ONTHEEXISTENCEOFLIPSCHITZSOLUTIONSTOSOME FORWARD-BACKWARDPARABOLICEQUATIONS By SeonghakKim ADISSERTATION Submittedto MichiganStateUniversity inpartialfullmentoftherequirements forthedegreeof Mathematics-DoctorofPhilosophy 2015 ABSTRACT ONTHEEXISTENCEOFLIPSCHITZSOLUTIONSTOSOME FORWARD-BACKWARDPARABOLICEQUATIONS By SeonghakKim Inthisdissertationwediscussanewapproachforstudyingforwar d-backwardquasilinear dusionequations.Ourmainideaismotivatedbyareformulationofsu chequationsas non-homogeneouspartialderentialinclusionsandreliesonaBaire 'scategorymethod.In thiswaytheexistenceofLipschitzsolutionstotheinitial-boundaryv alueproblemofthose equationsisguaranteedunderacertaindensitycondition.Finallywe studytwoimportant casesofanisotropicdusioninwhichsuchdensityconditioncanbere alized. TherstcaseisonthePerona-Maliktypeequations.In1990,P.Pe ronaandJ.Malik[35] proposedananisotropicdusionmodel,calledthePerona-Malikmode l,inimageprocessing u t =div Du 1+ j Du j 2 fordenoisingandedgeenhancementofacomputervision.Sincethe nthedichotomyof numericalstabilityandtheoreticalill-posednessofthemodelhasa ttractedmanyinterestsin thenameofthePerona-Malikparadox[28].Ourresultinthiscasepro videsthemodelwith mathematicallyrigoroussolutionsinanydimensionthatareevenree ctingsomephenomena observedinnumericalsimulations. TheothercasedealswiththeexistenceresultontheHolligtypeequ ations.In1983, K.Hollig[20]proved,indimension n =1,theexistenceofinnitelymany L 2 -weaksolu- tionstotheinitial-boundaryvalueproblemofaforward-backwardd usionequationwith non-monotonepiecewiselinearheatux,andthispiecewiselinearityw asmuchrelaxedlater byK.Zhang[45].Thework[20]wasinitiallymotivatedbytheClausius-Du heminequal- ityinthesecondlawofthermodynamics,wherethenegativeoftheh eatuxmayviolate themonotonicitybutshouldobeytheFourierinequalityatleast.Our resultinthiscase generalizes[20,45]toalldimensions. Formyfamily iv ACKNOWLEDGMENTS Firstofall,Iwouldliketodeeplythankmyadvisor,Dr.BaishengYanfo rhisencouragement andguidanceduringthiswork.Withouthissupportandendlesspatie nce,allofthiswork andmuchmorewouldhavebeenimpossible.Ibelievehehasshownmeth ebestexampleof anadvisoronecouldbeandhopeIcanresemblehimifIhaveachancet ohelpsomeonein thefuture. Ialsowouldliketoexpressmysinceregratitudetomycommitteememb ersDr.Casim Abbas,Dr.PeterW.Bates,Dr.KeithS.PromislowandDr.Zhengfa ngZhoufortheirtime, eortsandvaluablesuggestions. v TABLEOFCONTENTS LISTOFFIGURES ................................... viii Chapter1Introduction ............................... 1 1.1Reviewoftheliterature.............................5 1.1.1Youngmeasuresolutions.........................5 1.1.2Perona-Malikmodelandspatialregularizations............ 7 1.1.3Classicalsolutions.............................10 1.1.4Lipschitzsolutionsindimension n =1.................11 1.1.5Lipschitzsolutionsinalldimensions...................12 1.2Mainresults....................................13 1.2.1Perona-Maliktypeequations.......................14 1.2.2Holligtypeequations...........................14 Chapter2AnewapproachbyBaire'scategorymethod ........... 15 2.1Baire-onefunctions................................1 5 2.2Generalexistencetheorem........................... .19 2.3Existencetheoremsonanisotropicdusions............... ...27 2.3.1CaseI:Perona-Maliktypeequations..................28 2.3.2CaseII:Holligtypeequations......................30 2.3.3Radialandnon-radialsolutions.....................32 Chapter3Preliminaries ............................... 33 3.1Uniformlyparabolicequations..........................3 3 3.2Modcationofprolefunctions......................... 36 3.3Rightinverseofthedivergenceoperator................. ...37 Chapter4Perona-Maliktypeequations ..................... 41 4.1Geometryofrelevantmatrixset........................ .41 4.1.1Non-homogeneousderentialinclusionanditslimitation....... 41 4.1.2Geometryofthematrixset F 0 ......................43 4.2Relaxationof r ! ( z ) 2 F 0 ............................59 4.3Constructionofadmissibleset U .........................66 4.4CompletionofproofofTheorem2.3.2...................... 71 4.5ProofofTheorem2.3.5..............................77 Chapter5Holligtypeequations .......................... 79 5.1Geometryofrelevantmatrixset........................ .79 5.2Relaxationof r ! ( z ) 2 F 0 ............................93 5.3Constructionofadmissibleset U .........................97 5.4CompletionofproofofTheorem2.3.4...................... 102 vi BIBLIOGRAPHY .................................... 107 vii LISTOFFIGURES Figure1.1: CaseI: Perona-Maliktypeprole ˙ ( s ).................3 Figure1.2: CaseII: Holligtypeprole ˙ ( s )....................4 Figure1.3:Hollig'spiecewiselinearprole ˙ ( s )...................12 Figure2.1:Graphsoftwoproles ˙ ( s ). CaseI: Perona-Maliktype. CaseII: Holligtype.................................28 Figure2.2:Scale-spaceusinganisotropicdusionwithux A ( p )= p 1+ jp j2 =s 2 0 .ThreedimensionalplotofthebrightnessofFigure12in[35].(a) Originalimage,(b)aftersmoothingwithanisotropicdusion....31 Figure3.1: CaseI: Perona-Maliktypeprole ˙ ( s )andmodedfunction~ ˙ ( s ).37 Figure3.2: CaseII: Holligtypeprole ˙ ( s )andmodedfunction~ ˙ ( s )....38 viii Chapter1 Introduction Theevolutionprocessofmanyquantitiesinapplicationscanbemodele dbyadusion partialderentialequationoftheform u t =div( A ( Du ))in (0 ;T ),(1.1) where ˆ R n isaboundeddomain, T> 0isanyxednumber,and u = u ( x;t )isthedensity ofsomequantityatposition x andtime t ,with Du =( u x 1 ; ;u x n )and u t denotingits spatialgradientandrateofchange,respectively.Thevectorfu nction A :R n ! R n here representsthe diusionux oftheevolutionprocess.Theusualheatequationcorresponds tothecaseofisotropicdusiongivenbytheFourierlaw: A ( p )= kp ( p 2 R n ),where k> 0 isthedusionconstant. Forstandarddusionequations,theux A ( p )isassumedtobe monotone ;namely, ( A ( p ) A ( q )) ( p q ) 0( p;q 2 R n ) :Inthiscase,equation(1.1)isparabolicandcanbestudiedbythesta ndardmethodsof parabolicequationsandmonotoneoperators.Forexample,whent heux A ( p )isgivenby 1 A ( p )= Dp W ( p )forsomesmoothconvexfunction W :R n ! R satisfying j D2 p W ( p ) j ;n X i;j =1 W p i p j ( p ) ˘ i ˘ j j ˘ j 2 ( p;˘ 2 R n ) ;where ; arepositiveconstants,(1.1)canbeviewedandthusstudiedasace rtaingradient owgeneratedbythe convex energyfunctional I ( u )= Z W ( Du ( x )) dx inthecontextofnon-linearsemigrouptheoryandmonotoneopera tors;see,e.g.,Brezis[6]. Inregardtoclassicalsolutions,iftheux A ( p )satisestheuniformellipticitycondition j ˘ j 2 n X i;j =1 A i p j ( p ) ˘ i ˘ j j ˘ j 2 ( p;˘ 2 R n ) ;theexistenceandpropertiesofsolutionsto(1.1)canbeexaminedb yestablishingvari- ous apriori estimatesandappealingtotheLeray-Schauderxedpointtheore m;see,e.g., Ladyzenskaja etal. [29]andLieberman[30]. However,forsomeapplicationsoftheevolutionprocessincertainim portantphysical problems,underlyingdusionuxesmaynotbemonotone,yieldingno n-parabolicequations (1.1).Inthisdissertation,westudythedusionequation(1.1)with certainnon-monotone uxes A ( p )satisfyingFourier'sinequality: A ( p ) p 0( p 2 R n ).Wefocusontheinitial- 2 boundaryvalueproblem 8 > > > > > > > > < > > > > > > > > : u t =div( A ( Du ))in T ,A ( Du ) n =0on @ (0 ;T ), u = u 0 on f t =0 g ;(1.2) where T = (0 ;T ), n istheouterunitnormalon @ , u 0 = u 0 ( x )isagiveninitial datum,andtheux A ( p )isoftheform A ( p )= f ( j p j 2 ) p ( p 2 R n ) ;(1.3) givenbyafunction f :[0 ;1 ) ! R with prole ˙ ( s )= sf ( s 2 )havingoneofthegraphsin Figures1.1and1.2below.(Precisestructuralassumptionson ˙ ( s )= sf ( s 2 )willbegiven inChapter2.) s 0 ˙ ( s ) ˙ ( s 0 ) s 0 Figure1.1: CaseI: Perona-Maliktypeprole ˙ ( s ). ThetwocasesinFigures1.1and1.2correspondtotheapplicationsinim ageprocessing proposedbyPeronaandMalik[35]andinphasetransitionofthermod ynamicsstudiedby 3 s 0 ˙ ( s ) ˙ ( s 1 ) ˙ ( s 2 ) s 2 s 1 s 1 s 2 Figure1.2: CaseII: Holligtypeprole ˙ ( s ). Hollig[20],respectively.Forthesedusionequations(1.1),wehave ˙ 0 ( p s )= f ( s )+2 sf 0 ( s ) < 0forsomevaluesof s> 0. Inthesecases,thedusionis anisotropic sincethedusionmatrix( A i p j ( p )),where A i p j ( p )= f ( j p j 2 ) ij +2 f 0 ( j p j 2 ) p i p j ( i;j =1 ;2 ; ;n ) ;hastheeigenvalues f ( j p j 2 )(ofmultiplicity n 1)and f ( j p j 2 )+2 j p j 2 f 0 ( j p j 2 );hencethedusion coecientscouldbealsonegative.Insuchcases,problem(1.2)bec omes forward-backward parabolic .Moreover,setting W ( p )= Z jp j0 ˙ ( r ) dr;I ( u )= Z W ( Du ) dx; theinitial-boundaryvalueproblem(1.2)becomesa L 2 -gradientowoftheenergyfunctional I ( u );however, I ( u )is non-convex .Consequently,neitherthestandardmethodsofparabolic 4 equationsandmonotoneoperatorsnorthenon-linearsemigroupt heorycanbeappliedto study(1.2). Wenowintroducethenotionofaweaksolutiontoproblem(1.2)reec tingtheinitial andboundaryconditionsasfollows. Denition1.0.1. Wesaythatafunction u 2 W 1 ;1 ( T )isa Lipschitzsolution to(1.2) providedthatequality Z ( u ( x;s ) ( x;s ) u 0 ( x ) ( x; 0)) dx = Z s 0 Z ( u t A ( Du ) D ) dxdt (1.4) holdsforeach 2 C 1 ( T )andeach s 2 [0 ;T ]:Beforestatingthemainresultsofthisdissertation,webeginwithalit eraturereviewon forward-backwarddusionproblems. 1.1Reviewoftheliterature Inthisreview,wegenerallyassumetheux A ( p )isnon-monotone.However,weimpose atleast theFourierinequality: A ( p ) p 0forall p 2 R n ,whichisconsistentwiththe Clausius-Duheminequalityinthetheoryofthermalconductors. 1.1.1Youngmeasuresolutions A measure-valued or Youngmeasure solutiontoequation(1.1)isapair( u; )ofafunction u inasuitableSobolevspaceandaparametrizedfamily =( x;t ) ( x;t ) 2 T ofprobability 5 measureson R n generatedbythespatialgradientsofasequenceinthesamespac e,satisfying Z T 0 Z ( h ;A i D + u t ) dxdt =0 forall 2 C 1 c ( T ),where h ;A i = Z R n A ( p ) ( dp )a :e :in T :Inaddition,thepair( u; )isrequiredtosatisfy Du = h ; id i = Z R n p ( dp )a :e :in T ;whereid: R n ! R n istheidentityfunction. NotethatanyLipschitzsolution u 2 W 1 ;1 ( T )toequation(1.1)inthesenseofDe- nition1.0.1(withoutinitialandboundaryconditions)isaYoungmeasur esolutionwithits correspondingparametrizedfamily Du =( Du ( x;t ) ) ( x;t ) 2 T ofpointmassesat Du .TherehavebeenextensivestudiesonYoungmeasuresolutionstod usionequations(1.1) andtheirpropertiesunderderentassumptionsontheux A ( p )andDirichletorNeumann boundaryconditions.Twoearlyworkswereaccomplishedindepende ntlybySlemrod[39] andbyKinderlehrerandPedregal[27].In[39],equation(1.1)underD irichletorNeumann boundaryconditionsisapproximatedbyasequenceofregularands ingularlyperturbed problemswhosesolutionsareusedtogenerateaYoungmeasureso lution.Ontheother hand,thework[27]combinestheexplicitmethodsforsolutionstoev olutionequationswith variationalmethodsusedtoincorporatetheoscillatorybehavior. Suchcombinationthen leadstotheexistenceofYoungmeasuresolutionstoevolutionprob lemsthatmaybeof 6 forward-backwardtype.Howeverthederencesbetweenthes etwoworksaresubtle.In [39],theux A ( p )andinitialdatum u 0 areassumedtobesucientlysmooth, A ( p )has strictlysub-quadraticgrowth,and(1.1)issatisedinthesenseof distributions.In[27], A ( p )iscontinuousandoflineargrowth, u 0 2 H 1 0 (),and(1.1)issatisedin H 1 (). Followingtheapproachin[27],Demoulini[12]establishedtheexistence ofa unique Young measuresolutiontoequation(1.1)withux A ( p )asthegradientofsome C 1 potential ˚ ( p ) satisfyingacertaingrowthconditionandunderDirichletboundaryc ondition.Hermethod wasfurtherexploredbyYinandWang[43]toextendtheexistencer esultinvolvingother growthconditionson ˚ ( p ). FocusingonthePerona-Malikux A ( p )= p 1+ jp j2 (seebelow),existenceandproperties of innitelymany Youngmeasuresolutionstoproblem(1.2)hadbeenstudiedbyTaher iet al. [40]andbyChenandZhang[7]fordimensions n =1and n =2,respectively.Non- uniquenessofsolutionshereisinevitableduetotheintensityofforw ard-backwardnatureof thePerona-Malikux A ( p ).Thisisinsharpcontrasttotheuniquenessresultin[12]asa rathermildbackwardnessisinherentintheuxestreatedinthatpa per. 1.1.2Perona-Malikmodelandspatialregularizations IntheoriginalpaperofPeronaandMalik[35],theyproposedananiso tropicdusionmodel (1.2),calledthePerona-Malikmodel,fordenoisingandedgeenhance mentofacomputer vision,where ˆ R 2 isasquareandtheux A ( p )isgivenby(seeFigure1.1fortheshape ofprole ˙ ( s )) either A ( p )= p 1+ j p j 2 =s 2 0 or A ( p )=exp j p j 2 2 s 2 0 p (1.5) 7 withaxedthreshold s 0 > 0accordingtosomeexperimentalpurposes. Inthismodel, u ( x;t )representsanimprovedversionoftheinitialgraylevel u 0 ( x )ofa noisypicture.Theanisotropicdusiondiv( A ( Du ))isforwardparabolicinthe subcritical regionwhere j Du j s 0 :Theexpectationofthemodelisthatdisturbanceswithsmallgradie ntinthesubcritical regionwillbesmoothedoutbytheforwardparabolicdusion,whilesh arpedgescorre- spondingtolargegradientinthesupercriticalregionwillbeenhance dbythebackward parabolicequation.Suchexpectedphenomenologyhasbeenimpleme ntedandobservedin somenumericalexperiments;seee.g.,Esedoglu[13],showingstability andeectivenessof themodel.Ontheotherhand,manyanalyticalworkshaveshownth atthemodelishighly ill-posedwhentheinitialdatum u 0 is transcritical in;namely,therearesubregionsin where j Du 0 j s 0 ,respectively.Forsuchtranscriticalinitialdata,due tothebackwardparabolicity,evenapropernotionandtheexisten ceofwell-posedsolutions toproblem(1.2)haveremainedlargelyunsettled;seeKichenassamy [23]inthisregard. Therehavebeenmanyworkstryingtodeneasuitablenotionofwea ksolutionto problem(1.2)reectingexpectedphenomenologyofthemodel.One wayistostudyYoung measuresolutionsto(1.2)asin[40,7]explainedabove.Anotherway istoinvestigatethe nice solutionstoregularizedproblemsandthelimitingbehaviorsofsuchso lutionsasthe regularizationparameterapproaches0.Forthisdiscussion,letus take s 0 =1in(1.5)and focusontheux A ( p )= p 1+ jp j2 .Asaperturbationoftheoriginalproblem(1.2),amildregularizationw asproposedby Guidotti[19]includingaviscousterm( > 0): u t =div 1 1+ j Du j 2 + Du ;(1.6) 8 whichisstillofforward-backwardtypeatleastfor < 1 = 8.Itistheformalgradientowof theenergyfunctional I ( u )= 1 2 Z log(1+ j Du j 2 )+ j Du j 2 dx: Thisfunctionalhasanon-trivialconvexcationwhich uniquely determinesaYoungmea- suresolutionbymeansofapproximateweakYoungmeasuresolution s.Theconstruction oftheseapproximatesolutionswascarriedoutbyfollowingtheappr oachof[12].While Youngmeasuresolutionsin[40,7]arenotunique,thosein[12,19]ar euniqueduetothe reasonmentionedabove.Inadynamicalviewpoint,regularization( 1.6)seemstoreplace thestaircasingeectofthePeorna-Malikequationwithamicro-ram pingphenomenonby whichthecenterofmass(thatis,Youngmeasure)solutionisLipsch itzcontinuouswhileits gradientexhibitsamicro-structurecomposedofgradientsofsma llandlargesize.Oneof themainresultsofthisdissertationactuallyveressuchphenomen onfor exact Lipschitz solutionstothePerona-Maliktypeequationshavingproles ˙ ( s )asinFigure1.1(including thePerona-Malikequationitself)withoutregularization(1.6). Indimension n =1,fourthorderregularizationhasbeenstudiedbyBellettini etal. [3] andbyBellettiniandFusco[2].Thesepapersstudiedthesingularper turbation( > 0): u t = 2 u xxxx + u x 1+ u 2 x x whoseassociatedenergyfunctionalisgivenby I ( u )= 1 2 Z 1 0 u 2 xx +log(1+ u 2 x ) dx: 9 In[3],ithadbeenobservedthatinnitelymanyderentevolutionsma yariseunderthe sameinitialdatum u 0 byconsideringsequences u 0 ofinitialdatathatconvergeuniformlyto u 0 as ! 0 + .In[2],usingthe-limitconvergencetechniquewithappropriatesca ling,the authorscouldcapturethelongtimebehaviorofthePerona-Malikeq uationwithevolution ofpiecewiseconstantdata. 1.1.3Classicalsolutions Letusassumeforthemomentthat ˆ R n isabounded C 1 domainandthat A ( p )= p 1+ jp j2 .Givenapoint x 2 ,wesaythattheinitialdatum u 0 2 C 1 ( )is subcritical at x if j Du 0 ( x ) j < 1, supercritical at x if j Du 0 ( x ) j > 1,and critical at x if j Du 0 ( x ) j =1.The initialdatum u 0 is transcritical iniftherearetwopoints x;y 2 with j Du 0 ( x ) j < 1and j Du 0 ( y ) j > 1. Existenceofglobalorlocalclassicalsolutionstoproblem(1.2)depe ndsheavilyonthe initialdatum u 0 .KawohlandKutev[22]showedthataglobalclassicalsolutionexist s inanydimensionif u 0 issubcriticalon .However,inthiscase,theconvexityofis requiredtoguaranteesuchglobalexistenceaspointedoutbyKim[2 4].In[22],theyalso provedthat(1.2)cannotadmitaglobalclassicalsolutionfor n =1if u 0 istranscritical inundersometechnicalassumptions,andtheseassumptionswer ecompletelyremoved laterbyGobbino[17].ConcerningthePerona-Maliktypeequations,it hadbeenthegeneral beliefthatclassicalsolutionscanonlyexistiftheinitialdataaresmoo th,evenanalytic,at supercriticalpoints;thiswasformallystreamlinedbyKichenassamy [23].Inregardtothe classofsuitableinitialdataforclassicalsolutions,GhisiandGobbino [14]establishedthat for n =1,thesetofinitialdataforwhichproblem(1.2)hasalocalclassical solutionisdense in C 1 ( ). 10 Thesituationconcerningtheexistenceofaglobalclassicalsolution to(1.2)withatrans- criticalinitialdatumfor n 2turnsouttobequitederentfromthecase n =1 :Therst existenceresultofglobalclassicalsolutionswith u 0 transcriticalfor n 2wasobtainedby GhisiandGobbino[15],wheretheyconstructedaclassofglobalrad ial C 2 ;1 solutionswith suitablychosenradialinitialdatatranscriticalonanannuluscente redattheorigin;these solutionsalsohavethepropertyofnite-timeextinctionofsuperc riticalregion.Incontrast totheone-dimensionalresults[22,17]mentionedabove,theirres ultexhibitedaquiteder- entfeatureofthehigherdimensionalproblem.Ontheotherhand, intheradialcase,Ghisi andGobbino[16]alsoprovedthataglobal C 1 solutioncannotexistifthegradientofinitial datum u 0 isverylargeatapoint.Therefore,requirementofregularityfors olutions(e.g., classicalor C 1 )wouldpreventtheexistenceofsuchsolutionsiftheinitialdatasho uldbe arbitrarilygivenandtranscritical. 1.1.4Lipschitzsolutionsindimension n =1 Lettheux A ( p )begivenby(1.3)withitsprole ˙ ( s )asinFigures1.1or1.2.Whenthe initialdatum u 0 isanygivensmoothfunction(satisfyingcertaincompatibilityconditio ns on @ ),itseemsnaturaltolowertheexpectationontheregularityofs olutionsbynding plausibleweaksolutionstoproblem(1.2).Evenundertheloweringofr egularityhaveenor- mousdcultiesoccurredontheexistenceofsuitableweaksolutions aswediscussedabove. Toourbestknowledge,Zhang[44,45]wasthersttosuccessfully provethat,for n =1, thereareinnitelymanyLipschitzsolutionsto(1.2)foranygivennon -constantsmooth initialdata u 0 (withanextraassumptionwhentheprole ˙ ( s )isasinFigure1.2);his pivotalideawastoreformulatetheone-dimensionalPerona-Maliko rHolligtypeequations into2 2non-homogeneouspartialderentialinclusionsandthentoprov etheexistence 11 usingamodedmethodofconvexintegrationfollowingtheideasofKir chheim[28]andof MullerandSverak[32].BeforeZhang'swork[45],inthepioneeringwo rkofHollig[20],it wasprovedthatfor n =1,thereareinnitelymany L 2 -weaksolutionsto(1.2)whenthe prole ˙ ( s )ispiecewiselinearasinFigure1.3;however,themethodofHolligcann otbe appliedtogeneralizedproles ˙ ( s )asinFigure1.2. s 0 ˙ ( s ) ˙ ( s 1 ) ˙ ( s 2 ) s 2 s 1 Figure1.3:Hollig'spiecewiselinearprole ˙ ( s ). 1.1.5Lipschitzsolutionsinalldimensions Recently,KimandYan[25]extendedZhang'smethod[44]tostudyth ePerona-Maliktype equationsinalldimensions n forballs= f x 2 R n :j x j > > < > > > : ˆ R n isaboundeddomainwith @ of C 2+ ,u 0 2 C 2+ ( )isnon-constantwith Du 0 n j @ =0 ;(1.7) where 2 (0 ;1)isagivennumber. Althoughwewillrepeatthestatementsoftheconcreteexistence resultsinChapter2, weintroducethemhereasasummaryofthedissertation. 13 1.2.1Perona-Maliktypeequations Lettheux A ( p )beoftheform(1.3)withprole ˙ ( s )asinFigure1.1.Thenwehavethe following. Theorem1.2.1 (Perona-Maliktype) . Let and u 0 satisfy (1 :7) with convex,andlet T = (0 ;T ) foragiven T> 0 .ThenthereexistinnitelyLipschitzsolutions u to (1 :2) . AdetailedversionofthisresultisavailableinTheorem2.3.2thatprovide sthePerona- Malikmodelwithmathematicallyrigoroussolutionsreectingsomephe nomenaobservedin numericalsimulations.Notethatthisresultgeneralizesthoseof[44 ,26]. 1.2.2Holligtypeequations Lettheux A ( p )beoftheform(1.3)withprole ˙ ( s )asinFigure1.2.Thentheresultis asfollows. Theorem1.2.2 (Holligtype) . Let and u 0 satisfy (2 :10) with j Du 0 ( x 0 ) j2 ( s 1 ;s 2 ) for some x 0 2 ;andlet T = (0 ;T ) foragiven T> 0 .Thenthereexistinnitelymany Lipschitzsolutions u to (1 :2) . Thisistogeneralizetheresultsof[20,45]toHolligtypeproles ˙ ( s )illustratedinFigure 1.2foralldimensions. Precisestructuralassumptionsontheproles ˙ ( s )inTheorems1.2.1and1.2.2aregiven inChapter2.ThesetheoremsarecompletelyprovedinChapters4a nd5,respectively. Chapter3isreservedforpreliminaryresultsthatmaybeofindepen dentinterest. 14 Chapter2 AnewapproachbyBaire'scategory method Thepurposeofthischapteristodesignanew functional approachtostudyproblem(1.2), whichisbasedona Baire'scategorymethod .Asapreliminaryanalyticalbackground,we introduceaversionoftheBairecategorytheoremonBaire-onefu nctions.Wethenintroduce twoimportantclassesofnon-monotoneuxes A ( p )withwhichtheapproachcanbeapplied to(1.2).Indoingso,theconcreteexistenceresultsonthePeron a-MalikandHolligtype equationsarestatedalongwiththecoexistenceresultonradialan dnon-radialsolutionsfor thePerona-Maliktypewhenthedomainisaballandtheinitialdatum u 0 isradial. 2.1Baire-onefunctions Inthispreliminarysection,weintroduceaversionofthe Bairecategorytheorem on Baire- onefunctions followingtheexpositionof[9].Here,let X;Y denotemetricspaceswith correspondingmetrics d X ;d Y .Webeginwithbasicterminologies. Denition2.1.1. Let f :X ! Y .Wedenethe oscillation of f atapoint x 0 2 X by ! f ( x 0 )=lim ! 0 + sup x;y 2 B X ( x 0 ) d Y ( f ( x ) ;f ( y )) ;15 where B X ( x 0 ; )istheopenballin X withcenter x 0 andradius > 0.Thefunction ! f :X ! [0 ;1 ]iscalledthe oscillation of f .Denition2.1.2. Wesaythat f :X ! Y isa Baire-onefunction ifthereexistsasequence f f j g 1 j =1 ofcontinuousfunctionsfrom X into Y suchthat lim j !1 f j ( x )= f ( x )in Y ,8 x 2 X: Itisveryeasytoprovethefollowing;weskiptheproof. Proposition2.1.3. Let f :X ! Y .Then (i) f iscontinuousatapoint x 0 2 X ifandonlyif ! f ( x 0 )=0 ;(ii) 8 > 0 ,theset f := f x 2 X j ! f ( x ) < g isopenin X . Remark2.1.4. Let f :X ! Y .Let C f = f x 2 X j f iscontinuousat x g and D f = X nC f .ByProposition2.1.3,wehave D f = f x 2 X j ! f ( x ) > 0 g = [ j 2 N f x 2 X j ! f ( x ) 1 =j g ;whichisan F ˙ set.Also, C f = \ j 2 N f x 2 X j ! f ( x ) < 1 =j g isa G set. Somebasicdenitionsonmetricspacesareincludedhere. Denition2.1.5. (i)Aset N ˆ X iscalled nowheredense in X iftheclosure N of N containsnonon-emptyopensubsetof X ,thatis,theopenset N c = X n N isdensein X .(ii)Aset F ˆ X issaidtobeofthe rstcategory ifitisthecountableunionofnowhere densesubsetsof X .16 (iii)Aset S ˆ X thatisnotoftherstcategoryissaidtobeofthe secondcategory .TheBairecategorytheorembelowissostandardthatitiscontained inalmostallreal analysisbooks;seee.g.,[5]. Theorem2.1.6 (BaireCategoryTheorem:VersionI) . Let X becomplete.Thenanycount- ableintersectionofdenseopensubsetsof X isdensein X . Thetheorembelowisthemainpartofthissectionwhoseproofisprov idedforreader's convenience. Theorem2.1.7 (BaireCategoryTheorem:VersionII) . Let X becomplete.If f :X ! Y isaBaire-onefunction,then D f isoftherstcategory;so C f isdensein X . Proof. InviewofRemark2.1.4,itsucestoshowthatforeach > 0,theset F := f x 2 X j ! f ( x ) 5 g isnowheredensein X .Soxan > 0. Since f isaBaire-onefunction,wecanchooseasequence f f j g 1 j =1 ofcontinuousfunctions from X into Y suchthat lim j !1 f j ( x )= f ( x )in Y ,8 x 2 X: Foreach 2 N ,dene E = \ i;j f x 2 X j d Y ( f i ( x ) ;f j ( x )) g :Weshowthat E isclosedin X foreach 2 N .Todothis,let i;j 2 N .Thenitissucient tocheckthat x 7! d Y ( f i ( x ) ;f j ( x ))isacontinuousfunctionfrom X into[0 ;1 ).Let x 0 2 X 17 and > 0.Since f i ;f j arecontinuous(at x 0 ),thereexistsa = ( ;i;j ) > 0suchthat x 2 X;d X ( x 0 ;x ) < = ) d Y ( f i ( x 0 ) ;f i ( x )) <= 2 ;d Y ( f j ( x 0 ) ;f j ( x )) <= 2 = )j d Y ( f i ( x 0 ) ;f j ( x 0 )) d Y ( f i ( x ) ;f j ( x )) j d Y ( f i ( x 0 ) ;f i ( x ))+ d Y ( f j ( x ) ;f j ( x 0 )) <: Hencethefunction x 7! d Y ( f i ( x ) ;f j ( x ))iscontinuousat x 0 .Note E 1 ˆ E 2 ˆˆ X .Wenowcheck X = [ 2 N E .Chooseany x 0 2 X .Since f ( x 0 ) ! f ( x 0 )in Y as !1 ,thereisan N 2 N suchthat d Y ( f i ( x 0 ) ;f j ( x 0 )) 8 i;j N .Thus x 0 2 E N ˆ[ 2 N E :Thus X = [ 2 N E .Let I beanyclosedsetin X withinteriorint I 6= ; :Then I = I \ X = [ 2 N ( E \ I ) ;whereeach E \ I isclosedin X .Ifeach E \ I isnowheredensein X ,then I c = \ 2 N ( E \ I ) c isdensein X byTheorem2.1.6,andso I c \ int I 6= ; ,acontradiction.Sothereisanindex 0 2 N suchthat E 0 \ I isnotnowheredensein X ;thatis,thereexistsanon-emptyopen set J in X with J ˆ E 0 \ I: Thusforeach x 2 J ,wehave d Y ( f i ( x ) ;f j ( x )) forall i;j 0 ,andinparticular, d Y ( f 0 ( x ) ;f ( x )) ; since f j ( x ) ! f ( x )in Y .Bythecontinuityof f 0 ,foreach x 0 2 J ,thereisaneighborhood 18 I ( x 0 )of x 0 intheopenset J suchthat x 2 I ( x 0 )= ) d Y ( f 0 ( x 0 ) ;f 0 ( x )) = ) d Y ( f 0 ( x 0 ) ;f ( x )) d Y ( f 0 ( x 0 ) ;f 0 ( x ))+ d Y ( f 0 ( x ) ;f ( x )) 2 fromtheaboveinequality.Thus,foreach x 0 2 J ,wehave d Y ( f ( x ) ;f ( y )) 4 forevery x;y 2 I ( x 0 ),andso ! f ( x 0 ) 4 ;so x 0 62 F .Thisimplies J ˆ F c \ I .Puttingeverythingtogether,wecanconcludethatforanyclosed set I ˆ X withint I 6= ; ;thereisanon-emptyopenset J in X with J ˆ F c \ I .So F c \ O 6= ; foranynon-empty openset O in X .Therefore, F isnowheredensein X .2.2Generalexistencetheorem Tosetupageneralapproachforstudyingproblem(1.2),weassum e,inthissection,the domainhasaLipschitzboundary @ andtheinitialdatum u 0 2 W 1 ;1 ( T ).Without lossofgenerality,weassume Z u 0 ( x ) dx =0 ;(2.1) sinceotherwisewecansolve(1.2)forinitialdatum~ u 0 = u 0 u 0 with u 0 = 1 j jR u 0 dx .Ournewapproachismotivatedbythefollowingobservation. Proposition2.2.1. Suppose u 2 W 1 ;1 ( T ) issuchthat u ( x; 0)= u 0 ( x )( x 2 ) ;there existsavectorfunction v 2 W 1 ;2 ((0 ;T ); L 2 (; R n )) withweaktimederivative v t satisfying v t = A ( Du ) a.e.in T , (2.2) 19 andforeach 2 C 1 ( T ) andeach t 2 [0 ;T ];Z v ( x;t ) D ( x;t ) dx = Z u ( x;t ) ( x;t ) dx: (2.3) Then u isaLipschitzsolutionto (1 :2) . Proof. Toverify(1.4),givenany 2 C 1 ( T ),let g ( t )= Z u ( x;t ) ( x;t ) dx;h ( t )= Z u ( x;t ) t ( x;t ) dx ( t 2 [0 ;T ]) :Thenby(2.3),foreach 2 C 1 c (0 ;T ) ;Z T 0 t gdt = Z T 0 Z t v Ddxdt; Z T 0 hdt = Z T 0 Z v D t dxdt: Since v 2 W 1 ;2 ((0 ;T ); L 2 (; R n ))and v t = A ( Du )a.e.in T ,wehave Z T 0 Z ( D ) t vdxdt = Z T 0 Z A ( Du ) Ddxdt: As( D ) t = t D + D t ,combiningthepreviousequations,weobtain Z T 0 t gdt = Z T 0 h + Z A ( Du ) Ddx dt; whichprovesthat g isweaklyderentiablein(0 ;T )withitsweakderivative g 0 ( t )= h ( t ) Z A ( Du ( x;t )) D ( x;t ) dx a.e. t 2 (0 ;T ). Uponintegratingthis,(1.4)followsforeach s 2 [0 ;T ]:20 Condition(2.3)meansthatthefollowingconditionholdsinthesenseof distributionsin foreach t 2 [0 ;T ]:div v ( ;t )= u ( ;t ) ;v ( ;t ) n j @ =0 :Ifdimension n =1,thisconditiontogetherwith(2.2)implies v 2 W 1 ;1 ( T ;R 1 ) :However for n 2,sinceitisimpossibletobound k Dv k L 1 ( T ) intermsofdiv v ,thefunction v may notbein W 1 ;1 ( T ;R n );thisisthereasonweonlyassume v 2 W 1 ;2 ((0 ;T ); L 2 (; R n ))in Proposition2.2.1.Nevertheless,westilltrytoapproximatesuch v 'sbysomefunctionsin W 1 ;1 ( T ;R n ) :Tochoosesuitableapproximatingfunctions,werstintroducethe followingdenition. Denition2.2.2. Afunction=( u ;v ),with u 2 W 1 ;1 ( T )and v 2 W 1 ;1 ( T ;R n ), iscalleda boundaryfunction ifitsatises 8 > > > > > > > > < > > > > > > > > : u ( x; 0)= u 0 ( x ) ;x 2 ;div v ( x;t )= u ( x;t ) ;a.e.( x;t ) 2 T ;v ( ;t ) n j @ =0 ;t 2 [0 ;T ]:(2.4) Fixaboundaryfunction=( u ;v ).Wedenoteby W 1 ;1 u ( T ) ;W 1 ;1 v ( T ;R n )the usual Dirichletclasses withboundarytraces u ;v ;respectively.Wealsodenethefollow- ing. Denition2.2.3. Aclass Uˆ W 1 ;1 u ( T )iscalledan admissibleset providedthat U6 = ; isboundedin W 1 ;1 u ( T )andthatforeach u 2U ,thereexistsavectorfunction v 2 21 W 1 ;1 v ( T ;R n )satisfying div v = u a.e.in T ;k v t k L 1 ( T ) r ,where r> 0isaxednumber.Foranadmissibleset U andeach > 0,let U bethesetof all u 2U suchthatthereexistsafunction v 2 W 1 ;1 v ( T ;R n )satisfying div v = u a.e.in T ,k v t k L 1 ( T ) r ,Z T j v t ( x;t ) A ( Du ( x;t )) j dxdt j T j :Ournewapproachisthefollowing generalexistencetheorem underthepivotaldensity hypothesisof U in U ;whichisbasedonthe Bairecategorytheorem intheprevioussection. Theorem2.2.4. Let Uˆ W 1 ;1 u ( T ) beanadmissiblesetsatisfyingthe densityproperty: U isdensein U underthe L 1 -normforeach > 0 . (2.5) Then,givenany ' 2U ,foreach > 0 ;thereexistsaLipschitzsolution u 2 W 1 ;1 u ( T ) to (1 :2) satisfying k u ' k L 1 ( T ) <: Furthermore,if U containsafunctionwhichisnota Lipschitzsolutionto (1 :2) ;then (1 :2) itselfadmitsinnitelymanyLipschitzsolutions. Proof. Forclarity,wedividetheproofintoseveralsteps. 1.Let X betheclosureof U inthemetricspace L 1 ( T ) :Then( X ;L 1 )isanon-empty completemetricspace.Byassumption,each U isdensein X :Moreover,since U isbounded in W 1 ;1 u ( T ),wehave Xˆ W 1 ;1 u ( T ). 2.Let Y = L 1 ( T ;R n ).For h> 0,dene T h :X!Y asfollows.Givenany u 2X ,22 write u = u + w with w 2 W 1 ;1 0 ( T )anddene T h ( u )= Du + D( ˆ h w ) ;where ˆ h ( z )= h N ˆ ( z=h ),with z =( x;t )and N = n +1,isthestandard h -moerin R N ,and ˆ h w istheusualconvolutionin R N with w extendedtobezerooutside T :Then,for each h> 0,themap T h :( X ;L 1 ) ! ( Y ;L 1 )iscontinuous,andforeach u 2X ,lim h ! 0 + k T h ( u ) Du k L 1 ( T ) =lim h ! 0 + k ˆ h Dw Dw k L 1 ( T ) =0 :Therefore,thespatialgradientoperator D:X!Y isthepointwiselimitofasequence ofcontinuousfunctions T h :X!Y ;hence D:X!Y isa Baire-onemap .BytheBaire categorytheorem,Theorem2.1.7,thereexistsa residualset GˆX suchthattheoperator Discontinuousateachpointof G :Since XnG isofthe rstcategory ,theset G is dense in X .Therefore,givenany ' 2X ;foreach > 0,thereexistsafunction u 2G suchthat k u ' k L 1 ( T ) <: 3.Wenowprovethateach u 2G isaLipschitzsolutionto(1.2).Let u 2G begiven. Bythedensityof U in( X ;L 1 )foreach > 0,forevery j 2 N ,thereexistsafunction u j 2U 1 =j suchthat k u j u k L 1 ( T ) < 1 =j .Sincetheoperator D:( X ;L 1 ) ! ( Y ;L 1 )is continuousat u ,wehave Du j ! Du in L 1 ( T ;R n ) :Furthermore,fromthedenitionof U 1 =j ,thereexistsafunction v j 2 W 1 ;1 v ( T ;R n )suchthatforeach 2 C 1 ( T )andeach 23 t 2 [0 ;T ];Z v j ( x;t ) D ( x;t ) dx = Z u j ( x;t ) ( x;t ) dx; k ( v j ) t k L 1 ( T ) r; Z T j ( v j ) t A ( Du j ) j dxdt 1 j j T j :(2.6) Since v j ( x; 0)= v ( x; 0) 2 W 1 ;1 (; R n )and k ( v j ) t k L 1 ( T ) r ,itfollowsthatboth sequences f v j g and f ( v j ) t g areboundedin L 2 ( T ;R n ) ˇ L 2 ((0 ;T ); L 2 (; R n )) :Sowemay assume v j *v and( v j ) t *v t in L 2 ((0 ;T ); L 2 (; R n ))forsome v 2 W 1 ;2 ((0 ;T ); L 2 (; R n )) ;where * denotestheweakconvergence.Upontakingthelimitas j !1 in(2.6),since v 2 C 0 ([0 ;T ];L 2 (; R n ))and A 2 C 0 ( R n ;R n ),weobtain Z v ( x;t ) D ( x;t ) dx = Z u ( x;t ) ( x;t ) dx ( t 2 [0 ;T ]) ;v t ( x;t )= A ( Du ( x;t )) a:e: ( x;t ) 2 T :Consequently,byProposition2.2.1, u isaLipschitzsolutionto(1.2). 4.Finally,assume U containsafunctionwhichisnotaLipschitzsolutionto(1.2);hence G6 = U :Then G cannotbeaniteset,sinceotherwisethe L 1 -closure X = G = U wouldbe aniteset,making U = G :Therefore,inthiscase,(1.2)admitsinnitelymanyLipschitz solutions. Theproofiscomplete. Infact,onlywhenproblem(1.2)isnon-parabolic(thatis, A ( p )isnon-monotone)could Theorem2.2.4yieldthenon-uniquenessresult. Corollary2.2.5. Assumethedensityproperty (2 :5) holdsforsomeadmissibleset Uˆ W 1 ;1 u ( T ) :Suppose A :R n ! R n ismonotone.Thenanyfunction u 2U mustbea 24 Lipschitzsolutionto (1 :2); inthiscase, U containspreciselyonefunction. Proof. WefollowtheproofofTheorem2.2.4.Themonotonicityoftheux A ( p )implies thatthereexists atmost oneLipschitzsolutionto(1.2).Since U6 = ; ,wehave G6 = ; ,where everyfunctionin G isaLipschitzsolutionto(1.2).Thus U = G = f u g ,where u isthe unique Lipschitzsolutionto(1.2). WealsohavethefollowinggeneralpropertyforLipschitzsolutionst o(1.2)whentheux A ( p )satisesFourier'sinequality. Proposition2.2.6. Let A :R n ! R n satisfyFourier'sinequality: A ( p ) p 0 forall p 2 R n :ThenanyLipschitzsolution u to (1 :2) satises min u 0 u ( x;t ) max u 0 in T :(2.7) Proof. Let u 2 W 1 ;1 ( T )beanyLipschitzsolutionto(1.2).By(1.4),forall 2 C 1 ( T ), Z T u t ( x;t ) ( x;t ) dxdt = Z T A ( Du ) Ddxdt ;hencebyapproximation,thisequalityholdsforall 2 W 1 ;1 ( T ) :Taking ( x;t )= ˚ ( x;t ) ( t ) witharbitrary ˚ 2 W 1 ;1 ( T )and 2 W 1 ;1 (0 ;T ),wededucethat Z u t ( x;t ) ˚ ( x;t ) dx = Z A ( Du ( x;t )) D˚ ( x;t ) dx fora.e. t 2 (0 ;T )andall ˚ 2 W 1 ;1 ( T ).Nowtaking ˚ = u 2 k 1 with k =1 ;2 ; ,itfollows 25 fromtheFourierinequalityoftheux A ( p )thatfora.e. t 2 (0 ;T ) ;d dt Z u 2 k dx =2 k Z u t ˚dx = 2 k Z A ( Du ) D˚dx = 2 k (2 k 1) Z u 2 k 2 A ( Du ) Dudx 0 :Fromthiswededucethatthe L 2 k ()-normof u ( ;t )isnon-increasingon t 2 [0 ;T ];in particular, k u ( ;t ) k L 2 k () k u 0 k L 2 k () 8 t 2 [0 ;T ];k =1 ;2 ; :Letting k !1 ,weobtain k u ( ;t ) k L 1 () k u 0 k L 1 () ;hence k u k L 1 ( T ) = k u 0 k L 1 () :(2.8) Let m 1 =min u 0 ;m 2 =max u 0 ;anddene ~ A ( p )= A ( p ).Observethat ~ A ( p )alsosatisestheFourierinequality: ~ A ( p ) p 0 8 p 2 R n .Weshow m 1 u ( x;t ) m 2 forall( x;t ) 2 T tocompletetheproof.Weconsiderthree cases. Case1 :m 2 > 0and j m 1 j m 2 :Inthiscase, k u 0 k L 1 () = m 2 ;soby(2.8),forall ( x;t ) 2 T ;u ( x;t ) k u k L 1 ( T ) = k u 0 k L 1 () = m 2 :Toobtainthelowerbound,let~ u 0 = u 0 + m 2 + m 1 and~ u = u + m 2 + m 1 :Then~ u isa Lipschitzsolutionto(1.2)with A;u 0 replacedby ~ A; ~ u 0 ;respectively.Since m 1 ~ u 0 ( x ) 26 m 2 ,wehave~ u ( x;t ) m 2 asabove;hence u ( x;t ) m 1 forall( x;t ) 2 T :Case2 :m 2 > 0and m 1 < m 2 :Let~ u 0 = u 0 and~ u = u: Then~ u isaLipschitz solutionto(1.2)with A;u 0 replacedby ~ A; ~ u 0 ;respectively.Since m 2 ~ u 0 ( x ) m 1 , m 1 > 0,and j m 2 j = m 2 m 1 ;itfollowsfromCase1that m 2 ~ u ( x;t ) m 1 ;hence m 1 u ( x;t ) m 2 forall( x;t ) 2 T :Case3 :m 2 0 :Inthiscase, m 1 0 :If m 1 =0then m 2 =0andhence u 0 0;so, by(2.8), u 0.Nextassume m 1 < 0 :LetagainasinCase2~ u 0 = u 0 and~ u = u: Since m 2 ~ u 0 ( x ) m 1 and m 1 > 0 ;j m 2 j = m 2 m 1 ;itfollowsagainfromCase1 that m 2 ~ u ( x;t ) m 1 ;hence m 1 u ( x;t ) m 2 forall( x;t ) 2 T :2.3Existencetheoremsonanisotropicdiusions Inwhatfollows,westudyproblem(1.2)fornon-monotonedusion uxes A ( p )oftheform A ( p )= f ( j p j 2 ) p ( p 2 R n ) ;(2.9) where f :[0 ;1 ) ! R isafunctionwithprole ˙ ( s )= sf ( s 2 )havingoneofthegraphs inFigure2.1below.Precisestructuralassumptionson ˙ ( s )willbegiveninthefollowing subsections. Concerningthedomainandinitialdatum u 0 ,weassumethefollowinghereafter: 8 > > > < > > > : ˆ R n isaboundeddomainwith @ of C 2+ ,u 0 2 C 2+ ( )isnon-constantwith Du 0 n j @ =0 ;(2.10) where 2 (0 ;1)isagivennumber. 27 s 0 ˙ ( s ) ˙ ( s 0 ) s 0 s 0 ˙ ( s ) ˙ ( s 1 ) ˙ ( s 2 ) s 2 s 1 s 1 s 2 Figure2.1:Graphsoftwoproles ˙ ( s ). CaseI: Perona-Maliktype. CaseII: Holligtype. WeaimtoapplyTheorem2.2.4tostudytheexistenceofLipschitzsolut ionsto(1.2). Therestofthedissertationisdevotedtoconstructingsomeadmis siblesets U satisfyingthe densityproperty(2.5).Ofcourse,suchconstructionsdependo ntheinitialdatum u 0 and prole ˙ ( s )illustratedinFigure2.1. 2.3.1CaseI:Perona-Maliktypeequations Inthiscase,weassumethefollowingontheprole ˙ ( s )= sf ( s 2 ): Hypothesis(PM) (SeeFigure2.1.) (i)Thereexistsanumber s 0 > 0suchthat f 2 C 0 ([0 ;1 )) \ C 3 ([0 ;s 2 0 )) \ C 1 ( s 2 0 ;1 ) :28 (ii) ˙ 0 ( s ) > 0 8 s 2 [0 ;s 0 ), ˙ 0 ( s ) < 0 8 s 2 ( s 0 ;1 ),and lim s !1 ˙ ( s )=0 :WenowstatetheexistenceresultforthePerona-Maliktypeequat ions.Inthiscase,for each r 2 (0 ;˙ ( s 0 )),let s ( r ) 2 (0 ;s 0 )and s + ( r ) 2 ( s 0 ;1 )denotetheuniquenumberswith r = ˙ ( s ( r )). Theorem2.3.1 (Perona-Maliktype) . Let and u 0 satisfy (2 :10) with convex,andlet T = (0 ;T ) foragiven T> 0 .ThenthereexistinnitelyLipschitzsolutions u to (1 :2) . Dependingonthesizeof jj Du 0 jj L 1 () ,oursolutionssatisfyfurtherpropertiesasde- scribedinthetheorembelow;weprovethisdetailedversionthatimplie sTheorem2.3.1. Theorem2.3.2. InTheorem2.3.1,let M 0 = jj Du 0 jj L 1 () .Thenforeach r 2 (0 ;˙ ( M 0 )) , thereexistsanumber l = l r 2 (0 ;r ) suchthatforall ~ r 2 ( l;r ) andallbutatmostcountably many r 2 (0 ;~ r ) ,thereexisttwodisjointopensets 1 T ; 2 T ˆ T with j 1 T [ 2 T j = j T j and innitelymanyLipschitzsolutions u to (1 :2) satisfying u 2 C 2+ ; 1+ = 2 ( 1 T ) ;u t =div( A ( Du )) pointwisein 1 T ;j Du ( x;t ) j 0 ;where r 0 = f ( x; 0) j x 2 ;j Du 0 ( x ) j s 1 > 0suchthat f 2 C 0 ([0 ;1 )) \ C 1+ ([0 ;s 2 1 ) [ ( s 2 2 ;1 )) :(ii) ˙ 0 ( s ) > 0 8 s 2 [0 ;s 1 ) [ ( s 2 ;1 ), ˙ ( s 1 ) >˙ ( s 2 ) > 0,and ˙ 0 ( s ) 8 s 2 s 2 for someconstants > 0.Let s 1 2 (0 ;s 1 ) ;s 2 2 ( s 2 ;1 )denotetheuniquenumbers with ˙ ( s 1 )= ˙ ( s 2 ) ;˙ ( s 2 )= ˙ ( s 1 ) ;respectively. 30 Figure2.2:Scale-spaceusinganisotropicdusionwithux A ( p )= p 1+ jp j2 =s 2 0 .Threedimen- sionalplotofthebrightnessofFigure12in[35].(a)Originalimage,(b )aftersmoothing withanisotropicdusion. WestatetheexistenceresultfortheHolligtypeequations.Inthis case,foreach r 2 ( ˙ ( s 2 ) ;˙ ( s 1 )),let s ( r ) 2 ( s 1 ;s 1 )and s + ( r ) 2 ( s 2 ;s 2 )denotetheuniquenumberswith r = ˙ ( s ( r )). Theorem2.3.4 (Holligtype) . Let and u 0 satisfy (2 :10) with j Du 0 ( x 0 ) j2 ( s 1 ;s 2 ) for some x 0 2 ;andlet T = (0 ;T ) foragiven T> 0 .Thenthereexistinnitelymany Lipschitzsolutions u to (1 :2) . Chapter5dealswiththecompleteproofofthistheorem. 31 2.3.3Radialandnon-radialsolutions Weintroduceherethecoexistenceofradialandnon-radialLipsch itzsolutionstoproblem (1.2)whenthedomainisaballandtheinitialdatum u 0 isradial.Forconvenience,we focusonlyon CaseI: Perona-Maliktypeequations,althoughonecouldequallyjustifythe samefor CaseII: Holligtypeequations.Soweassumetheux A ( p )fulls Hypothesis (PM) .Let= B R (0)betheopenballin R n withcenter0andagivenradius R> 0.Letthe initialdatum u 0 2 C 2+ ; 1+ = 2 ( )satisfythecompatibilitycondition A ( Du 0 ) n =0on @ :Wesaythatafunction u denedin T ,resp.]is radial if u ( x;t )= u ( y;t ) 8 x;y 2 ;j x j = j y j ,8 t 2 (0 ;T )[ u ( x )= u ( y ) 8 x;y 2 ;j x j = j y j ,resp.]. Wenowhavethefollowing. Theorem2.3.5. Assume u 0 isradial.Thenthereareinnitelymanyradialandnon-radi al Lipschitzsolutionsto (1.2) . TheproofofthistheoremisgivenattheendofChapter4. 32 Chapter3 Preliminaries Thischapterpreparessomeessentialingredientsfortheproofs ofexistencetheorems,The- orems2.3.2and2.3.4. 3.1Uniformlyparabolicequations Werefertothestandardreferences[29,30]forsomenotations concerningfunctionsand domainsofclass C k + withaninteger k 0andanumber0 << 1. Assume ~ f 2 C 1+ ([0 ;1 ))isafunctionsatisfying ~ f ( s )+2 s ~ f 0 ( s ) 8 s 0 ;(3.1) where > 0areconstants.Thisconditionisequivalentto ( s ~ f ( s 2 )) 0 forall s 2 R ;hence, ~ f ( s ) forall s 0 :Let ~ A ( p )= ~ f ( j p j 2 ) p ( p 2 R n ) :Thenwehave ~ A i p j ( p )= ~ f ( j p j 2 ) ij +2 ~ f 0 ( j p j 2 ) p i p j ( i;j =1 ;2 ; ;n ;p 2 R n ) 33 andhencethe uniformellipticitycondition : j q j 2 n X i;j =1 ~ A i p j ( p ) q i q j j q j 2 8 p;q 2 R n :(3.2) Theorem3.1.1. Assume (2 :10) holds.Thentheinitial-Neumannboundaryvalueproblem 8 > > > > > > > > < > > > > > > > > : u t =div( ~ A ( Du )) in T ;@u=@ n =0 on @ (0 ;T ) ;u ( x; 0)= u 0 ( x ) for x 2 (3.3) hasauniquesolution u 2 C 2+ ; 1+ = 2 ( T ) .Moreover,if ~ f 2 C 3 ([0 ;1 )) and isconvex, thenthe gradientmaximumprinciple holds: k Du k L 1 ( T ) = k Du 0 k L 1 () :(3.4) Proof. Letusdividetheproofintothreesteps. 1.Asproblem(3.3)isuniformlyparabolicby(3.2),theexistenceofau niqueclassical solution u in C 2+ ; 2+ 2 ( T )followsfromthestandardtheory;see[30,Theorem13.24].To provethegradientmaximumprinciple(3.4),weassume ~ f 2 C 3 ([0 ;1 ))andisconvex. Notethat,since ~ A 2 C 3 ( R n ),astandardbootstrapargumentbasedontheregularitytheor y oflinearparabolicequations[29,30]showsthatthesolution u hasallcontinuouspartial derivatives u x i x j x k and u x i t within T for1 i;j;k n ;see[24]fordetails. 34 2.Let v = j Du j 2 :Then,within T ;wecompute v =2 Du D( u )+2 j D2 u j 2 ;u t =div( ~ A ( Du ))=div( ~ f ( v ) Du )= ~ f 0 ( v ) Dv Du + ~ f ( v ) u; Du t = ~ f 00 ( v )( Dv Du ) Dv + ~ f 0 ( v )( D2 u ) Dv + ~ f 0 ( v )( D2 v ) Du + ~ f 0 ( v )( u ) Dv + ~ f ( v ) D( u ) :Puttingtheseequationsinto v t =2 Du Du t ,weobtain v t L ( v ) B Dv = 2 ~ f ( j Du j 2 ) j D2 u j 2 0in T ,(3.5) whereoperator L ( v )andcoecient B aredenedby L ( v )= ~ f ( j Du j 2 ) v +2 ~ f 0 ( j Du j 2 ) Du ( D2 v ) Du; B =2 ~ f 00 ( v )( Dv Du ) Du +2 ~ f 0 ( v )( D2 u ) Du +2 ~ f 0 ( v )( u ) Du: Wewrite L ( v )= P n i;j =1 a ij v x i x j ;withcoecients a ij = a ij ( x;t )givenby a ij = ~ A i p j ( Du )= ~ f ( j Du j 2 ) ij +2 ~ f 0 ( j Du j 2 ) u x i u x j ( i;j =1 ; ;n ) :Notethaton T alleigenvaluesofthematrix( a ij )liein[ ; ]. 3.Weshow max ( x;t ) 2 T v ( x;t )=max x 2 v ( x; 0) ;35 whichproves(3.4).Weprovethisbycontradiction.Suppose M :=max ( x;t ) 2 T v ( x;t ) > max x 2 v ( x; 0) :(3.6) Let( x 0 ;t 0 ) 2 T besuchthat v ( x 0 ;t 0 )= M ;then t 0 > 0 :If x 0 2 ,thenthestrong maximumprincipleappliedto(3.5)wouldimplythat v isconstanton t 0 ;whichyields v ( x; 0) M on ,acontradictionto(3.6).Consequently x 0 2 @ andthus v ( x 0 ;t 0 )= M>v ( x;t )forall( x;t ) 2 T :WecanthenapplyHopf'sLemmaforparabolicequations [36]to(3.5)todeduce @v ( x 0 ;t 0 ) =@ n > 0 :However,aresultof[1,Lemma2.1](seeaslo[21, Theorem2])assertsthat @v=@ n 0on @ [0 ;T ];whichgivesadesiredcontradiction. 3.2Modicationofprolefunctions Thefollowingelementaryresultscanbeprovedinasimilarwayasin[44,4 5];weomitthe proofs(seeFigures3.1and3.2). Lemma3.2.1 ( CaseI: Perona-Maliktype) . Assume Hypothesis(PM) .Forevery 0 < > : = ˙ ( s ) ;0 s s ( r 1 ) ;<˙ ( s ) ;s ( r 1 ) 0 :Withsuchfunction ~ ˙ ,dene ~ f ( s )=~ ˙ ( p s ) = p s ( s> 0 )and ~ f (0)= f (0); then ~ f 2 C 3 ([0 ;1 )) fulllscondition (3 :1) . 36 s 0 ˙ ( s ) ˙ ( s 0 ) r 1 r 2 s + ( r 1 ) ~ ˙ ( s ) s + ( r 2 ) s 0 s ( r 1 ) Figure3.1: CaseI: Perona-Maliktypeprole ˙ ( s )andmodedfunction~ ˙ ( s ). Lemma3.2.2 ( CaseII: Cubic-liketype) . Assume Hypothesis(C) .Forevery ˙ ( s 2 ) > > > > < > > > > > : = ˙ ( s ) ;s 2 [0 ;s ( r 1 )] [ [s + ( r 2 ) ;1 ) ;<˙ ( s ) ;s ( r 1 ) ˙ ( s ) ;s + ( r 1 ) s 0 :Withsuchfunction ~ ˙ ,dene ~ f ( s )=~ ˙ ( p s ) = p s ( s> 0 )and ~ f (0)= f (0); then ~ f 2 C 1+ ([0 ;1 )) fulllscondition (3 :1) . 3.3Rightinverseofthedivergenceoperator Todealwithlinearconstraintdiv v = u ,wefollowanargumentof[4,Lemma4]toconstruct arightinverse R ofthedivergenceoperator:div R = Id (inthesenseofdistributionsin T ).Forthepurposeofthisdissertation,theconstructionof R isrestrictedto box domains, 37 s 0 ˙ ( s ) ˙ ( s 1 ) ˙ ( s 2 ) r 1 r 2 s 2 ~ ˙ ( s ) s 1 s 1 s ( r 1 ) s 2 s + ( r 2 ) s + ( r 1 ) Figure3.2: CaseII: Holligtypeprole ˙ ( s )andmodedfunction~ ˙ ( s ). bywhichwemeandomainsgivenby Q = J 1 J 2 J n ,where J i =( a i ;b i ) ˆ R isa niteopeninterval. Givenabox Q ,wedenealinearoperator R n :L 1 ( Q ) ! L 1 ( Q ;R n )inductivelyon dimension n asfollows.If n =1,for u 2 L 1 ( J 1 ),wedene v = R 1 u by v ( x 1 )= Z x 1 a 1 u ( s ) ds ( x 1 2 J 1 ) :Assume n =2.Let u 2 L 1 ( J 1 J 2 ) :Set~ u ( x 1 )= R b 2 a 2 u ( x 1 ;s ) ds for x 1 2 J 1 :Then ~ u 2 L 1 ( J 1 ) :Let~ v = R 1 ~ u ;thatis, ~ v ( x 1 )= Z x 1 a 1 ~ u ( s ) ds = Z x 1 a 1 Z b 2 a 2 u ( s;˝ ) d˝ds ( x 1 2 J 1 ) :Let ˆ 2 2 C 1 c ( a 2 ;b 2 )besuchthat0 ˆ 2 ( s ) C 0 b 2 a 2 and R b 2 a 2 ˆ 2 ( s ) ds =1 :Dene v = R 2 u 2 L 1 ( J 1 J 2 ;R 2 )by v =( v 1 ;v 2 )with v 1 ( x 1 ;x 2 )= ˆ 2 ( x 2 )~ v ( x 1 )and v 2 ( x 1 ;x 2 )= Z x 2 a 2 u ( x 1 ;s ) ds ~ u ( x 1 ) Z x 2 a 2 ˆ 2 ( s ) ds: 38 Notethatif u 2 W 1 ;1 ( J 1 J 2 )then~ u 2 W 1 ;1 ( J 1 );hence v = R 2 u 2 W 1 ;1 ( J 1 J 2 ;R 2 ) anddiv v = u a.e.in J 1 J 2 :Moreover,if u 2 C 1 ( J 1 J 2 )then v isin C 1 ( J 1 J 2 ;R 2 ) :Assumethatwehavedenedtheoperator R n 1 .Let u 2 L 1 ( Q )with Q = J 1 J 2 J n and x =( x 0 ;x n ) 2 Q ,where x 0 2 Q 0 = J 1 J n 1 and x n 2 J n :Set ~ u ( x 0 )= R b n a n u ( x 0 ;s ) ds for x 0 2 Q 0 :Then~ u 2 L 1 ( Q 0 ) :Bytheassumption,~ v = R n 1 ~ u 2 L 1 ( Q 0 ;R n 1 )isdened.Write~ v ( x 0 )=( Z 1 ( x 0 ) ; ;Z n 1 ( x 0 )) ;andlet ˆ n 2 C 1 c ( a n ;b n )be afunctionsatisfying0 ˆ n ( s ) C 0 b n a n and R b n a n ˆ n ( s ) ds =1 :Dene v = R n u 2 L 1 ( Q ;R n ) asfollows.For x =( x 0 ;x n ) 2 Q;v ( x )=( v 1 ( x ) ;v 2 ( x ) ; ;v n ( x ))isdenedby v k ( x 0 ;x n )= ˆ n ( x n ) Z k ( x 0 )( k =1 ;2 ; ;n 1) ;v n ( x 0 ;x n )= Z x n a n u ( x 0 ;s ) ds ~ u ( x 0 ) Z x n a n ˆ n ( s ) ds: Then R n :L 1 ( Q ) ! L 1 ( Q ;R n )isawell-denedlinearoperator;moreover, kR n u k L 1 ( Q ) C n ( j J 1 j + + j J n j ) k u k L 1 ( Q ) ;(3.7) where C n > 0isaconstantdependingonlyon n: Asinthecase n =2,weseethatif u 2 W 1 ;1 ( Q )then v = R n u 2 W 1 ;1 ( Q ;R n )and div v = u a.e.in Q .Also,if u 2 C 1 ( Q )then v = R n u isin C 1 ( Q ;R n ) :Moreover,if u 2 W 1 ;1 0 ( Q )satises R Q u ( x ) dx =0,thenonecaneasilyshowthat v = R n u 2 W 1 ;1 0 ( Q ;R n ) :Let I beaniteopenintervalin R .Wenowextendtheoperator R n toanoperator R on L 1 ( Q I )bydening,fora.e.( x;t ) 2 Q I ,( R u )( x;t )=( R n u ( ;t ))( x ) 8 u 2 L 1 ( Q I ) :39 Then R :L 1 ( Q I ) ! L 1 ( Q I ;R n )isaboundedlinearoperator. Wehavethefollowingresult. Theorem3.3.1. Let u 2 W 1 ;1 0 ( Q I ) satisfy R Q u ( x;t ) dx =0 forall t 2 I .Then v = R u 2 W 1 ;1 0 ( Q I ;R n ) , div v = u a.e.in Q I ,and k v t k L 1 ( Q I ) C n ( j J 1 j + + j J n j ) k u t k L 1 ( Q I ) ;(3.8) where Q = J 1 J n and C n isthesameconstantasin (3 :7) .Moreover,if u 2 C 1 ( Q I ) then v = R u 2 C 1 ( Q I ;R n ) :Proof. Given u 2 W 1 ;1 0 ( Q I ),let v = R u: Weeasilyverifythat v isLipschitzcontinuous in t andhence v t exists.Italsofollowsthat v t = R ( u t ) :Clearly,if R Q u ( x;t ) dx =0then v ( x;t )=0whenever t 2 @I or x 2 @Q .Thisproves v 2 W 1 ;1 0 ( Q I ;R n )andtheestimate (3.8)followsfrom(3.7).Finally,fromthedenitionof R u ,weseethatif u 2 C 1 ( Q I ) then v = R u 2 C 1 ( Q I ;R n ) :40 Chapter4 Perona-Maliktypeequations Inthischapter,wecompletelyprovetheexistenceresulton CaseI: Perona-Maliktype equations,thatis,Theorem2.3.2.Inordertoso,weassume Hypothesis(PM) throughout thischapter. 4.1Geometryofrelevantmatrixset Webeginthissectionbyintroducinganapproachforsolvingproblem( 1.2)thatturnsout tobeunsuccessful;however,itprovidesuswiththemainideaofsolv ing(1.2)inthecontext ofourmethod,Theorem2.2.4.Thenweembarkonanextensiveanaly sisof partial rank-one structureofsomerelevantmatrixsetthateventuallyyieldsTheor em4.1.6. 4.1.1Non-homogeneousdierentialinclusionanditslimit ation Let u 0 2 W 1 ;1 ().Assume=( u ;v ) 2 W 1 ;1 ( T ;R 1+ n )isaboundaryfunction,that is,itsatises 8 > > > > > > > > < > > > > > > > > : u ( x; 0)= u 0 ( x ) ;x 2 ;div v ( x;t )= u ( x;t ) ;a.e.( x;t ) 2 T ;v ( ;t ) n j @ =0 ;t 2 [0 ;T ]:41 Ifafunction w =( u;v ) 2 W 1 ;1 ( T ;R 1+ n )solvestheDirichletproblemof non-homogeneous derentialinclusion 8 > > > < > > > : r w ( x;t ) 2 K ( u ( x;t )) ;a.e.( x;t ) 2 T ,w ( x;t )=( x;t ) ;( x;t ) 2 @ T ;(4.1) thenitcanbeeasilyseenthat u isaLipschitzsolutionto(1 :2).Here, r w denotesthe space-timeJacobianmatrixof w thatliesin M (1+ n ) ( n +1) ,thespaceof(1+ n ) ( n +1) realmatrices,andforeach l 2 R ,K ( l )isthesubsetof M (1+ n ) ( n +1) denedby K ( l )= 8 > < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;c 2 R ;B 2 M n n ;tr B = l 9 > = > ; :(4.2) TheDirichletproblem(4.1)fallsintotheframeworkofgeneralnon-h omogeneouspartial derentialinclusionsthathavebeenstudiedbyDacorognaandMar cellini[10]usingBaire's categorymethodandbyMullerandSychev[34]usingtheconvexint egrationmethodfol- lowingtheworks[18,32,33];seealso[28].Recently,themethodsofd erentialinclusion havebeensuccessfullyappliedtootherimportantproblemsinpartia lderentialequations [8,11,31,38,41]. Wepointoutthattheexistenceresultof[34]isnotapplicabletoprob lem(4.1)even indimension n =1,ashasalreadybeennoticedin[44,45].Akeyconditioninthemain existencetheoremof[34],whenappliedto(4.1),wouldrequirethatt heboundaryfunction satisfy r ( x;t ) 2 U ( u ( x;t )) [ K ( u ( x;t )) a:e: ( x;t ) 2 T ;where U ( s ) ˆ M (1+ n ) ( n +1) ( s 2 R )areboundedsetsthatare reducibleto K ( s )inthe 42 sensethat,forevery s 0 2 R ;˘ 0 2 U ( s 0 ) ;> 0,andboundedLipschitzdomain G ˆ R n +1 ,thereexistapiecewiseanefunction w 2 W 1 ;1 0 ( G ;R 1+ n )anda > 0satisfying,for a.e. z =( x;t ) 2 G ,˘ 0 + r w ( z ) 2 \ js s 0 j< U ( s ) ;Z G dist( ˘ 0 + r w ( z ) ;K ( s 0 )) dz< j G j :Thesecondconditionwouldimplytr B 0 = s 0 foreach ˘ 0 = 0 B @ p 0 c 0 B 0 0 1 C A 2 U ( s 0 )and s 0 2 R ;butthen \ js s 0 j< U ( s )= ; ;whichmakestherstconditionimpossible. However,certainstructuresofset K (0)turnouttobestillquiteuseful,especiallywhenit comestotherelaxationof homogeneous derentialinclusion r ! ( z ) 2 K (0)with z =( x;t ) and ! =( '; ).Weinvestigatethesestructuresandestablishsuchrelaxationr esultthrough therestofthissectionandSection4.2. 4.1.2Geometryofthematrixset F 0 Fixanytwonumbers0 < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;j p j2 ( s ( r 1 ) ;s ( r 2 )) [ ( s + ( r 2 ) ;s + ( r 1 )) ;c 2 R ;B 2 M n n ;tr B =0 9 > = > ; :Wedecomposetheset F 0 intotwodisjointsubsetsasfollows: F = 8 > < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;j p j2 ( s ( r 1 ) ;s ( r 2 )) ;c 2 R ;B 2 M n n ;tr B =0 9 > = > ; ;43 F + = 8 > < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;j p j2 ( s + ( r 2 ) ;s + ( r 1 )) ;c 2 R ;B 2 M n n ;tr B =0 9 > = > ; :Inordertoextractmoredetailedinformationonsolutionsasstate dinTheorem2.3.2, wefocusonthehomogeneousderentialinclusion r ! ( z ) 2 F 0 ;thuswerstscrutinizethe rank-onestructureoftheset F 0 .Weintroducethefollowing: Denition4.1.1. Foragivenset E ˆ M (1+ n ) ( n +1) ,L ( E )isdenedtobethesetofall matrices ˘ 2 M (1+ n ) ( n +1) thatarenotin E butarerepresentableby ˘ = ˘ 1 +(1 ) ˘ 2 forsome 2 (0 ;1)and ˘ 1 ;˘ 2 2 E withrank( ˘ 1 ˘ 2 )=1,orequivalently, L ( E )= f ˘ 62 E j ˘ + t 2 E forsome t < 0 < > : 0 B @ pc B 1 C A c 2 R ;B 2 M n n ;tr B =0 ;( p; ) 2S 9 > = > ; (4.5) forsomeset S = S r 1 ;r 2 ˆ R n + n . Proof. Let( c;B ) ;( c 0 ;B 0 ) 2 R M n n besuchthattr B =tr B 0 =0,anddene S ( c;B ) = 8 > < > : ( p; ) 2 R n + n 0 B @ pc B 1 C A 2 R ( F 0 ) 9 > = > ; ;S ( c 0 ;B 0 ) = 8 > < > : ( p; ) 2 R n + n 0 B @ pc 0 B 0 1 C A 2 R ( F 0 ) 9 > = > ; :Itissucienttoshowthat S ( c;B ) = S ( c 0 ;B 0 ) =: S :Let( p; ) 2S ( c;B ) ,thatis, ˘ = 0 B @ pc B 1 C A 2 R ( F 0 ).Then ˘ := ˘ + t 2 F forsome t < 0 a ,thereexistsacontinuousfunction h ( a;b;c; ; ; ): I a;c =[0 ;a ) [0 ;1 ) [0 ;c ) ! [0 ;1 ) with h ( a;b;c; 0 ;0 ;0)=0 satisfyingthefollowing: Let 1 ; 2 and beanypositivenumberswith 0 s + ( r 2 ) s ( r 2 ) :(4.7) Next,considerthefunction F denedby F ( 0 ;q 0 ;s 0 ;p; )=( ˙ ( j p + s 0 q 0 j ) p + s 0 q 0 j p + s 0 q 0 j s 0 0 ;˙ ( j p + q 0 j ) p + q 0 j p + q 0 j 0 ; 0 q 0 ) 2 R n + n +1 forall 0 ;q 0 ;p; 2 R n and s 0 2 R with s ( r 1 ) < j p + s 0 q 0 j 0 ;where 0 2 [0 ;ˇ ]istheanglebetween p + s 0 q 0 and q 0 .Observeherethattheforwardpartof ˙ inthedenitionof F becomesessentialtoguaranteethat ˙ 0 ( j p + s 0 q 0 j ) > 0.Aftersome 55 elementarycolumnoperationstothelastmatrixfromtheaboverow operations,weobtain D( 0 ;q 0 ;s 0 ) F ! 0 B B B B B @ s 0 I n M s 0 N s 0 ! s 0 Os 0 M 1 M s 0 + N s 0 ! s 0 00 q 0 ! s 0 1 C C C C C A ;wherethe j thcolumnof N s 0 2 M n n is s 0 0 j + q 0 ( M s 0 ) j q 0 ! s 0 ! s 0 .So D( 0 ;q 0 ;s 0 ) F isinvertibleifand onlyifthe n n matrix M 1 1 s 0 M s 0 + 1 s 0 N s 0 isinvertible.Wecompute M 1 1 s 0 M s 0 + 1 s 0 N s 0 =( ˙ 0 ( j p + q 0 j ) ˙ ( j p + q 0 j ) j p + q 0 j ) p + q 0 j p + q 0 j p + q 0 j p + q 0 j + ˙ ( j p + q 0 j ) j p + q 0 j I n ( ˙ 0 ( j p + s 0 q 0 j ) ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j ) p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j I n + 1 q 0 ! s 0 ! s 0 ( 0 +( ˙ 0 ( j p + s 0 q 0 j ) ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j )( p + s 0 q 0 j p + s 0 q 0 j q 0 ) p + s 0 q 0 j p + s 0 q 0 j + ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j q 0 ) =( a 1 a s 0 ) I n +( b 1 a 1 ) p + q 0 j p + q 0 j p + q 0 j p + q 0 j ( b s 0 a s 0 ) p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j + 1 q 0 ! s 0 ! s 0 ! + s 0 ;andset(withanassumption a 1 6= a s 0 ) B = 1 a 1 a s 0 ( M 1 1 s 0 M s 0 + 1 s 0 N s 0 )= I n + b 1 a 1 a 1 a s 0 p + q 0 j p + q 0 j p + q 0 j p + q 0 j b s 0 a s 0 a 1 a s 0 p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j + 1 ( a 1 a s 0 ) q 0 ! s 0 ! s 0 ! + s 0 ;56 where a s 0 = ˙ ( jp + s 0 q 0 j) jp + s 0 q 0 j,b s 0 = ˙ 0 ( j p + s 0 q 0 j );then D( 0 ;q 0 ;s 0 ) F isinvertibleifandonlyifthe matrix B 2 M n n isinvertible. 4.ToclosetheargumentinStep2andthustonishtheproof,wech ooseasuitable l 2 = l r 2 2 (0 ;r 2 ),dependingon r 2 ,insuchawaythatforany r 1 2 ( l 2 ;r 2 ),thematrix B ,determinedthroughSteps2and3foranygiven( p 0 ; 0 ) 2S = S r 1 ;r 2 ,isinvertible. First,byHypothesis(PM),~ r 2 2 (0 ;r 2 )canbechosencloseenoughto r 2 sothat ˙ ( k ) k < ˙ ( l ) l 8 l 2 [s (~ r 2 ) ;s ( r 2 )] ;8 k 2 [s + ( r 2 ) ;s + (~ r 2 )] :Thendeneareal-valuedcontinuousfunction(toexpressthedet erminantofthematrix B fromStep3) DET( u;v;q; )=det I n + ˙ 0 ( j u j ) ˙ ( ju j) ju j˙ ( ju j) ju j ˙ ( jv j) jv ju j u j u j u j ˙ 0 ( j v j ) ˙ ( jv j) jv j˙ ( ju j) ju j ˙ ( jv j) jv jv j v j v j v j + 1 ( ˙ ( ju j) ju j ˙ ( jv j) jv j)(( ˙ 0 ( j v j ) ˙ ( jv j) jv j)( v jv j q ) 2 + ˙ ( jv j) jv j) ( ˙ 0 ( j v j ) ˙ ( j v j ) j v j )( v j v j q ) v j v j + ˙ ( j v j ) j v j q ( ˙ 0 ( j v j ) ˙ ( j v j ) j v j )( v j v j q ) v j v j + ˙ ( j v j ) j v j q + onthecompactset M ofpoints( u;v;q; ) 2 R n R n S n 1 R n with j u j2 [s + ( r 2 ) ;s + (~ r 2 )] ;j v j2 [s (~ r 2 ) ;s ( r 2 )] ;j j 1 :57 Set k = s + ( r 2 )and l = s ( r 2 );thenforeach q 2 S n 1 ;DET( kq; lq;q; 0)=det I n + ˙ 0 ( k ) ˙ ( k ) k + ˙ ( l ) l ˙ ( k ) k ˙ ( l ) l q q 6=0 ;since ˙ 0 ( k ) 6=0andhencethefractioninfrontof q q isderentfrom 1.So d :=min q 2 S n 1 j DET( kq; lq;q; 0) j > 0 :Next,chooseanumber > 0suchthatforall( u;v;q; ) ;(~ u; ~ v; ~ q; ~ ) 2M with j u ~ u j ;j v ~ v j ;j q ~ q j ;j ~ j < ,wehave j DET( u;v;q; ) DET(~ u; ~ v; ~ q; ~ ) j d= 2 > 0 :Theproofisnowcomplete. 4.2Relaxationof r ! ( z ) 2 F 0 Thefollowingresultisimportantfortheconvexintegrationwithlinear constraint;thefunc- tion ' determinedhereplaysasimilarroleasthetilefunction g usedin[44,45].Foramore generalcase,see[37,Lemma2.1]. 59 Lemma4.2.1. Let 1 ; 2 > 0 and 1 = 1 ; 2 = 2 with = 0 B @ qb 1 b q 1 C A ;j q j =1 ; q =0 ;b 6=0 :Let G ˆ R n +1 beaboundeddomain.Thenforeach > 0 ,thereexistsafunction ! = ( '; ) 2 C 1 c ( R n +1 ;R 1+ n ) with supp( ! ) ˆˆ G thatsatisesthefollowingproperties: (a) div =0 in G , (b) jf z 2 G jr ! ( z ) = 2f 1 ; 2 ggj <; (c) dist( r ! ( z ) ;[ 1 ; 2 ]) < forall z 2 G; (d) k ! k L 1 ( G ) <; (e) R R n ' ( x;t ) dx =0 forall t 2 R . Proof. Theprooffollowsasimpedversionof[37,Lemma2.1]. 1.Wedeneamap P :C 1 ( R n +1 ) ! C 0 ( R n +1 ;R 1+ n )bysetting P ( h )=( u;v ),where, for h ( x;t ) 2 C 1 ( R n +1 ), u ( x;t )= q Dh ( x;t ) ;v ( x;t )= 1 b ( q q ) Dh ( x;t ) :Weeasilyseethat P ( h )=( u;v ) 2 C 1 c ( R n +1 ;R 1+ n ) ;supp( P ( h )) ˆ supp( h ) ;div v 0, and R R n u ( x;t ) dx =0forall t 2 R ,forall h 2 C 1 c ( R n +1 ) :For h ( x;t )= f ( q x + bt )with f 2 C 1 ( R ), w =( u;v )= P ( h )isgivenby u ( x;t )= f 0 ( q x + bt )and v ( x;t )= f 0 ( q x + bt ) b ,andhence r w ( x;t )= f 00 ( q x + bt ) : Wealsonotethat P ( gh )= g P ( h )+ h P ( g )andhence rP ( gh )= g rP ( h )+ h rP ( g )+ B ( r g; r h ) 8 g;h 2 C 1 ( R n +1 ) ;(4.9) 60 where B ( r g; r h )isabilinearmapof r g and r h ;so jB ( r h; r g ) j C jr h jjr g j forsome constant C> 0 :2.Let G ˆˆ G beasmoothsub-domainsuchthat j G n G j <= 2 ;andlet ˆ 2 C 1 c ( G ) beacut-ofunctionsatisfying0 ˆ 1in G ,ˆ =1on G .As G isbounded, G ˆ f ( x;t ) j k 0,wecanndafunction f ˝ 2 C 1 c ( k;l )satisfying 1 f 00 ˝ 2 ;jf s 2 ( k;l ) j f 00 ˝ ( s ) = 2f 1 ; 2 ggj <˝; k f ˝ k L 1 + k f 0 ˝ k L 1 <˝: 3.Dene ! =( '; )= P ( ˆ ( x;t ) h ˝ ( x;t )) ;where h ˝ ( x;t )= f ˝ ( q x + bt ) :Then k h ˝ k C 1 C k f ˝ k C 1 C˝ ,! 2 C 1 c ( R n +1 ;R 1+ n ),supp( ! ) ˆ supp( ˆ ) ˆˆ G ,and(a)and(e)are satised.Notethat j ! jj ˆ jjP ( h ˝ ) j + j h ˝ jjP ( ˆ ) j C ˝; where C > 0isaconstantdependingon k ˆ k C 1 ( G ) .Sowecanchoosea ˝ 1 > 0sosmall that(d)issatisedforall0 <˝<˝ 1 :Notealsothat f z 2 G jr ! ( z ) = 2f 1 ; 2 gg ( G n G ) [f z 2 G j f 00 ˝ ( q x + bt ) = 2f 1 ; 2 gg :Since jf z 2 G j f 00 ˝ ( q x + bt ) = 2f 1 ; 2 gj N jf s 2 ( k;l ) j f 00 ˝ ( s ) = 2f 1 ; 2 ggj forsome constant N> 0dependingonlyonset G ,thereexistsa ˝ 2 > 0suchthat jf z 2 G jr ! ( z ) = 2f 1 ; 2 ggj 2 + N˝< 61 forall0 <˝<˝ 2 .Therefore,(b)issatised.Finally,notethat ˆ rP ( h ˝ ( x;t ))= ˆ f 00 ˝ ( q x + bt ) 2 [ 1 ; 2 ]in G and,by(4.9),forall z =( x;t ) 2 G ,jr ! ( z ) ˆ rP ( h ˝ ( x;t )) jj h ˝ jjrP ( ˆ ) j + jB ( r h ˝ ;r ˆ ) j C 0 ˝< forall0 <˝<˝ 3 ,where C 0 > 0isaconstantdependingon k ˆ k C 2 ( G ) ,and ˝ 3 > 0isanother constant.Hence(c)issatised.Taking0 <˝< min f ˝ 1 ;˝ 2 ;˝ 3 g ,theproofiscomplete. Wenowstatetherelaxationtheoremforhomogeneousderential inclusion r ! ( z ) 2 F 0 inaformthatismoresuitableforlateruse;werestricttheinclusiont oonly( p; ) components. Theorem4.2.2. Let 0 0 ,thereexistsa > 0 suchthatforeachbox Q I ˆ ~ Q ~ I ,point ( p; ) 2K ,andnumber ˆ> 0 sucientlysmall,thereexistsafunction ! =( '; ) 2 C 1 c ( Q I ;R 1+ n ) satisfyingthefollowingproperties: (a) div =0 in Q I , (b) ( p 0 + D' ( z ) ; 0 + t ( z )) 2S forall z 2 Q I and j ( p 0 ; 0 ) ( p; ) j ; (c) k ! k L 1 ( Q I ) <ˆ; (d) R Q I j + t ( z ) A ( p + D' ( z )) j dz< j Q I j = j ~ Q ~ I j ;(e) R Q ' ( x;t ) dx =0 forall t 2 I; (f) k ' t k L 1 ( Q I ) <ˆ: 62 Proof. ByTheorem4.1.6,thereexistnitelymanyopenballs B 1 ; ;B N ˆˆS covering K and C 1 functions q i : B i ! S n 1 , i : B i ! R n ,t i; : B i ! R (1 i N )with i q i =0 and t i; < 0 0,wechooseaconstant b i with 0 0 :Bycontinuity, H ˝ = S ( p ) 2 B j ;1 j N [˘ ˝ j; ( p; ) ;˘ ˝ j; + ( p; )]isacompactsubsetof R ( F 0 ), where R ( F 0 )isopeninthespace 0 = 8 > < > : 0 B @ pc B 1 C A tr B =0 9 > = > ; ;byLemma4.1.3andTheorem4.1.6.So d ˝ =dist( H ˝ ;@ j 0 R ( F 0 )) > 0,where @ j 0 isthe relativeboundaryin 0 .Let i; 1 = i; 1 i = 0 i (1 2 ˝ )( t i; + t i; ) i ; i; 2 = i; 2 i =(1 0 i )(1 2 ˝ )( t i; + t i; ) i on B i ,where i; 1 = ˝ ( t i; + )+(1 ˝ )( t i; ) > 0 ; i; 2 =(1 ˝ ) t i; + + ˝t i; > 0on B i ,and ˝> 0issosmallthat min 1 j N min B j j;k > 0( k =1 ;2) :ApplyingLemma4.2.1tomatrices i; 1 = i; 1 ( p; ) ; i; 2 = i; 2 ( p; )foraxed( p; ) 2 B i andagivenbox G = Q I ,weobtainthatforeach ˆ> 0,thereexistafunction ! = 64 ( '; ) 2 C 1 c ( Q I ;R 1+ n )andanopenset G ˆ ˆˆ Q I satisfyingthefollowingconditions: 8 > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > : (1)div =0in Q I ,(2) j ( Q I ) n G ˆ j <ˆ ;˘ i + r ! ( z ) 2f ˘ ˝ i; ;˘ ˝ i; + g forall z 2 G ˆ ,(3) ˘ i + r ! ( z ) 2 [˘ ˝ i; ;˘ ˝ i; + ]ˆ forall z 2 Q I; (4) k ! k L 1 ( Q I ) <ˆ ,(5) R Q ' ( x;t ) dx =0forall t 2 I ,(6) k ' t k L 1 ( Q I ) < 2 ˆ; (4.10) where[ ˘ ˝ i; ;˘ ˝ i; + ]ˆ denotesthe ˆ -neighborhoodofclosedlinesegment[ ˘ ˝ i; ;˘ ˝ i; + ]:Here,from (4.10.3),(4.10.6)followsas j ' t j < j c i; + c i; j + ˆ =( t i; + t i; ) j b i j + ˆ< 2 ˆ in Q I .Note(a),(c),(e),and(f)followfrom(4.10),where2 ˆ in(4.10.6)canbeadjustedto ˆ asin(f).Bytheuniformcontinuityof A on J = f p 0 2 R n jj p 0 j s + ( r 2 ) g ,wecannda 0 > 0suchthat j A ( p 0 ) A ( p 00 ) j < 3 j~ Q ~ I jwhenever p 0 ;p 00 2 J and j p 0 p 00 j < 0 :Wethen choosea ˝> 0sosmallthat C˝< 0 ;C j ~ Q ~ I j ˝< 3 :Next,wechoosea > 0suchthat < d ˝ 2 :If0 <ˆ< ,thenby(4.10.1)and(4.10.3),for 65 all z 2 Q I and j ( p 0 ; 0 ) ( p; ) j ,˘ i ( p 0 ; 0 )+ r ! ( z ) 2 0 ;dist( ˘ i ( p 0 ; 0 )+ r ! ( z ) ;H ˝ ) 0satisfying ( r 2 + M ˙ ) ˆ< jQ I j3 j~ Q ~ I j.Wehavevered(a){(f)forany( p; ) 2 B i and1 i N ,where > 0isindependent oftheindex i .Since B 1 ; ;B N cover K ,theproofisnowcomplete. 4.3Constructionofadmissibleset U Werstconstructasuitableboundaryfunction=( u ;v ) 2 W 1 ;1 ( T ;R 1+ n ).Assume and u 0 satisfy(2.10)withconvex.Let T = (0 ;T )foragiven T> 0and 66 M 0 = k Du 0 k L 1 () .Recallthatweassume(2.1);thatis, Z u 0 ( x ) dx =0 :(4.11) TotailorthedetailedresultofTheorem2.3.2intothegeneralexisten cetheorem,Theorem 2.2.4,weassumethefollowing:Let0 > > > > > > > < > > > > > > > > : u t =div( ~ A ( Du ))in T @u =@ n =0on @ (0 ;T ) u ( x; 0)= u 0 ( x ) ;x 2 (4.12) admitsauniqueclassicalsolution u 2 C 2+ ; 1+ = 2 ( T )satisfying j Du ( x;t ) j M 0 8 ( x;t ) 2 T :Fromconditions(2.10)and(4.11),wecanndafunction h 2 C 2+ ( )satisfying h = u 0 in ;@h=@ n =0on @ :Let v 0 = Dh 2 C 1+ ( ; R n )anddene,for( x;t ) 2 T ,v ( x;t )= v 0 ( x )+ Z t 0 ~ A ( Du ( x;s )) ds: (4.13) Thenitiseasilyseenthat:=( u ;v ) 2 C 1 ( T ;R 1+ n )satises(2.4);thatis, 8 > > > > > > > > < > > > > > > > > : u ( x; 0)= u 0 ( x )( x 2 ) ;div v = u a.e.in T ;v ( ;t ) n j @ =0 8 t 2 [0 ;T ]:(4.14) HenceisaboundaryfunctioninthesenseofDenition2.2.2. 68 Next,let F = f ( p;A ( p )) jj p j2 [0 ;s ( r 1 )] g :Thenwehavethefollowing: Lemma4.3.2. ( Du ( x;t ) ;v t ( x;t )) 2S[F8 ( x;t ) 2 T :Proof. Let( x;t ) 2 T and p = Du ( x;t );then j p j M 0 :If j p j s ( r 1 ),then ~ A ( p )= A ( p )andhenceby(4.13) ( Du ( x;t ) ;v t ( x;t ))=( p; ~ A ( p ))=( p;A ( p )) 2F :If s ( r 1 ) < j p j M 0 ,thenbyLemma4.3.1and(4.13) ( Du ( x;t ) ;v t ( x;t ))=( p; ~ A ( p )) 2S :Therefore( Du ;v t ) 2S[F in T .Denition4.3.3. Wesayafunction u is piecewise C 1 in T andwrite u 2 C 1 piece ( T )if thereexistsasequenceofdisjointopensets f G j g 1 j =1 in T suchthat u 2 C 1 ( G j ) 8 j 2 N ;j T n[ 1 j =1 G j j =0 :Notethatinthisdenition,wenecessarilyhave j @G j j =0forall j 2 N .(SelectionofinterfaceofmeasurezeroforclassicalandLi pschitzpartsof 69 solutions) Observethat jf ( x;t ) 2 T jj Du ( x;t ) j = s ( r ) gj > 0 for atmostcountablymany r 2 (0 ;~ r ).Wexany r 2 (0 ;~ r )with jf ( x;t ) 2 T jj Du ( x;t ) j = s ( r ) gj =0 ;andlet 1 T = f ( x;t ) 2 T jj Du ( x;t ) j s ( r ) g ;sothat 1 T and 2 T aredisjointsubsetsof T whoseunionhasmeasure j T j .Clearly, r 0 ˆ @ 1 T ,where r 0 isasinTheorem2.3.2. Let m = k u t k L 1 ( T ) +1.Wenallydenetheadmissibleset U asfollows: U = n u 2 C 1 piece \ W 1 ;1 u ( T ) u = u in 1 T ;k u t k L 1 ( T ) 0,let U bethesubsetof U givenby U = n u 2Uj9 v 2 C 1 piece \ W 1 ;1 v ( T ;R n )suchthatdiv v = u and ( Du;v t ) 2S[F a.e.in T ,and R T j v t A ( Du ) j dxdt j T j o :70 Remark4.3.4. From(4.14),Lemma4.3.2,andthedenitionof U ,itfollowsthat u 2U withitscorrespondingvectorfunction v ;so U isnon-empty.Also U isaboundedsubset of W 1 ;1 u ( T )as S[F isbounded.Moreover,by(i)ofTheorem4.1.6andthedenitionof F ,foreach u 2U ,itscorrespondingvectorfunction v satises k v t k L 1 ( T ) r 2 = r .Thus U isindeedanadmissiblesetinthesenseofDenition2.2.3withrespectto theboundary function=( u ;v ).Finally,notethat s ( r 1 ) < j Du j 0 , U isdensein U underthe L 1 -norm. Proof. Let u 2U ,> 0.Thegoalistoconstructafunction~ u 2U suchthat k ~ u u k L 1 ( T ) <: Forclarity,wedividetheproofintoseveralsteps. 1.Note k u t k L 1 ( T ) 0andthereexistsavectorfunction v 2 C 1 piece \ W 1 ;1 v ( T ;R n )suchthatdiv v = u and( Du;v t ) 2S[F a.e.in T :Sinceboth u and v arepiecewise C 1 in T ,thereexistsasequenceofdisjointopensets f G j g 1 j =1 in T with j @G j j =0suchthat u 2 C 1 ( G j ) ;v 2 C 1 ( G j ;R n ) 8 j 1 ;j T n[ 1 j =1 G j j =0 :71 2.Let j 2 N bexed,where N isthesetofpositiveintegers.Notethat( Du ( z ) ;v t ( z )) 2 S[F forall z =( x;t ) 2 G j andthat H j = f z 2 G j j ( Du ( z ) ;v t ( z )) 2 @ Sg isa(relatively) closedsetin G j withmeasurezero.So ~ G j = G j n H j isanopensubsetof G j with j ~ G j j = j G j j ,and( Du ( z ) ;v t ( z )) 2S[F forall z 2 ~ G j .3.Foreach ˝> 0,let G ˝ = f ( p; ) 2Sjj A ( p ) j >˝; dist(( p; ) ;@ S ) >˝ ) g ;then S =( [ ˝> 0 G ˝ ) [f ( p; ) 2Sj A ( p )= g as S isopen.Since A ( p )= 8 ( p; ) 2F ,wehave Z ~ G j j v t ( z ) A ( Du ( z )) j dz =lim ˝ ! 0 + Z f z 2 ~ G j j( Du ( z ) ;v t ( z )) 2G ˝ g j v t ( z ) A ( Du ( z )) j dz ;sowecannda ˝ j > 0suchthat Z F j j v t ( z ) A ( Du ( z )) j dz< 3 2 j j T j and j @O j j =0 ;(4.16) where F j = f z 2 ~ G j j ( Du ( z ) ;v t ( z )) = 2G ˝ j g and O j = ~ G j n F j isopen.Let J bethesetof allindices j 2 N with O j 6= ; .Thenfor j 62 J ,F j = ~ G j .4.Wenowxa j 2 J .Notethat O j = f z 2 ~ G j j ( Du ( z ) ;v t ( z )) 2G ˝ j g andthat K j := G ˝ j isacompactsubsetof S .Let ~ Q ˆ R n beaboxwith ˆ ~ Q and ~ I =(0 ;T ). ApplyingTheorem4.2.2tobox ~ Q ~ I; K j ˆˆS = S r 1 ;r 2 (recall r 2 = r;r 1 =~ r ),and 0 = j T j12 ,weobtainaconstant j > 0thatsatisestheconclusionofthetheorem.Bythe uniformcontinuityof A oncompactsubsetsof R n ,wecannda = ;r 1 > 0suchthat j A ( p ) A ( p 0 ) j < 12 (4.17) whenever j p j ;j p 0 j 2 s + ( r 1 )and j p p 0 j : Alsobytheuniformcontinuityof u ,v ,and 72 theirgradientson G j ,thereexistsa j > 0suchthat j u ( z ) u ( z 0 ) j + jr u ( z ) r u ( z 0 ) j + j v ( z ) v ( z 0 ) j + jr v ( z ) r v ( z 0 ) j < min f j 2 ; 12 ;;s + ( r 1 ) g (4.18) whenever z;z 0 2 G j and j z z 0 j j :Wenowcover O j (uptomeasurezero)byasequence ofdisjointopencubes f Q i j I i j g 1 i =1 in O j whosesidesareparalleltotheaxeswithcenter z i j anddiameter l i j < j :5.Fixan i 2 N andwrite w =( u;v ), ˘ = 0 B @ pc B 1 C A = r w ( z i j )= 0 B @ Du ( z i j ) u t ( z i j ) Dv ( z i j ) v t ( z i j ) 1 C A :Bythechoiceof j > 0inStep4viaTheorem4.2.2,since Q i j I i j ˆ ~ Q ~ I and( p; ) 2K j ,forallsucientlysmall ˆ> 0,thereexistsafunction ! i j =( ' i j ; i j ) 2 C 1 c ( Q i j I i j ;R 1+ n ) satisfying (a)div i j =0in Q i j I i j ,(b)( p 0 + D' i j ( z ) ; 0 +( i j ) t ( z )) 2S forall z 2 Q i j I i j andall j ( p 0 ; 0 ) ( p; ) j j ;(c) k ! i j k L 1 ( Q i j I i j ) <ˆ; (d) R Q i j I i j j +( i j ) t ( z ) A ( p + D' i j ( z )) j dz< 0 j Q i j I i j j = j ~ Q ~ I j ;(e) R Q i j ' i j ( x;t ) dx =0forall t 2 I i j ;(f) k ( ' i j ) t k L 1 ( Q i j I i j ) <ˆ: Here,welet0 <ˆ min f ˝ 0 ; j 2 C ; 12 C ; g ,where C n istheconstantinTheorem3.3.1and C is theproductof C n andthesumofthelengthsofallsidesof ~ Q .By(e),wecanapplyTheorem 3.3.1to ' i j on Q i j I i j toobtainafunction g i j = R ' i j 2 C 1 ( Q i j I i j ;R n ) \ W 1 ;1 0 ( Q i j I i j ;R n ) 73 suchthatdiv g i j = ' i j in Q i j I i j and k ( g i j ) t k L 1 ( Q i j I i j ) C k ( ' i j ) t k L 1 ( Q i j I i j ) j 2 :(by(f))(4.19) 6.As v t and A ( Du )areessentiallyboundedin T ,wecanselectaniteindexset Iˆ J N sothat Z S ( j;i ) 2 ( J N ) nI Q i j I i j j v t ( z ) A ( Du ( z )) j dz 3 j T j :(4.20) Wenallydene (~ u; ~ v )=( u;v )+ X ( j;i ) 2I ˜ Q i j I i j ( ' i j ; i j + g i j )in T .Asasideremark,noteherethatonly nitely manyfunctions( ' i j ; i j + g i j )aredisjointly patchedtotheoriginal( u;v )toobtainanewfunction(~ u; ~ v )towardsthegoaloftheproof. Thereasonforusingonlynitelymanypiecesofgluingisduetothelack ofcontroloverthe spatialgradients D( i j + g i j ),andovercomingthisdcultyisattheheartofourmethod. 7.Letusnallycheckthat~ u togetherwith~ v indeedgivesthedesiredresult.By construction,itisclearthat~ u 2 C 1 piece \ W 1 ;1 u ( T ),~ v 2 C 1 piece \ W 1 ;1 v ( T ;R n ) :By thechoiceof ˆ in(f)as ˆ ˝ 0 ,wehave k ~ u t k L 1 ( T ) s ( r ) :Thus, j L r; ~ r j > 0. 4.5ProofofTheorem2.3.5 Inthisnalsection,wecompletetheproofofTheorem2.3.5onthec oexistenceofradialand non-radialLipschitzsolutionstoproblem(1.2)whenisaballand u 0 isradial. ProofofTheorem2.3.5. UsingTheorem3.1.1,theexistenceofinnitelymany radial Lip- schitzsolutionsto(1.2)followsfrom[25].Weremarkthattheseradia lsolutionsarenot obtainedthroughtheexistenceresultofthisdissertation,Theor em2.3.2. Theexistenceofinnitelymany non-radial Lipschitzsolutionsto(1.2)canbeshownby modifyingtheproofofTheorem2.3.2.Weproceedtheproofasbelow .Itiseasytocheckthatthefunction u 2U constructedinSection4.3isradialin T .OurstrategyistoimitatetheprocedureofthedensityproofinSec tion4.4tothefunction 77 ( u ;v ) :Wechooseaspace-timeboxin T withpositivedistancefromthecentralaxisof T where v t issucientlyawayfrom A ( Du )in L 1 -sense.Thenasinthedensityproof, weperformthesurgeryon( u ;v )onlyinthisboxtoobtainafunction( u nr ;v nr )with membership u nr 2U maintained.Suchsurgerybreaksdowntheradialsymmetryof u ;hence, u nr isnon-radial.Notealsothatthis u nr cannotbeaLipschitzsolutionto(1.2). Supposethereisnonon-radialLipschitzsolutionto(1.2).Intheco ntextoftheproofof Theorem2.2.4,thismeansthatevery u 2G isaradialsolution.The L 1 -densityof G in X thenimpliesthateveryfunctionin X isradial.Thiscontradictstheexistenceofanon-radial function u nr in UˆX above.Thusthereexistsanon-radialLipschitzsolutionto(1.2). Supposethereareonlynitelymanynon-radialLipschitzsolutionst o(1.2).Thisforces thatthenon-radialfunction u nr shouldbethe L 1 -limitofsomesequenceofradialfunctions in G ,acontradiction.Therefore,thereareinnitelymanynon-radial Lipschitzsolutionsto (1.2). 78 Chapter5 Holligtypeequations Thischapterdealswiththeexistenceresulton CaseII: Holligtypeequations,thatis, Theorem2.3.4.Wethusassume Hypothesis(H) throughoutthischapter. 5.1Geometryofrelevantmatrixset ThissectionproceedsalmostinthesamewayasinSection4.1,andsow eskipmanydetails unlessthereshouldsomechangetobemade. Foreach l 2 R ,let K ( l )bethesubsetof M (1+ n ) ( n +1) denedby(4.2)withux A ( p ) withprole ˙ ( s )satisfyingHypothesis(H). Fixanytwonumbers ˙ ( s 2 ) < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;j p j2 ( s ( r 1 ) ;s ( r 2 )) [ ( s + ( r 1 ) ;s + ( r 2 )) ;c 2 R ;B 2 M n n ;tr B =0 9 > = > ; :Theset F 0 isdecomposedintotwodisjointsubsetsasfollows: F = 8 > < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;j p j2 ( s ( r 1 ) ;s ( r 2 )) ;c 2 R ;B 2 M n n ;tr B =0 9 > = > ; ;79 F + = 8 > < > : 0 B @ pc BA ( p ) 1 C A p 2 R n ;j p j2 ( s + ( r 1 ) ;s + ( r 2 )) ;c 2 R ;B 2 M n n ;tr B =0 9 > = > ; :Asin CaseI: Perona-Maliktypeequations,wefocusonthehomogeneousdere ntial inclusion r ! ( z ) 2 F 0 ;thuswerststudytherank-onestructureoftheset F 0 .Forthematrixset F 0 ,wedene R ( F 0 )= [ ˘ 2 F ;rank( ˘ + ˘ )=1 ( ˘ ;˘ + ) :Fromacarefulanalysis,onecanactuallydeduce L ( F 0 )= R ( F 0 ) :(5.1) Thisisadrasticderenceto(4.3)where L ( F + ) 6= ; .However,inthecurrentcase,asonly forwardpartsof ˙ areinvolvedin F 0 ,nosuchsetappearsin(5.1);soitisevenmorenatural toonlysticktotheanalysisoftheset R ( F 0 )towardstheexistenceresult,Theorem2.3.4. Weperformthestep-by-stepanalysisoftheset R ( F 0 ). 1. Alternateexpressionfor R ( F 0 ) . Theproofofthefollowinglemmajustfollowsthe linesofthatofTheorem4.1.2withminorchanges.Soweskiptheproof .Lemma5.1.1. Let ˘ 2 M (1+ n ) ( n +1) .Then ˘ 2 R ( F 0 ) ifandonlyifthereexistnumbers t < 0 < > : 0 B @ pc B 1 C A c 2 R ;B 2 M n n ;tr B =0 ;( p; ) 2S 9 > = > ; (5.2) forsomeset S = S r 1 ;r 2 ˆ R n + n . 3. Selectionofapproximatecollinearrank-oneconnectionsf or R ( F 0 ) . Werst equipwitha2-dimensionaldescriptionfortherank-oneconnection sofdiagonalcomponents ofmatricesin R ( F 0 )inageneralformwhoseproofissimilartothatofLemma4.1.4buthas severalminorchanges. Lemma5.1.3. Forallpositivenumbers a;b;c with b>a ,thereexistsacontinuousfunction h ( a;b;c; ; ; ): I a;b;c =[0 ;a ) [0 ;b a ) [0 ;c ) ! [0 ;1 ) with h ( a;b;c; 0 ;0 ;0)=0 satisfyingthefollowing: Let 1 ; 2 and beanypositivenumberswith 0 s + ( r 1 ) s ( r 2 ) :(5.4) Next,considerthefunction F denedby F ( 0 ;q 0 ;s 0 ;p; )=( ˙ ( j p + s 0 q 0 j ) p + s 0 q 0 j p + s 0 q 0 j s 0 0 ;˙ ( j p + q 0 j ) p + q 0 j p + q 0 j 0 ; 0 q 0 ) 2 R n + n +1 forall 0 ;q 0 ;p; 2 R n and s 0 2 R with s ( r 1 ) < j p + s 0 q 0 j 0 ;where 0 2 [0 ;ˇ ]istheanglebetween p + s 0 q 0 and q 0 .Observeherethattheforwardpartof ˙ inthedenitionof F becomesessentialtoguaranteethat ˙ 0 ( j p + s 0 q 0 j ) > 0.Aftersome elementarycolumnoperationstothelastmatrixfromtheaboverow operations,weobtain D( 0 ;q 0 ;s 0 ) F ! 0 B B B B B @ s 0 I n M s 0 N s 0 ! s 0 Os 0 M 1 M s 0 + N s 0 ! s 0 00 q 0 ! s 0 1 C C C C C A ;wherethe j thcolumnof N s 0 2 M n n is s 0 0 j + q 0 ( M s 0 ) j q 0 ! s 0 ! s 0 .So D( 0 ;q 0 ;s 0 ) F isinvertibleifand onlyifthe n n matrix M 1 1 s 0 M s 0 + 1 s 0 N s 0 isinvertible.Wecompute M 1 1 s 0 M s 0 + 1 s 0 N s 0 =( ˙ 0 ( j p + q 0 j ) ˙ ( j p + q 0 j ) j p + q 0 j ) p + q 0 j p + q 0 j p + q 0 j p + q 0 j + ˙ ( j p + q 0 j ) j p + q 0 j I n ( ˙ 0 ( j p + s 0 q 0 j ) ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j ) p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j 89 ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j I n + 1 q 0 ! s 0 ! s 0 ( 0 +( ˙ 0 ( j p + s 0 q 0 j ) ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j )( p + s 0 q 0 j p + s 0 q 0 j q 0 ) p + s 0 q 0 j p + s 0 q 0 j + ˙ ( j p + s 0 q 0 j ) j p + s 0 q 0 j q 0 ) =( a 1 a s 0 ) I n +( b 1 a 1 ) p + q 0 j p + q 0 j p + q 0 j p + q 0 j ( b s 0 a s 0 ) p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j + 1 q 0 ! s 0 ! s 0 ! + s 0 ;andset(withanassumption a 1 6= a s 0 ) B = 1 a 1 a s 0 ( M 1 1 s 0 M s 0 + 1 s 0 N s 0 )= I n + b 1 a 1 a 1 a s 0 p + q 0 j p + q 0 j p + q 0 j p + q 0 j b s 0 a s 0 a 1 a s 0 p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j p + s 0 q 0 j + 1 ( a 1 a s 0 ) q 0 ! s 0 ! s 0 ! + s 0 ;where a s 0 = ˙ ( jp + s 0 q 0 j) jp + s 0 q 0 j,b s 0 = ˙ 0 ( j p + s 0 q 0 j );then D( 0 ;q 0 ;s 0 ) F isinvertibleifandonlyifthe matrix B 2 M n n isinvertible. 4.ToclosetheargumentsinStep2andthustonishtheproof,wec hooseasuitable l 2 = l r 2 2 ( ˙ ( s 2 ) ;r 2 )insuchawaythatforany r 1 2 ( l 2 ;r 2 ),thematrix B ,determined throughSteps2and3foranygiven( p 0 ; 0 ) 2S = S r 1 ;r 2 ,isinvertible. First,byHypothesis(C),~ r 2 2 ( ˙ ( s 2 ) ;r 2 )canbechosencloseenoughto r 2 sothat ˙ ( k ) k < ˙ ( l ) l 8 l 2 [s (~ r 2 ) ;s ( r 2 )] ;8 k 2 [s + (~ r 2 ) ;s + ( r 2 )] :Thendeneareal-valuedcontinuousfunction(toexpressthedet erminantofthematrix B 90 fromStep3) DET( u;v;q; )=det I n + ˙ 0 ( j u j ) ˙ ( ju j) ju j˙ ( ju j) ju j ˙ ( jv j) jv ju j u j u j u j ˙ 0 ( j v j ) ˙ ( jv j) jv j˙ ( ju j) ju j ˙ ( jv j) jv jv j v j v j v j + 1 ( ˙ ( ju j) ju j ˙ ( jv j) jv j)(( ˙ 0 ( j v j ) ˙ ( jv j) jv j)( v jv j q ) 2 + ˙ ( jv j) jv j) ( ˙ 0 ( j v j ) ˙ ( j v j ) j v j )( v j v j q ) v j v j + ˙ ( j v j ) j v j q ( ˙ 0 ( j v j ) ˙ ( j v j ) j v j )( v j v j q ) v j v j + ˙ ( j v j ) j v j q + onthecompactset M ofpoints( u;v;q; ) 2 R n R n S n 1 R n with j u j2 [s + (~ r 2 ) ;s + ( r 2 )] ;j v j2 [s (~ r 2 ) ;s ( r 2 )] ;j j 1 :With k = s + ( r 2 )and l = s ( r 2 ),foreach q 2 S n 1 ;DET( kq; lq;q; 0)=det I n + ˙ 0 ( k ) ˙ ( k ) k + ˙ ( l ) l ˙ ( k ) k ˙ ( l ) l q q 6=0 ;since ˙ 0 ( k ) 6=0andhencethefractioninfrontof q q isderentfrom 1.So d :=min q 2 S n 1 j DET( kq; lq;q; 0) j > 0 :Next,chooseanumber > 0suchthatforall( u;v;q; ) ;(~ u; ~ v; ~ q; ~ ) 2M with j u ~ u j ;j v ~ v j ;j q ~ q j ;j ~ j < ,wehave j DET( u;v;q; ) DET(~ u; ~ v; ~ q; ~ ) j d= 2 > 0 :Theproofisnowcomplete. 5.2Relaxationof r ! ( z ) 2 F 0 Lemma4.2.1isincommonuseforbothTheorems4.2.2and5.2.1,andsowe donotrestate ithere. Westatetherelaxationtheoremforhomogeneousderentialinclu sion r ! ( z ) 2 F 0 ina formthatismoresuitableforlateruse;werestricttheinclusiontoo nly( p; )components. Althoughtheproofofthistheoremisquitesimilartothatofitscompa nionversion,Theorem 4.2.2,weincludeithereforthesakeofcompleteness. Theorem5.2.1. Let ˙ ( s 2 ) 0 ,thereexistsa > 0 suchthatforeach box Q I ˆ ~ Q ~ I ,point ( p; ) 2K ,andnumber ˆ> 0 sucientlysmall,thereexistsa function ! =( '; ) 2 C 1 c ( Q I ;R 1+ n ) satisfyingthefollowingproperties: (a) div =0 in Q I , (b) ( p 0 + D' ( z ) ; 0 + t ( z )) 2S forall z 2 Q I and j ( p 0 ; 0 ) ( p; ) j ; (c) k ! k L 1 ( Q I ) <ˆ; (d) R Q I j + t ( z ) A ( p + D' ( z )) j dz< j Q I j = j ~ Q ~ I j ;(e) R Q ' ( x;t ) dx =0 forall t 2 I; 93 (f) k ' t k L 1 ( Q I ) <ˆ: Proof. ByTheorem5.1.5,thereexistnitelymanyopenballs B 1 ; ;B N ˆˆS covering K and C 1 functions q i : B i ! S n 1 , i : B i ! R n ,t i; : B i ! R (1 i N )with i q i =0 and t i; < 0 0,wechooseaconstant b i with 0 0 :Bycontinuity, H ˝ = S ( p ) 2 B j ;1 j N [˘ ˝ j; ( p; ) ;˘ ˝ j; + ( p; )]isacompactsubsetof R ( F 0 ), where R ( F 0 )isopeninthespace 0 = 8 > < > : 0 B @ pc B 1 C A tr B =0 9 > = > ; ;byLemma5.1.2andTheorem5.1.5.So d ˝ =dist( H ˝ ;@ j 0 R ( F 0 )) > 0,where @ j 0 isthe relativeboundaryin 0 .Let i; 1 = i; 1 i = 0 i (1 2 ˝ )( t i; + t i; ) i ; i; 2 = i; 2 i =(1 0 i )(1 2 ˝ )( t i; + t i; ) i on B i ,where i; 1 = ˝ ( t i; + )+(1 ˝ )( t i; ) > 0 ; i; 2 =(1 ˝ ) t i; + + ˝t i; > 0on B i ,and ˝> 0issosmallthat min 1 j N min B j j;k > 0( k =1 ;2) :ApplyingLemma4.2.1tomatrices i; 1 = i; 1 ( p; ) ; i; 2 = i; 2 ( p; )withaxed( p; ) 2 B i andagivenbox G = Q I ,weobtainthatforeach ˆ> 0,thereexistafunction ! = 95 ( '; ) 2 C 1 c ( Q I ;R 1+ n )andanopenset G ˆ ˆˆ Q I satisfyingthefollowingconditions: 8 > > > > > > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > > > > > > : (1)div =0in Q I ,(2) j ( Q I ) n G ˆ j <ˆ ;˘ i + r ! ( z ) 2f ˘ ˝ i; ;˘ ˝ i; + g forall z 2 G ˆ ,(3) ˘ i + r ! ( z ) 2 [˘ ˝ i; ;˘ ˝ i; + ]ˆ forall z 2 Q I; (4) k ! k L 1 ( Q I ) <ˆ ,(5) R Q ' ( x;t ) dx =0forall t 2 I ,(6) k ' t k L 1 ( Q I ) < 2 ˆ; (5.6) where[ ˘ ˝ i; ;˘ ˝ i; + ]ˆ denotesthe ˆ -neighborhoodofclosedlinesegment[ ˘ ˝ i; ;˘ ˝ i; + ]:Here,from (5.6.3),(5.6.6)followsas j ' t j < j c i; + c i; j + ˆ =( t i; + t i; ) j b i j + ˆ< 2 ˆ in Q I .Note(a),(c),(e),and(f)followfrom(5.6),where2 ˆ in(5.6.6)canbeadjustedto ˆ as in(f).Bytheuniformcontinuityof A on J = f p 0 2 R n jj p 0 j s 2 g ,wecannda 0 > 0 suchthat j A ( p 0 ) A ( p 00 ) j < 3 j~ Q ~ I jwhenever p 0 ;p 00 2 J and j p 0 p 00 j < 0 :Wethenchoose a ˝> 0sosmallthat C˝< 0 ;C j ~ Q ~ I j ˝< 3 :Next,wechoosea > 0suchthat < d ˝ 2 :If0 <ˆ< ,thenby(5.6.1)and(5.6.3),forall z 2 Q I and j ( p 0 ; 0 ) ( p; ) j ,˘ i ( p 0 ; 0 )+ r ! ( z ) 2 0 ;dist( ˘ i ( p 0 ; 0 )+ r ! ( z ) ;H ˝ ) 0satisfying ( r 2 + M ˙ ) ˆ< jQ I j3 j~ Q ~ I j.Wehavevered(a){(f)forany( p; ) 2 B i and1 i N ,where > 0isindependent oftheindex i .Since B 1 ; ;B N cover K ,theproofisnowcomplete. 5.3Constructionofadmissibleset U Werstconstructasuitableboundaryfunction=( u ;v ) 2 W 1 ;1 ( T ;R 1+ n ).Assume and u 0 satisfy(2.10)and j Du 0 ( x 0 ) j2 ( s 1 ;s 2 )forsome x 0 2 .Let T = (0 ;T )for agiven T> 0and M 0 = k Du 0 k L 1 () .Recallthatweassume(2.1);hence, Z u 0 ( x ) dx =0 :(5.7) 97 Note M 0 j Du 0 ( x 0 ) j >s 1 .Wenowassumethefollowing:If M 0 > > > > > > > < > > > > > > > > : u t =div( ~ A ( Du ))in T @u =@ n =0on @ (0 ;T ) u ( x; 0)= u 0 ( x ) ;x 2 (5.8) 98 admitsauniqueclassicalsolution u 2 C 2+ ; 1+ = 2 ( T ). Noteherethatwemaynothavethegradientmaximumprinciple(3.4)f orthesolution u sincewedonotassumetheconvexityofinCaseII:Holligtypeequa tions.However,inthe casethatisconvex,suchgradientmaximumprincipleholds,anditinv okesadvantageous eectsontheexistenceresult,Theorem2.3.4,intwofolds: 1.Theprole ˙ ( s )inHypothesis(H)canbeallowedtohaveunboundedderivativefor largevaluesof s> 0. 2.Lipschitzsolutionstoproblem(1.2)canbechosentosatisfycert aingradientestimates intermsoftheinitialgradient Du 0 .Despiteoftheseadvantagescomingfromtheconvexityassumptio nonthedomain,we plannottopursuethosehere.Instead,weareincludingalargercla ssofdomainsforthe existenceresult. Fromconditions(2.10)and(5.7),wecanndafunction h 2 C 2+ ( )satisfying h = u 0 in ;@h=@ n =0on @ :Let v 0 = Dh 2 C 1+ ( ; R n )anddene,for( x;t ) 2 T ,v ( x;t )= v 0 ( x )+ Z t 0 ~ A ( Du ( x;s )) ds: (5.9) 99 Thenitiseasilyseenthat:=( u ;v ) 2 C 1 ( T ;R 1+ n )satises(2.4);thatis, 8 > > > > > > > > < > > > > > > > > : u ( x; 0)= u 0 ( x )( x 2 ) ;div v = u a.e.in T ;v ( ;t ) n j @ =0 8 t 2 [0 ;T ]:(5.10) HenceisaboundaryfunctioninthesenseofDenition2.2.2. Next,set M =max f s 2 +1 ;k Du k L 1 ( T ) g ,r = ˙ ( M )anddene F = f ( p;A ( p )) jj p j2 [0 ;s ( r 1 )] [ [s + ( r 2 ) ;M ]g :Thenwehavethefollowing: Lemma5.3.2. ( Du ( x;t ) ;v t ( x;t )) 2S[F8 ( x;t ) 2 T :Proof. Let( x;t ) 2 T and p = Du ( x;t );then j p j M: If j p j s ( r 1 )or s + ( r 2 ) j p j M ,then ~ A ( p )= A ( p )andhenceby(5.9) ( Du ( x;t ) ;v t ( x;t ))=( p; ~ A ( p ))=( p;A ( p )) 2F :If s ( r 1 ) < j p j 0,let U begivenby U = n u 2Uj9 v 2 C 1 piece \ W 1 ;1 v ( T ;R n )suchthatdiv v = u and ( Du;v t ) 2S[F a.e.in T ,and R T j v t A ( Du ) j dxdt j T j o :Remark5.3.3. From(5.10),Lemma5.3.2,andthedenitionof U ,itfollowsthat u 2U withitscorrespondingvectorfunction v ;so U isnon-empty.Also U isaboundedsubset of W 1 ;1 u ( T )as S[F isbounded.Moreover,by(i)ofTheorem5.1.5andthedenition of F ,foreach u 2U ,itscorrespondingvectorfunction v satises k v t k L 1 ( T ) r .Thus U isindeedanadmissiblesetinthesenseofDenition2.2.3withrespectto theboundary function=( u ;v ).Finally,notethat s ( r 1 ) < j Du j 0 , U isdensein U underthe L 1 -norm. Proof. Let u 2U ,> 0.Thegoalistoconstructafunction~ u 2U suchthat k ~ u u k L 1 ( T ) <: Forclarity,wedividetheproofintoseveralsteps. 1.Note k u t k L 1 ( T ) 0andthereexistsavectorfunction v 2 C 1 piece \ W 1 ;1 v ( T ;R n )suchthatdiv v = u and( Du;v t ) 2S[F a.e.in T :Sinceboth u and v arepiecewise C 1 in T ,thereexistsasequenceofdisjointopensets f G j g 1 j =1 in T with j @G j j =0suchthat u 2 C 1 ( G j ) ;v 2 C 1 ( G j ;R n ) 8 j 1 ;j T n[ 1 j =1 G j j =0 :2.Let j 2 N bexed.Notethat( Du ( z ) ;v t ( z )) 2 S[F forall z =( x;t ) 2 G j andthat H j = f z 2 G j j ( Du ( z ) ;v t ( z )) 2 @ Sg isa(relatively)closedsetin G j withmeasurezero. So ~ G j = G j n H j isanopensubsetof G j with j ~ G j j = j G j j ,and( Du ( z ) ;v t ( z )) 2S[F for all z 2 ~ G j .3.Foreach ˝> 0,let G ˝ = f ( p; ) 2Sjj A ( p ) j >˝; dist(( p; ) ;@ S ) >˝ ) g ;then S =( [ ˝> 0 G ˝ ) [f ( p; ) 2Sj A ( p )= g as S isopen.Since A ( p )= 8 ( p; ) 2F ,wehave Z ~ G j j v t ( z ) A ( Du ( z )) j dz =lim ˝ ! 0 + Z f z 2 ~ G j j( Du ( z ) ;v t ( z )) 2G ˝ g j v t ( z ) A ( Du ( z )) j dz ;102 thuswecannda ˝ j > 0suchthat Z F j j v t ( z ) A ( Du ( z )) j dz< 3 2 j j T j and j @O j j =0 ;(5.12) where F j = f z 2 ~ G j j ( Du ( z ) ;v t ( z )) = 2G ˝ j g and O j = ~ G j n F j isopen.Let J bethesetof allindices j 2 N with O j 6= ; .Thenfor j 62 J ,F j = ~ G j .4.Wenowxa j 2 J .Notethat O j = f z 2 ~ G j j ( Du ( z ) ;v t ( z )) 2G ˝ j g andthat K j := G ˝ j isacompactsubsetof S .Let ~ Q ˆ R n beaboxwith ˆ ~ Q and ~ I =(0 ;T ). ApplyingTheorem5.2.1tobox ~ Q ~ I; K j ˆˆS = S r 1 ;r 2 ,and 0 = j T j12 ,weobtaina constant j > 0thatsatisestheconclusionofthetheorem.Bytheuniformcont inuityof A oncompactsubsetsof R n ,wecannda = ;s 2 > 0suchthat j A ( p ) A ( p 0 ) j < 12 (5.13) whenever j p j ;j p 0 j 2 s 2 and j p p 0 j : Alsobytheuniformcontinuityof u ,v ,andtheir gradientson G j ,thereexistsa j > 0suchthat j u ( z ) u ( z 0 ) j + jr u ( z ) r u ( z 0 ) j + j v ( z ) v ( z 0 ) j + jr v ( z ) r v ( z 0 ) j < min f j 2 ; 12 ;;s 2 g (5.14) whenever z;z 0 2 G j and j z z 0 j j :Wenowcover O j (uptomeasurezero)byasequence ofdisjointopencubes f Q i j I i j g 1 i =1 in O j whosesidesareparalleltotheaxeswithcenter z i j anddiameter l i j < j :5.Fixan i 2 N andwrite w =( u;v ), ˘ = 0 B @ pc B 1 C A = r w ( z i j )= 0 B @ Du ( z i j ) u t ( z i j ) Dv ( z i j ) v t ( z i j ) 1 C A :103 Bythechoiceof j > 0inStep4viaTheorem5.2.1,since Q i j I i j ˆ ~ Q ~ I and( p; ) 2K j ,forallsucientlysmall ˆ> 0,thereexistsafunction ! i j =( ' i j ; i j ) 2 C 1 c ( Q i j I i j ;R 1+ n ) satisfying (a)div i j =0in Q i j I i j ,(b)( p 0 + D' i j ( z ) ; 0 +( i j ) t ( z )) 2S forall z 2 Q i j I i j andall j ( p 0 ; 0 ) ( p; ) j j ;(c) k ! i j k L 1 ( Q i j I i j ) <ˆ; (d) R Q i j I i j j +( i j ) t ( z ) A ( p + D' i j ( z )) j dz< 0 j Q i j I i j j = j ~ Q ~ I j ;(e) R Q i j ' i j ( x;t ) dx =0forall t 2 I i j ;(f) k ( ' i j ) t k L 1 ( Q i j I i j ) <ˆ: Here,welet0 <ˆ min f ˝ 0 ; j 2 C ; 12 C ; g ,where C n istheconstantinTheorem3.3.1and C is theproductof C n andthesumofthelengthsofallsidesof ~ Q .By(e),wecanapplyTheorem 3.3.1to ' i j on Q i j I i j toobtainafunction g i j = R ' i j 2 C 1 ( Q i j I i j ;R n ) \ W 1 ;1 0 ( Q i j I i j ;R n ) suchthatdiv g i j = ' i j in Q i j I i j and k ( g i j ) t k L 1 ( Q i j I i j ) C k ( ' i j ) t k L 1 ( Q i j I i j ) j 2 :(by(f))(5.15) 6.As v t and A ( Du )areessentiallyboundedin T ,wecanselectaniteindexset Iˆ J N sothat Z S ( j;i ) 2 ( J N ) nI Q i j I i j j v t ( z ) A ( Du ( z )) j dz 3 j T j :(5.16) 104 Wenallydene (~ u; ~ v )=( u;v )+ X ( j;i ) 2I ˜ Q i j I i j ( ' i j ; i j + g i j )in T .7.Letusnallycheckthat~ u togetherwith~ v indeedgivesthedesiredresult.By construction,itisclearthat~ u 2 C 1 piece \ W 1 ;1 u ( T ),~ v 2 C 1 piece \ W 1 ;1 v ( T ;R n ) :By thechoiceof ˆ in(f)as ˆ ˝ 0 ,wehave k ~ u t k L 1 ( T )