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Maxwell's problem on point charges in the plane

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Maxwell's problem on point charges in the plane
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Potential theory
Electrostatics
Bezout's Theorem
Algebraic curves
Harmonic functions
Dissertations, Academic -- Mathematics -- Masters -- USF   ( lcsh )
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Abstract:
ABSTRACT: This paper deals with approximating an upper bound for the number of equilibrium points of a potential field produced by point charges in the plane. This is a simplified form of a problem posed by Maxwell 4, who considered spatial configurations of the point charges. Using algebraic techniques, we will give an upper bound for planar charges that is sharper than the bound given in 6 for most general configurations of charges. Then we will study an example of a configuration of charges that has exactly the number of equilibrium points that Maxwell's conjecture predicts, and we will look into the nature of the extremal points in this case. We will conclude with a solution to the twin problem for the logarithmic potential, followed by a discussion of the conditions necessary for a degenerate case in the plane.
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Thesis (M.A.)--University of South Florida, 2008.
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by Kenneth Killian.
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Title from PDF of title page.
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Document formatted into pages; contains 37 pages.

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Maxwell's problem on point charges in the plane
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ABSTRACT: This paper deals with approximating an upper bound for the number of equilibrium points of a potential field produced by point charges in the plane. This is a simplified form of a problem posed by Maxwell [4], who considered spatial configurations of the point charges. Using algebraic techniques, we will give an upper bound for planar charges that is sharper than the bound given in [6] for most general configurations of charges. Then we will study an example of a configuration of charges that has exactly the number of equilibrium points that Maxwell's conjecture predicts, and we will look into the nature of the extremal points in this case. We will conclude with a solution to the twin problem for the logarithmic potential, followed by a discussion of the conditions necessary for a degenerate case in the plane.
538
Mode of access: World Wide Web.
System requirements: World Wide Web browser and PDF reader.
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Advisor: Dmitry Khavinson, Ph.D.
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Electrostatics
Bezout's Theorem
Algebraic curves
Harmonic functions
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Maxwell'sProblemonPointChargesinthePlane by KennethKillian Athesissubmittedinpartialfulllment oftherequirementsforthedegreeof MasterofArts DepartmentofMathematics CollegeofArtsandSciences UniversityofSouthFlorida MajorProfessor:DmitryKhavinson,Ph.D. CatherineBeneteau,Ph.D. YunchengYou,Ph.D. DateofApproval: June19,2008 Keywords:PotentialTheory,Electrostatics,Bezout'sTheorem,AlgebraicCurves, HarmonicFunctions c Copyright2008,KennethKillian

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Acknowledgments Iwouldliketothankallthosewhohavehelpedandsupportedmethrough theresearchanddevelopmentofthispaper.Myrstthanksgotomywife,whohas encouragedmetocontinuewhenitseemedtherewasnolightattheendofthetunnel. Iwouldliketoextendthankstomythesiscommitteemembersforalloftheirhelp andguidance.SpecialacknowledgmentandthanksgotomyadvisorDr.Khavinson forpresentingmewiththismostinterestingproblem,helpingmethroughsomany problems,andrespondingtoe-mailsbeginningwithIt's12:30amandIjusthada greatidea."Iwouldalsoliketothankmyfamilyforalloftheirprayersandsupport. IgreatlyappreciatethesupportbytheawardUSFThrustArea:Computational toolsfordiscovery.GrantforGraduateeducationandResearchinComputerVision andPatternrecognition,2007-2009.

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TableofContents ListofTablesii ListofFiguresiii Abstractiv 1Introduction1 2TwoPointChargeCase5 3AnUpperBoundViaBezout'sTheorem6 4ALookattheThreeChargeSystem7 5OtherCasesandConjectures9 6Proofs13 References37 i

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ListofTables Table1ComparingUpperBounds.............................6 ii

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ListofFigures Figure1Potentialeldgeneratedbyapositiveandanegativecharge......1 Figure2ForceeldgeneratedbythepotentialrepresentedinFigure1......2 Figure3Potentialofthreeequalchargesinequilateraltriangle.........8 Figure4Thecountourplotoffourequalcharges.................10 Figure5Thepotentialeldfrom2pointchargeswiththesamesign.......14 Figure6Graphof3 y 5 )]TJ/F15 11.9552 Tf 12.047 0 Td [(15 p 3 y 4 +54 y 3 )]TJ/F15 11.9552 Tf 12.047 0 Td [(30 p 3 y 2 +3 y + p 3...........22 Figure7Thegraphsof f 0 and )]TJ/F28 11.9552 Tf 9.298 0 Td [(g 0 ............................28 iii

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Maxwell'sProblemonPointChargesinthePlane KennethKillian ABSTRACT Thispaperdealswithapproximatinganupperboundforthenumberofequilibriumpointsofapotentialeldproducedbypointchargesintheplane.Thisisa simpliedformofaproblemposedbyMaxwell[4],whoconsideredspatialcongurationsofthepointcharges.Usingalgebraictechniques,wewillgiveanupperbound forplanarchargesthatissharperthantheboundgivenin[6]formostgeneralcongurationsofcharges.Thenwewillstudyanexampleofacongurationofchargesthat hasexactlythenumberofequilibriumpointsthatMaxwell'sconjecturepredicts,and wewilllookintothenatureoftheextremalpointsinthiscase.Wewillconcludewith asolutiontothetwinproblemforthelogarithmicpotential,followedbyadiscussion oftheconditionsnecessaryforadegeneratecaseintheplane. iv

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1. Introduction Inhisbook ATreatiseonElectricityandMagnetism [4],J.C.Maxwellsuggested thatgivenanynon-degeneratespatialcongurationofnpointcharges,themaximum numberofplacesthattheelectrostaticforcecanvanishis n )]TJ/F15 11.9552 Tf 12.859 0 Td [(1 2 .However,he oerednoproofofthisconjecture.Verylittleprogresshasbeenmadetoverifythis. In2007,Shapiro,Gabrielov,andNovikov[6]wereabletoestablishanupperbound of4 n 2 n 2 n foranynon-degeneratespatialcongurationofcharges.Forthecaseof threecharges,thisboundcanbesharpenedto12. Inordertoinvestigatethisconjecture,weneedanunderstandingofelectrostatic potentialandelectrostaticforce. Denition. Given n pointchargesinspace,eachofwhichhascharge q k andislocated atsomepoint p k = x k ;y k ;z k ,thepotential P generatedbythesepointchargesis .1 P = c n X k =1 q k j p )]TJ/F28 11.9552 Tf 11.956 0 Td [(p k j where c isCoulomb'sconstant [7] ,p.750. Figure1. Potentialeldgeneratedbyapositiveandanegative charge.Thepositivechargeisat-1,0,andthenegativechargeis at,0. Forourpurposes,wewilllettheconstant c equal1sinceitisnotrelevanttoour study. 1

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Denition. Theelectrostaticforce F generatedbythepotential P isgivenby F = r P = F x ~ i + F y ~ j + F z ~ k where .2 F x = )]TJ/F28 11.9552 Tf 10.494 8.087 Td [(@P @x = n X k =1 q k x )]TJ/F28 11.9552 Tf 11.956 0 Td [(x k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] 3 = 2 .3 F y = )]TJ/F28 11.9552 Tf 10.494 8.088 Td [(@P @y = n X k =1 q k y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k [ x )]TJ/F28 11.9552 Tf 11.956 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.956 0 Td [(z k 2 ] 3 = 2 and .4 F z = )]TJ/F28 11.9552 Tf 10.494 8.088 Td [(@P @z = n X k =1 q k z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 [ x )]TJ/F28 11.9552 Tf 11.956 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.956 0 Td [(z k 2 ] 3 = 2 Figure2. Thisistheforceeldgeneratedbythepotentialrepresented inFigure1 Itisimportanttonotethattheequations F x =0and F y =0arealgebraiccurves i.e.botharesumsofrationalfunctionswhosenumeratorsanddenominatorscanbe expressedaspolynomials.Thisimportantfactwillbeusedinestablishinganupper boundonthenumberofequilibriumpointsinSection3. Wewillonlyconsidercongurationsofpointchargesinthe xy -plane.Anyplace atwhichwehave F =0musthave F x =0and F y =0obviously, F z =0since allofthepointchargesareintheplane z =0.Ourgoalistondthemaximum numberofplacesatwhichtheforcewillvanishintheplanegiven n pointcharges. ThisisasimpliedformofMaxwell'sproblemwhichmayprovidesomeinsightonthe moregeneralnon-degeneratecases.Notethatanon-degeneratecongurationrefers 2

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toasituationinwhichtherearenitelymanyequilibriumpointspointswherethe electrostaticforcevanishes. Beforecontinuingontothe2-chargecase,rstrecallsomewell-knownresultsfrom classicalpotentialtheory. Denition. Asmoothfunction f x;y;z issaidtobeharmonicif f =0 ;thatis: @ 2 f @x 2 + @ 2 f @y 2 + @ 2 f @z 2 =0 [1] ,p.73. Proposition1.1. In R 3 ,thepotential P isaharmonicfunctionawayfromthe charges. Denition. Asmoothfunctionfx,yissaidtobesubharmonicif f 0 ;similarly, wesaythatfissuperharmonicif )]TJ/F28 11.9552 Tf 9.298 0 Td [(f issubharmonic. [5] ,pp.28,36 Proposition1.2. In R 2 ,thepotential P issubharmonicawayfromthechargesifall ofthechargesarepositive,andissuperharmonicifallofthechargesarenegative. In R 3 ,wehavethat P isalwaysharmonicawayfromthecharges,regardlessofthe signsofthecharges.Unfortunately,in R 2 ,nothingofthesortcanbesaidabout P ifthereisacombinationofpositiveandnegativecharges.Thiswillprovetobeone ofthegreatesthindrancestoshowingthatanon-degeneratecasecannotexistin R 2 Thefollowingtheoremisoneofthebasicresultsaboutharmonicfunctions.See[5], p.29,or[1],p.248foraproof. Theorem1.3.TheMaximumPrinciple :Iffisareal-valuedharmonicfunction inadomainD,thenfhasnolocallocalmaximumorlocalminimuminD,unlessf isaconstanteverywhere. Themaximumprincipleshowsthatzerosoftheforceareactuallysaddlepointsi.e. criticalpointsthatareneithermaximumnorminimumpointsin R 3 .Unfortunately, whenwerestrictourselvestotheplane,welosealloftheharmonicpropertiesofthe potential.InthespecialcasesmentionedinProposition1.2wehaveatbestthatthe potentialissubharmonicorsuperharmonic.Themaximumprincipledoesapplyto subharmonicfunctionsinthattheycannotattainalocalmaximum.However,there isnorestrictiononasubharmonicfunctionhavingalocalminimuminsideofhte domain.Similarly,superharmonicfunctionscannotattainaminimumbutcanhave alocalmaximum. 3

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Thepropertiesofharmonicfunctionsandalgebraiccurveswillbeinvaluabletools asweproceeswithourinvestigationofthevalidityofMaxwell'sconjecture.Wewill beginwiththesimplestcase:thetwo-chargecongurations.Thenwewillestablish anupperboundonthenumberofequilibriumpoints.Thiswillbefollowedwitha discussionofthree-chargecongurations,aswellasotherversionsofthisproblem andsomeconjectures.Section5alsocontainsadiscussionabouttheexistenceof degenratecongurationsofpointchargesintheplane. 4

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2. TwoPointChargeCase Maxwell'sconjecturepredictsthatforacongurationoftwopointcharges,there isatmostoneplaceatwhichtheforcewillvanish.Thefollowingtheoremshowsthis istrue. Theorem2.1. Giventwopointchargesintheplane,thereisatmostoneplaceat whichtheforcewillvanish. Remark :ItisinterestingtonotethattheproofgiveninSection6showsthat everycongurationoftwopointchargeswillvanishatexactlyonepoint,exceptwhen thetwochargesareofequalmagnitudebutoppositesign. Itwillbeusefultoobservewhattypeofextremathispointis.Inordertodothis, weneedtheSecondPartialsTest[3],p.889. Theorem2.2. Let f havecontinuoussecondpartialderivativesonanopenregion containingapoint a;b forwhich f x a;b =0 and f y a;b =0 .Totestforrelative extremaof f considerthequantity d = f xx a;b f yy a;y )]TJ/F15 11.9552 Tf 11.955 0 Td [([ f xy a;b ] 2 : 1.If d> 0 and f xx a;b > 0 ,then f hasarelativeminimumat a;b 2.If d> 0 and f xx a;b < 0 ,then f hasarelativemaximumat a;b 3.If d< 0 ,then a;b;f a;b isasaddlepoint. 4.Thetestisinconclusiveif d =0 : ApplyingtheSecondPartialsTesttothepotentialfunctionoftwopointcharges givesthefollowingresult. Corollary2.3. Anypointfromatwopoint-chargecongurationwheretheforce vanishesisasaddlepointforthepotential. 5

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3. AnUpperBoundViaBezout'sTheorem Wenowwishtondaneectiveupperboundforpointchargesintheplane. Onemightwishtousethepreviousresultasabasisstepforaninductiveproofto showMaxwell'sconjectureholds,butanexaminationofthestructureof F x =0and F y =0willbemoreuseful.RecallthatinSection1wenotedthattheseequations arealgebraiccurves.Theorderofanalgebraiccurveisthehighestdegreeofthe polynomial.Beforeestablishingthisupperbound,however,weneedthefollowing well-knownresultfromalgebrawhoseproofcanbefoundin[8],p.54. Theorem3.1. Bezout'sTheorem Twoplanealgebraiccurvesoforders m and n withnocommoncomponentshaveatmost mn intersections. Thealgebraiccurvesweareconsideringaredenedby F x =0and F y =0.Therefore,toapplyBezout'sTheorem,wemustassumethatwehaveanon-degenerate congurationofpointcharges.Thatis,if F x =0and F y =0shareacommonfactor, thenthereexistsacurvethroughtheplanealongwhicheverypointisaplacewhere theforcevanishes.InSection5wewillconsiderthisprobleminmoredepth. WithBezout'sTheorem,wearenowreadytogiveanupperboundonthenumber ofpointsthattheforcevanishes. Theorem3.2. Given n pointchargesintheplane,themaximumnumberofequilibriumpointsintheplaneis [2 n )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 [3 n )]TJ/F15 11.9552 Tf 11.955 0 Td [(2]] 2 ,assuminganon-degeneratecase. Table1. ComparingUpperBounds nTheorem3.2UpperBoundfrom[6] 264331,776 37841 : 393 10 11 46,4001 : 847 10 18 543,2646 : 493 10 26 6262,1445 : 463 10 36 Remark :Table1showsthattheupperboundgivenbyTheorem3.2providesa muchsmallerboundonthenumberofplaceswheretheforcevanishes.Thedrawback isthattheupperboundgivenin[6]considersspatialcongurationsofpointcharges, andtheupperboundfromTheorem3.2isrestrictedtocongurationsintheplane. 6

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4. ALookattheThreeChargeSystem Maxwell'sconjectureappliedtothethreechargesystempredictsthatthereare atmostfourpointswheretheforcewillvanish.Theorem1.5bof[6]givesanupper boundof12points.Thoughnocomputersimulationshavebeenabletoproducea congurationthatyields12pointsoreven5,itispossibletohave4pointswhere theforcevanishes.Wendthatoneofthesimplestcongurationsof3pointsgives thisresult. Proposition4.1. Giventhreeequalpointchargesontheplanearrangedinanequilateraltriangle,thereareexactlyfourplacesatwhichtheelectrostaticforcevanishes. Remark :Notethatthisproofshowsthatoneofthezerosoftheforceoccursat thecenterofmass"replacethepoint-chargeswithmasses,andthegravitational potentialandforceremainthesameofthissystem.Thismayleadonetoconclude thatacenterofmassofthesystemisalwaysoneofthelocationsofzeroforce. However,anobviouscounter-exampletothisistheearth-sunsystem.Thecenterof massofthesystemisinsideofthesun,butthepointwherethegravitationalpulls canceleachotherisclosetoearth. Theexistenceofacongurationthatproducesfourpointsofzeroforcebegsthe question:doesthereexistacongurationinwhichwehavemorethanfourpoints wheretheforcevanishes?Anexaminationofthepotentialofthisthreechargecongurationwillgivesomeinsightintothisproblem,aswellasshedsomelightonthe existenceofacurveintheplaneonwhichtheforcevanishes. Figure3showsthepotentialeldgeneratedbythreepositive,equalpointcharges placedinanequilateraltriangle.Weseethatthefourzerosoftheforceoccuratthe oneminimumandthethreesaddlepoints.Obviously,inordertohaveadditional zeros,eitheradditionalsaddlepointsorminimumpointsmustbegenerated.The followingtheoremgivessomeinsightintothebehaviorofthepotential. Theorem4.2. Givenacongurationofthreepositivecharges,if ~v isanyvector originatingontheboundaryofthetriangle,perpendiculartotheboundary,andis directedintotheinteriorofthetriangle,thenthepotentialtakesatmostoneminimum 7

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Figure3. Potentialofthreeequalchargesinequilateraltriangle.The chargesareplacedat )]TJ/F15 11.9552 Tf 9.299 0 Td [(1 ; 0, ; 0,and ; p 3 3 .Thereisoneminand threesaddlepoints. on ~v if ~v isanaltitudeofthetriangleandatmosttwominimaforallother ~v .The sameholdsforthreenegativechargesusingmaximumpoints. Oneinterestingfalloutofthisproofisthattheinwardnormalderivativeofthe potentialisalwayspositive.ThoughTheorem4.2doesnotprohibittheexistenceof multipleextremalpointsintheinteriorofthetriangle,itdoesshowthattheycannot lieonthesame ~v 8

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5. OtherCasesandConjectures Thusfar,wehaveonlyconsideredtheso-calledNewtoniankernelofthepotential, whichdenesthepotentialbyaninversedistancerule.Wecouldchangethedenition ofthepotentialandcreatenewkernelshavingthesamepropertiesastheNewtonian kernel,butwithsomesurprisingresultsintheforceequations.Onesuchexample islogarithmicpotential[2],pp.62-63,whichincomplexnotationisdenedby P = P n i =1 kq i log j 1 z )]TJ/F29 7.9701 Tf 6.587 0 Td [(z i j .Logarithmicpotentialisextremelyusefulwhenstudyinginnite cylindersorwiresthatcarrycharge.See[2],pp.62-63,formoredetails. Maxwell'sconjectureinthecontextoflogarithmicpotentialgivesthefollowing result: Theorem5.1. Theforce F producedby P = P n i =1 kq i log j 1 z )]TJ/F29 7.9701 Tf 6.586 0 Td [(z i j hasprecisely n )]TJ/F15 11.9552 Tf 12.257 0 Td [(1 equilibriumpoints,where F = r P: Anothercasewewishtoconsideristheexistenceofnon-degeneratecongurationof pointchargesintheplane.Thatis,doesthereexistacongurationsofpointcharges in R 2 thatproducesacurvein R 2 onwhichtheforcevanishesateverypoint?Itis possibleforchargesin R 2 toproducesuchacurvein R 3 .Forexample,considerthe potentialproducedbyplacinganevennumberofequalinmagnitudebutalternating chargesinaregular n -goncenteredattheorigin.Theneverypointofthe z -axisnot onlyhaszeropotentialbutalsozeroforce. Clearly,iftheforceistovanishonacurve,thenthepotentialmustbeaconstant onthatcurve.In R 3 ,weknowthatthepotentialisharmonicawayfromthecharges. Noticethatinboth R 2 and R 3 ,thelimitofthepotentialas x y ,or z goestoinnity iszero.Iftheforceistovanishonacurvethatgoestoinnity,thenthepotential mustbezeroonsuchacurve.However,whenwerestrictourselvesto R 2 ,wenolonger havethepotentialbeingaharmonicfunction,andhavebyProposition1.2thatthe potentialissubharmonicorsuperharmoniconlyifallofthechargeshavethesame sign.TheproofofTheorem5.2willhandlethepossibilityofacurvethroughthe planewhenallchargeshavethesamesign.Theonlyotherpossibilityofadegenerate congurationistheexistenceofaclosedlooponwhichtheforcevanishes. 9

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Thefollowingtheoremaddressestheexistenceofadegeneratecongurationofthree chargesofthesamesign. Theorem5.2. Giventhreechargesintheplanehavingthesamesign,thereisno equilibriumcurveintheplane. Theproofofthistheoremcanbeappliedtoanynumberofchargesofthesamesign; thatis,thethreepossiblepositionsofthesupposedcurvecannotexistregardlessofthe numberofcahrges.However,incongurationsoffourormorecharges,anadditional casearisesthatisnothandledbythisproof.Consideracongurationoffourcharges inwhichoneofthechargescallit q 4 inlocatedintheinteriorofthetriangleformed bytheotherthree.Onecouldthenarguethattherecouldexistanequipotentialcurve around q 4 onwhichtheforcevanishes.Bythemaximumprinciple,itisclearthat everypointofthecurvewouldhavetobeaminimumassumingallofthecharges arepositive. Figure4. Thecountourplotoffourequalcharges.Ifaminimum curveexisted,itwouldhavetobeononeofthelevelcurves.Inthis particularcase,thecomputersimulationscanonlyndthreeequilibriumpoints. Thoughitisstillanopenproblemwhethersuchacurvecanexistifallofthe chargeshavethesamesign,thefollowingtheoremgivesanecessaryconditionfor everypointonthecurve,ifitexists. Theorem5.3. Dene X = P n k =1 q k x )]TJ/F29 7.9701 Tf 6.587 0 Td [(x k 2 [ x )]TJ/F29 7.9701 Tf 6.587 0 Td [(x k 2 + y )]TJ/F29 7.9701 Tf 6.586 0 Td [(y k 2 ] 5 = 2 and Y = P n k =1 q k y )]TJ/F29 7.9701 Tf 6.586 0 Td [(y k 2 [ x )]TJ/F29 7.9701 Tf 6.586 0 Td [(x k 2 + y )]TJ/F29 7.9701 Tf 6.586 0 Td [(y k 2 ] 5 = 2 Givennitelymanysamesignchargesintheplane,thenecessaryconditionsforforce vanishonacurve S are 10

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1. P xx P yy = P xy 2 ateverypointof S ; 2.eitherboth P xx > 0 and P yy > 0 ,andtheinequality 1 2 X Y 2 X musthold; oratmostoneof P xx or P yy canequalzero. Incomputersimulations,theforcewillonlyvanishonaloopifthereisacharged wirethatformsaclosedloop,withanadditionalchargeintheinterioroftheloop. Thisleadsustothefollowingconjecture: Conjecture5.1. Givennitelymanypointchargesintheplane,theelectrostatic forcewillnevervanishonacurve. Thisconjecturealsoappliestocongurationsofmixedcharges.Forexample,considerapositiveandanegativechargeofequalmagnitureplacedat-1,0and,0, respectively.Weknowthatoneofthebasiccriteriafortheforcetovanishonacurve isthatthepotentialmustbeconstantonthatcurve.Inthiscase,theentirey-axis haszeropotential.However,the x -componentoftheforceisalwayspositiveonthe y-axis. Whenwelookedattheequilateraltriangle,wesawthattherearefourequilibrium points:threesaddlepointsandoneminimum.Intuitively,thisseemstobetheonly waytohavefourequilibriumpoints.Forinstance,consideracongurationofthree positivechargesinwhichweknowthataminimumpointexistssomewhereinthe triangle,butnotontheboundary.Let ~v beanyvectororiginatingatthatminimum point,andconsiderthebehaviorofthepotentialalong ~v .Since ~v originatesat aminimumpoint,thepotentialmustinitiallyincrease.However,weknowthat farenoughawayfromthechargesthepotentialwillbedecreasing.Thisimplies thatthepotentialtakesamaximumon ~v .However,weknowthatthepotentialis subharmonic,andsubharmonicfunctionscannothavemaximumpointsexceptonthe boundaryofthedomain.Sothismaximumweseeon ~v mustbeasaddlepoint,or issomepointonthe`ridge'joiningtwoofthecharges.Thisleadstothefollowing conjecture: Conjecture5.2. Giventhreepositivecharges,aminimumexistsintheinteriorof thetriangleifandonlyiftherearethreesaddlepoints. ThisconjecturewouldseemtoverifyMaxwell'sconjectureforthecaseofthree charges.Theonlymissingcomponentisthefollowingconjecture: 11

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Conjecture5.3. Foranycongurationofthreepositivecharges,thepotentialcan takeatmostoneminimumintheinteriorofthetriangle. Iftheminimumofthepotentialoccursontheboundaryofthetriangle,theforcewill notvanishattheminimumpoint.Thisfollowsfromthefactthattheinwardnormal derivativeofthepotentialisalwayspositive.Whenwelookedattheequilateral triangle,wesawthattherewerethreesaddlepoints,eachlocatednear"oneofthe legsofthetrianglei.e.thesaddlepointsdonotoccuronthelegsofthetriengle,but onecanidentifyalegthateachsaddlepointisgenerated"by,inthecontextofthe proofofTheorem5.2.Incomputersimulations,weseethatifallofthechargeshave thesamesignandareequalinmagnitude,thenthepotentialofallrightandobtuse congurationsofthechargeswillhaveaminimumonthelongestlegofthetriangle. However,thetwosaddlepointsnear"theshorterlegsremain. Conjecture5.4. Giventhreeequalchargesarrangedinarightorobtusetriangle, theforcewillonlyvanishattwosaddlepointsneartheshorterlegsofthetriangle. Foracongurationofthreemixedchargesitwouldseemthattheboundfor Maxwell'sconjectureisnotevenreached.Forinstance,giventwopositivecharges andonenegative,thepotentialisstrictlydecreasingonthelinesoriginatingatthe positivechargesanddirectedtowardsthenegativecharge.Thetwosaddlepoints thatmighthaveexistedneartheselinesinthecaseofallpositivechargescannot occur.However,thesaddlepointbetweenthetwopositivechargesmaystillexist, butwouldbelocatedoutsideofthetriangle. Computersimulationsonanequilibriumtrianglewithtwo+1chargesat-1,0and ,0andanegativechargeonthe y -axisat ; p 3showtwosaddlepointsinthe plane,bothonthe y -axis.Oneisatapproximately y = )]TJ/F28 11.9552 Tf 9.299 0 Td [(: 14629,andtheotherat y =6 : 20448.Onemightconjecturetheexistenceoftwosaddlepointsnotonthe plane.However,computersimulationsfailtondanyequilibriumpointsnotonthe plane.Thisleadstothefollowingconjecture: Conjecture5.5. Giventhreepointchargesintheplane,thenumberofequilibrium pointsforchargesofthesamestrengthattheverticesoftheequilibriumtriangleis maximizedwhenallchargeshavethesamesign. 12

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6. Proofs ProofofProposition1.1 :Weneedtoshowthat @ 2 P @x 2 + @ 2 P @y 2 + @ 2 P @z 2 =0.By denition,weknowthat F x = )]TJ/F29 7.9701 Tf 10.494 4.707 Td [(@P @x .Dierentiating )]TJ/F28 11.9552 Tf 9.298 0 Td [(F x withrespecttoxyields @ 2 P @x 2 @ 2 P @x 2 = @ @x )]TJ/F29 7.9701 Tf 17.356 14.944 Td [(n X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] 3 = 2 = )]TJ/F29 7.9701 Tf 17.357 14.944 Td [(n X k =1 q k [ )]TJ/F15 11.9552 Tf 9.298 0 Td [(2 x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.956 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] 5 = 2 Similarly,weseethat @ 2 P @y 2 = )]TJ/F29 7.9701 Tf 17.357 14.944 Td [(n X k =1 q k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] 5 = 2 and @ 2 P @z 2 = )]TJ/F29 7.9701 Tf 17.356 14.944 Td [(n X k =1 q k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 + z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k 2 ] 5 = 2 : Addingthesetogethergives @ 2 P @x 2 + @ 2 P @y 2 + @ 2 P @z 2 =0 : ProofofProposition1.2 :Weneedtoshowthat P 0,assumingall q k > 0. Usingthecalculationsofthepreviousproof,wehavethatin R 2 @ 2 P @x 2 = n X k =1 q k [2 x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [( y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 and @ 2 P @y 2 = n X k =1 q k [ )]TJ/F15 11.9552 Tf 9.299 0 Td [( x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 +2 y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 : Addingthese,weget @ 2 P @x 2 + @ 2 P @y 2 = n X k =1 q k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 3 = 2 > 0 : Similarly,weseethatifall q k < 0,then )]TJ/F28 11.9552 Tf 9.298 0 Td [(P willbesubharmonic,andthereforeP issuperharmonic. 13

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ProofofTheorem2.1 :Therearetwopossiblecasestoconsider.Intherst case,wehavebothpointchargeswiththesamesignseeFigure4,andinthesecond casewehaveonepositiveandonenegativechargeseeFigure1. Figure5. Thepotentialeldfrom2pointchargeswiththesamesign. Case1:Let q 1 and q 2 bepositiveandlocatedatpoints x 1 ;y 1 and x 2 ;y 2 ,respectively.Throughatranslationandrotation,thesepointscanbemovedsothatone pointisat ; 0,andtheotherisonthe x -axis.Withadilation,thesecondpoint canbemade ; 0withoutaectingthenumberofzerosoftheforce. Noticethatatanypoint p = x;y ontheplane,the y -componentoftheforceis givenby .1 F y = F 1 sin + F 2 sin where F 1 isthemagnitudeoftheforcegeneratedbythecharge q 1 locatedatthe origin, F 2 isthemagnitudeoftheforcegeneratedbythesecondcharge, istheangle between F 1 andthex-axis,and istheanglebetween F 2 andthex-axis.Noticethat F 1 and F 2 havethesamesignfromthedenitionofforce.2.Ifthepoint p isnot onthex-axis,then 6 =0and 6 =0,so F y 6 =0.Therefore,anypointsatwhichthe forcevanishesmustbeonthex-axis. Foranypoint p = x; 0,the x -componentoftheforceisgivenby .2 F x = F 1 cos + F 2 cos ': If p 2 ; 0 S ; 1 ,thenboth and willbe0or ,and F x cannotvanish.So weneedtoconsider p 2 ; 1. Noticethatif p 2 ; 1isapointwheretheforcevanishes,wehave =0and = ,so: F x = F 1 )]TJ/F28 11.9552 Tf 11.998 0 Td [(F 2 =0 F 1 = F 2 .Sincetheforceisinverselyproportionalto 14

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thesquareofthedistance,wegetthefollowingresult: q 1 x 2 = q 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(x 2 q 1 x 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 q 1 x + q 1 = q 2 x 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 x 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 1 x + q 1 =0 x = q 1 p q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 : Sinceboth q 1 and q 2 arepositive,thediscriminantisnon-zero,andwehavetwo distinctrootstothisequation. If q 1 >q 2 ,then q 1 + p q 1 q 2 >q 1 + p q 2 q 2 = q 1 + q 2 >q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 so q 1 + p q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 > 1 Since x 2 ; 1,thisrootcanbeignored.Considertheremainingroot: q 1 )]TJ 11.955 7.448 Td [(p q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 = p q 1 p q 1 )]TJ 11.955 7.448 Td [(p q 2 p q 1 + p q 2 p q 1 )]TJ 11.956 7.449 Td [(p q 2 = p q 1 p q 1 + p q 2 < 1 Therefore, x = q 1 )]TJ 11.955 7.448 Td [(p q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 istheonlysolutionin ; 1if q 1 >q 2 Similarly,if q 2 >q 1 ,then q 1 + p q 1 q 2 > 0,and q 1 )]TJ/F28 11.9552 Tf 11.836 0 Td [(q 2 < 0,sowehave q 1 + p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.587 0 Td [(q 2 < 0 andthisrootisnotin,1.Also, q 1 )]TJ 11.955 7.449 Td [(p q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.956 0 Td [(q 2 < q 1 )]TJ 11.955 7.449 Td [(p q 2 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 = q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 =1 : Therefore q 1 )]TJ 6.587 4.971 Td [(p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(q 2 istheonlysolutionin ; 1if q 2 >q 1 Also,notethatif q 1 = q 2 ,then F 1 = F 2 hasonlyonesolutionat x = 1 2 : Ifboth q 1 and q 2 arenegative,thenthenewpotentialeldwillbeidenticaltothe rst,justnegated.Theresultisthatthesignsoftheforceequationarereversed,but thesameresultsareobtained.Thus,Case1hasbeenproven. Case2:Let q 1 bepositiveand q 2 benegativeseeFigure1foragraphofthe potential.Again,througharigidmotionanddilation,thesechargescanbeplaced attheoriginand ; 0withoutaectingthenumberofzeros.Assumethat q 1 is 15

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positiveandthat q 2 isnegative.Thenwehaveagainthat F x = F 1 cos )]TJ/F28 11.9552 Tf 11.955 0 Td [(F 2 cos F y = F 1 sin )]TJ/F28 11.9552 Tf 11.955 0 Td [(F 2 sin Thereasonwewritethe F 2 termswithanegativesignisbecause q 2 isanegative charge.Bywritingtheforcethisway,wecanconsider q 2 positive.Take p = x;y where x 2 ; 1,andconsidertheforcesactingonthispoint.Since 2 )]TJ/F29 7.9701 Tf 10.494 4.707 Td [( 2 ; 2 and 2 2 ; 3 2 ,thenthehorizontalforcewillnotvanish.Ifthereexistsapoint p = x;y wheretheforcevanishes,then x 2 ; 0 S ; 1 .Considerwhathappensif y> 0 thesameargumentholdsfor y< 0and x 2 ; 1 .Obviouslywehaveboth ;'> 0.Noticethatinmagnitude,wehave F 1 = q 1 x 2 + y 2 and F 2 = q 2 x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 + y 2 Since x> 1,then x 2 + y 2 > x )]TJ/F15 11.9552 Tf 12.508 0 Td [(1 2 + y 2 .Ifthispoint p isapointatwhichthe electrostaticforcevanishes,thenwemusthavethat j q 1 j > j q 2 j andtherefore F 1 >F 2 Noticethat '> ,sosin '> sin .Combiningtheseinequalitiesgives F y = F 1 sin )]TJ/F28 11.9552 Tf 11.956 0 Td [(F 2 sin '>F 1 sin )]TJ/F28 11.9552 Tf 11.955 0 Td [(F 2 sin >F 2 sin )]TJ/F28 11.9552 Tf 11.955 0 Td [(F 2 sin =0 Thereforetheverticalforcewillnotvanish,hence y =0istheonlyway F y =0for x 2 ; 0 S ; 1 Take p = x; 0 ;x> 1.Thenboth ;' =0,so F x = F 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(F 2 =0givesus F 1 = F 2 q 1 x 2 = q 2 x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 : BytheproofofCase1,therewillbeonesolutionin0 ; 1andonesolutionin ; 1 Thesameargumentholdsfor x 2 ; 0.Noticethatif j q 1 j = j q 2 j ,thentheonly solutionto F 1 = F 2 is x = 1 2 ,andthereforethereisnosolution.Thiscompletesthe proof. ProofofCorollary2.3 :InordertoapplytheSecondPartialsTesttothepotential,weneed P xx P yy ,and P xy .Adirectcalculationshowsthatfor n pointcharges 16

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wehave P xx = n X k =1 q k [2 x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 )]TJ/F15 11.9552 Tf 11.956 0 Td [( y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 P yy = n X k =1 q k [ )]TJ/F15 11.9552 Tf 9.298 0 Td [( x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 +2 y )]TJ/F28 11.9552 Tf 11.956 0 Td [(y k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 and P xy = n X k =1 3 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 3 = 2 : UsingthesimplicationsfromtheproofofTheorem2.1,let x 1 ;y 1 = ; 0and x 2 ;y 2 = ; 0,sotheonlypoint p wheretheforcecanvanishmustbeoftheform p = x; 0.Thentheaboveequationssimplifyto P xx =2 2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 P yy = )]TJ/F26 7.9701 Tf 17.808 14.944 Td [(2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 P xy =0 : NowapplytheSecondPartialsTestbycalculatingthequantity d : d = P xx P yy )]TJ/F15 11.9552 Tf 11.955 0 Td [([ P xy ] 2 =[2 2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 ][ )]TJ/F26 7.9701 Tf 17.809 14.944 Td [(2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 ] )]TJ/F15 11.9552 Tf 11.955 0 Td [(0 = )]TJ/F15 11.9552 Tf 9.299 0 Td [(2[ 2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 ] 2 0 : Allthatremainsistocheckthat P 2 k =1 q k x )]TJ/F29 7.9701 Tf 6.586 0 Td [(x k 3 6 =0when x = q 1 p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.587 0 Td [(q 2 ,whichwasthe onlypointwheretheforcevanished.Recallthat x = q 1 )]TJ 6.587 4.971 Td [(p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(q 2 whenbothchargeshad thesamesignand x = q 1 + p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(q 2 whenthechargeshavedierentsigns.First,assume 17

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that q 1 6 = q 2 .Thensubstitutinggives 2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 = q 1 [ q 1 )]TJ 6.587 4.971 Td [(p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(q 2 ] 3 + q 2 [ q 1 )]TJ 6.587 4.971 Td [(p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.587 0 Td [(q 2 )]TJ/F15 11.9552 Tf 11.956 0 Td [(1] 3 = q 1 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 3 q 1 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 + q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 3 q 2 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 = q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 3 [ q 1 q 2 )]TJ 11.955 7.448 Td [(p q 1 q 2 3 + q 2 q 1 )]TJ 11.956 7.448 Td [(p q 1 q 2 3 q 1 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 q 2 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 ] = q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 3 [ q 1 q 3 2 )]TJ/F15 11.9552 Tf 11.956 0 Td [(4 q 1 q 2 2 p q 1 q 2 +6 q 2 1 q 2 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(4 q 2 1 q 2 p q 1 q 2 + q 3 1 q 2 q 1 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 q 2 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 ] = q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 3 p q 1 )]TJ 11.956 7.449 Td [(p q 2 4 q 1 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 q 2 )]TJ 11.955 7.449 Td [(p q 1 q 2 3 : Sinceweassumedthat q 1 6 = q 2 ,thenthisquantitycanneverequalzero,andtherefore x = q 1 )]TJ 6.587 4.971 Td [(p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(q 2 isasaddlepointifbothchargeshavethesamesign.Asimilar computationshowsthatwhen x = q 1 + p q 1 q 2 q 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(q 2 ,then 2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 3 = q 1 q 2 q 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(q 2 3 p q 1 + p q 2 4 q 1 + p q 1 q 2 3 q 2 + p q 1 q 2 3 : Again,thisonlyequalszeroif q 1 = q 2 .Therefore,when q 1 6 = q 2 ,thepointwherethe forcevanishesisasaddlepoint. Considerwhathappenswhen q 1 = q 2 .Obviously,theSecondPartialsTestcannot beappliedbecause P 2 k =1 q k x )]TJ/F29 7.9701 Tf 6.586 0 Td [(x k 3 = 0 0 .ByTheorem2.1,weknowthatthispoint p = x 0 ; 0wheretheforcevanishesmustbeonthelineconnectingthetwopoint charges.Sincebothchargesarepositive,then p cannotbeamaximumthepotential issubharmonic,andcanthereforeonlybeasaddlepointoraminimum.Assume that p isamin.Thenforall x;y inaballaround p ,wemusthave P x;y >P p Fix x 0 tobethe x -coordinatewheretheforcevanishes,andlet y> 0.Then P x 0 ;y = 2 X k =1 q k p x 0 )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 < 2 X k =1 q k x 0 )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k = P x 0 ; 0 = P p : 18

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Therefore, p cannotbeaminandmustbeasaddlepoint.Thesameargumentholds whenbothchargesarenegativebysimplyreversingtheinequalityandassumingthat p isamax. Therefore,thepointwheretheforcevanishesinthetwo-chargecaseisasaddle point. ProofofTheorem3.2 :Beforeprovingthistheorem,weneedthefollowing LemmaanditsCorollary. Lemma6.1. Anequationoftheform p p 1 p p 2 ::: p p n = k;k 6 =0 ,and p n > 0 forall n canberationalizedinthesensethatitcontainsnoradicaltermsbysquaring n times. Proof :Withoutlossofgenerality,considerthecase p p 1 + p p 2 + ::: + p p n = k .By induction,itistrivialtoseethisfor n =1.However,checkingthisresultfor n =2 willgivesomeinsightintothemethodofprovingthisforageneralnumberofterms. Wehave p p 1 + p p 2 = k .Squaringoncegives p 1 + p 2 +2 p p 1 p 2 = k 2 .Notethat theremainingradicalcanbeisolatedbymovingittotherightsideoftheequation andsubtracting k 2 tohave p 1 + p 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(k 2 = )]TJ/F15 11.9552 Tf 9.299 0 Td [(2 p p 1 p 2 : Squaringasecondtimegives p 2 1 + p 2 2 + k 4 +2 p 1 p 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(p 1 k 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(p 2 k 2 =4 p 1 p 2 : Observethatthiscanbere-writtenas p 2 1 + p 2 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 p 1 p 2 =2 p 1 k 2 +2 p 2 k 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(k 4 : InductiveStep:assumethat p p 1 + p p 2 + ::: + p p n = k canberationalizedby squaring n times. Needtoshowthat p p 1 + ::: + p p n + p p n +1 = k canberationalizedbysquaring n +1times.Subtract p p n +1 frombothsides,andlet j = k )]TJ 11.955 7.299 Td [(p p n +1 ,giving p p 1 + p p 2 + ::: + p p n = j: Assumethat j 6 =0sotheinductivestepstillappliesthecasewhere j =0ishandled intheCorollary.Bytheinductivestep,thiscanberationalizedbysquaring n times.Astheaboveexamplefor n =2shows,theequationresultingfromthese n 19

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`squarings'canbewrittensothatalltermswith j canbeisolatedtotheright-hand sideoftheequation.Since j isabinomial,anytermscontaining j raisedtosome powercanbeexpanded.Subtractfromtheright-handsidealltermsnotcontaining p p n +1 ,andthenfactor p p n +1 fromtheremainingterms.Squaringbothsidesonce moreeliminatesallradicals. Corollary6.2. Anequationoftheform p p 1 p p 2 ::: p p n =0 p i > 0 forall 1 i n ,canberationalizedbysquaring n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 times. Proof :Subtract p p n frombothsidesandlet k = )]TJ 9.299 7.449 Td [(p p n .ThenapplytheLemma. ProofofTheorem3.2 :Wehaveseenthat F x and F y areidenticalexceptforterms inthenumerator,whichallhavetheform q i x )]TJ/F28 11.9552 Tf 12.37 0 Td [(x i or q i y )]TJ/F28 11.9552 Tf 12.371 0 Td [(y i ,soconsideronly F x =0.Wehave n X i =1 q i x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x i [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x i 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y i 2 ] 3 = 2 =0 : TheLCMofthelefthandsideoftheequationis Q n j =1 [ x )]TJ/F28 11.9552 Tf 11.956 0 Td [(x i 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y i 2 ] 3 = 2 ,so multiplyingtheequationthroughbytheLCMtransforms F x =0into: n X i =1 q i x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x i [[ Y j 6 = i x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x i 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y i 2 ] 3 ] 1 = 2 =0 Theterms x )]TJ/F28 11.9552 Tf 12.081 0 Td [(x i 2 + y )]TJ/F28 11.9552 Tf 12.081 0 Td [(y i 2 eachhaveordertwo,so Q i 6 = j x )]TJ/F28 11.9552 Tf 11.956 0 Td [(x i 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y i 2 hasorder2 n )]TJ/F15 11.9552 Tf 11.751 0 Td [(1sincethereare n )]TJ/F15 11.9552 Tf 11.751 0 Td [(1terms.Cubingthisproductbringstheorder to6 n )]TJ/F15 11.9552 Tf 11.985 0 Td [(1.Distributingtheterm q i x )]TJ/F28 11.9552 Tf 11.984 0 Td [(x i intotheradicalrequiressquaring,soit addstwototheorderofeachtermintheradical. Nowwecanexpress F x as: p p 1 p p 2 ::: p p n =0 : Since p 1 ;p 2 ;:::;p n areoforder6 n )]TJ/F15 11.9552 Tf 12.879 0 Td [(1+2=2[3 n )]TJ/F15 11.9552 Tf 12.879 0 Td [(1+1],theneachtermin thisequationisoforder3 n )]TJ/F15 11.9552 Tf 12.407 0 Td [(1+1=3 n )]TJ/F15 11.9552 Tf 12.407 0 Td [(2afterapplyingthesquareroot.By theCorollary,thisequationmustbesquared n )]TJ/F15 11.9552 Tf 12.399 0 Td [(1timestoridtheequationofall radicals.Eachtimeitissquared,theorderoftheequationisdoubled,sotheresulting polynomialhasorder2 n )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 [3 n )]TJ/F15 11.9552 Tf 12.335 0 Td [(2].Noticethat F y hasthesameorderandnumber oftermsas F x .ByBezout'sTheoremthenumberofsolutionscannotexceedthe productoftheordersof F x and F y ,assumingthatthereisnocommonfactorbetween F x and F y Therefore,themaximumnumberofsolutionsis[2 n )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 [3 n )]TJ/F15 11.9552 Tf 11.955 0 Td [(2]] 2 20

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ProofofProposition4.1 :Withoutlossofgenerality,byapplyingarigidtransformationtogetherwithdilation,assumethatthethreepointcharges q 1 = q 2 = q 3 =+1 arelocatedat )]TJ/F15 11.9552 Tf 9.299 0 Td [(1 ; 0, ; 0,and ; p 3.Withthisconguration,wehaveanequilateraltrianglethathassymmetryaboutthey-axis.Becauseofthissymmetry,itis obviousthatthereisnohorizontalforceactingonanypointofthey-axisi.eforall p = ;y ;F x =0.Weneedtondanypoints y suchthat F y =0. Calltheforcesgeneratedbythechargesonthex-axis F 1 and F 2 ,andlet F 3 be theforcefromthechargeat ; p 3.Bysymmetry,foranypoint p onthey-axis,we have j F 1 j = j F 2 j inboththehorizontalandverticaldirections. If istheanglebetween F 1 andthex-axisandbysymmetry, F 2 aswell,then wehavethat F y = F 1 sin + F 2 sin )]TJ/F28 11.9552 Tf 12.334 0 Td [(F 3 istheverticalforceactingonapointon they-axis.Noticethatsince F 1 = F 2 ,thenatapointinwhichtheforcevanishes thisbecomes2 F 1 sin )]TJ/F28 11.9552 Tf 12.61 0 Td [(F 3 =0 2 F 1 sin = F 3 .Byconstruction,wehavethat sin = y p 1+ y 2 : Also,applyingthedenitionoftheforces,wegetthat F 1 = 1 1+ y 2 and F 3 = 1 p 3 )]TJ/F29 7.9701 Tf 6.587 0 Td [(y 2 : Combiningtheseyields 2 y + y 2 3 = 2 = 1 p 3 )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 2 2 y p 3 )]TJ/F28 11.9552 Tf 11.956 0 Td [(y 2 =+ y 2 3 = 2 4 y 2 p 3 )]TJ/F28 11.9552 Tf 11.956 0 Td [(y 4 =+ y 2 3 4 y 6 )]TJ/F15 11.9552 Tf 11.955 0 Td [(16 p 3 y 5 +72 y 4 )]TJ/F15 11.9552 Tf 11.955 0 Td [(48 p 3 y 3 +36 y 2 = y 6 +3 y 4 +3 y 2 +1 3 y 6 )]TJ/F15 11.9552 Tf 11.955 0 Td [(16 p 3 y 5 +69 y 4 )]TJ/F15 11.9552 Tf 11.956 0 Td [(48 p 3 y 3 +33 y 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(1=0 y )]TJ 13.151 17.977 Td [(p 3 3 y 5 )]TJ/F15 11.9552 Tf 11.955 0 Td [(15 p 3 y 4 +54 y 3 )]TJ/F15 11.9552 Tf 11.955 0 Td [(30 p 3 y 2 +3 y + p 3=0 : Clearly, ; p 3 3 isazerooftheforce.Computersoftwareisunabletondany factorsofthepolynomial3 y 5 )]TJ/F15 11.9552 Tf 12.764 0 Td [(15 p 3 y 4 +54 y 3 )]TJ/F15 11.9552 Tf 12.763 0 Td [(30 p 3 y 2 +3 y + p 3.However,if weset f y =3 y 5 )]TJ/F15 11.9552 Tf 12.733 0 Td [(15 p 3 y 4 +54 y 3 )]TJ/F15 11.9552 Tf 12.732 0 Td [(30 p 3 y 2 +3 y + p 3,noticethat f = p 3 and f p 3= )]TJ/F15 11.9552 Tf 9.299 0 Td [(32 p 3,sofyhasatleastonezeroon[0 ; p 3].Dierentiatinggives f 0 y =35 y 4 )]TJ/F15 11.9552 Tf 9.874 0 Td [(20 p 3 y 3 +54 y 2 )]TJ/F15 11.9552 Tf 9.875 0 Td [(20 p 3 y +1and f 00 y =12 y 3 )]TJ/F15 11.9552 Tf 9.875 0 Td [(15 p 3 y 2 +27 y )]TJ/F15 11.9552 Tf 9.874 0 Td [(5 p 3. Noticethat f 0 =3and f 0 p 3= )]TJ/F15 11.9552 Tf 9.299 0 Td [(96,so f y assumesatleastonemaximumon [0 ; p 3].However,since f 00 = )]TJ/F15 11.9552 Tf 9.298 0 Td [(60 p 3and f 00 p 3= )]TJ/F15 11.9552 Tf 9.299 0 Td [(96 p 3,weseethat f y is concavedownattheendpoints.Inorderfor f y tohaveadditionalzeroson[0 ; p 3], theremustbeadditionalmaximumandminimumpoints,implyingthat f y must 21

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changeconcavityon[0 ; p 3].Buttheonlyrealrootof f 00 y =0is 3 5 = 6 1 = 3 + p 10 2 = 3 +3 2 = 3 +5 2 = 3 + p 10 1 = 3 15+ p 10 1 = 3 3 : 94 > p 3 : Weseethat f y doesnotchangeconcavityon[0 ; p 3].Therefore,thereareno additionalextrama,and f y hasonlyonerooton[0 ; p 3]. Figure6. Graphof3 y 5 )]TJ/F15 11.9552 Tf 11.955 0 Td [(15 p 3 y 4 +54 y 3 )]TJ/F15 11.9552 Tf 11.955 0 Td [(30 p 3 y 2 +3 y + p 3 Figure5showstheonlyzeroof f y for y 2 ; p 3atapproximately0.2485.Since ; p 3 3 isthecenterofthetriangle,tworotationsof120degreesabout ; p 3 3 will givetwootherpointsatwhichtheforcevanishes,foratotaloffour. Observethatthethreealtitudesoftheequilateraltrianglepartitionitintosix congruentrighttriangles.Nowassumethattheforcevanishesatsomepoint p inthe interiorofoneoftheserighttriangles.Byreecting p aboutthe y -axis,andthen applyingthetworotations,weseethatassumingtheexistenceofonepoint p actually givesussixmoreequilibriumpointsintheforce,foratotaloften.By[6],weknow thatanycongurationofthreechargescannothavemorethantwelveequilibrium points;therefore,thereisatmostone p ineachoftheserighttriangles.Consider the x -componentoftheforcesat p inthetriangledeterminedby0
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Notethatthebound0 )]TJ/F28 11.9552 Tf 9.299 0 Td [(y> p 3 3 x )]TJ/F31 7.9701 Tf 13.109 11.318 Td [(p 3 3 .Observethen that x 2 + y )]TJ 11.955 10.473 Td [(p 3 2 = x 2 + y 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(2 p 3 y +3 = x 2 + y 2 +2 p 3 )]TJ/F28 11.9552 Tf 9.298 0 Td [(y +3 >x 2 + y 2 +2 p 3 p 3 3 x )]TJ 13.151 17.977 Td [(p 3 3 +3 = x 2 + y 2 +2 x +1 = x +1 2 + y 2 : So x 2 + y )]TJ 9.974 9.889 Td [(p 3 2 > x +1 2 + y 2 ,andtherefore x [ x 2 + y )]TJ 6.587 6.598 Td [(p 3 2 ] 3 = 2 < x x +1 2 + y 2 .Substituting thisresultyields .3 F x < 2 x +1 x +1 2 + y 2 + x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 [ x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 + y 2 ] 3 = 2 : Claim: 2 x +1 x +1 2 + y 2 < 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(x [ x )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 + y 2 ] 3 = 2 ProofofClaim:Noticethatalongtheline x =0,equalityholds.However,weare onlyconsidering0 x )]TJ/F15 11.9552 Tf 11.214 0 Td [(1 2 + y 2 .Applyingtheseinequalitiesgives @ @x 2 x +1 [ x +1 2 + y 2 ] 3 = 2 = )]TJ/F15 11.9552 Tf 9.298 0 Td [(4 x 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(5 x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1+2 y 2 [ x +1 2 + y 2 ] 5 = 2 < )]TJ/F15 11.9552 Tf 9.299 0 Td [(4 x 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(5 x )]TJ/F26 7.9701 Tf 13.15 4.707 Td [(1 3 [ x +1 2 + y 2 ] 5 = 2 < 2 x 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(4 x +2 )]TJ/F26 7.9701 Tf 13.15 4.707 Td [(1 3 [ x +1 2 + y 2 ] 5 = 2 < 2 x 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(4 x +2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 2 [ x +1 2 + y 2 ] 5 = 2 < 2 x )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 2 [ x )]TJ/F15 11.9552 Tf 11.956 0 Td [(1 2 + y 2 ] 5 = 2 = @ @x 1 )]TJ/F28 11.9552 Tf 11.955 0 Td [(x [ x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 + y 2 ] 3 = 2 : Since @ @x 2 x +1 [ x +1 2 + y 2 ] 3 = 2 < @ @x 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(x [ x )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 + y 2 ] 3 = 2 ,and 2 x +1 [ x +1 2 + y 2 ] 3 = 2 = 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(x [ x )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 + y 2 ] 3 = 2 when x = 0,thenatnopointinsidethetrianglecanequalityhold.Therefore 2 x +1 [ x +1 2 + y 2 ] 3 = 2 < 1 )]TJ/F29 7.9701 Tf 6.586 0 Td [(x [ x )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 + y 2 ] 3 = 2 ,andtheClaimisproven. Substitutingthisresultinto.3gives F x < 0ateverypointinsideofthetriangle determinedby0
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insidethistriangle;bysymmetry,noneofthesesixtrianglescanhaveadditional pointswheretheforcevanishes. Therefore,thereareexactlyfourpointequilibriumpoints. ProofofTheorem4.2 :Withoutlossofgenerality,assumethat q 1 q 2 ,and q 3 are threepositivecharges.Also,wecanthroughre-scalingandtranslationplace q 1 at theorigin, q 2 at ; 0,and q 3 intheupperhalf-plane.Let ~v beavectorhavingthe followingproperties: 1.Theoriginof ~v isonthelinejoining q 1 and q 2 2. ~v isperpendiculartothelinejoining q 1 and q 2 3. ~v passesthroughtheinteriorofthetriangle. Recallthatthepotentialisdenedby P x;y = P 3 k =1 q k p x )]TJ/F29 7.9701 Tf 6.586 0 Td [(x k 2 + y )]TJ/F29 7.9701 Tf 6.586 0 Td [(y k 2 .Theparametricequationsthatdescribethevector ~v are x t = c y t = t where0
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Substitutiongives @P @t = 3 X k =1 )]TJ/F28 11.9552 Tf 9.298 0 Td [(q k y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 3 = 2 = )]TJ/F28 11.9552 Tf 30.652 8.088 Td [(q 1 y [ x 2 + y 2 ] 3 = 2 )]TJ/F28 11.9552 Tf 48.093 8.088 Td [(q 2 y [ x )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 + y 2 ] 3 = 2 )]TJ/F28 11.9552 Tf 50.859 8.088 Td [(q 3 y )]TJ/F28 11.9552 Tf 11.956 0 Td [(y 3 [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x 3 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 3 2 ] 3 = 2 = )]TJ/F28 11.9552 Tf 29.845 8.088 Td [(q 1 t [ c 2 + t 2 ] 3 = 2 )]TJ/F28 11.9552 Tf 47.286 8.088 Td [(q 2 t [ c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 + t 2 ] 3 = 2 )]TJ/F28 11.9552 Tf 50.052 8.088 Td [(q 3 t )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 3 [ c )]TJ/F28 11.9552 Tf 11.955 0 Td [(x 3 2 + t )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 3 2 ] 3 = 2 : For t> 0,thersttwotermsofthisequationarenegative,butareincreasingto zero.Thisimpliesthatthersttwotermsofthepotentialaredecreasingtozero.For x;y insideofthetriangle, t )]TJ/F28 11.9552 Tf 12.217 0 Td [(y 3 < 0,sothethirdtermofthepotentialisstrictly increasing. Let P t = f t + g t ,where f t correspondstothersttwotermsofthepotential and g t isthe q 3 termofthepotential.Then f 0 t = )]TJ/F29 7.9701 Tf 23.97 5.256 Td [(q 1 t [ c 2 + t 2 ] 3 = 2 )]TJ/F29 7.9701 Tf 35.818 5.255 Td [(q 2 t [ c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 + t 2 ] 3 = 2 .In ordertoprovethistheorem,werstwanttoshowthat f 0 t hasonlyonemin.A directcalculationgivesthat .4 f 00 t = q 1 t 2 )]TJ/F28 11.9552 Tf 11.956 0 Td [(c 2 t 2 + c 2 5 = 2 + q 2 t 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [( c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 t 2 + c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 5 = 2 : Case1:Let c = 1 2 .Thensetting.4equaltozerobecomes q 1 t 2 )]TJ/F26 7.9701 Tf 13.151 4.707 Td [(1 4 t 2 + 1 4 5 = 2 + q 2 t 2 )]TJ/F26 7.9701 Tf 13.151 4.707 Td [(1 4 t 2 + 1 4 5 = 2 =0 q 1 + q 2 t 2 )]TJ/F26 7.9701 Tf 13.151 4.707 Td [(1 4 t 2 + 1 4 5 = 2 =0 2 t 2 )]TJ/F15 11.9552 Tf 13.151 8.088 Td [(1 4 =0 t 2 = 1 8 t = r 1 8 Sofor c = 1 2 ,whichistheperpendicularbisectorofthelinejoining q 1 and q 2 ,we seethat f 0 t hasonlyonemin. Case2:Assumethat c< 1 2 .Setting.4equaltozeroandrearrangingtermsgives q 1 q 2 2 t 2 )]TJ/F28 11.9552 Tf 11.955 0 Td [(c 2 2 t 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [( c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 = )]TJ/F15 11.9552 Tf 9.298 0 Td [([ t 2 + c 2 t 2 + c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 ] 5 = 2 : 25

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Let h t = q 1 q 2 2 t 2 )]TJ/F29 7.9701 Tf 6.586 0 Td [(c 2 2 t 2 )]TJ/F26 7.9701 Tf 6.587 0 Td [( c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 ,and j t = )]TJ/F15 11.9552 Tf 9.299 0 Td [([ t 2 + c 2 t 2 + c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 ] 5 = 2 .Observesomeobviousproperties ofthesefunctions: lim t !1 h t = q 1 q 2 : lim t !1 j t = )]TJ/F15 11.9552 Tf 9.299 0 Td [(1 : j t < 0forall t> 0 : h t =0at t = c p 2 : h t hasaverticalasymptoteat t = q c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 2 : If t>t then 2 t 2 )]TJ/F29 7.9701 Tf 6.586 0 Td [(c 2 2 t 2 )]TJ/F26 7.9701 Tf 6.587 0 Td [( c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 > c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 )]TJ/F29 7.9701 Tf 6.587 0 Td [(c 2 2 t 2 )]TJ/F26 7.9701 Tf 6.587 0 Td [( c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 = 1 )]TJ/F26 7.9701 Tf 6.587 0 Td [(2 c 2 t 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [( c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 > 0;sofor t>t ,wehavethat h t > 0.Also,weseethatlim t t + h t =+ 1 If t< c p 2 ,then 2 t 2 )]TJ/F29 7.9701 Tf 6.587 0 Td [(c 2 2 t 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [( c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 > 2 t 2 )]TJ/F29 7.9701 Tf 6.586 0 Td [(c 2 c 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [( c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 = 2 t 2 )]TJ/F29 7.9701 Tf 6.586 0 Td [(c 2 2 c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 > 0;sofor0 0. If c p 2 t 2 + 1 4 ,so t 2 + c 2 t 2 + c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 < 1,andtherefore t 2 + c 2 t 2 + c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 3 = 2 < 1.Therefore, h 0 t = q 1 q 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [(4 t )]TJ/F26 7.9701 Tf 6.587 0 Td [(2 c t 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [( c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 2 < 26

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q 1 q 2 )]TJ/F26 7.9701 Tf 6.587 0 Td [(4 t )]TJ/F26 7.9701 Tf 6.587 0 Td [(2 c t 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [( c )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 2 2 t 2 + c 2 t 2 + c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 3 = 2 .Inordertohave h 0 t p 5+ q q 1 q 2 p 5 )]TJ/F34 7.9701 Tf 6.587 8.207 Td [(q q 1 q 2 c )]TJ/F15 11.9552 Tf 12.506 0 Td [(1 2 ,whichholdsforall t sincetheright handsideisnegative.If p 5 )]TJ/F33 11.9552 Tf 11.768 12.327 Td [(q q 1 q 2 > 0,thenthisgives t 2 < p 5+ q q 1 q 2 p 5 )]TJ/F34 7.9701 Tf 6.586 8.208 Td [(q q 1 q 2 c )]TJ/F15 11.9552 Tf 11.768 0 Td [(1 2 .Since p 5+ q q 1 q 2 p 5 )]TJ/F34 7.9701 Tf 6.586 8.207 Td [(q q 1 q 2 > 1,thenweseethat p 5+ q q 1 q 2 p 5 )]TJ/F34 7.9701 Tf 6.586 8.207 Td [(q q 1 q 2 c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 > c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 > c )]TJ/F26 7.9701 Tf 6.586 0 Td [(1 2 2 Soforall t suchthat t 2 < p 5+ q q 1 q 2 p 5 )]TJ/F34 7.9701 Tf 6.586 8.207 Td [(q q 1 q 2 c )]TJ/F15 11.9552 Tf 12.438 0 Td [(1 2 ,wehavethat h 0 t
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Thisshowsthatforall c< 1 = 2,thepotentialonthesegmentofthelinedenedby vector ~v assumesatmostoneminimumexcludingendpoints.However,byreecting thechargesabouttheline x = 1 2 ,weseethatthisresultholdsforall c Adirectcalculationgivesus f 000 t = )]TJ/F15 11.9552 Tf 10.494 8.088 Td [(3 q 1 t t 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(3 c 2 c 2 + t 2 7 = 2 )]TJ/F15 11.9552 Tf 13.151 8.088 Td [(3 q 2 t t 2 )]TJ/F15 11.9552 Tf 11.955 0 Td [(3 c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 [ c )]TJ/F15 11.9552 Tf 11.955 0 Td [(1 2 + t 2 ] 7 = 2 andweseethatalthough f 000 =0,for t> 0sucientlysmallwehave f 000 t > 0, soinitially f 0 t isconcaveup. Also,notethat g 00 t = q 3 [2 t )]TJ/F29 7.9701 Tf 6.586 0 Td [(y 3 2 )]TJ/F26 7.9701 Tf 6.586 0 Td [( c )]TJ/F29 7.9701 Tf 6.587 0 Td [(x 3 2 ] [ t )]TJ/F29 7.9701 Tf 6.587 0 Td [(y 3 2 + c )]TJ/F29 7.9701 Tf 6.586 0 Td [(x 3 2 ] 5 = 2 ,sothemaximumof g 0 t musthave q 3 [2 t )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 3 2 )]TJ/F15 11.9552 Tf 11.956 0 Td [( c )]TJ/F28 11.9552 Tf 11.955 0 Td [(x 3 2 ]=0 t )]TJ/F28 11.9552 Tf 11.955 0 Td [(y 3 2 = c )]TJ/F28 11.9552 Tf 11.955 0 Td [(x 3 2 2 t = y 3 c )]TJ/F28 11.9552 Tf 11.955 0 Td [(x 3 p 2 Clearly,oneofthesesolutionsisgreaterthan y 3 ,so g 0 t hasatmostonemaximum on ~v Figure7. Thegraphsof f 0 and )]TJ/F28 11.9552 Tf 9.298 0 Td [(g 0 .Forthisparticularexample,all chargesareequal,and q 3 islocatedat 1 4 ; 3. G 1 uses c = 1 2 ,while G 2 uses c = 1 4 ,thealtitudeofthetriangle. Wenowneedthefollowinglemmas. Lemma6.3. Let F x < 0 forall x> 0 F =0 lim x !1 F x =0 F x has exactlyonemin,initially F x isconvexup, G < 0 G x isconcavedownand decreasingforall x> 0 F x and G x arebothsmoothfunctions.Thenthegraphs of F x and G x intersectatmosttwice. 28

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ProofofLemma6.3 :Let x 0 betheminof F ,andlet x 1 betherstintersection of F and G .Itisclearthatif F and G intersectatsome x>x 0 ,thenbecause F isincreasingand G isdecreasingwemusthave G>F approachingthisintersection point.Butbytheinitialconditionswehave G 0.Itfollowsthenthatfor x>x 1 ,eventhoughboth F 0 < 0and G 0 < 0,wehave F 0 isincreasingto0.Thereforethisistheonlypointofintersection. Next,assumethat G istransversalto F .Because F 0andsince G isdecreasing, itfollowsfromtheIntermediateValueTheoremthat F and G willintersectatleast oncemore.Callthispoint x 2 .Assumethat x 2 >x 0 .Thenitistrivialthat x 2 isthe onlypointofintersectionbecause F isincreasingand G isdecreasing. Next,assumethat x 2 x 0 .Again,bytheTaylorserieswehavethat G 00 < 0, F 00 > 0, F 0 isincreasingto0,so x 2 istheonlypointofintersection. Thereforethereisatmosttwopointsofintersection. Lemma6.4. Let F x < 0 forall x> 0 F =0 lim x !1 F x =0 F x has exactlyonemin,initially F x isconvexup, F x switchesconcavityonceafterits min, G < 0 forall 0 0 F x and G x arebothsmoothfunctions.Thenthegraphsof F x and G x intersectatmosttwice. ProofofLemma6.4 :Bythepreviouslemma, F and G canintersectatmosttwo timesbefore F takesitsminimumat x 0 .Assumethisisthecase,andtherefore G
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At t =0,wehave f 0 t =0and g 0 t > 0,soinitiallythepotentialwillincrease. Since ~v isorthogonaltotheboundary,weseethattheinwardnormalderivativeof thepotentialisalwayspositive. If g 0 t > j f 0 t j forall t ,thenthepotentialismonotonicallyincreasingon ~v However,ifthereexistsa t suchthat j f 0 t j >g 0 t ,since f 0 t takesonlyone minimumweseethatthereareatmosttwointervalsonwhichthepotentialwill decrease.Thisfollowsfromthelemmas.Ifwelet F = f 0 and G = )]TJ/F28 11.9552 Tf 9.299 0 Td [(g 0 ,thenifwe choose ~v tobeanaltitudeofthetrianglewehavetheconditionsofLemma6.3.For anyother ~v wehavetheconditionsofLemma6.4. Sincethepotentialisinitiallyincreasing,therstpointsuchthat f 0 t = )]TJ/F28 11.9552 Tf 9.299 0 Td [(g 0 t mustbeamaximum.If ~v waschosentobeanaltitudeofthetriangle,thenby Lemma6.3thereisexactlyoneminmum.Foranyother ~v considerthefollowing: sincethepotentialisdecreasingon ~v at y 3 ,wemusthavethatthelastcriticalpoint isamaximum.Iftherearethreecriticalpoints,weseethattherearetwomaximum andoneminimum.Thereisnowaythatwecanhavefourcriticalpointsbecause thesecondandthirdwouldhavetobeminimums,whichisimpossible.Fivecritical pointsimpliesthreemaximumsandtwominimums. Thesameproofappliestonegativecharges. ProofofTheorem5.1 :Withoutlossofgenerality,let k =1.Also,since log j 1 z )]TJ/F29 7.9701 Tf 6.586 0 Td [(z k j = )]TJ/F15 11.9552 Tf 11.291 0 Td [(log j z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k j ,weonlyneedtoconsiderthezerosoflog j z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k j Notethatlog j z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z 0 j = 1 2 log z )]TJ/F28 11.9552 Tf 11.955 0 Td [(z 0 + 1 2 log z )]TJ/F15 11.9552 Tf 12.191 0 Td [( z 0 = 1 2 log x + iy )]TJ/F28 11.9552 Tf 11.955 0 Td [(z 0 + 1 2 log x )]TJ/F28 11.9552 Tf -422.702 -19.527 Td [(iy )]TJ/F15 11.9552 Tf 14.67 0 Td [( z 0 .So P x;y = P n k =1 )]TJ/F29 7.9701 Tf 10.494 5.256 Td [(q k 2 log x + iy )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k )]TJ/F29 7.9701 Tf 13.151 5.256 Td [(q k 2 log x )]TJ/F28 11.9552 Tf 11.955 0 Td [(iy )]TJ/F15 11.9552 Tf 14.307 0 Td [( z k .Since F = r P = )]TJ/F29 7.9701 Tf 10.494 4.707 Td [(@P @x )]TJ/F28 11.9552 Tf 11.955 0 Td [(i @P @y ,and @P @x = )]TJ/F15 11.9552 Tf 10.494 8.087 Td [(1 2 2 X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k )]TJ/F15 11.9552 Tf 14.308 0 Td [( z k x + iy )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(iy )]TJ/F15 11.9552 Tf 14.308 0 Td [( z k and @P @y = )]TJ/F28 11.9552 Tf 11.423 8.087 Td [(i 2 n X i =1 q k iy )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k )]TJ/F15 11.9552 Tf 14.307 0 Td [( z k x + iy )]TJ/F28 11.9552 Tf 11.955 0 Td [(z k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(iy )]TJ/F15 11.9552 Tf 14.308 0 Td [( z k ; thenweseethat F = P n i =1 q i z )]TJ/F26 7.9701 Tf 8.725 0 Td [( z k 30

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Soif P = P n i =1 q i log j 1 z )]TJ/F29 7.9701 Tf 6.587 0 Td [(z i j ,then F = r P reducesto n X k =1 q k z )]TJ/F15 11.9552 Tf 14.307 0 Td [( z k = q 1 z )]TJ/F15 11.9552 Tf 14.113 0 Td [( z 1 + q 2 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 2 + ::: + q n z )]TJ/F15 11.9552 Tf 14.566 0 Td [( z n = q 1 z )]TJ/F15 11.9552 Tf 14.113 0 Td [( z 2 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 3 z )]TJ/F15 11.9552 Tf 14.566 0 Td [( z n + q 2 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 1 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 3 z )]TJ/F15 11.9552 Tf 14.566 0 Td [( z n z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 1 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 2 z )]TJ/F15 11.9552 Tf 14.566 0 Td [( z n + ::: + q n z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 1 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 2 z )]TJ/F15 11.9552 Tf 19.976 0 Td [( z n )]TJ/F26 7.9701 Tf 6.587 0 Td [(1 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 1 z )]TJ/F15 11.9552 Tf 14.114 0 Td [( z 2 z )]TJ/F15 11.9552 Tf 14.566 0 Td [( z n : Sincethenumeratoristheconjugateofapolynomialofdegree n )]TJ/F15 11.9552 Tf 11.972 0 Td [(1,thenbythe FundamentalTheoremofAlgebra, F hasexactly n )]TJ/F15 11.9552 Tf 11.955 0 Td [(1solutions. ProofofTheorem5.2 :Beforeprovingthistheorem,weneedthefollowing lemmasanddenition. Lemma6.5. Green'sTheorem:Let D beasimplyconnectedregionwithapiecewise smoothboundary C ,orientedcounterclockwise.If F x;y = M x;y ~ i + N x;y ~ j ,and M and N havecontinuouspartialderivativesin D ,then R C F ~nds = RR D r FdA where ~n istheoutwardunitnormalvector. Foraproof,see[3],p.1025.Noticethat r F = r )]TJ/F29 7.9701 Tf 10.494 4.707 Td [(@P @x ~ i )]TJ/F29 7.9701 Tf 13.151 4.707 Td [(@P @y ~ j = )]TJ/F15 11.9552 Tf 9.298 0 Td [( P Denition. Asingularpointofafunction f isapoint a;b wherethegradientof f andpossiblyhigherorderderivativesvanish;thatis, f x a;b = f y a;b =0[8] Itisclearthatiftheforcevanishesonacurve,theneverypointofthiscurveisa singularpoint. ThefollowingtheoremofPuiseauxfoundin[8],pp.95,111,showsthatina neighborhoodofasingularpoint,thecurve,throughaproperchangeofcoordinates, hasaparametrizationthatishomeomorphicinsomepossiblysmallerneighborhood to y p = x q ;p;q 2 Z Lemma6.6. Inasuitablecoordinatesystemanygivenparametrizationisequivalent tooneofthetype x = t n ;y = a 1 t n 1 + a 2 t n 2 + ::: where 0
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thenthecurvehasacuspatthesingularpoint,butdoesnotterminate.Thesingular pointmaybeapointwherethefunctionbranchesconsidertheorigininthegraph of x 2 + y 2 3 )]TJ/F15 11.9552 Tf 12.419 0 Td [(4 x 2 y 2 =0,butthebranchesformloopsorgotoinnity.Itisnot possibletohavethefunctionterminateatasingularpoint.Therefore,thecurves wheretheforcevanishescaneithergoto 1 orareclosedloops. ProofofTheorem5.2 :Withoutlossofgenerality,assumethatallofthechargesas positive.Assumethat C isanequipotentialcurveonwhichtheforcevanishes.There arethreecasestoconsider: Case1: C isacurvethatgoestoinnity. Sincethelim x;y !1 P =0,thenbythemaximumprinciplewemusthave P =0at everypointon C .Otherwise,assumethat P = k> 0on C ,andchoosean p = x;y on C sucientlyfarfromthethreecharges.Fix "> 0,andlet B p beaball ofradius centeredat p .Let p 0 beanyotherpointin B p .If p 0 ison C ,then P p 0 = k .Otherwise,sincelim x;y !1 P =0wehavethat P p 0 0forall p 0 not on C ,so P p 0
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ateverypointof C .ApplyingGreen'sTheorem,weget 0= Z C F ~nds = ZZ D r FdA = ZZ D )]TJ/F15 11.9552 Tf 9.298 0 Td [( PdA < 0 : Thiscontradictionshowsthat C cannotbeaclosedloopintheinteriorofthetriangle. ProofofTheorem5.3 :Assumeforcevanishesalongacurve S intheplane,then ateverypointon S wewouldhave F x =0and F y =0.Consider y tobeafunction of x ,andrecallthat F x = )]TJ/F29 7.9701 Tf 10.494 4.707 Td [(@P @x ,andsoimplicitdierentiationon F x =0weget )]TJ/F28 11.9552 Tf 9.298 0 Td [(P xx )]TJ/F28 11.9552 Tf 11.955 0 Td [(P xy y 0 1 =0 : Similarly,implicitdierentiationon F y =0gives )]TJ/F28 11.9552 Tf 9.298 0 Td [(P yx )]TJ/F28 11.9552 Tf 11.955 0 Td [(P yy y 0 2 =0 : Solvingeachofthesefor y 0 gives y 0 1 = )]TJ/F28 11.9552 Tf 10.494 8.087 Td [(P xx P xy and y 0 2 = )]TJ/F28 11.9552 Tf 10.494 8.088 Td [(P yx P yy : Alongacurvewheretheforcevanishes,wemusthavethat y 0 1 = y 0 2 ,which,after re-arrangingtermsandusingthefactthat P xy = P yx ,becomes .5 P xx P yy = P xy 2 : Noticethenthatalong S ,wegettherelationthattheproduct P xx P yy mustbenonnegative.Obviously,onewaythisconditioncanbesatisedisif P xy =0atevery pointof S Wehavealreadyseenthatadirectcalculationgives P xx = n X k =1 q k [2 x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 )]TJ/F15 11.9552 Tf 11.956 0 Td [( y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 33

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P yy = n X k =1 q k [ )]TJ/F15 11.9552 Tf 9.298 0 Td [( x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 +2 y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 and P xy =3 n X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 2 : Dene X = n X k =1 q k x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 [ x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.956 0 Td [(y k 2 ] 5 = 2 and Y = n X k =1 q k y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 [ x )]TJ/F28 11.9552 Tf 11.956 0 Td [(x k 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y k 2 ] 5 = 2 : Wecanwrite P xx =2 X )]TJ/F28 11.9552 Tf 12.337 0 Td [(Y and P yy = )]TJ/F28 11.9552 Tf 9.298 0 Td [(X +2 Y .Noticethatboth X and Y are strictlypositive. Wewishtondunderwhatconditions P xx P yy 0. Case1:Whenallchargesarepositiveweknowthatthepotentialfunctionis subharmonic;thatis, .6 P = P xx + P yy 0 : Itistrivialthentoseethat P xx > 0and P yy > 0willsatisfythesubharmonic condition.Notice,though,thatifwewrite P xx =2 X )]TJ/F28 11.9552 Tf 12.653 0 Td [(Y> 0,thenwegetthat Y< 2 X .Also, P yy = )]TJ/F28 11.9552 Tf 9.298 0 Td [(X +2 Y> 0givesthat Y> 1 2 X .Soweseethat 1 2 X 0and P yy < 0,but j P xx j P xx tomaintainthesubharmonicinequality.But .5saysthattheproduct P xx P yy 0.Therefore,atnopointof S canwehavethat P xx > 0and P yy < 0,orviceversa.If P xx < 0atsomepointof S ,thenatthatpoint wemusthavethatboth P yy = P xy =0inorderfor.5tohold.However,if P xx < 0 and P yy =0,then.6isviolated.Therefore,atnopointof S caneither P xx > 0or P yy > 0. Case3:Assumethat P xx = P yy =0.Thenwehavethesystemofequations 2 X )]TJ/F28 11.9552 Tf 11.955 0 Td [(Y =0 )]TJ/F28 11.9552 Tf 9.298 0 Td [(X +2 Y =0 : Theonlysolutiontothissystemis X = Y =0,whichisnotpossiblebecause X and Y weredenedtobestrictlypositive.Therefore, P xx and P yy cannotbothequal zeroatanypointof S 34

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Numericalmethodforndingzerosofthepotential :Thecalculationsfor thispaperweredoneusingMaple11.Herewediscusstheprogramwhichndszeros intheforceoftheequilateraltriangle.Withslightmodication,thisprogramcanbe usedtondzerosofanycongurationofanynumberofpointchargesintheplane orinspace. Inordertoplotandanalyzeapotentialeld,thepositionsofthepointcharges andtheircorrespondingvaluesmustbeestablished.Fortheequilateraltriangle,the chargesallhadvalue+1,andwerelocatedat )]TJ/F15 11.9552 Tf 9.298 0 Td [(1 ; 0, ; 0,and ; p 3.Toplace thesechargesontheplaneusingMaple,usethefollowingcommand: q [1]:=1; q [2]:=1; q [3]:=1; x [1]:= )]TJ/F15 11.9552 Tf 9.298 0 Td [(1; y [1]:=0; x [2]:=1; y [2]:=0; x [3]:=0; y [3]:= sqrt ; Note:semicolonscanbereplacedwithcolonstosuppressoutput. Todenethepotentialasafunction,usethecommand: P := x;y )]TJ/F28 11.9552 Tf 12.619 0 Td [(>sum q [ i ] =sqrt x )]TJ/F28 11.9552 Tf 11.955 0 Td [(x [ i ] 2 + y )]TJ/F28 11.9552 Tf 11.955 0 Td [(y [ i ] 2 ;i =1 :: 3; TheforceequationsaredenedbytakingthederivativeofPx,ywithrespectto thetwovariables,andthennegating. Fx := )]TJ/F28 11.9552 Tf 9.298 0 Td [(diff P x;y ;x ; Fy := )]TJ/F28 11.9552 Tf 9.299 0 Td [(diff P x;y ;y ; Nowweneedtheprogramtosolvethesystemofequations Fx =0 Fy =0 : Unfortunately,Maple'sbuilt-in solve commandissomtimesunabletosolvethissystemforallsolutions.Insomecases,itmayreturnsomesolutions,butoftenitwill generateawarningthatsolutionsmayhavebeenlost.However, fsolve isaprocedurethatwillattempttondasolutionoveraspeciedrangeofthevariables. Thedrawbackisthat fsolve willmanytimesonlyreturntherstsolutionitnds. If fsolve isunabletondanysolutions,itwillreturnthecallingcommand. Inordertohave fsolve ndallsolutions,weneedtopartitiontheportionofthe planewewishtostudyintosmallersquarestowhich fsolve canbeapplied.Giventhat 35

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ourchargesarelocatedat )]TJ/F15 11.9552 Tf 9.298 0 Td [(1 ; 0, ; 0,and ; p 3,thesquareregion[ )]TJ/F15 11.9552 Tf 9.299 0 Td [(1 ; 1] [0 ; 2] containstheentiretriangleandthereforeallpossiblesolutions. Choose n tobethenumberofpartitions.Let d =2 =n bethelengthofthepartitionedsquaresthe2comesfromthelengthofthesquareregionthatisbeing partitioned.Thefollowingcodewillhave fsolve checkeachofthesquaresforsolutionsandoutputtheminaset.Thefollowingexampleuses n =20forthenumber ofpartitions. n:=20: SolutionSet:= fg : d:=2/n: forifrom0ton-1do forjfrom0ton-1do p:=fsolve [ Fx=0,Fy=0 ] ,x=-1+i*d..-1+ i+1 *d,y=j*d.. j+1 *d : ifmember f x=-1+i*d..-1+ i+1 *d,y=j*d.. j+1 *d g ,p then next: else SolutionSet:=SolutionSetunion f p g : endif: enddo: enddo: SolutionSet; Line7isnecessaryforsuppressingunwantedoutput.TheBooleanprocedure member checksiftheoutputcontainedaportionofthecallingcommand.Ifitdoes,then fsolve wasunabletondasolution,andLine8iteratestheloop.IfLine7returned false",thenLine10addsthesolutiontothe SolutionSet 36

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References [1]S.Fisher,ComplexVariables,SecondEdition,DoverPublicationsInc,1999. [2]O.Kellogg,FoundationsofPotentialTheory,DoverPublications,Inc.1953. [3]R.Larson,R.Hostetler,B.Edwards,Calculus,SixthEdition,HoughtonMiinComplany, 1998. [4]J.C.Maxwell,ATreatiseonElectricityandMagnetism,vol.1,Republicationofthe3rdrevised edition,DoverPublications,Inc.,1954. [5]T.Ransford,PotentialTheoryintheComplexPlane,CambridgeUniversityPress,1995. [6]A.Gabrielov,D.Novikov,B.Shapiro,Mysteryofpointcharges.Proc.Lond.Math.Soc.95 ,no.2,443-472. [7]P.TiplerandG.Mosca,PhysicsforScientistsandEngineers,5thed,FreemanandCompany, 2004. [8]R.Walker,AlgebraicCurves,DoverPublications,1962. 37