetermined.WecheckinrthefollowingthatLh&0inΣ.DirectcalculationyieldsLh=e?α|x|2??????????4α2n??i,j=1n??i,j=1aij(x)xixj?2αaij(x)xixj?2αn??i=1n??i=1aii(x)?2αn??n=1??2bi(x)xi+c?ce?αr≥e?α|x|24α2??[aii(x)+bi(x)xi]+c.Byellipticityassumption,wehaven??i,j=1aij(x)xixj≥λ|x|≥λ2??r??22&0inΣ.Soforαlargeenough,weconcludeLh&0inΣ.Withsuchh,wehaveLv=Lu+εLh&0inΣforanyε&0.ByLemma1.1,vcannotattainitsnonnegativemaximuminsideΣ.29Nextweproveforsomesmallε&0vattainsatx0itsnonnegativemaximum.Considervontheboundary?Σ.(i)Forx∈?Σ∩B,sinceu(x)&u(x0),sou(x)&u(x0)?δforsomeδ&0.Takeεsmallsuchthatεh&δon?Σ∩B.Henceforsuchεwehavev(x)&u(x0)forx∈?Σ∩B.(ii)OnΣ∩?B,h(x)=0andu(x)&u(x0)forx=x0.Hencev(x)&u(x0)onΣ∩?B\{x0}andv(x0)=u(x0).Thereforeweconcludev(x0)?v(x0?tν)≥0foranysmallt&0.tHenceweobtainbylettingt→0?h1liminf[u(x0)?u(x0?tν)]≥?ε(x0).t→0t?νByde?nitionofh,wehave2?????h?h????(x0)=(x0)n?ν=?2αre?αrn?ν&0.?ν?nThis?nishestheproof.ˉsatisfyLu≥0Theorem1.4(StrongMaximumPrinciple).Letu∈C2(?)∩C(?)ˉcanbeassumedonlyonwithc(x)≤0in?.Thenthenonnegativemaximumofuin???unlessuisaconstant.ˉSetΣ={x∈?;u(x)=M}.ItProof.LetMbethenonnegativemaximumofuin?.isrelativelyclosedin?.WeneedtoshowΣ=?.Weprovebycontradiction.IfΣisapropersubsetof?,thenwemay?ndanopenballB??\ΣwithapointonitsboundarybelongingtoΣ.(Infactwemaychooseapointp∈?\Σsuchthatd(p,Σ)&d(p,??)?rstandthenextendtheballcenteredatp.IthitsΣbeforehitting??.)Supposex0∈?B∩Σ.ObviouslywehaveLu≥0inBandu(x)&u(x0)foranyx∈Bandu(x0)=M≥0.?u(x0)&0wherenistheoutwardnormaldirectionatx0totheTheorem1.3impliesballB.Whilex0istheinteriormaximalpointof?,henceDu(x0)=0.Thisleadstoacontradiction.??ˉsatis?esLu≥0Corollary1.5(ComparisonPrinciple).Supposeu∈C2(?)∩C(?)in?withc(x)≤0in?.Ifu≤0on??,thenu≤0in?.Infact,eitheru&0in?oru≡0in?.Inordertodiscusstheboundaryvalueproblemwithgeneralboundarycondition,weneedthefollowingresult,whichisjustacorollaryofTheorem1.3andTheorem1.4.30ˉCorollary1.6.Suppose?hastheinteriorspherepropertyandthatu∈C2(?)∩C1(?)satis?esLu≥0in?withc(x)≤0.Assumeuattainsitsnonnegativemaximumatˉ.Thenx0∈??andforanyoutwarddirectionνatx0to??x0∈??u(x0)&0?νˉ.unlessuisconstantin?Application.Suppose?isboundedinRnandsatis?estheinteriorsphereproperty.ConsiderthethefollowingboundaryvalueproblemLu=f(*)?u+α(x)u=??nin?on??ˉand?∈C(??).Assumeinadditionthatc(x)≤0in?andα(x)≥0forsomef∈C(?)ˉifc≡0oron??.Thentheproblem(*)hasauniquesolutionu∈C2(?)∩C1(?)ˉupα≡0.Ifc≡0andα≡0,theproblem(*)hasuniquesolutionu∈C2(?)∩C1(?)toadditiveconstants.Proof.SupposeuisasolutiontothefollowinghomogeneousequationLu=0?u+α(x)u=0?nin?on??.Case1.c≡0orα≡0.Wewanttoshowu≡0.ˉIfu≡const.&0,thiscontradictsSupposethatuhasapositivemaximumatx0∈?.?u(x0)&0bytheconditionc≡0in?orα≡0on??.Otherwisex0∈??andCorollary1.6,whichcontradictstheboundaryvalue.Thereforeu≡0.Case2.c≡0andα≡0.Wewanttoshowu≡const.ˉisassumedonlyon??Supposeuisanonconstantsolution.Thenitsmaximumin??ubyTheorem1.4,sayatx0∈??.AgainCorollary1.6implies(x0)&0.Thisisacontradiction.Thefollowingtheorem,duetoSerrin,generalizesthecomparisonprincipleundernorestrictiononc(x).ˉsatis?esLu≥0.Ifu≤0in?,theneitherTheorem1.7.Supposeu∈C2(?)∩C(?)u&0in?oru≡0in?.Proof.Method1.Supposeu(x0)=0forsomex0∈?.Wewillprovethatu≡0in?.31Writec(x)=c+(x)?c?(x)wherec+(x)andc?(x)arethepositivepartandnegativepartofc(x)respectively.Henceusatis?esaijDiju+biDiu?c?u≥?c+u≥0.Sowehaveu≡0byTheorem1.4.Method2.Setv=ue?αx1forsomeα&0tobedetermined.ByLu≥0,wehaveaijDijv+[α(a1i+ai1)+bi]Div+(a11α2+b1α+c)v≥0.Chooseαlargeenoughsuchthata11α2+b1α+c&0.Thereforevsatis?esaijDijv+[α(a1i+ai1)+bi]Div≥0.HenceweapplyTheorem1.4tovtoconcludethateitherv&0in?orv≡0in?.NextresultisthegeneralmaximumprinciplefortheoperatorLwithnorestrictiononc(x).ˉsatisfyingw&0in?ˉandTheorem1.8.Supposethereexistsaw∈C2(?)∩C1(?)ˉsatis?esLu≥0in?,thenucannotassumein?Lw≤0in?.Ifu∈C2(?)∩C(?)uuitsnonnegativemaximumunless≡const.If,inaddition,assumesitsnonnegativeumaximumatx0∈??and≡const.,thenforanyoutwarddirectionνatx0to??thereholds???u??(x0)&0?νwif??hastheinteriorspherepropertyatx0.Proof.Setv=u.Thenvsatis?esaijDijv+BiDiv+(Lw)v≥0wwhereBi=bi+2aijDijw.WemayapplyTheorem1.4andCorollary1.6tov.Remark.IftheoperatorLin?satis?estheconditionofTheorem1.8,thenLhasthecomparisonprinciple.Inparticular,theDirichletboundaryvalueproblemLu=fin?u=?on??hasatmostonesolution.Nextresultistheso-calledmaximumprinciplefornarrowdomain.32Proposition1.9.Letdbeapositivenumberandebeaunitvectorsuchthat|(y???x)?e|&dforanyx,y∈?.Thenthereexistsad0&0,dependingonlyonλandthesup-normofbiandc+,suchthattheassumptionsofTheorem1.8aresatis?edifd≤d0.??ˉliesin{0&x1&d}.AssumeinProof.Bychoosinge=(1,0,???,0)wesuppose?addition|bi|,c+≤NforsomepositiveconstantN.Weconstructwasfollows.SetˉBydirectcalculationwehavew=eαd?eαx1&0in?.??Lw=?(a11α2+b1α)eαx1+c(eαd?eαx1)≤?(a11α2+b1α)+Neαd.Chooseαsolargethata11α2+b1α≥λα2?Nα≥2N.HenceLw≤?2N+Neαd=N(eαd?2)≤0ifdissmallsuchthateαd≤2.Remark.Someresultsinthissection,includingProposition1.pareProposition1.9withTheorem4.8.§2.APrioriEstimatesInthissectionwederiveaprioriestimatesforsolutionstoDirichletproblemandNeumannproblem.Suppose?isaboundedandconnecteddomaininRn.ConsidertheoperatorLin?Lu≡aij(x)Diju+bi(x)Diu+c(x)uˉWeassumethataij,biandcarecontinuousandhenceboundedforu∈C2(?)∩C(?).ˉandthatLisuniformlyellipticin?,i.e.,in?aij(x)ξiξj≥λ|ξ|2foranyx∈?andanyξ∈Rnwhereλisapositivenumber.WedenotebyΛthesup-normofaijandbi,i.e.,max|aij|+max|bi|≤Λ.??ˉsatis?esProposition2.1.Supposeu∈C2(?)∩C(?)??Lu=fin?u=?on??ˉand?∈C(??).Ifc(x)≤0,thenthereholdsforsomef∈C(?)|u(x)|≤max|?|+Cmax|f|foranyx∈????33whereCisapositiveconstantdependingonlyonλ,Λanddiam(?).Proof.Wewillconstructafunctionwin?suchthat(i)L(w±u)=Lw±f≤0,orLw≤?fin?;(ii)w±u=w±?≥0,orw≥??on??.DenoteF=max|f|andΦ=max|?|.Weneed???Lw≤?Fin?w≥Φon??.Supposethedomain?liesintheset{0&x1&d}forsomed&0.Setw=Φ+(eαd?eαx1)Fwithα&0tobechosenlater.Thenwehavebydirectcalculation?Lw=(a11α2+b1α)Feαx1?cΦ?c(eαd?eαx1)F≥(a11α2+b1α)F≥(α2λ+b1α)F≥Fbychoosingαlargesuchthatα2λ+b1(x)α≥1foranyx∈?.Hencewsatis?es(i)and(ii).ByCorollary1.5(thecomparisonprinciple)weconclude?w≤u≤win?,inparticularsup|u|≤Φ+(eαd?1)F?whereαisapositiveconstantdependingonlyonλandΛ.ˉsatis?esProposition2.2.Supposeu∈C2(?)∩C1(?)??Lu=f?u??in?on??+α(x)u=?wherenistheoutwardnormaldirectionto??.Ifc(x)≤0in?andα(x)≥α0&0on??,thenthereholds??|u(x)|≤Cmax|?|+max|f|foranyx∈??????whereCisapositiveconstantdependingonlyonλ,Λ,α0anddiam(?).Proof.Specialcase:c(x)≤?c0&0.Wewillshow|u(x)|≤11F+Φc0α034foranyx∈?.De?nev=10F+10Φ±u.Thenwehave????11Lv=c(x)F+Φ±f≤?F±f≤0in?c0α0?????v11+αv=αF+Φ±?≥Φ±?≥0on??.?nc0α0ˉthenvattainsiton??byTheorem1.2,sayatIfvhasanegativeminimumin?,?????vx0∈??.Thisimplies(x0)≤0forn=n(x0),theoutwardnormaldirectionatx0.Thereforeweget?????v+αv(x0)≤αv(x0)&0?nˉinparticular,whichisacontradiction.Hencewehavev≥0in?,|u(x)|≤11F+Φforanyx∈?.c0α0Notethatforthisspecialcasec0andα0areindependentofλandΛ.GeneralCase:c(x)≤0foranyx∈?.ˉConsidertheauxiliaryfunctionu(x)=z(x)w(x)wherezisapositivefunctionin?tobedetermined.Directcalculationshowsthatwsatis?es????aijDijz+biDizfaijDijw+BiDiw+c+w=in?zz?????w1?z?+α+w=on???nz?nz1ˉsuchthatwhereBi=(aij+aji)Djz+bi.Weneedtochoosethefunctionz&0in?thereholdinc+aijDijz+biDiz≤?c0(λ,Λ,d,α0)&0in?z1?z1α+≥α0on??,z?n2oraijDijz+biDiz≤?c0&0in?z??????1?z??1??????z?n??≤2α0on??.35Supposethedomain?liesin{0&x1&d}.Choosez(x)=A+eβd?eβx1forx∈?forsomepositiveAandβtobedetermined.Directcalculationshows????1β2a11+βb1(β2a11+βb1)eβx11?aijDijz+biDiz=≥≥&0,zA+eβd?eβx1A+eβdA+eβdifβischosensuchthatβ2a11+βb1≥1.Thenwehave??????1?z??????≤βeβd≤1α0??z?n??A2ifAischosenlarge.Thisreducestothespecialcasewejustdiscussed.Thenewextra?rstordertermdoesnotchangetheresult.Wemayapplythespecialcasetow.Remark.Theresultfailsifwejustassumeα(x)≥0on??.Infact,wecannotevengettheuniqueness.§3.GradientEstimatesThebasicideainthetreatmentofgradientestimates,duetoBernstein,involvesdi?erentiationoftheequationwithrespecttoxk,k=1,...,n,followedbymultiplicationbyDkuandsummationoverk.Themaximumprincipleisthenappliedtotheresultingequationinthefunctionv=|Du|2,possiblywithsomemodi?cation.Therearetwokindsofgradientestimates,globalgradientestimatesandinteriorgradientestimates.Wewillusesemi-linearequationstoillustratetheidea.Suppose?isaboundedandconnecteddomaininRn.Considertheequationaij(x)Diju+bi(x)Diu=f(x,u)in?foru∈C2(?)andf∈C(?×R).Wealwaysassumethataijandbiarecontinuousandˉandthattheequationisuniformlyellipticin?inthefollowinghenceboundedin?senseaij(x)ξiξj≥λ|ξ|2foranyx∈?andanyξ∈Rnforsomepositiveconstantλ.ˉsatis?esProposition3.1.Supposeu∈C3(?)∩C1(?)(1)aij(x)Diju+bi(x)Diu=f(x,u)in?ˉandf∈C1(?ˉ×R).Thenthereholdsforaij,bi∈C1(?)sup|Du|≤sup|Du|+C???36whereCisapositiveconstantdependingonlyonλ,diam(?),|aij,bi|C1(?)ˉ,M=|u|L∞(?)and|f|C1(?ˉ×[?M,M]).Proof.SetL≡aijDij+biDi.WecalculateL(|Du|2)?rst.NoteDi(|Du|2)=2DkuDkiuand(2)Dij(|Du|2)=2DkiDkju+2DkuDkiju.Di?erentiating(1)withrespecttoxk,multiplyingbyDkuandsummingoverk,wehaveby(2)aijDij(|Du|2)+biDi(|Du|2)=2aijDkiuDkju?2DkaijDkuDiju?2DkbiDkuDiu+2Dzf|Du|2+2DkfDku.Ellipticityassumptionimplies??i,j,kaijDkiuDkju≥λ|D2u|2.ByCauchyinequality,wehaveL(|Du|2)≥λ|D2u|2?C|Du|2?CwhereCisapositiveconstantdependingonlyonλ,|aij,bi|C1(?)ˉand|f|C1(?ˉ×[?M,M]).Weneedtoaddanothertermu2.WehavebyellipticityassumptionL(u2)=2aijDiuDju+2u{aijDiju+biDiu}≥2λ|Du|2+2uf.ThereforeweobtainL(|Du|2+αu2)≥λ|D2u|2+(2λα?C)|Du|2?C≥λ|D2u|2+|Du|2?Cifwechooseα&0large,withCdependinginadditiononM.Inordertocontroltheconstanttermwemayconsideranotherfunctioneβx1forβ&0.Henceweget????????L|Du|2+αu2+eβx1≥λ|D2u|2+|Du|2+β2a11eβx1+βb1eβx1?C.37Ifweputthedomain??{x1&0},theneβx1≥1foranyx∈?.Bychoosingβlarge,wemaymakethelasttermpositive.Therefore,ifwesetw=|Du|2+α|u|2+eβx1forlargeα,βdependingonlyonλ,diam(?),|aij,bi|C1(?)ˉ,M=|u|L∞(?)and|f|C1(?ˉ×[?M,M]).thenweobtainLw≥0in?.Bythemaximumprinciplewehavesupw≤supw.???This?nishestheproof.Similarly,wecandiscusstheinteriorgradientbound.Inthiscase,wejustrequiretheboundofsup|u|.?Proposition3.2.Supposeu∈C3(?)satis?esaij(x)Diju+bi(x)Diu=f(x,u)in?ˉandf∈C1(?ˉ×R).Thenthereholdsforanycompactsubsetforaij,bi∈C1(?)??????sup|Du|≤C???whereCisapositiveconstantdependingonlyonλ,diam(?),dist(???,??),|aij,bi|C1(?)ˉ,M=|u|L∞(?)and|f|C1(?ˉ×[?M,M]).∞Proof.Weneedtotakeacut-o?functionγ∈C0(?)withγ≥0andconsidertheauxiliaryfunctionwiththefollowingformw=γ|Du|2+α|u|2+eβx1.Setv=γ|Du|2.ThenwehaveforoperatorLde?nedasbeforeLv=(Lγ)|Du|2+γL(|Du|2)+2aijDiγDj|Du|2.RecallaninequalityintheproofofProposition3.1L(|Du|2)≥λ|D2u|2?C|Du|2?C.HencewehaveLv≥λγ|D2u|2+2aijDkuDiγDkju?C|Du|2+(Lγ)|Du|2?C.38Cauchyinequalityimpliesforanyε&0|2aijDkuDiγDkju|≤ε|Dγ|2|D2u|2+c(ε)|Du|2.Forthecut-o?functionγ,werequirethat|Dγ|2≤Cγin?.Thereforewehavebytakingε&0small????2|Dγ|Lv≥λγ|D2u|21?ε?C|Du|2?Cγ1≥λγ|D2u|2?C|Du|2?C.2Nowwemayproceedasbefore.Intherestofthissectionweusebarrierfunctionstoderivetheboundarygradientestimates.Weneedtoassumethatthedomain?satis?estheuniformexteriorsphereproperty.ˉsatis?esProposition3.3.Supposeu∈C2(?)∩C(?)aij(x)Diju+bi(x)Diu=f(x,u)ˉandf∈C(?ˉ×R).Thenthereholdsforaij,bi∈C(?)|u(x)?u(x0)|≤C|x?x0|foranyx∈?andx0∈??in?whereCisapositiveconstantdependingonlyonλ,?,|aij,bi|L∞(?),M=|u|L∞(?),2ˉ|f|L∞(?×[?M,M])and|?|C2(?)with?=uon??.ˉforsome?∈C(?)Proof.Forsimplicityweassumeu=0on??.AsbeforesetL=aijDij+biDi.ThenwehaveL(±u)=±f≥?Fin?wherewedenoteF=sup?|f(?,u)|.Now?xx0∈??.WewillconstructafunctionwsuchthatLw≤?Fin?,w(x0)=0,w|??≥0.Thenbythemaximumprinciplewehave?w≤u≤w39in?.Takingnormalderivativeatx0,wehave???????u????(x0)??≤?w(x0).???n???nSoweneedtobound?w(x0)independentlyofx0.ˉR(y)∩?ˉ={x0}.De?ned(x)asthedistanceConsidertheexteriorballBR(y)withBfromxto?BR(y).Thenwehave0&d(x)&D≡diam(?)foranyx∈?.Infact,d(x)=|x?y|?Rforanyx∈?.Considerw=ψ(d)forsomefunctionψde?nedin[0,∞).Thenweneedψ(0)=0(=?w(x0)=0)ψ(d)&0ford&0(=?w|??≥0)ψ??(0)iscontrolled.?Fromthe?rsttwoinequalities,itisnaturaltorequirethatψ??(d)&0.NoteLw=ψ????aijDidDjd+ψ??aijDijd+ψ??biDid.Directcalculationyieldsxi?yi|x?y|δij(xi?yi)(xi?yi)Dijd(x)=?|x?y||x?y|3Did(x)=whichimply|Dd|=1andwithΛ=sup|aij|aijDijd=aiiaijnΛλnΛ?λnΛ?λ?DidDjd≤?=≤.|x?y||x?y||x?y||x?y||x?y|RThereforeweobtainbyellipticitynΛ?λLw≤ψ????aijDidDjd+ψ??+|b|R????nΛ?λ≤λψ????+ψ??+|b|R40????ifwerequireψ????&0.HenceinordertohaveLw≤?Fweneed????nΛ?λλψ????+ψ??+|b|+F≤0.RTothisend,westudytheequationforsomepositiveconstantsaandbψ????+aψ??+b=0whosesolutionisgivenbyC1C2?adb?eψ(d)=?d+aaaforsomeconstantsC1andC2.Forψ(0)=0,weneedC1=C2.HencewehaveforsomeconstantCbCψ(d)=?d+(1?e?ad)aawhichimplies????bbψ??(d)=Ce?ad?=e?adC?eadaaψ????(d)=?Cae?ad.baDe.Sinceψ??(d)&0ford&0,soInordertohaveψ??(d)&0,weneedC≥ψ(d)&ψ(0)=0foranyd&0.ThereforewetakebbaDψ(d)=?d+2e(1?e?ad)aa????b1aD=e(1?e?ad)?d.aaSuchψsatis?esalltherequirementsweimposed.This?nishestheproof.§4.Alexandro?MaximumPrincipleSuppose?isaboundeddomaininRnandconsiderasecondorderellipticoperatorLin?L≡aij(x)Dij+bi(x)Di+c(x)wherecoe?cientsaij,bi,careatleastcontinuousin?.Ellipticitymeansthatthecoe?cientmatrixA=(aij)ispositivede?niteeverywherein?.WesetD=det(A)41andD?=DsothatD?isthegeometricmeanoftheeigenvaluesofA.Throughoutthissectionweassume0&λ≤D?≤ΛwhereλandΛaretwopositiveconstants,whichdenote,respectively,theminimalandmaximaleigenvaluesofA.Beforestatingthemaintheoremwe?rstintroducetheconceptofcontactsets.Foru∈C2(?)wede?neΓ+={y∈?;u(x)≤u(y)+Du(y)?(x?y),foranyx∈?}.ThesetΓ+iscalledtheuppercontactsetofuandHessianmatrixD2u=(Diju)isnonpositiveonΓ+.Infactuppercontactsetcanalsobede?nedforcontinuousfunctionubythefollowingΓ+={y∈?;u(x)≤u(y)+p?(x?y),foranyx∈?andsomep=p(y)∈Rn}.Clearly,uisconcaveifandonlyifΓ+=?.Ifu∈C1(?),thenp(y)=Du(y)andanysupporthyperplanemustthenbeatangentplanetothegraph.NowweconsidertheequationofthefollowingformLu=fforsomef∈C(?).ˉ∩C2(?)satis?esLu≥fin?withthefollowingTheorem4.1.Supposeu∈C(?)conditions|b|f,?∈Ln(?),andc≤0in?.?DDThenthereholdsf?+supu≤supu+C?????Ln(Γ+)D???whereΓ+istheuppercontactsetofuandCisaconstantdependingonlyonn,diam(?)band????Ln(Γ+).Infact,Ccanbewrittenas??????2n?2??bnd?exp{????++1}?1ωnnnD?Ln(Γ)withωnasthevolumeoftheunitballinRn.Remark.TheintegraldomainΓ+canbereplacedbyΓ+∩{x∈?;u(x)&supu+}.??1in?parewiththeHopf’smax-imumprincipleinSection1.Weneedalemma?rst.42n2ˉLemma4.2.Supposeg∈L1loc(R)isnonnegative.Thenforanyu∈C(?)∩C(?)thereholds????g≤g(Du)|detD2u|BM?(0)Γ+?=(supu?supu+)/dwithd=diam(?).whereΓ+istheuppercontactsetofuandM???Remark.Foranypositivede?nitematrixA=(aij)wehave1det(?D2u)≤D???aijDijun??nonΓ+.HencewehaveanotherformforLemma4.2???????aijDiju??ng(Du).g≤?nD+BMΓ?(0)Remark.Aspecialcasecorrespondstog=1:?supu≤supu++?d?|detD2u|?1??ωnΓ+?1???n????daijDiju?≤supu++1?.??nD??ωnΓ+??1?NotethisistheTheorem4.1ifbi≡0andc≡0.ProofofLemma4.2.Withoutlossofgeneralityweassumeu≤0on??.Set?+={u&0}.Bythearea-formulaforDuinΓ+∩?+??,wehave????(1)g≤g(Du)|det(D2u)|,Du(Γ+∩?+)Γ+∩?+where|det(D2u)|istheJacobianofthemapDu:?→Rn.Infactwemayconsiderχε=Du?εId:?→Rn.ThenDχε=D2u?εI,whichisnegativede?niteinΓ+.Hencebychangeofvariableformulawehave????g=g(χε)|det(D2u?εI)|,χε(Γ+∩?+)Γ+∩?+43whichimplies(1)ifweletε→0.++n?NowweclaimBM?(0)?Du(Γ∩?),i.e.,foranya∈Rwith|a|&Mthereexistsx∈Γ+∩?+suchthata=Du(x).Wemayassumeuattainsitsmaximumm&0at0∈?,i.e.,u(0)=m=supu.??)Considerana?nefunctionfor|a|&m/d(≡ML(x)=m+a?x.ThenL(x)&0foranyx∈?andL(0)=m.Sinceuassumesitsmaximumat0,thenDu(0)=0.Hencethereexistsanx1closeto0suchthatu(x1)&L(x1)&0.Notethatu≤0&Lon??.Hencethereexistsanx?∈?suchthatDu(?x)=DL(?x)=a.Nowwemaytranslateverticallytheplaney=L(x)tothehighestsuchposition,i,e.,thewholesurfacey=u(x)liesbelowtheplane.Clearlyatsuchpoint,thefunctionuispositive.ProofofTheorem4.1.WeshouldchoosegappropriatelyinordertoapplyLemma4.2.Noteiff≡0andc≡0then(?aijDiju)n≤|b|n|Du|nin?.Thissuggeststhatweshouldtakeg(p)=|p|?n.Howeversuchfunctionisnotlocallyintegrable(atorigin).Hencewewillchooseg(p)=(|p|n+μn)?1andthenletμ→0+.FirstwehavebyCauchyinequality?aijDij≤biDiu+cu?f≤biDiu?fin?+={x;u(x)&0}≤|b|?|Du|+f?????1?n1n?2(f)nn,≤|b|n+?(|Du|+μ)?(1+1)μninparticular?????????fn(?aijDiju)n≤|b|n+(|Du|n+μn)?2n?2.μg(p)=ByLemma4.2wehave??BM?(0)Nowwechoose1.|p|+μ2n?2g≤nn??Γ+∩?+|bn|+μ?n(f?)n.D44Weevaluatetheintegralinthelefthandsideinthefollowingway??g=ωnBM?(0)0???M?n+μn?nrn?1ωnMωnMdr=log=log(n+1).rn+μnnμnnμThereforeweobtain????n?2?????????2bfn?nn?n≤μnexpM?????????1.n(Γ+∩?+)+μn++LωnnDDL(Γ∩?)fIff≡0,wechooseμ=????Ln(Γ+∩?+).Iff≡0,wemaychooseanyμ&0andthenletμ→0.?InthefollowingweuseTheorem4.1andLemma4.2toderivesomea-prioriestimatesforsolutionstoquasilinearequationsandfullynonlinearequations.Inthenextresultwedonotassumeuniformellipticity.ˉ∩C2(?)satis?esProposition4.3.Supposethatu∈C(?)Qu≡aij(x,u,Du)Diju+b(x,u,Du)=0whereaij∈C(?×R×Rn)satis?esaij(x,z,p)ξiξj&0forany(x,z,p)∈?×R×Rnandξ∈Rn.nnSupposethereexistnon-negativefunctionsg∈Lnloc(R)andh∈L(?)suchthatin?|b(x,z,p)|h(x)≤forany(x,z,p)∈?×R×Rn?nDg(p)and??hn(x)dx&?Rn??gn(p)dp≡g∞.Thenthereholdssup|u|≤sup|u|+Cdiam(?),???whereCisapositiveconstantdependingonlyongandh.Example.Theprescribedmeancurvatureequationisgivenby(1+|Du|2)??u?DiuDjuDiju=nH(x)(1+|Du|2)453forsomeH∈C(?).Wehaveaij(x,z,p)=(1+|p|2)δij?pipj?D=(1+|p|2)n?1b=?nH(x)(1+|p|2).Thisimplies+2|b(x,z,p)||H(x)|(1+|p|2)2n≤=|H(x)|(1+|p|)n?1?2nD(1+|p|)33andinparticularg∞=??gn(p)dp=Rn??Rndp(1+|p|2)=ωn.ˉ∩C2(?)satis?esCorollary4.4.Supposeu∈C(?)(1+|Du|2)??u?DiuDjuDiju=nH(x)(1+|Du|2)in?forsomeH∈C(?).Thenif??H0≡?3|H(x)|ndx&ωnwehavesup|u|≤sup|u|+Cdiam(?)???whereCisapositiveconstantdependingonlyonnandH0.ProofofProposition4.3.Weproveforsubsolutions.AssumeQu≥0in?.Thenwehave?aijDiju≤bin?.Notethat{Diju}isnonpositiveinΓ+.Hence?aijDiju≥0,whichimpliesb(x,u,Du)≥0inΓ+.TheninΓ+∩?+thereholdsb(x,z,Du)h(x)≤.nD?g(Du)WemayapplyLemma4.2tognandget???????aijDiju??n≤gn≤gn(Du)nD?BM?(0)Γ+∩?+??gn(Du)Γ+∩?+??b??nnD?????hn≤hn(&???gn).Rn≤Γ+∩?+46ThereforethereexistsapositiveconstantC,dependingonlyongandh,suchthat?≤C.ThisimpliesMsupu≤supu++Cdiam(?).???NextwediscussMonge-Amp′ereequations.ˉ∩C2(?)satis?esProposition4.5.Supposeu∈C(?)det(D2u)=f(x,u,Du)in?nforsomef∈C(?×R×Rn).Supposethereexistnonnegativefunctionsg∈L1loc(R)andh∈L1(?)suchthat|f(x,z,p)|≤and??h(x)g(p)forany(x,z,p)∈?×R×Rn??h(x)dx&?Rng(p)dp≡g∞.Thenthereholdssup|u|≤sup|u|+Cdiam(?),???whereCisapositiveconstantdependingonlyongandh.TheproofissimilartothatofProposition4.3.Therearetwospecialcases.The?rstcaseisgivenbyf=f(x).Wemaytakeg≡1andhenceg∞=∞.Soweobtainˉ∩C2(?)satisfyCorollary4.6.Letu∈C(?)det(D2u)=f(x)ˉ.Thenthereholdsforsomef∈C(?)1????diam(?)?sup|u|≤sup|u|+|f|n?.1???ωn?in?SecondcaseisabouttheprescribedGaussiancurvatureequations.47ˉ∩C2(?)satisfyCorollary4.7.Letu∈C(?)det(D2u)=K(x)(1+|Du|2)ˉ.ThenifforsomeK∈C(?)K0≡?n+2in???|K(x)|&ωnwehavesup|u|≤sup|u|+Cdiam(?)???whereCisapositiveconstantdependingonlyonnandK0.We?nishthissectionbyprovingamaximumprincipleinadomainwithsmallvolume,whichisduetoVaradham.ConsiderLu≡aijDiju+biDiu+cuin?where{aij}ispositivede?nitelypointwiselyin?and|bi|+|c|≤Λanddet(aij)≥λforsomepositiveconstantsλandΛ.ˉ∩C2(?)satis?esLu≥0in?withu≤0on??.Theorem4.8.Supposeu∈C(?)Assumediam(?)≤d.Thenthereisapositiveconstantδ=δ(n,λ,Λ,d)&0suchthatif|?|≤δthenu≤0in?.Proof.Ifc≤0,thenu≤0byTheorem4.1.Ingeneralwritec=c+?c?.ThenaijDiju+biDiu?c?u≥?c+u(≡f).ByTheorem4.1wehavesupu≤c(n,λ,Λ,d)??c+u+??Ln(?)?≤c(n,λ,Λ,d)??c+??L∞|?|?supu≤?11supu2?if|?|issmall.Hencewegetu≤0in?.<parethiswithProposition1.9,themaximumprinciplefornarrowdomain.§5.MovingPlaneMethodInthissectionwewillusethemovingplanemethodtodiscussthesymmetryofsolutions.Thefollowingresultwas?rstprovedbyGidas,NiandNirenberg.48ˉ1)∩C2(B1)isapositivesolutionofTheorem5.1.Supposeu∈C(B?u+f(u)=0inB1u=0on?B1wherefislocallyLipschitzinR.ThenuisradiallysymmetricinB1andforx=0.?u(x)&0TheoriginalproofrequiresthatsolutionsbeC2uptotheboundary.Herewegiveamethodwhichdoesnotdependonthesmoothnessofdomainsnorthesmoothnessofsolutionsuptotheboundary.Lemma5.2.Supposethat?isaboundeddomainwhichisconvexinx1directionandˉ∩C2(?)isapositivesymmetricwithrespecttotheplane{x1=0}.Supposeu∈C(?)solutionof?u+f(u)=0in?u=0on??wherefislocallyLipschitzinR.Thenuissymmetricwithrespecttox1andDx1u(x)&0foranyx∈?withx1&0.Proof.Writex=(x1,y)∈?fory∈Rn?1.Wewillprove(1)??u(x1,y)&u(x?1,y)foranyx1&0andx1&x1withx1+x1&0.Thenbylettingx?1→?x1,wegetu(x1,y)≤u(?x1,y)foranyx1.Thenbychangingthedirectionx1→?x1,wegetthesymmetry.Leta=supx1for(x1,y)∈?.For0&λ&a,de?neΣλ={x∈?;x1&λ}Tλ={x1=λ}Σ??λ=there?ectionofΣλwithrespecttoTλxλ=(2λ?x1,x2,???xn)forx=(x1,x2,???xn).InΣλwede?newλ(x)=u(x)?u(xλ)forx∈Σλ.ThenwehavebyMeanValueTheorem?wλ+c(x,λ)wλ=0inΣλwλ≤0andwλ≡0on?Σλ.49wherec(x,λ)isaboundedfunctioninΣλ.Weneedtoshowwλ&0inΣλforanyλ∈(0,a).Thisimpliesinparticularthatwλassumesalong?Σλ∩?itsmaximuminΣλ.ByTheorem1.3(HopfLemma)wehaveforanysuchλ∈(0,a)??????????Dx1wλ??=2Dx1u??&0.??x1=λx1=λForanyλclosetoa,wehavewλ&0byProposition1.9(themaximumprinciplefornarrowdomain)orTheorem4.8.Let(λ0,a)bethelargestintervalofvaluesofλsuchthatwλ&0inΣλ.Wewanttoshowλ0=0.Ifλ0&0,bycontinuity,wλ0≤0inΣλ0andwλ0≡0on?Σλ0.ThenTheorem1.4(strongmaximumprinciple)implieswλ0&0inΣλ0.Wewillshowthatforanysmallε&0wλ0?ε&0inΣλ0?ε.Fixδ&0(tobedetermined).LetKbeaclosedsubsetinΣλ0suchthat|Σλ0\K|&δ/2.Thefactwλ0&0inΣλ0implieswλ0(x)≤?η&0foranyx∈K.Bycontinuitywehavewλ0?ε&0inK.Forε&0small,|Σλ0?ε\K|&δ.WechooseδinsuchawaythatwemayapplyTheorem4.8(maximumprinciplefordomainwithsmallvolume)towλ0?εinΣλ0?ε\K.Hencewegetwλ0?ε(x)≤0inΣλ0?ε\KandthenbyTheorem1.7wλ0?ε(x)&0inΣλ0?ε\K.Thereforeweobtainforanysmallε&0wλ0?ε(x)&0inΣλ0?ε.Thiscontradictsthechoiceofλ0.50CHAPTER3WEAKSOLUTIONS,PARTIGUIDEThe?rstsectionprovidesomegeneralknowledgeofCampanatoandBMOspacesthatareneededinbothChapter3and4.Sections3.2,3.3andsections5.3,5.4canbeviewedasperturbationtheory(fromconstantco?cientsequations).Theformerdealswithequationsofdivergencetype,andthelatterisfornondivergencetypeequations.TheclassicaltheoryofSchauderestimatesandLpestimatesarealsocontainedinthelattertreatment.Notewedidnotusetheclassicalpotentialestimates.HeretwopapersbyCa?arelli[C1,2]andthebookofGiaquinta[G]aresourcesforthefurtherreadings.Inthischapterandthenextwediscussvariousregularityresultsforweaksolutionstoellipticequationsofdivergenceform.Inordertoexplainideasclearlywewilldiscusstheequationswiththefollowingformonly?Dj(aij(x)Diu)+c(x)u=f(x).Weassumethat?isadomaininRn.Thefunctionu∈H1(?)isaweaksolutionifitsatis?es????????1aijDiuDj?+cu?=f?forany?∈H0(?),??whereweassume(i)theleadingcoe?cientsaij∈L∞(?)areuniformlyelliptic,i.e.,forsomepositiveconstantλthereholdsaij(x)ξiξj≥λ|ξ|2nforanyx∈?andξ∈Rn;2n(ii)thecoe?cientc∈L(?)andnonhomogenuoustermf∈L(?).NotebySobolevembeddingtheorem(ii)istheleastassumptiononcandftohaveameaningfulequation.Wewillprovevariousinteriorregularityresultsconcerningthesolutionuifwehavebetterassumptionsoncoe?cientsaijandcandnonhomogenuoustermf.Basicallytherearetwoclassofregularityresults,perturbationresultsandnonperturbationre-sults.The?rstisbasedontheregularityassumptionontheleadingcoe?cientsaij,whichareassumedtobeatleastcontinuous.Undersuchassumptionwemaycomparesolutionstotheunderlyingequationswithharmonicfunctions,orsolutionstoconstantcoe?cientequations.Thentheregularityofsolutionsdependsonhowclosetheyaretoharmonicfunctionsorhowclosetheleadingcoe?cientsaijaretoconstantcoe?cients.TypesetbyAMS-TEX51InthisdirectionwehaveSchauderestimatesandW2,pestimates.InthischapterweonlydiscusstheSchauderestimates.Forthesecondkindofregularity,thereisnocon-tinuityassumptionontheleadingcoe?cientsaij.Hencetheresultisnotbasedontheperturbation.TheiterationmethodsintroducedbyDeGiorgiandMoseraresuccessfulindealingwithnonperturbationsituation.Theresultsprovedbythemarefundamentalforthediscussionofquasilinearequations,wherethecoe?cientsdependonthesolu-tions.Infactthelinearlityhasnobearingintheirarguments.Thispermitsanextensionoftheseresultstoquasilinearequationswithappropriatestructureconditions.Onemaydiscussboundaryregularitiesinasimilarway.Weleavethedetailstoreaders.§1.GrowthofLocalIntegralsLetBR(x0)betheballinRnofradiusRcenteredatx0.Thewell-knownSobolevoldercontinuouswiththeoremstatesthatifu∈W1,p(BR(x0))withp&nthenuisH¨exponentα=1?n/p.Inthe?rstpartofthissectionweproveageneralresult,duetoS.Campanato,whichcharacterizesH¨oldercontinuousfunctionsbythegrowthoftheirlocalintegrals.Thisresultwillbeveryusefulforstudyingtheregularityofsolutionstoellipticdi?eren-tialequations.Inthesecondpartofthissectionweprovearesult,duetoJohnandNirenberg,whichgivesanequivalentde?nitionoffunctionsofboundedmeanoscillation.Let?beaboundedconnectedopensetinRnandletu∈L1(?).ForanyballBr(x0)??,de?ne??1ux0,r=u.|Br(x0)|Br(x0)Theorem1.1.Supposeu∈L2(?)satis?es??|u?ux,r|2≤M2rn+2α,foranyBr(x)??,Br(x)forsomeα∈(0,1).Thenu∈Cα(?)andforany??????thereholds????|u(x)?u(y)|sup|u|+sup≤cM+??u??L2(?)α????|x?y|?x,y∈?x=ywherec=c(n,α,?,???)&0.Proof.DenoteR0=dist(???,??).Foranyx0∈???and0&r1&r2≤R0,wehave|ux0,r1?ux0,r2|2≤2(|u(x)?ux0,r1|2+|u(x)?ux0,r2|2)52andintegratingwithrespecttoxinBr1(x0)|ux0,r1????2?ux0,r2|2≤{|u?ux0,r1|2+|u?ux0,r2|2}ωnr1Br1(x0)Br2(x0)fromwhichtheestimate(1)?nn+2αn+2α|ux0,r1?ux0,r2|2≤c(n)M2r1{r1+r2},follows.ForanyR≤R0,withr1=R/2i+1,r2=R/2i,weobtain|ux0,2?(i+1)R?ux0,2?iR|≤c(n)2?(i+1)αMRαandthereforeforh&k|ux0,2?hR?ux0,2?kR|≤αMR2(h+1)αc(n)k?h?1??i=01c(n,α)α≤MR.22Thisshowsthat{ux0,2?iR}?RisaCauchysequence,henceaconvergentone.Itslimitu?(x0)isindependentofthechoiceofR,since(1)canbeappliedwithr1=2?iRandˉwhenever0&R&Rˉ≤R0.Thuswegetr2=2?iRu?(x0)=limux0,rr↓0with(2)|ux0,r?u?(x0)|≤c(n,α)Mrαforany0&r≤R0.Recallthat{ux,r}converges,asr→0+,inL1(?)tothefunctionu,bytheLebesguetheorem,sowehaveu=u?a.e.and(2)impliesthat{ux,r}convergesuniformlytou(x)in???.Sincex→ux,riscontinuousforanyr&0,u(x)iscontinuous.By(2)weget|u(x)|≤CMRα+|ux,R|foranyx∈???andR≤R0.Henceuisboundedin???withtheestimateαsup|u|≤c{MR0+??u??L2(?)}.???53FinallyweprovethatuisH¨oldercontinuous.Letx,y∈???withR=|x?y|&R0/2.Thenwehave|u(x)?u(y)|≤|u(x)?ux,2R|+|u(y)?uy,2R|+|ux,2R?uy,2R|.The?rsttwotermsontherightsidesareestimatedin(2).Forthelasttermwewrite|ux,2R?uy,2R|≤|ux,2R?u(ζ)|+|uy,2R?u(ζ)|andintegratingwithrespecttoζoverB2R(x)∩B2R(y),whichcontainsBR(x),yields????2|ux,2R?uy,2R|2≤{|u?ux,2R|2+|u?uy,2R|2}|BR(x)|B2R(x)B2R(y)≤c(n,α)M2R2α.Thereforewehave|u(x)?u(y)|≤c(n,α)M|x?y|α.For|x?y|&R0/2weobtain|u(x)?u(y)|≤2sup|u|≤c{M+???1??u??L2}|x?y|α.R0This?nishestheproof.TheSobolevtheoremisaneasyconsequenceofTheorem1.1.InfactwehavethefollowingresultduetoMorrey.1Corollary1.2.Supposeu∈Hloc(?)satis?es??|Du|2≤M2rn?2+2α,foranyBr(x)??,Br(x)forsomeα∈(0,1).Thenu∈Cα(?)andforany??????thereholds????|u(x)?u(y)|sup|u|+sup≤cM+??u??L2(?)????|x?y|?x,y∈?x=ywherec=c(n,α,?,???)&0.Proof.ByPoincar′einequality,weobtain????|u?ux,r|2≤c(n)r2Br(x)Br(x)|Du|2≤c(n)M2rn+2α.ByapplyingTheorem1.1,wehavetheresult.Thefollowingresultwillbeneededinsection2.54Lemma1.3.Supposeu∈H1(?)satis?es??|Du|2≤Mrμ,foranyBr(x0)??,Br(x0)forsomeμ∈[0,n).Thenforany??????thereholdsforanyBr(x0)??withx0∈???????|u|2≤c(n,λ,μ,?,???){M+u2}rλBr(x0)?whereλ=μ+2ifμ&n?2andλisanynumberin[0,n)ifn?2≤μ&n.Proof.AsbeforedenoteR0=dist(???,??).Foranyx0∈???and0&r≤R0,Poincar′einequalityyields????|u?ux0,r|2≤cr2|Du|2dx≤c(n)Mrμ+2.Br(x0)Br(x0)Thisimpliesthat??Br(x0)|u?ux0,r|2≤c(n)MrλwhereλisasintheTheorem1.3.Forany0&ρ&r≤R0wehave??????u2≤2|ux0,r|2+2|u?ux0,r|2Bρ(x0)Bρ(x0)??Bρ(x0)≤c(n)ρn|ux0,r|2+2|u?ux0,r|2Br(x0)????ρ??n≤c(n)u2+MrλrBr(x0)whereweused??c(n)|ux0,r|2≤nr??u2.Br(x0)Hencethefunctionφ(r)=(1)rforsomeλ∈(0,n).IfwemayreplacethetermMrλintherightbyMρλ,wearedone.Infact,wewouldobtainthatforany0&ρ&r≤R0thereholds????ρ??λ??(2)u2≤c{u2+Mρλ}.rBρ(x0)Br(x0)Chooser=R0.Thisimplies????u2≤cρλ{u2+M}foranyρ≤R0.Bρ(x0)?φ(ρ)≤c(n){Br(x0)??ρ??nu2satis?estheinequalityφ(r)+Mrλ},forany0&ρ&r≤R0Inordertoget(2)from(1),weneedthefollowingtechnicallemma.55Lemma1.4.Letφ(t)beanonnegativeandnondecreasingfunctionon[0,R].Supposethat??ρ??αφ(ρ)≤A[+ε]φ(r)+Brβrforany0&ρ≤r≤R,withA,B,α,βnonnegativeconstantsandβ&α.Thenforanyγ∈(β,α),thereexistsaconstantε0=ε0(A,α,β,γ)suchthatifε&ε0wehaveforall0&ρ≤r≤R??ρ??γφ(ρ)≤c{φ(r)+Bρβ}rwherecisapositiveconstantdependingonA,α,β,γ.Inparticularwehaveforany0&r≤Rφ(R)φ(r)≤c{rγ+Brβ}.RProof.Forτ∈(0,1)andr&R,wehaveφ(τr)≤Aτα[1+ετ?α]φ(r)+Brβ.Chooseτ&1insuchawaythat2Aτα=τγandassumeε0τ?α&1.Thenwegetforeveryr&Rφ(τr)≤τγφ(r)+Brβandthereforeforallintegersk&0φ(τk+1r)≤τγφ(τkr)+Bτkβrβ≤τ(k+1)γφ(r)+Bτkβrβ≤τ(k+1)γk??j=0τj(γ?β)Bτkβrβφ(r)+.1?τγ?βChoosingksuchthatτk+2r&ρ≤τk+1r,thelastinequalitygives1??ρ??γBρβ.φ(ρ)≤γφ(r)+2βτrτ(1?τγ?β)Intherestofthissectionwediscussfunctionsofboundedmeanoscillation(BMO).ThefollowingresultisprovedbyJohnandNirenberg.56Theorem1.5(John-NirenbergLemma).Supposeu∈L1(?)satis?es??|u?ux,r|≤Mrn,foranyBr(x)??.Br(x)ThenthereholdsforanyBr(x)????eBr(x)p|u?ux,r|≤Crn,forsomepositivep0andCdependingonlyonn.Remark.FunctionssatisfyingtheconditionofTheorem1.5arecalledfunctionsofboundedmeanoscillation(BMO).WehavethefollowingrelationL∞?BMO.=Thecounterexampleisgivenbythefollowingfunctionin(0,1)?Ru(x)=log(x).Forconvenienceweusecubesinsteadofballs.WeneedtheCalderon-Zygmunddecomposition.Firstweintroducesometerminology.TaketheunitcubeQ0.Cutitequallyinto2ncubes,whichwetakeasthe?rstgeneration.Dothesamecuttingforthesesmallcubestogetthesecondgeneration.Continuethisprocess.Thesecubes(fromallgenerations)arecalleddyadiccubes.Any?,whichiscalledthe(k+1)-generationcubeQcomesfromsomek-generationcubeQpredecessorofQ.??Lemma1.6.Supposef∈L1(Q0)isnonnegativeandα&|Q0|?1Q0fisa?xedconstant.Thenthereexistsasequenceof(nonoverlapping)dyadiccubes{Qj}inQ0suchthatf(x)≤αa.e.inQ0??∪jQjand1α≤|Qj|??fdx&2nα.Qj??Proof.CutQ0into2ndyadiccubesandkeepthecubeQif|Q|?1Qf≥α.Forothers???1keepcuttingandalwayskeepthecubeQif|Q|f≥αandcuttherest.Let{Qj}Qbethecubeswehavekeptduringthisin?niteprocess.Weonlyneedtoverifythatf(x)≤αa.e.inQ0??∪jQj.57LetF=Q0??∪jQj.Foranyx∈F,fromthewaywecollect{Qj},thereexistsasequenceofcubesQicontainingxsuchthat??1f&αi|Q|Qianddiam(Qi)→0asi→∞.ByLebesguedensitytheoremthisimpliesthatf≤αa.e.inF.ProofofTheorem1.5.Assume?=Q0.Wemayrewritetheassumptionintermsofcubesasfollows??|u?uQ|&M|Q|QforanyQ?Q0.Wewillprovethatthereexisttwopositiveconstantsc1(n)andc2(n)suchthatforanyQ?Q0thereholds??c??2|{x∈Q;|u?uQ|&t}|≤c1|Q|exp?t.MThenTheorem1.5followseasily.???1|u?uQ0|dx.AssumewithoutlossofgeneralityM=1.Chooseα&1≥|Q0|Q0ApplyCalderon-Zygmunddecompositiontof=|u?uQ0|.Thereexistsasequenceof(1)(nonoverlapping)cubes{Qj}∞j=1suchthat??1|u?uQ0|&2nαα≤(1)|Qj|Q(1)j|u(x)?uQ0|≤αa.e.x∈Q0??∪∞j=1Qj.Thisimplies??j(1)1(1)|Qj|≤α1??Q0|u?uQ0|≤1|Q0|α??Qj(1)|uQ(1)?uQ0|≤j(1)|Qj||u?uQ0|dx≤2nα.De?nitionofBMOnormimpliesforeachj??1|u?uQj(1)|dx≤1&α.(1)(1)|Qj|Qj58Applydecompositionprocedureabovetof=|u?uQ(1)|inQj.Thereexistsasequencej(1)of(nonoverlapping)cubes{Qj}in∪jQj∞??(2)(1)suchthat1??(2)|Qj|≤αjj=1??11??(1)|Qj|≤2|Q0||u?uQ(1)|≤(1)jαjαQj(1)(2)andj|u(x)?uQ(1)|≤αa.e.x∈Qjwhichimplies??∪Qj,(2)|u(x)?uQ0|≤2?2nαa.e.x∈Q0??∪jQj.Continuethisprocess.Foranyintegerk≥1thereexistsasequenceofdisjointcubes(k){Qj}suchthat??(k)1|Qj|≤k|Q0|,αjandThus|{x∈Q0;|u?uQ0|&2kα}|≤n|u(x)?uQ0|≤k2nαa.e.x∈Q0??∪jQj.∞??j=1(k)|Qj|(k)1≤k|Q0|.αForanytthereexistsanintegerksuchthatt∈[2nkα,2n(k+1)α).Thisimpliesα?k=αα?(k+1)=αe?(k+1)logα≤αeα?logt.This?nishestheproof.§2.H¨olderContinuityofSolutionsInthissectionwewillproveH¨olderregularityforsolutions.Thebasicideaistofreezetheleadingcoe?cientsandthentocomparesolutionswithharmonicfunctions.Theregularityofsolutionsdependsonhowclosesolutionsaretoharmonicfunctions.Henceweneedsomeregularityassumptionontheleadingcoe?cients.Supposeaij∈L∞(B1)isuniformlyellipticinB1=B1(0),i.e.,λ|ξ|2≤aij(x)ξiξj≤Λ|ξ|2,foranyx∈B1,ξ∈Rn.59Inthefollowingweassumethataijisatleastcontinuous.Weassumethatu∈H1(B1)satis?es????1(?)aijDiuDj?+cu?=f?forany?∈H0(B1).B1B1ThemaintheoremwewillprovearethefollowingH¨olderestimatesforsolutions.ˉ1),c∈Ln(B1)andTheorem2.1.Letu∈H1(B1)solve(?).Assumeaij∈C0(Bf∈Lq(B1)forsomeq∈(n/2,n).Thenu∈Cα(B1)withα=2?n/q∈(0,1).1andr≤R0Moreover,thereexistsanR0=R0(λ,Λ,τ,??c??Ln)suchthatforanyx∈Bthereholds??????2+??u??|Du|2≤Crn?2+2α??f??2H1(B1)Lq(B1)Br(x)whereC=C(λ,Λ,τ,??c??Ln)isapositiveconstantwith|aij(x)?aij(y)|≤τ(|x?y|),foranyx,y∈B1.Remark.Inthecaseofc≡0,wemayreplace??u??H1(B1)with??Du??L2(B1).Theideaoftheproofistocomparethesolutionuwithharmonicfunctionsandusetheperturbationargument.Lemma2.2.(BasicEstimatesforHarmonicFunctions.)Suppose{aij}isaconstantpositivede?nitematrixwithλ|ξ|2≤aijξiξj≤Λ|ξ|2foranyξ∈Rnforsome0&λ≤Λ.Supposew∈H1(Br(x0))isaweaksolutionof(1)aijDijw=0inBr(x0).Thenforany0&ρ≤r,therehold?}