Module 4M12: Partial Differential Equations and Variational Methods IndexNotationandVariationalMethods

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1 Module 4M12: Partial Differential Equations and ariational Methods IndexNotationandariationalMethods IndexNotation(2h) ariational calculus vs differential calculus(5h) ExamplePaper(1h) Fullinformationat www2.eng.cam.ac.uk/ jl305/4m12/4m12.html induecourse. Thereisnodata sheet for this course. The lecturenotes are kindly provided by Dr. Garth Wells. 1

2 1 Indexnotation The summation convention Supposexandyarevectors,andAandBarematrices.Writeafewcommoncombinations in terms of their components: Dot product n=3 x y= x i y i i=1 Matrix vector multiplication [Ax] i = n A ij x j j=1 Matrix matrix multiplication [AB] ij = n A ik B kj k=1 Notice the curious thing: Everysumgoeswithanindexwhichisrepeatedtwice. Non-repeated indices are not summed. We can use a simplified notation by adopting the summation convention(due to Einstein), Donotwritethesummationsymbol. Arepeatedindex implies summation. (Anindexmaynotappearmorethantwiceononesideofan equality.) 1 Indexnotationisalsoknownas suffixnotation. 4M12 PAD/JL(jl305) 2

3 Using the summation convention, x y=x i y i [Ax] i =A ij x j [AB] ij =A ik B kj Summary Ifanindexoccurs once, itmustoccuronceineverytermoftheequation, andthe equationistrueforeachseparatevalueofthisindex. Ifanindexappearstwice itis summedoverallvalues. Itdoesnotmatterwhatthisiscalled: itisa dummyindex whosenamecanbechangedatwill.ifanindexappearsthreeormoretimesinanygiven terminanequation,itiswrong! Thismayseemaverypeculiartrick,withnoobviousbenefit.However,itwillturnout to be surprisingly powerful, and make many calculations involving vector identities and vector differential identities much simpler. 1.2 TheKroneckerdeltaδ ij Twoadditionalpiecesofnotationareneeded. Thefirstisawaytowritetheidentity matrix I, I= We define the Kronecker delta as 1 i=j δ ij =, 0 i j Weknowthat 4M12 PAD/JL(jl305) 3

4 Iy=y and δ ij y j =y i Inotherwords ifoneindexofδ ij issummed,theeffectistoswapthistotheother index. 1.3 Thepermutationsymbolǫ ijk Anothernecessaryingredientisawaytowritethecrossproductoftwovectorsinindex notation, e 1 e 2 e 3 ] x y= x 1 x 2 x 3 = [x 2 y 3 x 3 y 2, x 3 y 1 x 1 y 3, x 1 y 2 x 2 y 1 y 1 y 2 y 3 wheree i arethebasisforthevectors.wehaveassumedthate i aretheunitvectorsfor Cartesiancoordinates(youmayhaveseenthebasisvectorswrittenasi,jandk). To expressthecrossproductinindexnotation,wewillusethepermutationsymbolǫ ijk. Thepermutationsymbolǫ ijk isdefinedas 1 if(ijk)isanevenpermutationof(1,2,3) ǫ ijk = 1 if(ijk)isanoddpermutationof(1,2,3) 0 otherwise For example, 4M12 PAD/JL(jl305) 4

5 ǫ 112 =0 (anyrepeatedindex) ǫ 312 =1 (evennumberofinversionsofpairs) ǫ 132 = 1 (oddnumberofinversionsofpairs) The permutation symbol is also known as the alternating symbol or the Levi-Civita symbol. Using the permutation symbol, we can write the cross product of two vectors as: [x y] i =ǫ ijk x j y k Toprovethis,foreachisumoverjandk.Thepermutationsymbolpossessesanumber of symmetries, ǫ ijk =ǫ kij =ǫ jki (cyclicpermutation) = ǫ jik = ǫ ikj = ǫ kji (switchpairij,switchpairki,switchpairjk) 1.4 Theǫ ijk δ ij identity Thereisanimportantidentity( contractedepsilonidentity)relatingǫ ijk andδ ik : ǫ ijk ǫ klm }{{} sumoverk =δ il δ jm δ im δ jl Theproofissimple(butsomewhattedious!),justcheckeverycase. Wesumoverk, whichleavesfourfreeindicesandeachindexruns1 3,thereforethereare4 3 =81 cases. Here are two examples: 4M12 PAD/JL(jl305) 5

6 i=1,j=2,l=1,m=3 ǫ ijk ǫ klm =ǫ 121 ǫ 113 +ǫ 122 ǫ 213 +ǫ 123 ǫ 313 =0+0+0=0 δ il δ jm δ im δ jl =δ 11 δ 23 δ 13 δ 21 =0+0=0 i=1,j=2,l=1,m=2 ǫ ijk ǫ klm =ǫ 121 ǫ 112 +ǫ 122 ǫ 212 +ǫ 123 ǫ 312 =0+0+1=1 δ il δ jm δ im δ jl =δ 11 δ 22 δ 12 δ 21 =1+0=1 Example A vector identity [a (b c)] i =ǫ ijk a j (b c) k =ǫ ijk a j ǫ klm b l c m =(δ il δ jm δ im δ jl )a j b l c m =b i a m c m c i a j b j =[(a c)b (a b)c] i 4M12 PAD/JL(jl305) 6

7 1.5 A trick: symmetry and anti-symmetry Weexpectthea a=0. In index notation, [a a] i =ǫ ijk a j a k Theterma j a k issymmetricinj,k.thatis,a j a k =a k a j. Thetermǫ ijk isanti-symmetricinj,k.thatisǫ ijk = ǫ ikj. ǫ ijk a j a k = 1 2 (ǫ ijka j a k +ǫ ijk a j a k ) = 1 2 (ǫ ijka j a k +ǫ ikj a k a j ) = 1 2 (ǫ ijka j a k ǫ ijk a j a k ) =0, as expected. Symmetry and anti-symmetry are often useful to simplify expressions. The permutation symbol is anti-symmetric in any of its two indexes. Some other symmetric expressions include a i a j foranyvectora TheKroneckerdeltaδ ij ThematrixBwhen[B] ij =[A] ij +[A] ji The partial second derivative of a scalar function φ, 2 φ 4M12 PAD/JL(jl305) 7

8 1.6 ector derivatives The real power of index notation is revealed when we look at vector differential identities. Thevectorderivativesknownasthegradient,thedivergenceandthecurlcanallbe writtenintermsoftheoperator, = [ x 1, x 2, x 3 where[x 1,x 2,x 3 ]arethecomponentsofthepositionvectorx. Forthescalarfunctionφandavectorfunctionu,bothofwhichareafunctionsof positionx,wecanwrite ], Gradient Divergence gradφ: [ φ] i = φ [ φ =, φ, φ ] x 1 x 2 x 3 i divu: u= u i = u 1 x 1 + u 2 x 2 + u 3 x 3 Curl(recallthat[a b] i =ǫ ijk a j b k ) curlu: [ u] i =ǫ ijk u k The machinery we have developed for index notation can be used directly to manipulate quantities without having to be constantly thinking about the complicated physical meanings of div or curl, for example. 4M12 PAD/JL(jl305) 8

9 Example: Product rule for div and curl Divergence (φu)= (φu i ) = φ u i +φ u i (productrulefordifferentiation) =u φ+φ u Curl [ (φu)] i =ǫ ijk (φu k ) x ( j φ =ǫ ijk u k +φ u ) k =[ φ u+φ u] i Example: curl(gradφ) [ ( φ)] i =ǫ ijk =ǫ ijk 2 φ x k ( ) φ x k =0 (bysymmetryandanti-symmetryinj,k) 1.7 Derivatives of the position vector If the position vector x itself appears in a vector expression which is being manipulated inindexnotation,thenitisusefultonoticethat =δ ij, 4M12 PAD/JL(jl305) 9

10 which follows directly from the definition of a partial derivative: x 1 x 1 =1, x 1 x 2 =0, etc. So for example: x= =δ ii =3 [ x] i =ǫ ijk x k =ǫ ijk δ jk =0 (bysymmetry/anti-symmetryinj,k) 1.8 Magnitude of a vector Sometimes the magnitude(or modulus) of a vector appears in an expression which you wishtodifferentiateinsomeway.thetrickhereistoforceitintotheformofascalar product, even though this may make the expression look more complicated than when you started. Index notation allows scalar products to be handled easily, and the normal rules of calculus will allow the calculation to be done without difficulty. Example: Whatis x? [ x ] i = ( (x j x j ) 1/2) = 1 2 (x kx k ) 1/2 2x j Thereisadangerofjappearingfourtimesinthesameexpressionwhichisnotpermitted andwouldindicateanerror. Recallthatjwasadummyindex,summedoverallvalues. Soinoneofthetwoexpressions,jhasbeenreplacedbyk(oranyotherindexyoulike, apart from i or j). Continuing: 1 2 (x kx k ) 1/2 2x j =(x k x k ) 1/2 x j δ ij =(x k x k ) 1/2 x i 4M12 PAD/JL(jl305) 10

11 or in vector notation, x = x x, whichisjusttheunitvectorinthedirectionofx. 1.9 Integral theorems Index notation allows the divergence theorem and Stokes s theorem to be written in a way which makes them look familiar. Recall this result for ordinary integration and differentiationofafunctionf(x)ontheinterval(a,b): b a f x dx=f b a =f(b) f(a) Inotherwords,theintegralofthederivativeofafunctionisequaltothefunction,and canbeevaluatedfromthevalueofthefunctionontheboundariesoftheintegration interval. This general description will turn out to apply just as well to volume integrals (via the Divergence Theorem) and to surface integrals(via Stokes s theorem) Divergence theorem The divergence theorem states: fd = S f nds, wherefisavectorfunctionofposition,isavolumeenclosedbythesurfaces(s= ) andnistheunitoutwardnormalvectorons(itpointsoutwardsfrom). n s 4M12 PAD/JL(jl305) 11

12 Wecanalsowrite da=nds,orusingindexes da i =n i ds. For the divergence theorem, we can derive a number of other results(corollaries). For example, what can we say about the gradient of a scalar function, integrated over the volume? To examine this, consider the divergence theorem using index notation, f i d = f i n i ds. S Nowsupposethewehaveascalarfunctionφ. Wecancreateavectorfunctionf by multiplyingitbyavectora, f =φa, whereaisanyfixedvector(itdoesnotdependonx).substitutingthisintothedivergence theorem, φa i d = φa i n i ds S Sinceaisfixeditcanbetakenoutoftheintegral, ( ) φ a i d φn i ds =0 Thevectoraisarbitrary,aslongasitisfixed.Thereforethe above expression must hold for all fixed vectors. This implies S that and therefore ( ) φ d φn i ds =0 S φ d = φn i ds S 4M12 PAD/JL(jl305) 12

13 In vector notation, we have φd = S φnds. Amoregeneralresultwhichcanbeprovedusingthe fixedvector typeapproachisthat ( )d = ( )nds, where( ) is any index notation expression. We could, for example, insert S ǫ jik f k and deduce a curl theorem (check it). We now have a nice generalisation for relating volume integrals of vector derivatives to surface integrals. The integral of any regular function over a volume can be transformed into a surface integral. When in doubt, one can always return to index notation to check and develop the necessary relationships. Transforming surface integrals into volume integrals is the key to deriving many equations used in physics and mechanics which come from conservation laws. We can start byposingabalanceintermsofwhatishappeningontheboundary,andthentransform the expression of conservation into a differential equation Stokes theorem Stokes theorem is stated as: fda= f dl, S C wheresisasurface(possiblycurved), da=ndsisanareavectorelementandnis theunitnormalvectortothesurfaces,c= Sisthecurvethatboundsthesurface and dl=sdcisthevectorelementrunningalongthecurveandsistangentialtothe boundary. The direction of s is important! 4M12 PAD/JL(jl305) 13

14 C s n s ds As for the Divergence theorem, we can write Stokes theorem using indices, f k ǫ ijk n i ds= f i s i dc andwecanusethetrickwiththefixedvectortoget ( ) ǫ ijk n i ds= ( )s i dc S S The expression can be manipulated further using another expression which involves anotherǫsymbol,andthenusingtheǫ δidentitytopushtheǫtotheothersideof the equation. However, it is not possible to eliminate the ǫ entirely, therefore Stokes s theorem always appears to be more complicated that the divergence theorem. C C Integration by parts An important tool, particularly for variational methods, is integration by parts. Starting in one dimension, and then rearranging, b a d(uv) dx b a dx= b a v du b dx dx+ u dv a dx dx, v du b dx dx= u dv a dx dx+uv b a. Wewantnowtogeneralisethistotwoandthreedimensions.Thedothis,considerthe divergenceofatermφu, (φu)d = φ ud+ φ ud 4M12 PAD/JL(jl305) 14

15 where the product rule for differentiation has been applied. Applying the divergence theorem for the term on the left-hand side, φu nds= φ ud+ φ ud S The operation can also be performed using index notation, Product rule for differentiation (φu i ) φ d = u i d+ φ u i d Apply divergence theorem to LHS φ φu i n i ds= u i d+ S φ u i d 1.10 Directional derivative We will make frequent use of the directional derivative when working with variational problems. The directional derivative of a function f(u) is defined as Df(u)[v]= df(u+ǫv) dǫ Itmeans thederivativeoff withrespecttouinthedirectionofv. Incomputingthe directional derivative, we first compute the derivative with respect to ǫ, and then set ǫ=0.forexample ǫ=0 4M12 PAD/JL(jl305) 15

16 Forf=u u,whereuissomevectorfunction,thedirectional derivative is Df(u)[v]= d dǫ ((u+ǫv) (u+ǫv)) ǫ=0 =2((u+ǫv) v) ǫ=0 =2u v 1.11 Tensors Themanipulationofobjectsusingindiceshasbeenpresentedonthepremiseofafew basicrules.however,muchofwhathasbeenpresentedcarriesovertothericherfieldof tensors. Some tensor basic tensor operations will be considered, but only scratching the surfacesothatwecantacklemoreproblemsofpracticalrelevance 2. Wearealreadyfamiliarwiththeideaofvectors. However,youmightfindithardto give a formal definition of a vector, although you have though been exposed to aspects of the formal definition. For example, vectors are often expressed as a=a 1 i+a 2 j+a 3 k, or using modern notation as a=a i e i =a 1 e 1 +a 2 e 2 +a 3 e 3, The mathematical definition is in two parts. First, a vector a is quantity which, with respecttoaparticularcartesiancoordinatesystem,hasthreecomponents(a 1,a 2,a 3 ), orequivalentlya i, i=1,2,3.butthatisnotthewholestory.wealsoneedtoexpress thefactthatifwechoosetorotateourcoordinateaxes,thevector(the arrow which youmightdrawinadiagram)remainsthesame. Thismeansthatavectoralsohasan associatedbasis.intheaboveexample,thebasisise i.wemaychangethecomponents 2 Basicconsiderationoftensorsisnewin M12 PAD/JL(jl305) 16

17 andthebasisofavectorsuchthatthevectorremainsthesame(itstillpointsinthe same direction and still has the same length). A second-order(or second-rank) tensor is like a matrix but has a basis associated, and can be expressed using the notation A=A ij e i e j. Luckily, whenworkingwithafixedorthonormalbasis, wecanworkwitha ij, usually manipulatingitjustaswewouldamatrix. You have in fact considered second-order tensors in Part IA without being told(and most likely without the lecturer realising). Recall that matrix can be rotated into a new coordinate system via whereitturnsoutthat A ij=r ik A kl R jl R ij =e i e j. Common second-order tensors include stress and strain. Derivatives and second order tensors Thetwooperationsthatwewishtoconsiderarethegradientofavectorandthedivergence of a second-order tensor. The gradient of a vector produces a second-order tensor: a= a = a i e i e j. Forourpurposes,itsufficestojustrecall[ a] ij = a i /. Justtomakethingsmore confusing, in fluid mechanics, the gradient is often defined as the transpose of what is defined here. Another important operation is the divergence of a second-order tensor. It leads to a vector: A= A ij e i Similartothegradient,itsufficesforourpurposestorecall[ A] i = A ij /.Justas foravector,thedivergencetheoremcanbeappliedtoasecondordertensor: Ad = AndS 4M12 PAD/JL(jl305) 17

18 Tensors, in combination balance laws, can be used to systematically derive many differential equations of importance in physics and engineering. Example: Equilibriumequationofabodywithmassdensityρ:Iftheforceperunitarea acting on a surface is denoted by the vector t(the traction vector), equilibrium(forces sum to zero) requires that tds+ ρgd =0. Thestress(asecondordertensor)atapointsatisfiesbydefinitiont=σn,wheren istheoutwardunitnormalvectortoasurface. Usingindices,t i =σ ij n j. Insertingthe expression for σ, σnds+ ρgd =0. Applying the divergence theorem, σd+ ρgd =0. Sinceequilibriummustapplytoanysub-bodyof,theequationcanbelocalised, σ+ρg=0. Theisbalanceoflinearmomentum.Itcanbeshownthatangularmomentumbalanceis satisfiedifσ=σ T. 4M12 PAD/JL(jl305) 18