ERRATAPage19: In Problemx-4 xshould read x-4 -x.Page 69: Replace dsby dsoin parenthetical remark.Page102: The correct reference to Ohm's Law is Eqn.(5.4.20).Page116: In Problem 6.1.4, uv -4 2uv.Page190: In the fifthline from bottom, W-4 V.Page 276: Problem 9.6.6. Ignore last sentence and answer at the back.Page 286: Problem 9.7.8: e1xl-4 e-1xl.Page 311: In Problem1part (i),42D -42D.Page 331: In line "corner" replaces "edge".The following corrections apply tothe answersgivenat the endProblem 3.1.5: d=4/.;5.Problem 3.2.8: V= 1287r.Problem 4.2.4: In part (ii)r= Problem 4.3.1: Converges forIxl>1.Problem 5.2.4: (ii)-4 (iv)Izi = '(},= I-f6 - i V3!V2.Problem 5.3.2: Argument of Zl is 2tan-1 Problem 6.2.3: cosy -4 coshy.Problem 6.2.14: (i) i repeated twice.P bl 9 2 6 - 9+i3+9i - 8ro em .. . a - V2' JJ = - V2' "Y - Problem 9.6.4: The saddle point is atx = 0, y = 1.Problem 9.7.3:-4 i.Problem 9.7.13: Replace8 by 32.Problem 9.7.14: Replaceby 2:.Problem 9.9.2: Part (b): The degenerate eigenvalueis Problem10.4.9: Part (ii)y=+x64P bl 10 5 8 ( ) - 42n' 2mrx [2mrYJro em .. , ux, y -;;:L..odd4n2-1 sm -y;- exp - L .BasicTrainingin MathematicsR. Shankar0-306-45035-6(Hardbound)0-306-45036-4(Paperback)PlenumPress, NewYork, 1995Basic Training in MathematicsA Fitness Program for Science StudentsBasic Training in MathematicsA Fitness Program for Science StudentsR.SHANKARYaleUniversityNew Haven, ConnecticutPLENUM PRESS. NEW YORK AND LONDONLibrary of Congress Cataloging-in-PublicationDataOn fileISBN 0-306-45035-6 (Hardbound)ISBN0-306-45036-4 (Paperback)1995Plenum Press, New YorkA Division of Plenum Publishing Corporation233 Spring Street, New York, N. Y. 1001310987654321All rights reservedNo part of this book may be reproduced, stored in a retrievalsystem, or transmitted in any formor by any means, electronic, mechanical, photocopying, microfilming, recording, or otherwise,without writtenpermission from thePublisherPrinted in theUnited States of AmericaForUMAPREFACEThis book is based on a course I designed a fewyears ago andhave been teachingatYale ever since. Itis a required course for physics majors, and students wishingto skip it have to convince the Director of Undergraduate Studies of their familiaritywithits contents. Althoughit is naturally slanted towardphysics, I cansee a largepart of it servingtheneedsof anyoneinthephysical sciencessince, forthemostpart, only very basicphysicsideasfromNewtonianmechanicsareemployed. Theraisond 'etrefor thisbook andthe course areidenticaland asfollows.While teaching many ofthe core undergraduate courses, I frequently had to digressto clear up some elementary mathematical topicthat bothered some part of the class.For instance, I recall the time I was trying to establish how ubiquitous the hannonicoscillator was by showing that the Taylor series of any potential energy functionat astationary point was given to leading order by a quadratic function of the coordinate.At this point somestudents wanted toknow what a Taylor series was. A digressiontodiscussTaylorseriesfollowed. At thenext stage, whenItriedtoshowthat ifthepotential involvedmany coordinates, the quadraticapproximationtoitcould bedecoupled into independent oscillators by a change of coordinates, I was forced to usesomefonn of matrixnotationand elementary matrixideas, andthatbotheredsomeother set of students. Once again we digressed. Now, I was not averse to the idea thatin teaching physics, one would also have to teach some new mathematics. For example,the course on electricity and magnetism is a wonderful context in which to learn aboutLegendre polynomials. On the other hand, it is not the place to learn for the first timewhat a complex exponential likeeimmeans. Likewise, in teaching special relativityone does not want to introduce sinh and cosh, one wants to use them and to admirehow naturally they serve our purpose. To explain what these functions are at this pointis like explaining a pun. In other words, some of the mathematicaldigressions weresimply not desirable and quite frustrating for the teacher and student alike.Now, this problem was, of course, alleviated as the students progressed throughthe system, since they were taking first-rate courses in the mathematics departmentin the meantime and could soon tell you a surprising thing or two about the edge-of-the-wedge theorem. But one wished thestudentswouldhave a grasp of the basicsof each essential topic at some rudimentary level from the outset, so that instructorscould get on with their job with the least amount of digressions. From the student'spoint of view, thisallowedmoretimetothinkaboutthesubjectproper andmorefreedomtotake advancedcourses.Whenthis issue was raisedbefore the faculty, mysentiments were sharedbymany. It was thereforedecidedthat I woulddesignandteachacoursethatwoulddeal withtopicsindifferential calculus of oneor morevariables(includingVIIviii Prefacetrigonometric, hyperbolic, logarithmic, and exponential functions), integral calculusof oneand many variables, power series, complex numbers and function of a com-plexvariable, vector calculus, matrices, linear algebra, andfinallytheelementsofdifferential equations.Incontrast tothemathematical methods coursestudents usuallytake inthesenior year,thisone would dealwith each topicinits simplest form. For example,matriceswouldbetwo-by-two, unlessa bigger onewasabsolutely necessary(say,to explain degeneracy). On the other hand, the treatment of this simple case wouldbethoroughandnot superficial. Thecoursewouldlast onesemesterandbeself-contained. It wasmeantforstudentsusually inthesophomoreyear, thoughit hasbeen taken by freshmen, upper-class students, and students from other departments.Thisbookisthat course.Each department has to decide if it wants to devote a course in thesophomoreyear tothistopic. Myownview(basedonourexperienceat Yale) isthat sucha preventive approach, whichcosts one course for just one semester, is worthhoursof curinglateron. Hourforhour, Icanthinkof noothercoursethat willyielda higher payofffor the beginningundergraduateembarkedona career inthephysical sciences, sincemathematics isthechosenlanguageof nature, whichpervadesall quantitative knowledge. The differencebetween strengthor weaknessin mathematics will subsequently translateinto the differencebetween success andfailureinthesciences.Asismypractice, I directly addressthestudent, anticipatingtheusual ques-tions, imagining heor sheisinfront of me. Thusthebookisideal forself-study.For this reason, even a department that does not have, as yet, a course at this level,candirect studentstothisbookbeforeorduringtheirsophomoreyear. Theycanturn to it whenever they runinto troublewith themathematical methods employedin various courses.AcknowledgmentsI ampleasedtothankall thestudentswhotookPhysics30 Iafor theirinput andIlya Gruzberg and SenthilTodari for comments on themanuscript.As always,it has been a pleasure to work with the publishing team at Plenum.Myspecial thankstoSenior Editor AmeliaMcNamara, her assistant KenHowell,andSenior Production Editor JosephHertzlinger.I thankMeera andAJShankar for theirhelpwith theindex.But mygreatest debt istomywifeUma. OvertheyearsmychildrenandIhave been able to flourish, thanks toher nurturing efforts, rendered at great cost toherself. This book is yet another example of what she has made possible through hertirelesscontributionsasthefamilymuse. It isdedicated toher andwill hopefullyserveasone tangiblerecord of her countless efforts.R. ShankarYaleUniversityNewHaven, ConnecticutNOTE TO THE INSTRUCTORIf youshouldfeel, as Ido, that it isnot possibletocoverall thematerial inthebookinonesemester, here aresome recommendations. Tobeginwith, youcanskipanytopicinfineprint. Ihavetriedtoensurethat doingsowill notdisrupt continuity. Thefineprint isforstudentswhoneedtobechallenged, or for astudent who, longafter thecourse, beginstowonder about somesubtlety, or runsintosome of thismaterial ina latercourse, andreturnstothebookfor clarification. At that stage, thestudentwill have the timeandinclination toreadthe fineprint. Theonlychapterthat onecanskipwithout anyseriousimpactonthesub-sequent ones, is that onvector calculus. It will beapityif this route istaken; but it isbettertoleaveout atopicentirelyratherthanrushthrougheverything. Moremoderatesolutions, likeomittingsomesections, arealsopossible. Nothing teachesthestudent asmuchasproblem solving. I havegivenalotof problems and wish I could have given more. When I say more problems, Ido not mean more that are isomorphic to the ones given, except for a changeof parameters, but genuinely new ones. As for problems that are isomorphic,youcangenerateanynumber(sayforatest) andhavethemcheckedbyaprogram like Mathematica. Whilethiscourseisforallphysicalscientists, itisgenerallyslanted towardphysics. Ontheotherhand, mostof thephysicsideasarefromelementaryNewtonianmechanicsandmust befamiliartoanyonewhohastakena cal-culuscourse. You may stillhave to customize some of the examples to yourspecialty.I welcome your feedback.ixNOTE TO THESTUDENTIn American parlance the expression "basic training" refers to the instruction givento recruits in the armed forces. Its purpose is to ensure that the trainees emerge withthefitnessthat willbe expected of them when they embark on their main mission.In this sense the course provides basic training to one like yourself, wishing toembarkon a program of study in thephysicalsciences. It hasbeen my experiencethat incomingstudentshaveawidespectrum of preparationandmost haveareasthat needtobestrengthened. If thisisnotdoneat theoutset, it isfoundthattheresultsarepainful for theinstructorandstudent alike. Conversely, if youcoverthebasicmaterial inthisbookyoucanlookforwardtoasmoothentryintoanycourseinthephysicalsciences. Of course, youwill learnmore mathematics whilepursuing your major and through courses tailored to your specialization, as well asincoursesofferedbythemathematicsdepartment. This c o u r s ~ isnot asubstituteforany of that.But this course is unlikeaboot campinthat youwill not beaskedtodothingswithout question; noinstructorwill barkat youto"hit that deskandgivemefiftyderivativesof eX." Youareencouragedto question everything, and asfaraspossible everything youdo will begivenalogicalexplanation and motivation.The course will be like a boot camp in that youwill be expected to work hardandstruggle often, andwill emerge proud of your mathematical fitness.I have done mybest tosimplifythis subject as much as possible(but nofurther), aswill your instructor. But finally it is up to you to wrestle with the ideasandstrugglefortotal mastery of thesubject. Otherscannot dothestrugglingforyou, any more than they can teach you to swim if you won't enter the water. Hereisthemost important rule: doasmanyproblemsasyoucan! Readthematerialbeforeyoustartontheproblems, instead of starting onthe problems and jumpingbacktothe texttopickupwhatever youneed tosolve them. Thisleadsto patchyunderstanding andpartialknowledge. Start withthe easy problems andwork yourwayup. Thismayseemtoslowyoudownat first, but youwill comeoutahead.Lookat other books if youneed to do more problems. One I particularly admire isMathematicalMethodsinthe Physical Sciences, by M. Boas, publishedby WileyandSons, 1983. It ismoreadvancedthanthisone, butisvery clearly writtenandhaslots of problems.Behonest withyourself andconfront your weaknesses beforeothers do, astheyinvariablywill. Stayontopof thecoursefromdayone: inmathematics,morethananythingelse, yourearlyweaknesseswill returntohaunt youlaterinthecourse. Likewise, anyweaknessinmathematical preparationwill troubleyouduring the rest of you career. Conversely, the mental muscles you develop here willstand youingood stead.XICONTENTS1 DIFFERENTIALCALCULUS OF ONE VARIABLE1.1. Introduction . . . . . . . . . .1.2. Differential Calculus .1.3. Exponential andLogFunctions1.4. TrigonometricFunctions . . . .1.5. PlottingFunctions .1.6. MiscellaneousProblemsonDifferentialCalculus1.7. Differentials1.8. Summary .2 INTEGRAL CALCULUS2.1. Basics of Integration .2.2. Some Tricks of theTrade2.3. Summary .3 CALCULUS OF MANY VARIABLES3.1. Differential Calculus of Many Variables.3.1.1. Lagrangemultipliers .....3.2. Integral Calculus of Many Variables.3.2.1. Solid angle.3.3. Summary ...4 INFINITESERIES4.1. Introduction4.2. TestsforConvergence4.3. PowerSeriesinx4.4. Summary. . . . . .5 COMPLEX NUMBERS5.1. Introduction . . . .5.2. Complex NumbersinCartesianForm5.3. Polar Form of ComplexNumbers5.4. AnApplication.5.5. Summary .6 FUNCTIONS OF A COMPLEX VARIABLE.6.1. AnalyticFunctions. . . . . . ....xiii111519232529303333444951515561687275757780878989909498104107107XIV6.1.1. Singularities of analyticfunctions . .6.1.2. Derivatives of analyticfunctions ...6.2. AnalyticFunctions Defined by Power Series6.3. Calculus of AnalyticFunctions ...6.4. TheResidue Theorem . . . . . . . .6.5. Taylor Series for AnalyticFunctions6.6. Summary. . . . . .7 VECTOR CALCULUS.7.1. Review of Vectors Analysis7.2. Time Derivatives of Vectors7.3. Scalar and Vector Fields ..7.4. Line andSurface Integrals .7.5. Scalar Field and the Gradient7.5.1. Gradient in other coordinates7.6. Curl of a Vector Field .7.6.1. Curlin other coordinate systems .7.7. TheDivergence of a Vector Field ....7.7.1. Divergence in other coordinate systems7.8. DifferentialOperators.....7.9. Summary of IntegralTheorems . . . .7.10. Higher Derivatives . . . . . . . . . . .7.10.1. Digression on vector identities.7.11. Applications fromElectrodynamics7.12. Summary .8 MATRICES ANDDETERMINANTS.8.1. Introduction . .8.2. Matrix Inverses .8.3. Determinants. . . . . . . . . . . .8.4. Transformations on Matrices andSpecialMatrices8.4.1. Action on row vectors8.5. Summary. . . . . . . . .9 LINEAR VECTOR SPACES.9.1. Linear Vector Spaces: Basics9.2. Inner ProductSpaces. .9.3. Linear Operators .....9.4. SomeAdvanced Topics .9.5. TheEigenvalueProblem.9.5.1. Degeneracy ...9.6. Applications of Eigenvalue Theory9.6.1. Normal modes of vibration: two coupled massesContents111114116126132139144149149156158159167171172181182186187188189190193202205205211214219223227229229237247253255263266267Contentsxv9.6.2. Quadraticfonns .9.7. FunctionSpaces .9.7.1. Generation of orthononnalbases9.7.2. Fourier integrals .9.7.3. Nonnal modes: vibrating string9.8. Some Tenninology ....9.9. Tensors: AnIntroduction9.10. Summary. . . . . . . . .274277279287289 294294 30010DIFFERENTIAL EQUATIONS10.1. Introduction .10.2. ODEswithConstant Coefficients10.3. ODEs withVariable Coefficients: First Order.10.4.ODEswithVariableCoefficients: Second Order andHomogeneous10.4.1. Frobeniusmethodforsingular coefficients .. . . .10.5.Partial Differential Equations . . . . . . . . . . . . . . . .10.5.1. The wave equationinone andtwospace dimensions10.5.2. Theheat equationinoneandtwo dimensions.10.6.Green'sFunctionMethod10.7. Summary.ANSWERS.305 305307315318324329329336344347351INDEX .. . 3591DIFFERENTIAL CALCULUSOF ONEVARIABLE1.1. IntroductionStudents taking the course on mathematicalmethods generally protested vigorouslywhentoldthat weweregoingtostart withareviewof calculus, onthegroundsthatthey knewit all. Now, thatproved to be the casefor some, whileformany itwassomewhat different: either they once knew it,or thought they once knewit, oractuallyknewsomeonewhodid, andsoon. Sinceeverythinghingesoncalculusrather heavily, we will play it safe and review it in the first three chapters. However,tokeeptheinterestlevel up, thereviewwill be brief andonly subtleties related todifferential and integral calculus will be discussed at any length. The main purposeof the reviewis to go through results you probably know and ask where they comefrom and how they are interrelated and also to let you know where you really stand.If youfind anyportionwhereyouseemtobeweak, youmust findabookoncalculus, workout manymoreproblemsthanareassignedhere, andremedythedefectatonce.1.2. Differential CalculusLetusassumeyouknowwhat a function f (x) is: a machine that takesina valueof x andspitsoutavaluef whichdependsonx. For examplef(x) = x 2+ 5isafunction. Youput inx =3andit givesbackthevalue f =32+ 5 =14. Werefer to f asthe dependent variable andxastheindependent variable. Note thatinsomeotherproblemx canbethelocationof a particleattimet inwhichcasex (t) isthe dependent variable and t is theindependent variable.We will assume thefunctioniscontinuous: thismeans that you candrawthegraph of f without taking the pen off the paper. More formally:2 Chapter IDefinition1.1. A functionI (x) iscontinuousat x= a ifforanyE>O. howeversmall. we can find absuchthat II (x)- I (a) I (2.1.38)(2.1.39)and so on.After thisset of rules, we tumto a fewtricks of the trade.Problem2.1.2. Findtheintegral of In x usingintegrationby partsbyrewritingIn x= 1 . In x. Makethe right choice forwhich factoristobeFand which istobeg.Problem2.1.3. Consider the functionF(n)= 100xne-xdxwherenisanon-negativeinteger. Showusing integrationby partsthat F (n) =nF(n- 1) andthat F(n) = nL The gammafunction is definedbyr(n)F(n - 1). What isO! as defined by thisintegral?Problem2.1.4. Consider Jo1rcos2xdx. Givearguments for whythis must equalJo1rsin2xdx. By adding thetwointegrals and averaging, and using a well-knowntrigonometricidentity, show that11rcos2xdx= 11rsin2xdx= 7T /2. (2.1.40)Thisisa veryuseful result.(2.2.1)(2.2.2)44 Chapter 22.2. Some Tricks of the TradeThetricksoneusestoevaluateintegralsaresonumerousthatwecannot hopetocover them all here. There arehowever two fundamental ploysthatarefrequentlyemployed. These are Substitution or change of variable. Differentiating withrespect to a parameter.Asanexample of theformer considerlX2jdxXllX2x2dxXl (x 3+4)2'Let ussay we only knowtointegrate powers of x. Theintegrandj heredoesnothave thatform. We willnow bring it to that form by a change of variable. Let(2.2.3)Our original goal was to plotj as a function of x, chop the region betweenx 1andX2into segments of width dx,evaluate the productsj(x)dxover the segments andadd them all up in the appropriate limit. Our hopeis that the job will be simpler interms of u. Supposeweexpressj interms of uandplot j(u). Alongtheu-axisthepoints Ul =x ~ andU2=x ~ definethelimits. Everyvaluethat j took asafunction ofxit will now take as afunction ofu at the corresponding point u= x3.We donot however want the areaunder j(u)betweenUl andU2., i.e., we donotwant Jj(u)du. We want Li j(ui)dx, which is the old area we began with. Thusgiventwonearbypointsseparatedbydu, wewant toformtheproduct j(u)dx,wheredx isthecorresponding separation along thex-axis. It isclear thatsince xisa functionof udxdx=-duo (2.2.4)duThusthe integrandforthe u-integration isj (u )dx/ du, where the factordx / duistobe expressed interms of u. Going back to our problemududxdxduj(x)dx1 13x2= 3u2/ 3du duj(x(u)). 3u2/ 3 = 3(u +4)2(2.2.5)(2.2.6)(2.2.7)(2.2.8)Integral CalculuslX2f(x)dxXl45(2.2.9)Switching oncemoretov= u +4 we finallyobtain(2.2.10)It isnow trivial tointegrate1/v2betweenthegivenlimits.Let us recall the mainpoint intheabove manipulations. Givenadifficultintegrand, wehopethat goingtoanewvariable, wecanendupwiththesimplerintegrand. Thenewintegrandishowevernot justtheoldoneexpressedintermsof thenewvariable, but that timesthe JacobianThus(X) dxJ - =-.u du(2.2.11)(2.2.12)lX2f(x)dx= rU(X2)f(x(u))J (:.)duXl }u(xd uwiththelimitsexpressedinterms of thecorresponding variables.Our mission is accomplished onlyifthe newintegrandis simple. In ourexample, the denominator ofthe original integranddidsimplifyupon goingtov=x3+ 4, but the numerator didnot. But fortunatelyit got canceledbytheJacobian. Conversely, the substitutionofv wouldnot have beenveryeffectivewithout theX 2inthenumerator.Problem2.2.1. Evaluate JX2 h byswitchingto e definedbyx = a sin e.Xl a-xAssume 0 0 dxJo X 2+ 11f2Chapter 2(2.2.27)(2.2.28)and wanted to evaluate the integral with (x 2+1)2in the denominator. We have noparameter todifferentiate. Thenwemustintroduceone! Firstweviewthegivenintegralasroo dxJo x2+a2evaluatedat a=1. Next wecanchangetou =ax anddeduceEqn. (2.2.26).Thereafter weproceedasbeforeintakingparametricderivatives. At theendweset a= 1. The notion of embedding a zero parameter problem into a parametrizedfamilyisvery powerfulandhasmany uses.Problem2.2.10. Considerrlt - 11 = Jo h;t.Think of thet int - 1 asthea =1limit of ta. Let1(a) bethecorrespondingintegral. Take the a derivative ofboth sides (using ta= eaIn t) and evaluate dl/ daby evaluating the corresponding integral by inspection. Given dl / daobtain1byperformingtheindefiniteintegral of bothsideswithrespecttoa. Determinetheconstant of integrationusing yourknowledgeof 1(0). Showthat the originalintegral equals In 2.Problem2.2.11. Givenevaluate1000xe-axsin kxdxand 1000xe-axcos kxdx.Integral Calculus2.3. SummaryHere are the key ideastoremember at all times: TheequationlX2j(x )dx= F(x)- F(xo)Xl49canbeviewedintwoways. First. it gives thechangeinafunction F (x)between Xo andx giventhat its rateofchange is j(x), interms of thedefiniteintegralover j between the saidlimits,assuming the integralcan befoundsomehow, sayastheareaunderthegraph. Conversely, if weknowanyintegral F of j,thisequation expressesthedefiniteintegral interms ofthelatter. The members of the family of integralsdiffer by constantswhichdrop outin the difference computed in the definiteintegral.lX2j(x)dx = _lxIj(x)dxXl X2lX2F(x)g(x)dx= F G I ~ ~ _lX2j(x)G(x)dxXl XllX2lU(X2) dx(u)j(x)dx= j(x(u))--du,Xl U(XI) duwhere d ~ ~ u ) = J( ~ ) is the Jacobian.IflX2F(a) = j(x, a)dxXlthen,dF(a) = lX2aj(x, a) dx.da Xl aa3CALCULUSOF MANY VARIABLESWenowtumtodifferential andintegral calculusof manyvariables, inthatorder.Most of our discussion will be limited to the case of two variables since it illustratesmost of thenewideas. We willhowever occasionally discussthreevariablessincewe live andsolve problemsinthreedimensions.3.1. Differential Calculus of Many VariablesLetusbeginwithafunctionf (x, y) of twovariables. Forexample(x, y) couldlabel points intheplaneandf couldbesomefunctionsuchasthetemperatureT (x, y) or h(x, y), theelevationabovesealevel.The partial derivativewithrespect to, say x, is definedasofax= f - I' f (x + ~ x , y) - f (x, y)- x - 1m .Ax---+O ~ x(3.1.1 )Thus, tofindthepartial derivativealongx, weimaginemovinginfinitesimallyinjust the x-direction and measuring the rate of change. Operationally this means thatwhile taking the x-partial derivative, we treatyasa constant since it is indeed heldconstant. Asimilarset of relationsexist for f y, thederivativeinthey-direction.Consider forexample thefollowingfunctionanditspartialderivatives:f(x,y)fxf yx 3+2xy23x2+ 2y 24xy.(3.1.2)(3.1.3)(3.1.4)Conversely thex-partialderivativedeterminesthechangeinthex-direction:off(xo +~ x , YO) = f(xo,YO) + -aI x o , y o ~ X + ...x(3.1.5)where(xo, YO) is somepoint intheplaneandthepartial derivativeisevaluatedthere. Theellipsisrefertohigherorderterms. What if wewantedtomovetoa5152point displacedin both thex andydirections? We do thisin twostages:Chapter 3f(xo +f::t.x,yO +f::t.y)- f(xo,yo)f(xo + f::t.x,Yo + f::t.y) - f(xo + f::t.x, Yo) + f(xo + f::t.x, Yo) - f(xo, Yo)ofl ofl- f::t.y+- f::t.x8y xo+Ax,yo ox Xo,Yo8fI f::t.y +of I f::t.x + ...oy oxXo,Yo X o , ~where ingoingtothelast equationwehaveignoredthechange inthe y-partialderivativeas we move from(xo, Yo) to(xo + f::t.x, yo). Thisis because we wish towork to just first order in the infinitesimals. (That is to say, the y-partialderivative,itself afunctionof twovariables, assumedtobedifferentiable, will changeby anamount proportionaltof::t.x. Due to a prefactorf::t.y multiplying thisderivative,theneglected termis of orderf::t.x f::t.y.)What about aTaylorserieswithmoretermsasinthecaseof onevariable?Forthisweneedtolookat higherderivatives. Therearefourof themat secondorder andarelistedbelow, theexpressionsinparenthesesbeingthevaluesfortheexample f = .r3+ 2xy2:fxx02foof(= 6x) (3.1.6) --ox2oxoxfyy02foof(= 4x) (3.1.7) ---oy2oyoyfxy02foof(= 4y) (3.1.8) ----oxoy oxoyfyx82f oof(= 4y) (3.1.9) ----oyox oy oxThelast two mixed derivativesarealwaysequal toeach other foranynonsingularfunction.Problem 3.1.1. Establishtheequality of thetwomixed derivatives. Gobacktotheir definitions(givenabove) and show that bothare givenby the limitsfxy fyx= lim f(x+Ax,y+Ay)- f(x+Ax,y)- f(x,y+Ay)+f(x,y)Ax,Ay---+O AXAy .TheTaylor seriesnow goes asfollows:of off(xo + f::t.x,Yo + f::t.y) - f(xo, Yo) = -f::t.x + -f::t.yox oy1[ 2 2 ]+2 fxx(f::t.x) + fyy(f::t.y) + fxyf::t.xf::t.y + fyxf::t.yf::t.x + ...Calculus of Many Variables 53whereit isunderstood that all derivativesare takenat (xo,YO) and where we havenotusedthefreedomtolumpthelast twotermsusingtheequalityof themixedderivatives.Problem 3.1.2. Findthepartial derivatives uptosecondorder forthe functionf(x,y) = x 3+x 2y 5 +y4. Observe theequality of the mixed derivatives.Let us now consider the question of maxima and minima for functions of manyvariables. Let us recall howit goesforthecaseof onevariable. Weplot f onone axisandx alongtheother. Wefollowtheundulationsof f aswevaryx. Astationary point is one where a changeinx doesnot produce a changef).f to firstorder inf).x. Thus the first derivative, which is the ratio of the former with respectto thelatter, vanishesat thispoint. This point couldbea maximum, minimum, ora point of inflexion. Todecidebetweenthesealternativeswecomputethesecondderivative. Thethreecaseslistedabovecorrespondtothesecondderivativebeingnegative, positive, andzero, respectively. Wearegenerally moreinterestedinthecase of a maximum or minimum.Considernowafunctionof twovariables, f (x, y). Let us plot aboveeachpointinthex - y planethevalueof f measuredinthezdirection. Consider forexamplethefunction f = JR2- x 2- y2, withlxi,Iyl < R andthepositivebranch of theroot. Itisclear that theprofile of.f is just theupper hemisphere ofthe spherex 2+ y2 + .f2= R 2. It isalso clear that thenorthpole, situated ontopof (0, 0) isa maximum. How doesthiscome out of a calculation?Once again we look for stationary points, defined to be points where a changeinx ory producesno changeinf to firstorder ineitherf).x orf).y. Thischangeis given by(3.1.10)Eventhoughit is thesumwhichhas tovanishat thestationarypoint. wecanarguethat eachpiecemust separatelyvanish. Thisisbecausethefirst variationhas tovanishfor anychoiceof the f).x or f).y. If wechoosejust oneof themto benonzero, thecorrespondingpartial derivativemust v a n i s h ~ thereisnoroomfor cancellationsbetweenthe twoterms. Consequently both thepartialderivativesmust vanishat a stationary point:f x= f y = 0 atstationary point.In our example f = JR2- x 2- y2, wefindthe partialderivatives(3.1.11)(3.1.12)(3.1.13)indeed vanish at the origin. To further diagnose this as a maximum, we must clearlylook at thesecondderivatives. A simple calculation showsthat at the origin1fxx= f yy= - R54with themixedderivatives vanishing. Going back tothe Taylor seriesChapter 3(3.1.14)toquadratic order. Notice that we have droppedthe first derivatives sinceweareat astationarypoint. It isclearthatnomatterwhichdirectionwemovein,AfD.) Wehadnotroubleheresince Afhad a unique sign which could be seen by inspection. This in tum was because themixedderivativevanished, andtheinfinitesimals(Ax)2and(Lly)2werepositivedefinite: if theybothmultipliedpositive(negative) secondderivatives, wehadaminimum (maximum).Considernext aslightlymoregeneral stationarypoint for somef suchthatthemixedderivativesarestillzero, but fxx and f yyhaveoppositesigns, say withf xx>D. Thisis a case where moving alongxcauses f to increase, while movingalongycausesit todecrease. Thisiscalledasaddle point. Thisisbecausetheshape of the functionresemblesthat of a saddle in thevicinity of itscenter: aswemovealong thehorsethefunctionrises, whileaswemovetransversetothehorseit falls.Consider finally a function with all three second order partial derivativesnonzero. It is not possible now to say what is happening byjust looking at the signso/the second derivatives, because Af has cross terms AxLlyin its expansionandthis factor is ofindefinite sign. Thus there is no simple relation between the sign ofLlfand the signs of the partial derivatives. We shall see how to solve this problemwhenwe learnaboutmatrices. For the present let usnote the following. Considerthe case where near the origin(3.1.15)where Aand Bare both positive. We clearly are at a minimum since f rises for alldisplacements. Let us rotate our axes by 450 and usenew coordinatesx = X!;!.In terms of these(3.1.16)Nowweknowthat despite the cross terms, thisis just a minimum indisguise. Toseethis, wesimplyhavetogobacktotheoldcoordinatesinwhichcross termsvanish. But whatabout someother problem with crossterms? Will therebenicecoordinates in which cross terms disappear so that the nature of the stationary pointcan be read off by inspection? Matrix theory answers this in the affirmative as youwill learnlaterinChapter 9.Calculus of Many Variables 553.1.1. Lagrange multipliersThefinal topic inthis sectionpertains to locatingthe maxima or minima ofafunctionwithina submanifold determined by a constraint. For exampleletf(x, y) = x 2- xy + y2 (3.1.17)be the temperature distribution in the plane. Let some bug be restricted tolive in acircle of radius5 givenby the constraintequation9(X,y) = x 2+ y2- 25 =O. (3.1.18)Whatisthehottestpointinthisbug'sworld? Sincethispointneednotbearealmaximum of the function in the unrestricted plane, it is not necessarily characterizedbythe vanishingof all its first derivatives. Thesolutiontothis problemlies inunderstanding this pointinsomedetail.As the bug walks around the circle measuring the temperature, it willfind thatas it approaches a maximum (minimum) on the circle, the temperature willfirstgoup(down) andthenstart goingdown(up). At eitherextremum, thevariationtofirstorder will indeedvanish:df = fxdx +fydy= O. (3.1.19)We cannot however conclude that each partial derivativevanishes since dxanddyare not independent, they are chosen so that the displacement isalong the (tangentto the)circle. Inother words,they are chosen such that9remains zero beforeandafter the displacement:d9dy9xdx + 9ydy= 09x--dx.9ywhichimplies (3.1.20)(3.1.21)If we feedthisinto Eqn. (3.1.19) we findthe following condition for the stationarypoint:(3.1.22) at stationary point.fx _ f y9x 9yWe can now find the two coordinates of the stationary point given the above and theconstraintequation9 = O. Turning tothegiventemperaturedistribution, equation(3.1.22)becomes2x - y2x2y- x2y(3.1.23)which impliesx=y. (3.1.24)56 Chapter 3This, alongwiththeconstraint x 2+y2 = 25, tells us therearefour stationarypoints:(x.y) = (5/v'2,5/v'2) (3.1.25)Of coursewemust domoreworktoseewhich of thesestationary pointsisreallya maximum as comparedtoa minimum or saddle point.Problem 3.1.3. (Very important). Ifyou want to really check Eqn. (3.1.25), sim-ply eliminatey in favor of xin f using the constraint. Find the stationary pointsof this functionofjust onevariable. Alternativelywritex = 5 cos e, y = 5 sin eand vary withrespect toe.Thereisan equivalent way to writeEqn. (3.1.22)forthestationary point:fxf yg(x,y)(3.1.26)(3.1.27)(3.1.28)whereAiscalledtheLagrangemultiplier. Thethreeequations abovedeterminethetwocoordinatesandoneLagrangemultiplier. (If thereareseveral stationarypoints, therecanbeadifferent A foreach.) Inthisformtheresult generalizestomorevariables.Notice thatwhether we arrive atthisresult by eliminating one of dxordyinfavorof theother, thefinal equationsabovearesymmetricbetweenthevariables.Lagrangeinvented aclever trick by whichthe symmetry betweentheconstrainedvariablesisretained at all stagesinthe calculationand the procedure for findingthese equationsisreduced to findingthe minimum of some other function:Fwithno constraints. Hereis howit works. Consider thebug problem. Let ususepolarcoordinates rand erelated tox andybyxyr cos er sin e.(3.1.29)(3.1.30)Thesecoordinatesarechosenbecausetheconstraint issimply9 = r2- 25= O.Moving on this fixed rcurve, we satisfy the constraint. The other coordinate ede-scribes variations within the constrained space. We can write in these coordinatesdf = fr dr + lode (3.1.31)fora general displacement. At themaximum onthe circler = 5, dl = 0, but thisonlyimplies10 = 0butsaysnothingabout I r sincethevariationdr =O. Let usnowintroducea newfunction:F = I - A9, (3.1.32)Calculus of ManyVariableswhere). isa constant to be chosen shortly. Itsvariationis57dF df - )'dg(J9- ).g9)dO + (Jr- ).gr)dr(J9)dO + (Jr- ).gr)dr,(3.1.33)(3.1.34)(3.1.35)whereinthelaststepwehaveusedthefact that 9doesnot vary with0sincethelatterischosentoparametrizetheconstrainedsurface,q = r2- 25= O. Thusfand Fhave the samederivativesalong theconstraint surfacenomatter). is. Inparticular when/9vanishessodoesF9. At thispoint thederivative of F normalto the surface equalsfr- ).gr. Let us now use our freedom in choosing). to ensurethat this combinationvanishes. For this choice, thepoint inquestionbecomesstationary withrespect toall variations(radialand tangential). Thisstatementisof course true evenin the cartesian coordinates. Thus the point wearelooking foris the solution todF o(Jx- ).gx)dx + (Jy- ).gy)dy(3.1.36)(3.1.37)withnoconstraintsonthevariations. Consequently thefollowingthreeequationslocate thestationary point and thelagrange multiplier:fxfyg(x,y)(3.1.38)(3.1.39)(3.1.40)whichis what wehadearlier inEqns. (3.1.26-3.1.28).To summarize:To findthestationary points of /. a functionof Nvariables, subject totheconstraint 9=0, find the extrema ofF= f - ).gwith no constraint. The extremalpointand). aredeterminedbytheresultingN equations (the vanishing partialderivatives of F)and the equation9= O.Asanexample consider theextremization off(x,y) = x2+ 2xysubject to the constraintg(x,y) = x 2+ y2- 4 = O.ThusF(x, y) = x 2+ 2xy- ).(x2+ y2- 4)(3.1.41)(3.1.42)(3.1.43)58andthestationary points obey2x + 2y- 2AX=02x- 2Ay=04.Chapter 3(3.1.44)(3.1.45)(3.1.46)Themiddleequation tellsusx = Ay, whichuponfeedingintothefirst wefindol V52(3.1.47)(3.1.48)It is now straightforward to plug these valuesin and to obtain the stationary pointsl [J2(1 + V5), J8)]J5+V5l [J2(1 - V5). J8)]J5- J5(3.1.49)(3.1.50)ShowninFig. 3.1 isthecontourplot of .f andtheconstraint surface. Notethatat theextremal points, thecontoursof constant I aretangential totheconstraintcircle. Thiswillbe takenupin Chapter 7.Problem 3.1.4. Repeat thetemperatureproblemdoneearlierusingaLagrangemultiplier. Thistime find the value of A aswell foreach stationary point.If you have followed these arguments, you will see that the method generalizesinanobviousway to morecoordinatesandmoreconstraints.Forexampleinthecasewhere I and9 dependon(x, y, z ), youmayshowthat byeliminatingsaydz, that thefollowingfour equations determinethethreecoordinates of the stationary point andone Lagrangemultiplier:IxAgx(3.1.51)IyAgy(3.1.52)IzAgz(3.1.53)g(x,y,z) - O. (3.1.54)Or youcouldmoreeasily get theequationsby minimizing :F=I - Ag.Problem3.1.5. Find the shortest distance fromthe originto any point onthe linex +2y = 4 by using Lagrange multipliers. Check this by more elementary means:by first .finding theequation forthelinewhichis perpendiculartothe givenlineand passing throughthe origin.Calculus of Many Variables 59yxFigure 3.1. Contours of constant f(x,y)=x2+2xy and the constraint circle x2+y2=4on whichwewant toextremize f. Thedensityof lines reflect theheight of thefunction. Thusthestationarypointsinthefirstandthirdquadrants aremaxima, while theother twoareminima. Notethetangencyof the constraintcircle and theconstant f curves at the extrema.Finally let usask whathappens if there are threeindependent variablesx, y,z andtwo constraints91 (x,y, z) = 092 (x,y, z) = O. If wewant, we canstill eliminatetwo of the differentials, say dzand dyin favor of the remaining one. It is howeverwiser to followLagrange'sidea andusedF = (Jx-A191x-A292x)dx+(Jy-AW1y-A292y)dy+(fz-A191z-A292z)dz = 0(3.1.55)to obtainIx- A191x- A292xI y- A191y- A292yIz- A191z- A292z91oooo(3.1.56)(3.1.57)(3.1.58)(3.1.59)60920,Chapter 3(3.1.60)which detennine the three coordinates of the stationary point and the twoLagrangemultipliers. Noticethat the final set of equationstreatsall threevariablessym-metrically. Inmany casesit is possible to findthe coordinates of the point withoutsolvingfor themultipliers. (Recall thebugexample.) Onethenmayormaynotbother tofindthelatter.Problem 3.1.6. ShowusingLagrangemultipliersthatamongall rectangles of agiven perimeter, the square has the greatest area.Problem 3.1.7. AStatistical MechanicsInterlude. This exampleshowshowLa-grangemultipliersappear inStatistical Mechanics. Consider Nparticles inabox. According to quantummechanics, the energies of the particles are quantizedto some set of values or energy levels, e1, e2,.... Let ni be the number ofparticlesinlevel i with energyei. The multiplicity or number of distinct rearrangements ofthe particles consistent with any givendistributionni, is givenby(3.1.61)Forexample if all the particles areinthelowest energy level, wehave nl = N,rest ofthe ni= 0 and W=1. (Recall thatO! =1.) It'one particle is in the secondlevel and the rest are still inthe lowest, vV= N! / (N- I)! = N, the multiplicityreflecting the Nwaysto choose the one who goesup. The questionis this: whichdistribution ofparticles, subject to the constraint that the total number equal Nandthetotal energy equal E, givesthe biggest W?Proceed to findthis as follows: Work withS= In W. Argue thatIn n! ':::::n In n - nforlarge n by approximating the suminvolved inIn n!by anintegral. Writethe constraints onthe ni's duetototal number Nand energy E. Treat all ni ascontinuousvariables, introduce Lagrangemultipliersaand13 for Nand Eand maximizeS. Derive the Boltzmanndistributionni =e-o-{3c, .Themultipliers may thenbe found by setting thetotal number and energy comingfromthis distributionto Nand E, respectively. But thisis not our problemhere.Calculus of Many Variablesx61Figure 3.2. Integrationof afunctionof twovariables. ThedomainDisshownonthex- yplane.The integral equals the volume of the cylindricalobject standing on top of the domain. Wehave chosento slice thevolumeinplanes paral1el tothex- f plane.3.2. Integral Calculus of Many VariablesWebeginbyrecallinghowwe integrate a function of just one variable. Firstwe plot f (x) alonganaxisperpendiculartothex-axis, thenweselect adomainof integrationboundedbythelimits Xl andX2, thenwetakeathininterval ofwidthAxat x, formtheproduct f (x) Ax, whichcorrespondedtotheareaof arectangle "standing" on top of that intervaland sum over such areasin thelimit ofvanishingAx. Geometricallythiscorrespondedtothearea of thefigureboundedby thetwovertical lines x = x I, andx = X2, thex-axis, andthe function f.There was however an algebraic waytodo it, which did not involve actuallymeasuringand addingareas. For example inthe casef (x) = x2, we simplyevaluated [F(x) = x3/3]::. Theequivalence of thetwoschemesstemsfromthefact that inthealgebraicschemewearelookingforanF (x) whosederivativeisthe given f (x), andarea constructionis the geometricsolution tothisproblem.Consider now the following straightforward extension of this notion to f (x, y).We firstchoosea domainDin thex- y-plane, asshownin Fig. 3.2.We want to calculate thevolume bounded below by thex- y-plane,boundedabove by the function f (x,y ), and in the sides by a cylindrical surface whose crosssection in thex- y-plane isthe domainD. We denotethisbyF(D) = JLf(x,y)dx dy.(3.2.1)Thenotationalsotellshowweplantogoaboutcomputingthis. First wetakealittle patch in D, of size Axby Ay at the point (x, y ), multiply it by the value of fthere to obtain thevolume of therectangular solid withbasegiven by theselectedpatch and height f. We then do a sum over such volumes in the limit of vanishinglysmall patches and call it F (D). The subscript Din the right-hand side tells us thatthe little patchesmust coverDnomore, noless. Weusetwointegrationsymbolsto tellusweareintegrating a functionover a two dimensionaldomain.62 Chapter 3Thisisthegeneralization of a definiteintegral F (x I, X2) inonevariable. Inthecaseof onevariable, wealsohadtheoptionof callingX2 as x andstudyingthedependenceof F onthis x, callingtheresultingfunction(defineduptoanadditiveconstant) theindefiniteintegral of f. All thiswas possiblebecausetheone-dimensional domainhad disjoint lower and upper boundaries so we could holdtheformerfixedandvarythelatter. Inhigher dimensionsanatural generalizationof thisidea doesnot exist.Tofindtheintegral algebraicallyweproceedasfollows. First wedecidetofindthe volume of theslab that liesbetween the two verticalplanesat distancesyandy + fly fromthe f - x plane, asshowninFig. 3.2. Thisisclearlygivenbythe area of either sheet lying within the solid timesfly. As for the former, it is justthe integral of f (x, y)with respect to x, between the limits x I (y) andX2 (y) showninthefigure. ThelimitsthusdependonthedomainDandthepresent valueofy. Thevariableyistreatedasa constantduringthisx-integration, asit indeediswithintheslab. The volume of thisslabisa functionof y. Wefinallydothesumover such slabs, or equivalently the integraloverybetween the limitsYI andY2 asshownin thefigure. InsummarylY2[lX2(Y) ]F(D) = f(x,y)dx dy.YI Xl (y)(3.2.2)We thus seethat thetwo-dimensional integral canbereducedtoa sequenceofone-dimensional integrals. Note that we could just as easily have sliced the volumeintoslabsbounded by planes of constant x.Letustry asimple example: findV (R), thevolume of a sphere of radiusR,centeredatthe origin. Let usfindH (R), thevolume of thehemisphereabovethex-y-planeanddouble the answer. The domain of integrationis clearly a circle:(3.2.3)On top of the point (x,y ), the height of the hemisphere is f (x,y)JR2-x 2_y2. At a given y, the range for x is Xl = _JR2_y2, X2J R2- y2, asyoushouldconvinceyourself bydrawingasketch of Dorbyex-amining theaboveequationforD. Thus(3.2.4)Next wearguethat theintegrationoverthefull circleis fourtimeswhat wegetfrom the first quadrant where both xand yare positive. This is clear if you imaginethesphere withits symmetries. It is also clear even if youhave no imagination butcansee that theintegrandisan even function of both variablesandtheintegrationregionissymmetricabout theoriginof coordinates. ThuseverypatchinsidetheCalculus of Many Variables 63chosenquadrant hasidentical counterpartsintheotherthreequadrantsrelatedbychanging thesign of x,of y, or both. ThusLet usnowdothex-integralby the followingchange of variables:x JR2_y2sinOo < 0::; Jrj2,whichleads to(3.2.5)(3.2.6)(3.2.7)(7r/2io JR2- y2cosOJR2- y2cosOdO(R2- y2)Jr j 4. (3.2.8)where the repeated factorsin the integrand come once from rewriting f in terms of() and oncefromchangingfromdxtodO. It isnoweasy todothey-integral andto multiply theanswer by 2 togetV (R) = ~ J r R3.Problem3.2.1. Fillinthe missing steps.Problem 3.2.2. By doingtheintegral of f = lover thesame domainobtainthearea of a circle.Consider now thefollowing problem: tointegrate f (x, y) over a square of side2acentered at theorigin. TheanswerisI(D) = 1: [1: f(x, Y)dX]dy.(3.2.9)The simplification we noticeis that the range of x-integrationisindependent of y.Thus cartesiancoordinatesarevery convenient touse if Disboundedby lines ofconstantxorycoordinates. Thingsgeteven easier iff(x,y)= X(x)Y(y), (3.2.10)that is, if f factorizesinto a product of a functionX(x) that depends on just xandY (y) that dependson justy. LetDstillberectangularinshapesothat limitsonxare independent of y. ThenlY2lx2I(D) = Y(y)dy X(x)dx,YI Xl(3.2.11)64yf------ y + t. yyxx x+t.xCartesian CoordinatesChapter 3Polar CoordinatesFigure 3.3. Cartesianandpolar coordinate systems.that is, I (D) becomes a product oftwo one-dimensional integrals. This is aseasyas the probleminhigher dimensions canget: factorize intoa product ofone-dimensional integrals. But noticethat theverysameproblemcanlookverynasty in some other coordinate systems. For example suppose we switched to polarcoordinates(r.O)l showninFig. 3.3They aredefinedbyor theinverse relationsrx r cosOLl t -1yu = an -xy = r sinO(3.2.12)(3.2.13)In this coordinate system, neither willtheintegrand factorizenor willthe limits oneach variable beindependent of thevalue takenby the other.By the same token, a problem that looksnasty in one coordinate system couldpossibly be simplified by going to a different system. Let us go back to the volumeof thesphere. Theintegrand and domainwereVR2- x 2- y2integrandR2domain.(3.2.14)(3.2.15)Itisvery clearthattheproblem will bealot simpler if weusepolarcoordinates.Indeedwefind:f(r,O) VR2- r2 integrandr < Rdomain.1Sometimes thepolar coordinates arereferredto as(p, 0), (p, ), (r, 0). or(r, ).(3.2.16)(3.2.17)Calculus of Many Variables65Noticehow f inthenewcoordinateshassimplified: not onlyisitaproduct of afunctionofr times afunctionof e, thelatteris just unity! Next, thedomainofintegrationis described entirely by one of the coordinates, so that nomatter e is, rwill gooverthisrange. Howdoweactually dotheintegral inthesecoordinates?As inonedimension, it is not enoughto just rewritetheintegrandandlimits inthenewsystem, wemust includetheJacobian. Inotherwords, theareaelementdxdy does not simply get replaced by drde. Indeed the latter does not even have thedimensions of an area. What we need is a JacobianJ (r. e)that needs to be insertedto ensure thatwe arestill computing thesame object(the volume)asbefore.Thefirst stepistobetter understandthe cartesiansystem. Givenaplane, wepickanorigin. Thenwedrawlinesof constant x andy, theformer areparalleltothey-axisandthelatterareparallel tothex-axis, asshowninFig. (3.3). Theintersection of two suchlines of constant coordinatesdefines a point. For examplethelines x = 3andy = 4meet at thepoint wecall (3,4). Toperformatwo-dimensionalintegral, we draw two constant-xlines with coordinatesxandx + and two constant-ylines with coordinatesyandy + Their intersection enclosesa region of sizewhich we then multiply by f(.1:, y) in doing the integration.We finally take the limit---+0 in defining the integral. All this is understoodwhenwewrite Jdxdy. Figure3.3summarizes all this. Let us repeat this withpolar coordinates. Tolocateapoint, wefirst drawacurve of constant coordinater, whichis justacircleof radiusr. Thenwedrawacurveof constante, whichis just araygoingfromthe originto infinityat anangle e with respect tothex-axis. Theintersectionof thesetwo curvesdefinesthepointcalled(r, e). Todointegrals, we draw another constant r curve at r +and another constant e curveat() +andmultiply the enclosed area by f (r. e). What is theenclosed area? Itis clear fromthefigurethat the enclosedarea hastheshape of arectangle of sidesrb.eand(All right, it isnot quitearectanglesincetheconstant e linesarenot parallel andthe constant r linesare curved. But theseobjectionsvanishinthelimit of infinitesimalpatches.) Noticethattheareaissimplygivenby theproductof the twolengthsrand r because r, e are orthogonal coordinates: curves ofconstant rand emeet at right angles. To summarize, what wehave learned is thatdxdy ---+ r dr deJ(r,O) r.(3.2.18)(3.2.19)Let usnoticethatwiththeJacobianinplace, theintegrationmeasurehastheright dimensions. The factor rwhich converts the infinitesimal change in e(at fixedr)to thecorresponding displacementrdeiscalled thescalefactorcorrespondingto e, and is often denoted by h(}. The scale for r is unity since dris itself the actual Likewise the scale factorsforthe cartesian coordinatesisunity. TheJacobianisthe product of the scale factors inorthogonal coordinates.If we go to three dimensions,there are two popular coordinate systems besides66zzFigure 3.4. The cylindrical coordinatesystem.Chapter 3the cartesian. One is called thecylindrical coordinate systemwithcoordinates:Pd>z z.(3.2.20)(3.2.21 )(3.2.22)Thesystemisshownin Fig. 3.4.Thesurface of constantpisa cylinder of radiuspwiththezaxisasitsaxissincepmeasuresthedistancefromthez axis. Thesurfaceof constant $ is likeadoor hingedonthe z axis, makinganangle $ withthe x - z plane. Thesetwosurfaces intersect ona vertical linewhere the door penetrates the cylinder.Finally thesurfaceof constant z isa planeadistancez abovethex - y plane. Itintersects this verticalline at one point, called the point (p, $. z). Note that this toois an orthogonal coordinate system. You should prove that the scale factors(factorswhichconverta changeinone coordinate withtheother heldfixed) andJacobian(product of scalefactorsforanorthogonal system suchasthisone)arehpJ(p,$,z)1 h= phz= 11lp(3.2.23)(3.2.24)Thismeansfor examplethat at fixedpandz, aninfinitesimal changedt 2zThefunctionalsohasa pole at z= -iwitharesidue1R(z = -i) = lim,(z + i)j(z) = -.z--->t - 2z(6.1.28)(6.1.29)Functions of aComplexVariable 113In the present problem we can obtain the above results simply by using partialfractionsinEgn. (6.1.27):1 1[1 1]l(z)= =- --- .(z +i)(z - i) 2i z- i z + i(6.1.30)More generally, if a functionhasasimple pole at zo, itsresiduethereisR(zo)= lim(z - zo)l(z).Z-+ZO n-thorder poleWesay that f hasann-th order pole at zoif1(z) --+ R (zo )(z- zo)71.asz --+ zoo(6.1.31)(6.1.32)Then-thorderpoleis calledaremovablesingularity: bymultiplyingthefunctionby(z- zo) n, wecaneliminatethesingularity. Essential singularityAfunctionhasanessential singularity at apoint zoifit haspoles of arbitrarily high order which cannot be eliminated by multipli-cationby (z- zo) n forany finitechoice of n. Anexampleisthefunction00 1l(z)=L ~o z n.which haspoles of arbitrarily highorder at the origin.(6.1.33) Branch point Afunctionhasa branch point at zoif, uponencirclingzoandreturningtothestartingpoint, thefunctiondoes not returntothe startingvalue. Thusthefunctionis multiple valued. Anexampleis(6.1.34)Notice that as we increase() from 0 to271",at some fixedr, the functiongoesfrom r! to r! ei7l" = - r!. Wesay 1hasa branch point at theorigin.Youmight thinkthat amonganalytic functions, those withno singularitiesanywhereare speciallyinteresting. This is not so, becausetheonlyexampleofsuchafunction is aconstant. (Thiscanbeproven, thoughwewill not proveithere.) Even polynomialsmisbehaveasz --+00;they blow up.In thischapter wewill mainly emphasizemeromorphic functions:Definition 6.5. A function1(z) ismeromorphic if itsonly singularities for finitezare poles.114 Chapter 6Problem6.1.7. Locateandnamethesingularities of f1/(z4 + 2z2+ 1).1/(1 +z4) and f =Problem6.1.8. Showthatdz =(dr + irdO)ei8givenz = rei8. Interpret thetwofactorsindz, in particular therole of theexponential factor. Equating the deriva-tivesintheradial and angular directions. findtheeREinterms qf Ur, U8, Vr V8.Beginnowwith f(r,B) = f(rei8)as thedefinitionof analyticity, relate rand 0derivatives of f, and regaintheCREin polar form.Problem 6.1.9. Confirmthe analyticity ofx- zyx2 + y2'sin x cosh y + i cos x sinh y.Youcaneither usetheCRE or findf(z).6.1.2. Derivatives of analyticfunctionsWhenU andvpossesscontinuous partialderivativesthatobeytheCREwemaydefine the z-derivative of f =u+ivas follows. If the function is known explicitlyin terms of z, say f = z2, we take derivatives aswe didwitha real variablex:(6.1.35)(6.1.36) 2z.1).fdfdzSuppose we have the functionin terms of x andyand only know that it obeys theCRE, sothat inprincipleitcanbewritteninterms of just z. Let usvary bothxandytoobtain to first orderdfof of-dx+ -dyOx oy'(ux + ivx)dx + (u y + iVy)dy(ux + ivx)dx + (-iuy + vy)idy.(6.1.37)(6.1.38)(6.1.39)If wenowinvoketheCREweseethatbothbracketsareequalandthechangeinf isproportional todx + idy = dz. It followsthat the derivativeis(6.1.40)Functions of a Complex Variablein accordance withEqn. (6.1.18).Thus forexample thederivative off = x2- y2 + 2ixy.IS115(6.1.41)df 8f- = - = 2x + 2iy = 2z. (6.1.42)dz 8xNoticethat thez-derivative of ananalytic functionisindependent of thedi-rectioninwhichwe make theinfinitesimal displacementthat entersthedefinitionof thederivative. Youcanchoosedz =dx or dz =idy or anygeneral casedz = dx + idyandget thesameanswer. Thefollowingexerciselooksatthisinmore detail.Problem 6.1.10. (Important). Consider df. the first order changeinresponsetoa change dxand dy inthecoordinates. of a function f = u + ivwhereuand vhavecontinuousderivatives whichdonothowever obeytheCRE. Theshiftinxandycorrespondstochangingz bydz =dx + idyandz*bydz*=dx- idy.Inother words, aswe move inthe complex plane labeled by x, y, we change bothzandz*. Showthatdf generallyhas parts proportional tobothdz and dz*byreexpressing dxand dy interms of the former. Showthatasaresult thesymbol1zmakes no sense in general: it is like trying to define df (x, y)/ dx for a functionof twovariables, whenall one candefine isthe partial derivative. If however, thefunctionof x andyhappened tohavenodependenceony, wecoulddefinethetotal derivative with respect to x. That is what is happening with analytic functionsf whichdepend only onz.It follows fromtheprecedingdiscussionthat if f isanalyticat apoint, 1zexistsina neighborhood of that point.From the eRE,we can easily show, by taking derivativesthatu xx +U yy =0 vxx +Vyy =o.(6.1.43)(6.1.44)Note that so far we have only made assumptions about the continuity of first deriva-tives ofu andv. Theaboveequationassumes the secondderivatives are alsowell-defined. Weshall seelater that this isnotanadditional assumption: it willbe shown towards the end of the chapter, (see discussion followingEqn. (6.4.22,that if f (z) hasafirst derivativeina domain, it hasall higher derivativesinthatdomain.The equation obeyed by u82u 82u-+-=08x28 y2(and similarly v)is called Laplace's equationand uand vare said to be harmonicfunctions.116Problem6.1.11. Prove that uand vare harmonic giventheeRE.Chapter 6Problem6.1.12. Youare giventhat u= x 3- 3xy2is the real part of ananalyticfunction. (i) Findv. Reconstruct f (z ). Verifythat 1L and'V areharmonic. (ii)Repeat/or the case v = e-Ysin x. (iii) Youare givenu = x3- 3x2yand asked to.find v. Yourunintoreal problems. Why?Problem6.1.13. Prove that (::2+ :;2 )If(z)12= 4If'(z)12.Problem 6.1.14. Given that(u, v) forma harmonic pair, show without brute/orcethat uvand u2- 1)2are also sucha pair.6.2. Analytic Functions Definedby Power SeriesWe can build any number of analytic functions by simply writing do\\'n any expres-sion thatinvolves just z. Takefor example the functionf(Z)=Z2+Z (6.2.1)which assigns definite values to each complex number z. For example at the pointz = 3 + 4i, f (z) = - 4 + 28i.Thisis a special case of n-thorder polynomialsPn(z)(6.2.2)which will obey the eRE for any finitez. Equivalently they will have well-definedderivativesforall finitez. They willhowever misbehave aswe approachinfinity:the n-th order polynomial will (mis)behave aszn= rneinBblowing up in differentwaysin different directions.Now we tum to a very important notion: analyticfunctionsdefined by infiniteseries. As a prelude we must discuss the notion of convergence of complex series.Definition 6.6. Theinfinite series of complex termsan,(6.2.3)is said to converge (( its real and imaginary parts. i.e.. the series that sum the realand imaginary parts of an. converge.Functions of aComplexVariable 117Since we already to know how to test realseries for convergence, no new tools areneeded. Consider forexample00 00 00(6.2.4) s=L(2-n+ie-n) = L2-n+iLe-n.000The ratio rest onthe real andimaginaryparts showthat bothcOliverge(beinggeometricserieswithr 1 theroots are real and their product isunity. Thus both roots must lie onthe real axis,withoneinsideandoneoutsidetheunitcircle. Usingthecontribution fromtheformerobtainthe above result. Thecasea ---> --->dr = er dr +e(J rdO (7.5.18)---> --->hasanangular component rdOand notdOand we want to writedcf; as V' cf;. dr. Itisnowclear uponinspectionthat(7.5.19)---> B 1 Bcf; --->V' cf; = -er+ --e(J.Br r BONoticethat thesignificance of thegradientisstill thesame: it isthesum over or-thogonal unit vectors times the spatial rate of change in the corresponding direction.Inthecase of other coordinatesor threedimensionsthegradientisgivenby(7.5.20)where hi is the scale factor whichconverts the coordinate shift dUi tothe cor-responding displacement in space ande: is the unit vector in the directionofincreasing Ui.--->Problem7.5.4. Writedown V' 'Ij; incylindrical and spherical coordinatesinthreedimensions. Showthat x. y, zareanorthogonal system of coordinatesby evalu-atingtheir gradients(afterexpressing theminterms of cylindrical and sphericalcoordinates). (Thisis just the converse of Problem(7. 5. 3.)).172 Chapter 7Problem7.5.5. (i) Giventhat = x 2y + xyfindthegradient incartesianand- polar coordinates. (ii)Evaluate I V' 1 at (x= 1, y = 1) inbothcoordinatesandmake suretheyagree. (iii)Find therate of changeatthis point inthedirection- - - parallel to2i + 3 j . (iv) I n ~ e g r a t e V' fromthe originto(x =1, y =1)alonga450 line. What should this equal?Did you get that result?Problem 7.5.6. Showthat thework doneindisplacinga pendulumof lengthL,bob mass m, by anangle efromthe vertical is W=mgL(l - cos e), 9being theaccelerationdueto gravity.Problem 7.5.7. You are on a hot volcanic mountain where the temperature is givenby T(x, y) =x 2+xy3. (i) Ifyou are located at(x =1, y =1), in which directionwill youruntobeat the heat?(ii) Ifyour steps are1/10 units long, by how muchwill thetemperature dropafter the firststep?(Work to .first order.)Problem 7.5.8. Find the directional derivative of the following fieldsinthe radialdirection at the point (1, 1, 1): (i) f =x 2+y2 +z2, (ii) f = eX+Y+z, (iii)xyz. (iv)- - - Repeat forthe direction parallel to i +j - 2 k . Can youinterpret the results?7.6. Curl of a Vector FieldFor a general vector field, the line integral around a closed path will not be zero andiscalledthecirculationaroundthat loop. Thecirculationmeasuresthetendencyof thefieldtogoaroundsomepoint. Considerfor exampletheflowof wateras- youdrainthetub: thevelocityvector V has alargecomponent intheangular- direction. Let us say it is V = Vre: + Ve e/t. Consider a contour that goes aroundtheoriginat fixedradiusR. Thend;; = Rdee/t andthecirculation, Jo27rVeRde- will be nonzero. Likewise the magnetic fieldB due to an infinitely long conductorcarryingcurrentI coilsaroundthewire: if thecurrentisalongthez-axisandweusecylindrical coordinates,- _fLOIB = e-, (7.6.1)21rpwherepand arecoordinatesintheplaneperpendicular tothez-axis andfLO=41r10-7Newtons/Amp2 is aconstant calledthe permeability offreespace. The- line integral of B along a contour encircling the wire and lying in the planeVector CalculusyIV"x,y..IIx+dx,yx173Figure 7.8. The tiny contour forthe curl formula.perpendiculartothewirewill benonzero. Weshall havemoretosayabout thislater.---+Incontrast consider a field W = W r e; inthe x - y plane. This purelyradial fieldhasnotendencytogoaroundanypoint andthecirculationarounda---+ ---+circular path centered at the origin vanishes (since rvanddr are perpendicular)inagreement.---+Consider aloopCin thex- yplane and thelineintegral of Wtakeninthecounterclockwise direction. Just as in the case of the complex integral (Fig. 6.1) wecan argue that this integral is the sum over integrals around smaller nonoverlappingtilesor plaquettesthat fullycover theareaS enclosedby thegivencontour: thecontributionsfromedgessharedbyanytwoadjacent tilescancel, leavingbehindjust the exposed edges, whose union is just C. Consider therefore the tiny rectanglefromFig. 7.8The contributions from edges I and III are computed as in the complex plane:I + IIIlX+dxlx+dxx Wx(x,y)dx- x Wx(x.y + dy)dxlx+dx( oW( ))_ x x, Y d dx(j Yx yoWx(x, y) ( b')- 0 dxdy +0 CU IC .Y(7.6.2)(7.6.3)(7.6.4)Adding now thesimilar contributionfromsidesI I and IV, wefindfI + I I I + I I + IV= (7.6.5)174 Chapter 7(7.6.6)( 8Wy8Wx)-- - -- dxdy.8.e 8yNotethat thefinal contributionisproportional tothearea of theplaquetteandtheneglectedtennsareof higherorder. Let usrecall what wehavedonesofar. Wewantedthelineintegral around a macroscopiccontour C. Wearguedthat itisthesum of integralsaroundtinyplaquettesthat tiletheareaboundedby C. Wenowfind that each plaquette makes a contribution equal to its area times(a::::v_ aJ:;/ ).Consequently asweincrease thenumber of tilesenclosed and decreasethearea ofeachone, wewill obtaininthelimit thedouble integral of (a::::v- a';):;x) overthe enclosed area. The neglected tenns(called "cubic" in the derivationaboveandproportional tohigherpowersof theinfinitesimals) will not survive inthelimit.Thereasonmustbe quitefamiliar fromearlier discussions of thesamequestion.It is clear thatwherei---; ---t 11 (8Wy8Wx) ~ t . dr = --- -- dxdy. c=as s 8x 8yC= 88(7.6.7)(7.6.8)means that C is theboundaryof the region 8. This result is called Green'sTheorem. It allowsusto express thelineintegralof a vector fieldaround a closedloopastheareaintegral of theobject a::::y- a';):;x calleditscurl. Inparticular ifthe vector fieldis conservative, thisintegralmustvanishon any region8whichintummeansthat---t . (8Hly 8Wx )~ v ISconservative - -- - -- =O.8x 8y(7.6.9)(7.6.10)Considerfor exampleour friend F=2xy2"7 + x2j. This better not beconservative since its line integral from the origin to (1, 1) was found to be differenton different paths. Wefindindeed(8:Xy- 8 ~ x ) = 2x- 4xy,On the other hand we have seen that any gradient is a conservative vector fieldsince its integral depends only on the difference of the scalar field at the two ends ofthe path. It must then be curl-free. Ifwe apply Eqn. (7.6.9) to W="7~ ~ + j ~wefind82 82-- - -- == O. (7.6.11)8x8y 8y8xInother wordsthecurl of a gradient vanishesidentically. Weshall havemoretosay onthislater.Vector Calculus 175Problem7.6.1. Considerthecomplexintegral f(z)dz, wheref = u + ivanddz= dx + idyover a contour that lies in the domain of analyticity of f, Write thereal and imaginary partsascirculations of tworeal vector fields and verifythattheCRE ensure that both fieldsare conservative.--+Problem7.6.2. Test each ofthefollowingfields to see ifit is conservative: (i)W=i y sinx + --; cosy, (U) TV= i2xy3 + --;3y2x 2, (iii) TV=i sinx + --; cosy,--+ --+ --+(iv)W= i cosh x cosh y +j sinh x sinh y.--+Wesawthatthevectorfield \7 isconservative. Theconverse isalsotrue:if a field is conservative, it may be written as a gradient of some scalar field. Thecorrespondingscalar fieldis not unique: givenonemember, wecangenerateawhole family by adding a constant without changing the gradient. Tofindanyone---+memberof this familycorrespondingtoagivenconservativefield ~ F , picksomepoint 1 andset (1) = O. Assign to1V' I. If youmoveinanyotherdirection, therateof changewill bediminished~by thecosine of theanglebetween V' andthedisplacement vector. Likewise, if~ ~youknowV' xWat somepoint, youknowthat thecirculationperareawill be---+ -->greatest around a loop that lies in the plane perpendicular to V' xW(i.e., the area~ ~vector isparallel to V' xW)andthatthecirculationwill equal theproduct of the---+ -->area and1V' xWI. The circulation around any other differently oriented areawill~ -->bereduced by thecosine of the anglebetween the areavector and V' xW.Let ustakeone more look at Eqn. (7.6.19). It says that the circulation aroundatinyloop is the product of the areaofthe looptimes the component ofcurlparallel totheareavector. Conversely. thecomponent of curlinanydirectionisthe circulation per unit areaina plane perpendicular tothat direction.Consider nowasurfaceSinthreedimensions. For definitenessimaginethisis a hemisphere. TheboundaryC = aS ofthis surface is the equator. IfweVector Calculus179Figure7.9. Anexample of Stokes' Theorem. Thelineintegral of avector fieldalongthe equator ofthe hemisphereis the sum of lineintegrals around the plaquettes that tile it. The latter sum becomes thesurfaceintegral of the curl inthelimit of vanishingly small plaquettes.nowtilethehemispherewithlittleareasit followsthat thecirculationaroundtheequatoristhesumof circulationsaroundthetiles, as showninFig. 7.9(Onlyafewrepresentativetilesareshown.)Sincethelatter aregivenby Eqn. (7.6.19)we obtain1 ---4 ---+----t ----t-----+Jr _ WdrL-\7 xW. dB.C-8S tiles(7.6.22)(7.6.23)We donot usetheequalitysignsinceas longas thetiles areof finite size, thecontourcanonlybeapproximatedbythetilesandthequestionof whereineachtiletoevaluate thevariousderivativeswill plagueus. However if weconsider thelimit of infinitelymanytilesof vanishingsize, theresult becomesexact. Inthis----t ---4limit theright-handside becomes thesurface integral of\7 xWover thesurfaceSand1 }C=8S iswhichiscalled Stokes' Theorem.ThusGreen'sTheoremisStokes' Theoremrestrictedtotheplane. NotethatStokes' Theoremrelates thelineintegral of avectoraroundaclosedlooptothesurfaceintegralof itscurl over anysurfaceboundedbytheclosedloop. Thereisno Stokes' Theorem for theline integralover anopen curveP12withboundaries1and2.Notealsothat for ageneral vector, the surface integral candependonthedetails of thesurface given a fixedboundary. On the other hand, if thevector fieldis the curl of another vector field, the answer depends on just the boundary C =/3 B.Thisshould remind you of line integrals of vector fieldsthat are generally sensitiveonly to thepath anddepending on the boundary points of the path when the vectorfieldisagradient. Weshall havemoreto say onthislater.180 Chapter 7Let us consider two examples ofGreen's Theorem. Suppose we want tointegrate----+ ---t ---t 2F = i xy + j x y (7.6.24)counterclockwisearoundsome contourinthex-yplane. Thetheorem tellsusf F .d: = JJdxdy (iJ(;:Y) - = JJdxdy(2xy- T) (7.6.25)Say the contouris aunitsquare centered atthe origin. Then thedoubleintegral iszerosince thefunctionsbeingintegratedareoddinx ory or both. Consider thena unitsquarein thefirst quadrant, withitslowerleft-hand comer attheorigin. Inthat case.f .f dxdy(2xy- x) = .i1dx 11dy(2xy- x) =-= 0(7.6.26)whereweget zeroduetoacancellationbetweenthetwoterms. Hadthesquarebeen of side two, theresult wouldhavebeen(7.6.27)---t ----+As thesecondexampleconsiderthetwodimensional field F = 2xy2 i +x 27whose curl was2x - 4xy. (See Eqn. (7.6.10) in the worked example.) Sincethecurl is nonzero, not onlydoweknowtheanswer is path-dependent, wecanalso relate the difference (in the line integral) between any two paths to thesurfaceintegral of thecurl over theareatrappedbetweenthepaths. Let usrecall thattheline integral was1 if we went along the x-axis to (L 0)and then upto(1, 1), whileit wasequal to5/6if onewent alongx = y. Itfollowsthat if wewent to(1, 1)alongthefirst routeandcamebackalongthesecond, thecontributionfromthisclosedcounter-clockwisepathwouldbe1 - 5/6= 1/6. Let usverifythat thisisindeed what we get by integrating the curl within the triangular region between thepaths:11 1x 2 1 1dx dy[2x- 4xy] = - - - = -.o 0 3 2 6(7.6.28)----+ ----+ ----+Problem7.6.9. Recall that the line integral of F = 2xy2 i +:r2j betweentheoriginand(1. 1) wasequal to 1 along paths that first went upandthentotheright and viceversa. Integratethecurl over theregioninquestionand reconcilethis factwithGreen'sTheorem. Suppose we goto(0, 1), thenstraight up to(1. 1)andbackalong:r = y2. Fromtheworkedexample(in theintroduction tolineintegrals)deducewhat we get forthis path. ReconcilewithGreen's Theorembyintegrating thecurl over theappropriatearea. Finallyreconcilethe fact that thelineintegral hasthesamevalue on pathsy = x2and x = y2 byintegrating thecurlinthe regionbetweenthe two curves.Vector Calculus 181Problem 7.6.10. Showthat the line integral of W= - ~ (y - ~ sin 2y) i +x cos2Y j counterclockwisearoundacircleof radius 11" centeredat thepoint(x= e, y= e7l") equals11"3.~ ~ ~Problem 7.6.11. (i) Findtheworkdonebythe force Fix + j y as onemovesalong thesemicircle of radius2 centered attheoriginstarting from(1,0)andendingat (-1,0). (ii) Howmuchworkisdoneinreturningtothestartingpoint along the diameter? (iii) What doesall thissuggest about the force? Verifyyour conjecture.Problem 7.6.12. Showthat theareaenclosedbyacounterclockwisecurveCinthe plane is givenby ~ fc(xdy- ydx).Problem 7.6.13. Evaluatef(ydx- xdy) clockwise uround thecircle(x- 2)2 +y2= 2 any way youwant.~ ~ ~ ~Problem 7.6.14. Show that the line integral of F = i 2y +j x +kz2 fromtheoriginto(1,1,1) alongthe pathx2+ y2 = 2z, x = y is ll. Visualizethe pathand parametrize it using x. Repeat the integral along the straight line joining thepoints. Showthat thisvector fieldisnot curl free. How dowe reconcile thiswiththe above tworesults?Problem 7.6.15. Evaluatef ((x- 2y)dx- xdy) around theunit circle first asaparametrized lineintegral and thenby theuse of Green'stheoremand show thatit equals 11".7.6.1. Curl in other coordinate systemsLet us sayU1, U2, u3 arethreeorthogonal coordinates, suchas thespherical co-ordinates. Let hI, h2, h3bethescalefactors that convert thedifferentials dUi todistances. Thecurlcanthenbewrittenasfollows:(7.6.29)182 Chapter 7Let ustrytounderstandthis. Considerthecomponentof curl along e;. This isgivenbythecirculationperunit areaonaloopthat liesintheplanespannedbycoordinates 112, U3. Considertheroughlyrectangular loopboundedbythe lines112,112 +dU2, U3, U3 +dU3. By pairing together line integrals along opposite edges,wewill get asincartesian coordinates,(7.6.30)wherethescalefactors multiplyWi becausethecorrespondinginfinitesimal lineelements are hidui. We must then divide this by the area of the loop,h2du2h3du3,to get thecurl.Problem 7.6.16. Writethe expression forthe curlintwodimensional polar coor-dinates, spherical and cylindrical coordinates.Problem 7.6.17. Consider once againf((x - 2y)dx - xdy)around the unit circlefromProblem(7. 6.15.). Giventhat this is the line integral of a vector field, extractthevector .field incartesiancoordinatesand thenexpressit in polar coordinates.(Start by writingdown-; and -; in terms of e: andeo.) Nowdo the lineintegral in polar coordinates. Check using Green'stheoremin polar coordinates.Repeatfor the following: f(xydx +x dy) counter-clockwise around the unit circlecentered at the origin.7.7. The Divergence of a Vector FieldImagineaclosedsurfaceS whichistheboundary of somevolumeV: S = 8V.--7Our goal isnowtocompute thesurfaceintegral of Wover S. ImaginefillingthevolumeV withtiny parallelepipeds, calledcubestosavespaceandavoidspellingmistakes. (If they are infinitesimal, they can fillany shape.) Each cube has its areavectorpointingoutwardoneachface. Thusadjacent faces of neighboringcubeshavetheir areavectors oppositely directed. Suppose weknowhowto compute thesurfaceintegralover any such tiny cube. Then we can express thesurfaceintegralover Sasthe sum of surface integrals over the tiny cubes that fill it. The reasonisjust likeinStokes' Theorem. Consider anycubeintheinterior. Everyone of itssixfaces is shared by some other cube. Consequently the surface integral over thiscube isannulled by thecontribution fromthesevery same faceswhenconsideredas parts of the surrounding cubes. The reason is clear: we are integrating the samevector fieldinthetwocases, but withoppositelypointingnormals: theoutwardVector Calculusz183yFigure7.10. Theinfinitesimal cube atthe originonwhosefaceswecomputethesurfaceintegral.pointingnonnal onour cubeisinwardpointing according toitsneighbors. If theneighboring cubesthemselveshaveother neighbors, they will experience asimilarcancellationandall wewill haveintheendisthecontributionfromtheexposedfaces, Le., thesurfaceS.Tosummarize, thesurfaceintegralof thevectorfieldover theclosed surfaceSisthesum of surface integrals over tiny cubes that fill the enclosed volume Vintheusual limit of infinitecubesof vanishingsize. Soourfirst jobistocomputethesurfaceintegral over aninfinitesimal cubeshownin Fig. 7.10.It is centered at the origin (which is a convenience but not a necessity) and haslinear dimensions dx, dy,and dz. Its surface is the union of six planar areas. Eachis represented by a vector nonnal to the plane and pointing outward. Let us imagine--tthat this cube isimmersed in thevector fieldWandcalculate the fluxcoming out--tof thiscubeby adding thefluxonall sixfaces, i.e., dothesurfaceintegralof W--tover thiscube. First imaginethat Wisconstant. Thenthetotal fluxwill vanish.This is because whatever we get from one face will be exactly canceled by the faceoppositetoit sincethetwononnalsareoppositeindirectionwhilethevectorisconstant. If thefluxwereindeedduetoflowof particles, this wouldmeanthatwhatever number flowsinto the cubefromone faceflowsout of the opposite face.Thus, forustoget a nonzeroflux, thefieldwill havetobe nonunifonn.Consider twoof thecube'sfacesthat lieinthey- z plane. Theirareasare184 Chapter7---+givenbyi d,/jdz, wherethe plus/minus signs applytothe faces with .1' > 0and,I: due to charge ql at thelocation of charge q2whichwill cause charge q2to experience a force q2 Ewhenplacedthere. Thusthefieldisa result ofjust onechargeandit runscountertothesimplernotionthat ittakes twochargesforanythingtohappen. The fieldisa forcewaitingtohappen. Butwe giveit a meaning evenbefore the other chargeisputthere. To get a forceoutofthefield, youhavetointroduce the other charge q2.)Theobject D issimilar. Whereassintakesina number and spitsout a number, D takesinafunctionand spitsout afunction. Thefunctionit spitsoutisthe derivative of theinputfunction.ThusDacting onf = df.dxThusDactingonthefunctionx2givesthefunction2x; actingonthefunctionsin x givescos xandsoon. OnecallsDthederivativeoperator. It isusuallywrittenas d ~ tomakeitsactionverytransparent. Youcanuse anysymbol youlike, but what it doesiswhat it does: take the derivativeof what youfeedit.The derivative operator isa linear operator:D(af(x) +bg(x)) = aDf(x) +bDg(x). (7.8.8)That is to say, the action on a linear combinationis the sum of the action on the pieces. By contrast,sinisnot linearbecausesin(x + y) :f; sin x + sin y. However if youwerealwaysgoingtolet xandybesmall, orif youweresomehowrestrictedtosmall xandy, thenyoucouldcometotheconclusion that sinis a linear operator since then sinx:= xand siI1(x+y):= x+y. Many relationsinnaturethat arereallynonlinearappear linearat first sight sinceexperimentsinitiallyprobethesituationina limitedrange.Thevectordifferential operator, "del," inEqn. (7.8.2) is withinourgraspnow: actingona--->scalar it gives a vector field V' , actingona vector fieldviaa dotproduct, it gives a scalar, thedivergence, while actingviathe crossproduct, it gives another vector field, the curl.Problem7.8.1. Canyoudefine the operator D2?Isit linear?Thevector differential operator canbe writteninother orthogonal coordinatesystemsas---> ---> 1 a ---> 1 a ---> 1 aV' = el- -- + e2- -- + e3- --.hI aUI h2 aU2 h3 aU3(7.8.9)7.9. Summary of Integral TheoremsLet usrecall the threeintegraltheoremsestablishedinthepreceding pages:lap=(2)- (1),J A . d: (Stokes' Theorem),}p=asi----+A . dS(Gauss'sTheorem).s=av(7.9.1 )(7.9.2)(7.9.3)Note thatin all threecasesVector Calculus 189 Ontheleft-handsidesomesortof derivativeof afieldisintegratedoveraone-, two-, or three-dimensional region. On theright-hand side is the contribution of the fieldfrom just the boundaryof theregioninquestionintheleft hand side.In general, the line, surface, and volume integrals will not receive their contributionfrom just the boundaries of the integration regions. This isso only for the gradient,curl, or divergenceinthethreerespectivecasesshownabove. Itfollowsfor thesespecial fields that if theintegrationregionontheleft-handsideis overaregionwith no boundary, we will get zero. This willlead to some identitieswe discuss inthenextsection.7.10. Higher DerivativesSofarwehavecomputedsomefirst-orderderivativesof vectorandscalar fields:---'> ---'> ---'> ---'>---'>\7 dJ, \7. Wand\7x W. Let us move on to higher-order derivatives of fields. Forexample,wecan explore thedivergence of the curlor the gradient of a divergenceand so on. We will examine a fewsecond-order derivatives. We start with two thatareidentically zero:---'> ---'> ---'>\7\7xWo0,(7.10.1)(7.10.2)whichyoumayeasilyverify by straightforward differentiation. All youwill needistheequality of mixed partialderivatives: lij= iji,wherei, j =x, yor z.Problem7.10.1. VerifYthese identities.Wecanunderstandtheseidentitiesinadeeperway. Considerthelineintegral ofa gradient. It isthesameonany twopathswiththesame end pointssince all thecontributioncomesfromtheend points. Thefact thatwe get thesame answer ontwopaths joining thesame end pointsmeansthatthedifference, theintegral onaclosed path(obtained by going up one path and coming back counter to the other)is zero. But by Green's or Stokes'Theorems, the surface integral of the curl (of thisgradient) hadtovanishontheareaboundedby thisclosedpath. Sincethepathsand hence enclosed area were arbitrary, and could be infinitesimal,the integrand ofthesurfaceintegral, whichisthe curl of thegrad, had tovanish everywhere.Consider now the surface integral of a curl. It is the same on any two surfaceswiththesame boundary since theanswer comesonlyfromtheboundary. Inotherwords thedifferenceof thetwosurfaceintegralsiszero. Thisdifferenceis then190 Chapter 7asurfaceintegral over aclosed surface. Let usmakesureweunderstandthelastpoint, by considering anexample. Consider a unit circleonthex - yplane, withitsedgesrunningcounterclockwiseasseenfromabovethex - yplane. Itistheboundary of theunitdiscintheplaneandalso of aunithemispheresittingabovethe x - y plane. In computing surface integrals, we are instructed by the orientationof the loop to choose the outward normal on the hemisphere and the upward normalon the disc. In the difference of integrals, we will use the downward normal for thedisc. Thedifferenceisthenthesurfaceintegral overaclosedsurfaceformedbythehemisphereanddisc, withall normalspointingoutward. Withthis fact clear,wenowuseGauss'sLaw torelatetheintegral over theclosedsurface of thecurltotheintegralof the divergence(of the curl)over theenclosed volume. Since thisis to vanish for any closed volume, even an infinitesimalone, the divergence of thecurl mustvanish.7.10.1. Digression on vector identitiesThereis yet another way tounderstandtheseidentities,another way that tells youtheidentitieshadtobetrue.2Asa prelude tothiswerequire the followingtheoremstatedwithoutproof:The boundary of aboundaryis zero.Here are some examplestoillustrate this. Consider a pathP12inthe x - yplane goingfrompoint 1topoint 2. Itsboundariesareitsendpoints. But thepathitself isnot theboundaryofanything. But now, let the two extremities be brought together makingit a closedcurve. Two thingshave simultaneouslyhappened: Thepathhasbecometheboundaryofa twodimensional region. (Intwodimensionsthisregionisunique. Inthreedimensions therearemany surfacesboundedby thispath. Thinkof the closedpathas therimof a butterflynet. Asthenetwigglesinthree dimensions, thatrimcontinues tobeitsboundary.) Thepathnolongerhas a boundary.We summarizethisby sayingIf PaP =aaSoStheno.(7.10.3)(7.10.4)Consider next the case of S, the butterfly net. It has a boundary oS =P, where Pis therim.Shrink theboundarytoa point. Two thingshappennow: Thesurface S isnow closedandisthe boundary of a volume W: S = aV. The surface nolonger has a boundary!Once againweseeIf SoS =aaV2Thissectionispurely foryour amusement.=aVtheno.(7.10.5)(7.10.6)Vector CalculusAll thesecasesare summarizedby one powerful identity:r;2 =0.191(7.10.7)If you want to go beyond our intuitive treatment of this result, youmust lookit up under the headingHomology.It turns out that the identitiesEqn. (7.10.1,7.10.2)are related to thisidentity. Let us beginwith12--> -->1 "il >. dr =1>(2) - >(1) =>18P (7.10.8)(where 1 and 2 are short for rJ. and r2respectively) which states that the contribution to the integralcomes entirely fromthe boundary of theregionof integration. Letus now jointhe endpointssoastoobtaina closedpath. Let Sbeany surface withthis closedpathasits boundary. Once thepathis closed, two thingshappen: The endpoint contributioniszerosince thereisno endpoint. WecannowuseStokes' Theoremrelatingthelineintegral tothesurfaceintegral overtheenclosedsurface.Thus1 v>.-;;,}P=8S>18P=88s=o0.(7.10.9)(7.10.10)(7.10.11)(7.10.12)---+NowSisanysurfacewitha boundary. Wecantakethena verytinyinfinitesimal surfacedSon---> -->whichthevector field("il x "il isessentiallyconstant. Inthat casethesurfaceintegral isjust-; -; ---1' ---+ ----t'"il x "il >. dS. If thisis to vanishfor any orientation of the surface, ("il x "il1 mustitself vanishgivingustheidentityEqn. (7.10.1).Consider next Stokes' Theorem{VxW.dS= 1 W.(l;,Js }P=8Swhereonceagaintheentirecontributioncomesfromtheboundaryof S. Let usnowshrinktheboundaryP=oStoa point. Thisclosesthesurfaceandmakesit theboundary of a volume V:S = oV. Evaluating the surfaceintegral aboveusing Gauss's Theoremwe find{---> ---> -->Jv"il. "il xWdxdydz1 VxW.dS}S=8Vi--->--->WdrP=8S=88V=O0.(7.10.13)(7.10.14)(7.10.15)Inthelast stepall wearesayingisthatthecirculationona closedsurface(suchasthesurfaceofa sphere)must vanishsince every tiny tileisfully surroundedbyother tilesso that the contributionfromanyedgeofanytiletothelineintegral iscanceledbyanadjoiningtilewithwhichit sharesthatedge.Since the volumeintegral must vanishfor any volume, theintegranditself mustvanish,givingustheidentity Eqn. (7.10.2).Tosummarize, inEqn. (7.10.9)we are taking thesurfaceintegral of a curl (of a gradient). Sotheanswer comesentirelyfromtheboundary, asthelineintegral of thegradient. Thelineintegral192 Chapter 7of thegradient oncegivesall itscontributionat theboundaryoftheboundary, which iszero. InEqn. (7.10.13)we are doing thisagain. We areintegrating the divergence (of a curl)over a volume.Sothecontributioncomesfromtheboundarysurface. But ontheboundaryweareintegratingacurl, whichgives all its contributionfromtheboundary of thebounding surface, whichiszero.Noticethat inEqns. (7.10.9-7.10.11)andEqns. (7.10.13-7.10.15)we start withtheintegral ofa (scalar orvector)fieldthat hasbeendifferentiatedtwice. At thefirst step we tradeonederivativeonthefieldfor one aactingontheintegrationregion, givingitsboundary. Thenwedoit again,tradingonemorederivative ontheintegrandandforonemore aactingontheintegrationregion.Butthisis one toomany boundary operations, for twostrikes andyouare out inthis game.The advantage of thelastwayof thinkingabout vectoridentitiesisthat it isverygeneral andapplies toall kindsof manifoldsinall dimensions.Wenowreturnto our study of higher-order operators.A second derivative that doesnot vanish andplaysa significant roleiscalledtheLaplacian. Itisdefined asfollows:A functionthat obeys82cb 82 82-+-+-8:['2 8y2 8z2\J2.(7.10.16)(7.10.17)(7.10.18)issaidtoobeyLaplace'sequation, or tobeahannonicfunction. (Recall thatthereal and imaginary parts of analyticfunctionare hannonic in two dimensions.) Thedifferential operator{7.1O.19)iscalled theLaplacian.By combining the expressionsfor gradient and divergence in other orthogonalsystems, Eqns. (7.5.19, 7.7.10),we can obtain the Laplacian in other coordinates:There areonly twoother second derivatives:->-> ->\J(\JW),--+ --+ --+\Jx\JxW(7.10.21 )(7.10.22)The first has no name and rarely comes up. The second equation is true in cartesiancoordinates and willbeinvokedlater in thischapter.Vector Calculus 193Problem 7.10.2. Verifythat therearenoothersecondderivatives wecan formstarting with scalar and vector fieldsand the"del" operator.7.11. Applications fromElectrodynamicsInthissectionwewill discusssomeideasfromelectrodynamicsthat will illustratehowtheideasdevelopedabove arisenaturallyinphysics.We begin with the flow of electric charge. Let p be the density of charge and 17its velocity atsome point. First consider a wire along the x-axis andignore the vectorial character of the velocity.The currentthroughthewire, I, isgivenby theamount of chargecrossingthe crosssectionof thewire per unittime. Byargumentsgivenearlierinthischapter,I = pvS (7.11.1 )where Sis the cross-sectional area of the wire. The current density j is givenas the current per unitareaandis givenbyj = pv. (7.11.2)(7.11.3)(7.11.4)Ingeneral;~ -J =pvisthe current density at a point where the charge densityisp andvelocityis17.Consider somevolumeV boundedbyasurfaceS. Thetotal chargeleavingthesurfaceisgivenbythe surfaceintegral of j:1-->chargeleaving W= Is j . dS.Nowit turns out that thetotal charge intheuniverse is conserved, i.e.. does not changewithtime. Furthermore, theconservation islocal: wedonot seechargedisappear inoneregionandinstantaneously reappear elsewhere. Instead we see a more detailed accounting: any charge decreasein a volume V is accounted for by the flow out of that volume. Indeed without thislocalaccountingof charge, itsconservationisquiteemptyandunverifiable: if somechargeissuddenlylost inourpart of the universe, it is pointless to argue over whether this chargeis really gone or hiding insomeremote andinaccessible part of our universe. Using the fact that the total chargein Vis the volumeintegral of p, andthe above equationfor the flow out of V, we may express charge conservationasfollows:(7.11.7)a r 3at Jvp(x, y, z, t)dr-:Is T .dB (7.11.5)-1Tl . Td3r (Gauss'sTheorem) (7.11.6)where d3r is shorthandfor dxdydzinthis case andfor whatever isthe element of volumein othercoordinatesystems, suchasr2sin ()d()dinthespherical system. Sincethevolumeintheaboveequationis arbitrary, theintegrands onbothsidesmustbe equal, givingusa - -atp(x, y, z, t) +'\7. j =0,whichiscalledthecontinuity equation. Thisequationalsoappearsinother contextssuchasfluidmechanics, whereit expressesthe conservationof matter.194 Chapter 7Wenow turnto Maxwell's electrodynamics. The cause of every fieldisthe electric charge, atrest orinmotion. The effect isalsofelt by thecharges, at rest orinmotion. Butweseeit asa twostage process: charges produce fieldsandfieldsleadto forcesfeltby charges. Asmentionedearlier,the electrostaticforce--+F = ql q2 --+er- - - " - ~ 2 (7.11.8)41Tcorbetweentwostaticchargesisseenasa twostage process: the firstchargeproducesa field--+ ql--+E = er (7.11.9)41Tcor2atthelocationof the secondcharge andthisfieldcausesa forceonthesecondgivenby--+ ---.F = q2 E. (7.11.1 0)Electric charges in motionalso experience another force, called the magnetic force, whichwasdiscussed earlier. Thusthetotal forceona chargeis givenbytheLorentzforcelaw:---+ --+ ---+ --+F = q( E+v xB). (7.11.11)Tofindtheelectricandmagnitudefieldsat a point, wefirst put a staticchargethereandmeasure--+the force(bymeasuringtheaccelerationtimesmass), givingus E. Thenweput movingcharges--+withdifferent velocities there andmeasure the forcethey feel tofindout B.Problem7.11.1. Give aset of measurements that will determine B.This thenis one half of the story: how the chargesrespondto the fields. The other half specifiestherelationbetweenthefieldsandthe chargesthat produce them.Let usbeginwiththe electricfieldduetoa staticpoint charge qat the origininMKSunits:(7.11.12)(7.11.16)--+where e: isa unitradial vector. If wenow calculate thefluxof E overa sphere ofradiusRwefindiE'dB i q 2 e:. e:dS(7.11.13)s41TcoRq 1R2do'(7.11.14) = -- ---41Tco S R2q(7.11.15) =cOwherewehave usedthe fact that theinfinitesimal area vector ona sphere of radiusr is everywhereradial andhasa magnitude r 2sin ()d()d R.concentric withthe chargedistribution, we getJ B.dSJs=av=J e;"tf(r). e;"tdSJs=av47l'r2fer)f .!!-d3rJvcO47l'R3--p3eo(7.11.25)(7.11.26)(7.11.27)(7.11.28)(7.11.29)fromwhichfollows41rR3--pfer) = 4 32'1rr cOwhichis the fieldyouwouldget if all the chargeinthe ball concentratedat the origin. Youshouldconvince yourself that Theresult canbe generalizedevenif p =per).----> If p has angular dependence,so will E and givenits integral over a sphere,we cannot workour way to thevaluesat all pointsonit. Gauss'slaw givesusjust one equationper surfaceandif only one thingisunknownaboutthefield(suchasitsmagnitude at some distance)itmay be determined.Problem7.11.2. Consideraninfinitewirealongthez-axis withchargeAper unit length. Giveargumentsfor why thefieldwill be radially outward and afunctiononly of p, the axial distancetothewireincylindrical coordinates. Byconsidering acylindricalsurface of lengthL, coaxial with----> ----> ).,thewire, show that E = ep-2--.1rrcoProblem7.11.3. Consider aninfinite plane of chargewithdensity aper unit area. By consideringa cylinder that piercesthe plane withaxis normal to the plane, show that thefieldis perpendiculartoit, hasnodependence ondistance fromthe plane and is of magnitude E= 2;0 .----> ---->Problem7.11.4. Considertheradial field E =~ e r . Showthat itsdivergenceiszeroat any1rcorpoint other thanthe origin by using the divergence formulain spherical coordinates. Next calculateVector Calculus 197thesurfaceintegral onasphereof radiusr. Arguethat if thedivergenceiszeroattheoriginaswell Gauss'sTheoremwill beviolated. Thusthedivergencemustbe givenby afunctionthathassupport only attheorigin, but have so muchpunchtherethatitsintegral isfinite. Afunctionwiththis propertyiscalled Dirac delta function. Itis denoted by the symbol 83(r> - TO) and ithas thefollowingproperties:(7.11.30)(7.11.31 )Thus 83(r> - TO) is nonzero only at TObut so singular there that its volume integral is unity. Express--->the divergence of E inthis problem in terms of the Dirac deltafunctions. We shall returnto a fullerstudy of thisfunctioninChapter9.--->Consider next themagnetic fieldB. Itis anempirical factthat there are no magnetic chargesproducingradial magneticfields. (Howthendowehaveanymagneticfields? Becausemovingelectriccharges, i.e.,currents, canproduce a magneticfield. More onthisshortly.) Theabsence ofmagneticchargesis expressedby thesecondMaxwell equation:---> --->V' . B =0 (Maxwell II) (7.11.32)(7.11.34)Now we ask about the magnetic field producedby currents. Consider aninfinite wire carryinga current 1upthez-axis. Incylindrical coordinates, wecansummarizetheexperimentsbytheequationB= e ~ : ~ (7.11.33)where110 is a constant determined by measuring magnetostatic (time-independent) phenomena. Thelineintegral of Barounda circle of radiusp (i.e., (f;. = epd)centeredonthe wireandlyinginthe x- yplaneis givenbyf---> --->1211' 1101pdB . dr = =1101.o 2npNoticethat theresult is independent of thecircleradius. Indeedit is independent of thedetailedshape of the contour, aslongasit enclosesthe z-axis.Tosee this, first deformthe circleasfollows(seeFig. 7.11): withinaninterval d, cut out thearc, pull itradially outward, rescalingit soi