The Euler-Lagrange equation - KAIST

1 Chapter2 TheEuler-LagrangeequationIn thischapter,we willgive necessaryconditionsforanextremumof a functionof thetypeI(x) =ZbaF(x(t); x0(t); t)dt;withvarioustypes of in theformof a di erentialequationthattheextremalcurve shouldsatisfy, andthisdi erentialequationis beginwiththesimplesttype of boundaryconditions,wherethecurves areallowed to varybetweentwo formulatedas follows:LetF( ; ; ) be a functionwithcontinuous rstandsecondpartialderivatives withrespectto( ; ; ). Then ndx2C1[a; b] such thatx(a) =yaandx(b) =yb, andwhich is anextremumforthefunctionI(x) =ZbaF(x(t); x0(t); t)dt:( )In otherwords,thesimplestoptimisationproble mconsistsof ndinganextremumof a functionof theform( ),wheretheclassof admissiblecurves comprisesallsmoothcurves joiningtwo xedpoints; willapplythenecessaryconditionforanextre mum(establishedin )to thesolve thesimplestoptimisationproblemdescribed [a; b]jx(a) =yaandx(b) =ybg, andletI:S!

2 Rbe afunctionof theformI(x) =ZbaF(x(t); x0(t); t)dt:IfIhasanextremumatx02S, thenx0satis estheEuler-Lagrangeequation:@F@ (x0(t); x00(t); t) ddt @F@ (x0(t); x00(t); t) = 0;t2[a; b]:( ) :Possiblepathsjoiningthetwo xedpoints (a; ya) and(b; yb).ProofTheproof is longandso we divideit into allwe notethatthesetSis nota vectorspace(unlessya= 0 =yb)! notapplicabledirectly. Hencewe introducea newlinearspaceX, andconsidera newfunction~I:X!Rwhich is de nedin termsof [a; b]jx(a) =x(b) = 0g;withtheinducednormfromC1[a; b]. Thenforallh2X,x0+hsatis es(x0+h)(a) =yaand(x0+h)(b) =yb. De ning~I(h) =I(x0+h), forh2X, we notethat~I:X!Rhasa localextremumat 0. It ~I(0)= now calculateD~I(0).We have~I(h) ~I(0)=ZbaF((x0+h)(t);(x0+h)0(t); t)dt ZbaF(x0(t); x00(t); t)dt=Zba[F(x0(t) +h(t); x00(t) +h0(t); t)dt F(x0(t); x00(t); t)]dt:RecallthatfromTaylor'stheorem,ifFp ossessespartialderivatives of order2 in a ballBof radiusraroundthepoint ( 0; 0; 0) inR3, thenforall( ; ; )2B, thereexistsa 2[0;1] such thatF( ; ; )=F( 0; 0; 0) + ( 0)@@ + ( 0)@@ + ( 0)@@ F ( 0; 0; 0)+12!

3 ( 0)@@ + ( 0)@@ + ( 0)@@ 2F ( 0; 0; 0)+ (( ; ; ) ( 0; 0; 0)):Henceforh2 Xsuch thatkhkis smallenough,~I(h) ~I(0)=Zba @F@ (x0(t); x00(t); t)h(t) +@F@ (x0(t); x00(t); t)h0(t) dt+12!Zba h(t)@@ +h0(t)@@ 2F (x0(t)+ (t)h(t);x00(t)+ (t)h0(t);t)dt:It canbe checked thatthereexistsaM>0 such that 12!Zba h(t)@@ +h0(t)@@ 2F (x0(t)+ (t)h(t);x00(t)+ (t)h0(t);t)dt Mkhk2;1 Notethatby the0 in theright handsideof theequality, we meanthezeromap,namelythecontinuouslinear mapfromXtoR, which is de nedbyh7!0 ~I(0)is themaph7!Zba @F@ (x0(t); x00(t); t)h(t) +@F@ (x0(t); x00(t); t)h0(t) dt:( ) show thatif themapin ( )is thezeromap,thenthisimpliesthat( ) neA(t) =Zta@F@ (x0( ); x00( ); )d :Integratingby parts,we ndthatZba@F@ (x0(t); x00(t); t)h(t)dt= ZbaA(t)h0(t)dt;andso from( ),it followsthatD~I(0)= 0 impliesthatZba A(t) +@F@ (x0(t); x00(t); t) h0(t)dt= 0 forallh2 , ,we obtain A(t) +@F@ (x0(t); x00(t); t) =kforallt2[a; b]:Di erentiatingwithrespecttot, we obtain( ).

4 Thiscompletestheproof of onlyanecessaryconditionfortheexistenceof anextremum( ).However,in many cases,theEuler-Lagrangeequationby itselfis enoughto give a completesolutionof fact,theexistenceofanextremumis sometimesclearfromthecontextof in such scenarios,thereexistsonlyonesolutionto theEuler-Lagrangeequation,thenthissoluti onmusta fortioribe thepointforwhich theextremumis [0;1]jx(0)= 0 andx(1)= 1g. ConsiderthefunctionI:S!Rgiven byI(x) =Z10 ddtx(t) 1 2dt:We wishto ndx02 SthatminimizesI. We proceedas haveF( ; ; ) = ( 1)2, andso@F@ = 0 and@F@ = 2( 1). ( )is now given by0 ddt(2(x00(t) 1))= 0forallt2[0;1] , we obtain2(x00(t) 1) =C, forsomeconstantC, andsox00=C2+ 1 = ,we havex0(t) =At+B, determinedby usingthatfactthatx02S, andsox0(0)= 0 andx0(a) = 1.

5 Thus we haveA0 +B=0;A1 +B=1;which yieldA= 1 andB= theEuler-LagrangeequationinSisx0(t) =t,t2[0;1]; we arguethatthesolutionx0indeedminimizesI. Since(x0(t) 1)2 0 forallt2[0;1],it followsthatI(x) 0 forallx2C1[0;1].ButI(x0) =Z10(x00(t) 1)2dt=Z10(1 1)2dt=Z100dt= 0:AsI(x) 0 =I(x0) forallx2S, it followsthatx0minimizesI. De theEuler-Lagrangeequation( )arecalledcritical in generala secondorderdi erentialequation,butin somespecialcases,it canbe reducedto a rstorderdi erentialequationor whereitssolutioncanbeobtainedentirelyby indicatesomespecialcasesin Exercise3 onpage31,wherein each instance,Fis independent of oneof LetS=fx2C1[0;1]jx(0)= 0 =x(1)g. ConsiderthemapI:S!Rgiven byI(x) =Z10(x(t))3dt;x2 , ndthecriticalcurvex02 SforI.

6 DoesIhave a localextremumatx0?2. WritetheEuler-LagrangeequationwhenFis given by(a)F( ; ; ) = sin ,(b)F( ; ; ) = 3 3,(c)F( ; ; ) = 2 2,(d)F( ; ; ) = 2 2+ 3 Prove that:(a)IfF( ; ; ) does notdependon , thentheEuler-Lagrangeequationbecomes@F@ (x(t); x0(t); t) =c;whereCis a constant.(b)IfFdoes notdependon , thentheEuler-Lagrangeequationbecomes@F@ (x(t); x0(t); t) = 0:(c)IfFdoes notdependon andifx0is twice-di erentiablein [a; b], thentheEuler-LagrangeequationbecomesF(x( t); x0(t); t) x0(t)@F@ (x(t); x0(t); t) =C;whereCis a : Whatisddt F(x(t); x0(t); t) x0(t)@F@ (x(t); x0(t); t) ?4. Findthecurve which hasminimumlengthbetween(0;0) and(1;1).5. LetS=fx2C1[0;1]jx(0)= 0 andx(1)= 1g. Findcriticalcurves inSforthefunctionsI:S!

7 R, whereIis given by:(a)I(x) =Z10x0(t)dt(b)I(x) =Z10x(t)x0(t)dt(c)I(x) =Z10(x(t) +tx0(t)) Findcriticalcurves forthefunctionI(x) =Z21t3(x0(t))2dtwherex2C1[1;2] withx(1)= 5 andx(2)= Findcriticalcurves forthefunctionI(x) =Z21(x0(t))3t2dtwherex2C1[1;2] withx(1)= 1 andx(2)= Findcriticalcurves forthefunctionI(x) =Z10 2tx(t) (x0(t))2+ 3x0(t)(x(t))2 dtwherex2C1[0;1] withx(0)= 0 andx(1)= Findcriticalcurves forthefunctionI(x) =Z10 2(x(t))3+ 3t2x0(t) dtwherex2C1[0;1] withx(0)= 0 andx(1)= 1. Whatifx(0)= 0 andx(1)= 2? miningcompany mentionedin futuremoneyisdiscountedcontinuouslyat a constant rater, thenwe canassessthepresent valueof pro tsfromthisminingoperationby introducinga factorofe rtin theintegrandof ( ).Supposethat = 4, = 1,r= 1 andP= 2.

8 Finda criticalminingoperationx0such thatx0(0)= 0andx0(T) = variations:someclassicalproblemsOptimisa tionproblemsof thetype consideredin theprevioussectionwerestudiedin variousspecialcasesby many leadingmathematiciansin varioustechniques,andthesegave riseto thebranch of mathematicsknownas the`calculusof variations'.Thenamecomesfromthefactthato ftentheprocedureinvolved thecalculationof the`variation'in thefunctionIwhenitsargument (which was typicallya curve) was changed, thissection,we mentiontwo classicalproblems,andindicatehow thesecanbe variationsoriginatedfroma problemposedby theSwissmathematicianJohannBernoulli(166 7-1748).Herequiredtheformof thecurve joiningtwo xedpointsAandBin averticalplanesuch thata bodyslidingdownthecurve (undergravity andnofriction)travelsfromAtoBin nothave a trivialsolution;thestraight linefromAtoBis notthesolution(thisis alsointuitivelyclear,sinceif theslope is highat thebeginning,thebodypicks upa highvelocity andso itsplausiblethatthetravel timecouldbe reduced)andit canbe veri edexperimentallyby slidingbeadsdownwiresin variousshapes.

9 PSfragreplacementsA(0;0)B(x0; y0) posetheproblemin mathematicalterms,we introducecoordinatesas shownin ,so thatAis thepoint (0;0),andBcorrespondsto (x0; y0). Assumingthattheparticleis releasedfromrestatA, conservationof energygives12mv2 mgy= 0, wherewe have takenthezeropotentialenergylevel aty= 0, andwherevdenotesthespeedof thespeedisgivenbyv=dsdt=p2gy, wheresdenotesarclengthalongthecurve. ,we seethatanelement of arclength, sis given approximatelyby (( x)2+ ( y)2) variations:someclassicalproblems33 PSfragreplacements y x :Element of descent is given byT=Zcurvedsp2gy=1p2gZy00vuut1 + dxdy 2ydy:Ourproblemis to ndthepathfx(y); y2[0; y0]g, satisfyingx(0)= 0 andx(y0) =x0, whichminimizesT, thatis, to determinetheminimizerforthefunctionI:S!

10 R, whereI(x) =1p2gZy00 1 + (x0(y))2y 12dy;x2S;andS=fx2C1[0; y0]jx(0)= 0 andx(y0) =x0g. Here2F( ; ; ) =q1+ 2 is independent of , andso theEuler-Lagrangeequationbecomesddy x0(y)p1 + (x0(y))21py!= 0:Integratingwithrespecttoy, we obtainx0(y)p1 + (x0(y))21py=C;whereCis a constant. It canbe shownthatthegeneralsolutionof thisdi erentialequationisgiven byx( )=12C2( sin )+~C;y( )=12C2(1 cos );where~Cis anotherconstant. Theconstants arechosenso thatthecurve passesthroughthepoints(0;0) and(x0; y0).PSfragreplacements(0;0)(x0; y0) :Thecycloidthrough(0;0) and(x0; y0).Thiscurve is knownas acycloid, andin factit is thecurve described by a pointPin a circlethatrollswithoutslippingonthexaxis ,in such a way thatPpassesthrough(x0; y0); ,theFheredoesnotsatisfythedemandsmadein ,withsomeadditionalargument, thesolutiongiven herecanbe fullyjusti revolutionTheproblemof minimumsurfaceareaof revolutionis to ndamongallthecurves joiningtwogivenpoints (x0; y0) and(x1; y1), theonewhich generatesthesurfaceof thesurfaceof revolutiongeneratedby rotatingthecurveyaboutthexaxisisS(y) = 2 Zx1x0y(x)p1 + (y0(x))2dx:Sincetheintegranddoes notdependexplicitlyonx, theEuler-LagrangeequationisF(y(x); y0(x); x) y0(x)@F@ (y(x); y0(x); x) =C.