Methods of Theoretical Physics I
Methods of Theoretical Physics I PHY 581
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1 Analysis of Complex Functions Let7s jump right in and begin with some de nitions De nition 11 The complecc plane C is a representation in R2 of the set 2 lt 00 and the eptended complecc plane is CUoo where 00 is the single point at in nity corresponding to the origin under the transformation 2 a 2 1 De nition 12 An argument ofz is any one of the numbers 6 2am where z re and m is an integer The principal value of argz satis es 77139 lt argz S 7139 De nition 13 A region D C C is a non empty connected open subset ofC De nition 14 P C D is a simple closed curve scc if is a continuous image of Slz lzl1 De nition 15 A function of a complecc variable 2 is a mapping from D to C Fundamental to a discussion of complex functions is the issue of differentiation We say that a function f D a C is di erentiable at the point 20 E D if f2 7 fzogt we 2 jgngolt Z 7 20 lt1 exists and is unique We then say that a single valued function fz is complep analytic regular or holomorphic in D if f z exists and is continuous at each point in D Finally7 fz is entire if it is holomorphic everywhere in C7 or meromorphic if its only singularities in C are poles see later Theorem 11 Ifw u iv fz iy is di erentiable at 20 0 iyo then u and v are di erentiable as functions of two real variables u uz7y v v7y and Lu 61 3p 7 3y 3U 31 a a 7 2 and are known as the Cauchy Riemann equations PHY581 Methods of Theoretical Physics I Fall 2005 2 Corollary 11 If f is at least twice di erentiable in D then the real and imaginary parts are both harmonic functions 0 3x2 3y i 321 321 g 672 0 3 Thus7 u and i cannot have maxima or minima in D and so any stationary points must be saddle points Hence7 their biggest and smallest values are attained only on the boundary 3D of D As with functions of a real variable7 we may consider power series expansions of complex functions For now we7ll look at holomorphic functions7 but later we7ll extend our discussion to include functions with singularities As with real functions7 if fz 2200 anz 7 then the derivative of this function is given by f z 2201 nanzn l Each power series has a radius of convergence7 R7 such that 2200 anz converges for lt R and diverges for gt R We may use power series expansions to de ne complex versions of some of our favourite real functions oozn 22 expz Z 1zi n0 39 2 i 23 25 s1nz z 7 7 22 24 cosz 17 7 4 In all these cases the radius of convergence is R 00 Note that7 with this de nition of expz7 the usual result exp21 22 exp21 exp22 holds As an introduction to more complicated functions7 consider trying to de ne the complex logarithm7 logz We de ne it to be a logarithm of 2 if z expw 5 If we write it u in then 2 e cosi isini 6 and therefore7 in particular7 e Formally7 we can write it logz logz iargz i27rk 7 k E N 7 PHY581 Methods of Theoretical Physics I Fall 2005 3 where we have de ned 77139 lt argz S 7139 With argz restricted to this region each choice for the integer k speci es a branch of the logarithm function The principal value of logz is obtained by setting h 0 Note that given some domain D with 0 D then once we specify the value of log20 for some 20 E D then logz is uniquely de ned on D This is equivalent to specifying the branch However it is impossible to de ne logz in this way in any domain which contains a simple closed curve which encircles the origin since log0 is unde ned For any holomorphic function fz which is many valued ie has several branches a point 20 which behaves in the same way as the origin for logz is referred to as a branch point In such a situation f20 may or may not be de ned There exists a power series expansion for the logarithm given by n1 z 22 23 727777 s 1eua i4gt n7 2 3 n1 with radius of convergence R 1 In addition all points on this circle also converge except for z 1 Another example of a function with branches is provided by the function it z with a E C This can be seen by w z expalogz emmmmmr 9 Here we have set 2 rem It is the choice of range for 0 for example 77139 lt 6 S 7139 that speci es the branch Note that the choice becomes unnecessary if a E Z because in this case the choice of a different branch merely changes 2 by ezmm 1 This ts with our elementary de nition of 2 Example 11 i expilogi 7 exp i 227k exp 7 lt2k 7T 10 This takes in nitely many distinct real values PHY581 Methods of Theoretical Physics I Fall 2005 4 zplane wplane 7 Example 12 w 2 where pq is a rational number in lowest terms In this case we obtain afunetion with q branches having 2 0 as a branch point Consider the special case w Write z re 0 S 6 lt 27139 note a different more convenient choice of range here Then w pew with 0 S t lt 7139 where p 77 and t 92 Geometrically the whole complex plane with the positive real line removed is mapped to the upper half plane see below As we have de ned f w2 2 it de nes a branch of V3 with z 0 as a branch point We must have a cut along the positive real line because we cannot possibly have continuity there since f4 2 but f4ei9 a 72 as 6 a 27139 from below 11 Integration and Cauchy s Theorem We shall consider curves 0 in the z plane parameterized by a single real variable t that is to say 0 70 W wt l t6 071 7 11 where t and 7 are continuously differentiable functions of t A particularly useful example is when G is a union of line segment or in other words a polygon It is useful to be able to de ne the length of such a curve In general yt is called reeti able if supElength of inscribed line segments exists and if so then this quantity is referred to as the length of the curve It is clear that a union of line elements is recti able and that the length is equal to the sum of the lengths of the included line segments PHY581 Methods of Theoretical Physics I Fall 2005 5 Proposition 11 Subject to the conditions introduced so far 1 fzdz uio iiZdt 12 C 0 Theorem 12 Cauchy7s Theorem Let D be a simply connected domain in C and let G be a piecewise di erentiable simple closed curve entirely contained in the interior of D then if f z epists for each point z E D fz dz 0 13 0 Note that simple connectivity of D is used to ensure the existence of a domain D1 C D whose boundary is C From the point of view of physicists7 this is one of the most important theorems of complex analysis The proof is somewhat involved7 and I will not go into it here Rather7 1711 just give a simple demonstration that the theorem is true under the stronger assumption that fz is holomorphic in D ie f z exists and is continuous in D Proof 13 fz dz udx 7 ydy i odd udy 14 C C C lfC bounds the domain D1 we may apply Green s theorem do do do do 7 77dd 777dd 731lt6p6ygt zyz 7319x 3y zy 0 7 by the Cauchy Riemann equations Of equal importance to physicists is a result that allows us to calculate the values of integrals As well see7 Cauchy7s theorem is required to prove this Theorem 14 Cauchy7s Integral Formula Let f z eccist inside the open disc B207r Then for each point a with la 7 20 lt r we have 1 z fa 7 lt 2m 3327a dz 16 here 3B 2 l 2 7 20 PHY581 Methods of Theoretical Physics I Fall 2005 6 Proof 15 f2 7 2 AB 2 7 a dz 7 33a 3 dz f2fad u lim z 7 0 3Bas 2 7 a 3Bas 2 7 a dz 17 The rst term on the right hand side can be made arbitrarily small because the integrand is bounded and the length of the simple closed curve 7 0 As for the second term substitute yt a 6e 7 t E 07 27f 18 and the nal answer then becomes 27r i q ieeitdt 0 8equot 7 2mm 7 lt19 as required This must seem like a lot of heavy handed mathematical machinery7 but 17d like to present one more important result before we get down to some examples of how these theorems are applied Theorem 16 Power Series Expansions Let f U 7 C U open be compler di erentiable and let B207p Q U Then 3 a unique power series f2 Z Cn2 20 7 20 n0 with positive radius of convergence 2 p in some neighbourhood of 20 Furthermore 1 f2 n 7 d 7 21 C 2m 27mm 2 7 20 1 Z for 0 lt r lt p This power series7 if it exists7 must be unique7 since we can use Cauchy7s integral formula to show that 1 cn Hf zo 22 Thus7 we have formally shown the existence of a power series that we mentioned earlier PHY581 Methods of Theoretical Physics I Fall 2005 7 12 Singularities and Laurent Series Thus far we have been concerned with functions that are well de ned everywhere We now turn to the various ways in which they can misbehave De nition 16 A function fz is singular at 20 if there is no neighbourhood of 20 in which it is holomorphic fz has a singularity at z 00 if gi7 de ned by gi7 E fz and 77 E 2 1 has a singularity at 77 0 De nition 17 A singularity at 20 is isolated iff is holomorphic on B20820 There are three types of isolated singularities 1 20 is removable if with a suitable de nition of f20 we obtain a function holo rnorphic on all of B20 e 2 20 is called a pole of f if for some in 2 1 the function gz z 7 20mfz has a removable singularity at 20 The smallest value of m for which this condition holds is called the order of the pole at 20 3 Otherwise 20 is called an essential singularity For an example of a non isolated singularity see fz cscz 1 De nition 18 A function fz is meromorphic on the domain D if all its singularities in D are isolated and non essential Lemma 11 Iffz is meromorphic then locally fz is eppressible as a quotientgzhz where g and h are holomorphic Conversely ifg and h are holomorphic on a domain D and h f 0 then is meromorphic perhaps with removable singularities Proof 17 i is clear from de nition just take hz z 7 20 39 If h20 31 0 then is holomorphic near 2 20 If h20 0 then since h is holomorphic we may use a power series eccpansion about 20 to eppress hz z 7 20mh20 with h20 31 0 Furthermore gz z 7 20kg20 for some h gt 0 it may be that g and h vanish at the same point 20 in which case gh has a removable singularity at 20 We can now de ne 23 PHY581 Methods of Theoretical Physics I Fall 2005 8 Note how the following de nition parallels that of a pole and its order De nition 19 Let f U 7 C be a holornorphic function de ned on some open subset U C C and let f20 0 We say that 20 is a zero off and de ne its order k by the condition fltzogt 7 f ltzogt 7 7 fk 1zo 7 0 f s 0 24 If no such nite value k exists ie fi20 0 V 0 S i lt oo7 we say that 20 is a zero of in nite order In this case7 the assumption of holornorphy irnplies f20 V 2 in some neighborhood of 20 121 Laurent Series well now generalize our power series discussion to functions with singularities Recall the expansion we derived earlier for holornorphic functions7 and compare to the following If fz is holornorphic in the annulus R1 lt l2 7 20 lt R2 note7 f not required to be holornorphic at 20 then we may express 0 fz Z anz 7 20 25 n7oo and this series is referred to as a Laurent series for f This can be decomposed as 00 71 fz Z anz 7 20 Z anz 7 20 26 n0 n7oo where the rst term is referred to as the subsidiary part and the second term as the principal part Both parts are unique7 and7 as earlier7 7 f2 aniCWdz 27 with C a simple closed curve containing 20 in the domain of f The Laurent series converges absolutely7 and uniformly in any closed subset of the annulus PHY581 Methods of Theoretical Physics I Fall 2005 9 13 Calculus of Residues and Contour Integration After all the waiting7 well get down to many applications of what we7ve learned in this section De nition 110 Residue The coe cient any of z 7 20 1 in the Laurent expansion is called the residue of fz at z 20 Note that Cauchy7s integral formula gives c e i fzdz 2s 1 7 2m 0 where C is a small circle of radius 8 say7 about 20 This leads us to the most important result of complex analysis as far as well be concerned in this course Theorem 18 Residue theorem Cauchy If fz is meromorphic in some do main D which contains the subdomain D1 bounded by the simple closed curve C with no singularities on C then C fz d2 2m R7 7 29 where R is the residue at the isolated singularity at z 27 j 17 27 7 N in D1 Proof 19 Easy The problem with this result is actually to calculate the residues Rj Two hints 1 Use the Laurent expansion about 27 if this is easily calculated 2 Assume that 27 0 is a pole of order m for Then Rj is given by 71 Rj m 7 1 i133 dzm l lzm zll 30 Note that for a simple pole in 1 Bi limzn0zfz Proof 110 Assume that f2 i wquot 7 31 n7m and di erentiate term by term PHY581 Methods of Theoretical Physics I Fall 2005 10 Now the main point of many of the recent results is that they allow us to analyti cally evaluate many de nite integrals which would seem almost impossible without the techniques we7ll learn here Although these methods will apply to complex integrals we7ll see that they provide an excellent method for the evaluation of real integrals In all the examples that follow the procedure is the same We have a de nite real integral to evaluate and we do this by rst making the integral complex and including the range of integration eg 7a a where a frequently tends to in nity inside some suitable simple closed curve C in C With minor adaptation we suppose that the complex integrand has no singular points on C We then apply the residue theorem and arrange that all contributions to the integral other than that on the real axis are vanishingly small Warning Watch out for branching Recall that this occurs for example when f contains a factor x where a is non integral or logz We shall consider various different cases 1 Integrals of the form 27r Rcos6sin6 d6 32 0 Method write 22 1 cost9 sint9 2 271 33 ie z 6w dz 22 d0 and take 0 to be de ned by 1 E0 fz meromorphic with nite number of poles in the region gt 0 and no poles on the real axis Assume that 22f2 S M whenever 2 0 and gt R say Label the poles as 11 ak Then 00 k f dx 2m 2 Resfaj 34 00 F1 To see this integrate around a contour consisting of an upper semicircle of radius R together with the real interval 7PM B As R a 00 we obtain the above result 00 The same general method applies to fxeim dab 35 700 subject to the weaker assumption that limzH00 fz 0 for 2 0 m E R PHY581 Methods of Theoretical Physics I Fall 2005 r 11 Lemma 12 J0rdan7s Lemma Let P be art upper semicircle of radius R eeri tered at 0 Let fz have a riite number of poles or removable singularities iri the upper half plarie gt 0 arid let limgTSo fz 0 Theri 6imzf2d2gt0 as R 007 36 1 form gt 0 Proof 111 Choose B so large that lt e V 2 oh P Now leimzl lexpimRcos 6 isin exp7mRsin0 37 Therefore F fzeimz dz 07r fRei9 expimRei9Rei9 d0 lt 50 e mRsmeRdQ 2R5 AWZemRsmedo lt 21750W2e 2mmW do 717 WE in 7T8 lt77 m lt38 as required Note7 that we can distinguish between real and imaginary parts7 and thereby replace eimz by cosm2 or sinm2 We can adapt this general technique to allow for a nite number of singular points on the real axis Diagramatically7 we replace our original contour by In this form7 the following lemma can be useful PHY581 Methods of Theoretical Physics I Fall 2005 U 12 Lemma 13 Suppose that fz has a simple pole at z 0 and we integrate round a circular are of radius r between the angles 91 and 02 Then 9 9 fz dz residue at 0 gtlt i02 7 01 er 7 39 r52 24mg2 2 where er a 0 as r a 0 Proof 112 Eppand fz as residue at 02 regular function and use Jor dan s lemma Branching Problems Let fz be rational7 and 22f21 S M outside a large semicircle Assume that fz has at worst a simple pole at z 07 and that there are no other poles on the real axis If 0 lt a lt 17 we can determine A glam clip 40 as follows Cut C along the non negative real axis Then 2 rem is uniquely de ned with 0 lt r and 0 lt 6 lt 27139 Write the logarithm over this domain as 2 a logr i67 and de ne 20 em With these conventions we have xed the branches of logz and 20 with which we work in this problem Choose 8 gt 0 so small and R gt 0 so large that the contour encloses all poles except the one that may exist at z 0 B PHY581 Methods of Theoretical Physics I Fall 2005 13 Then R limj z fz dz z fxdz lim z fz dz 5H0 7m R fezma z fz dz 7 41 because on 75 we are about to cross over to another branch of the integrand Therefore7 lim lim z fz dz 1 7 627m z fz dz aU USU39y 0 Race 5410 2m Resj 42 6 With logz rather than z in the integrand7 it may be more convenient to replace the given contour by 7 For some integrands it may be useful to use a large rectangle7 for example if a trigonometric or hyperbolic function appears in the denominator 7 in 15 in 14 Worked Examples Contour Integration 141 Example 1 0 dz 0 z2 12x2 4 PHY581 Methods of Theoretical Physics I Fall 2005 14 As with all our examples we choose a contour and extend the integral around the contour Here consider f d2 43 1 22 1222 4 7 where P is de ned to be the contour consisting of that part of the real axis from 7R to R R gt 07 and the semicircle7 radius R7 center 0 in the upper half plane iR Then dx 13 dr dz R lt42 41W 4 4 1 o lt22 1222 4 4 Now let R a 00 Clearly the second term in the above goes to zero in this limit We now evaluate the left hand side using the calculus of residues The integrand has a pole of order 2 at z 2397 and a pole of order 1 at z 22397 which lie within the contour P Now d 1 R 39 l 7 es 2 lltz4ltz4lt224l i 2 22 lim 7 H39 lltz 4322 4 4 lt2 44W 4 42l 7 42 i 2239 7 323 92 239 i 7 44 Similarly Res272239 lim T T 242439 221222239 239 7 45 Therefore f dz 2 z z m iiii r2212224 36 36 7r i 46 9 PHY581 Methods of Theoretical Physics I Fall 2005 15 So we obtain ltgtltgt dz 7 7T 47 700 z212z24 9 39 Finally7 since the integrand is an even function of z7 this implies 00 dz 7T 7 i 48 0 z2 12z2 4 18 142 Example 2 7agt0 cosz dz 700 z2 a2 This requires a little more cunning Consider f eizdz 1 z2 a2 with P the same contour as in example 1 We then obtain z2a2 7R z2a2 7R z2a2 6 dz R cosz dz R z sinz dz 6 dz 7i c z2a2 Obviously7 the nal term goes to zero as R a 007 thus f 6 dz 0 cosz dz 0 sinz dz 7 7 Z 7 rz2a2 foo z2a2 foo z2a2 The integrand on the left hand side has simple poles at z iz39a However7 since a gt 0 we have chosen the above contour7 and hence only require the pole at z 211 enclosed by the contour So Reszz39a lim 6 zaia z 20 6 04 49 2m Therefore 00 cosz dz 0 sinz dz 6 7T 7 7 2 7 a 50 L00 z2a2 2Lm z2a2 m 2m 16 Finally7 taking the real parts of both sides we obtain 00 cosz dz 67a 39 51 700 z2a2 a PHY581 Methods of Theoretical Physics I Fall 2005 16 143 Example 3 00 z 7 sinz d2 52 700 23 We7ll begin by integrating by parts twice7 to put the integral into a form that is more simple to handle by complex methods 00 z 7 sinx x 7 sinz 0 1 00 17 cosx LOT TWWWT 71 7 cosx 100 s1nx Ch 22 700 2 700 x l 00 i 7 s1nz d7 53 2 700 2 since both terms in the square brackets vanish Now to use our complex variable machinery Consider eiz 7d 54 AZ 27 lt gt where P is the contour shown Now7 this integral is zero7 by Cauchy7s theorem We may write it as 12 i2 7 ix 12 R iz 67d267d2 67d2 67d2 67d2 55 r 2 C 2 7R 2 0 2 39r 2 Now7 as R 7 00 the integral around the large semicircle7 0 becomes zero Thus7 7 67d2 dz7 67d2 56 700 2 39r 2 0 2 Now7 since 6 7 12 has a removable singularity at the origin7 we have 7 ix 0 ix 1 67d2 dz 7 7d2 700 2 39r 2 0 2 0 1 ew d0 7 7r rem m39 57 PHY581 Methods of Theoretical Physics I Fall 2005 17 Letting r 7 07 we have 00 eim 7 dz m 58 700 x Taking imaginary parts we nally obtain mwdzn7 59 700 so that by our initial integration by parts f 7 mm dz 3 60 700 p3 2 2 Exact and Approximate Evaluation of Sums and Integrals De nition 21 An asymptotic sequence is a set offunctions such that n1z o nz 7 as 2 7 20 61 Usually we take on 2 and 20 00 De nition 22 If nz is an asymptotic sequence then the asymptotic expansion for afunction fz is 00 f2 N Earwz 7 62 r0 provided that n71 f2 EWMZ 00 0r 0 n71 7 63 r0 as 2 7 20 ie the remainder after n terms is smaller than the last included term or the same order as the rst neglected term Some important properties of asymptotic expansions are Here consider f N Earz l always 1 Asymptotic expansions depend on the sector ie argz For example7 1 1 7T 1 7 7 lt 7 64 6 2 2 7 larglt2gtl 2 ltgt no more terms7 since e is smaller than any power of But7 1 1 37139 7139 1 7 7 7 lt lt 7 65 e sz 7 2 larglt2gtl 2 ltgt lfthe asymptotic expansion of fz is different in different sectors7 we say it exhibits Stokes phenomenon PHY581 Methods of Theoretical Physics I Fall 2005 18 Theorem 21 If fz is single valued and holornorphie for 2 a and fZV EarZ T 7 66 r0 is valid for all argz ie doesn t emhibit Stokes phenomenon then the series is in fact oonuergent ie fz ITZTT 67 n Proof 22 f is single valued holornorphie for 2 a therefore fz 23 00 enz with 1 f z n 7 7 d 68 C 2m 02 Z Choose 0 to be a large circle radius R Then 1 27 1 lt47 waka w loLQWO e R ltgt Now since fz N 23 aTZ T f a a0 as a 00 Therefore we can nd M such that lfl lt M for large enough This implies that M M lt E n gt 0 70 But R can be as large as we like so on 0 for n gt 0 Also an 0 since asymptotic empansions are unique see newt property E0 For a given range of argz the asymptotic expansion of fz is unique To see this let fz N 23 anz as 2 a 00 in a given sector Then f N a0 as 200 and in m an as 2 a 00 Similarly n71 f 7 Z aw z a an 72 0 as 2 a 00 Thus the coef cients an are uniquely de ned Note that the converse does not hold 3 Asymptotic expansions can be added and multiplied as if they were convergent Let7s now see how we might calculate asymptotic expansions for several different classes of functions PHY581 Methods of Theoretical Physics I Fall 2005 19 21 Watson s Lemma and Laplace s Method Lemma 21 Watson Let MmeM m wa25gt0 am 0 with b 23 bnt for M lt R Then FZ m It is important to note here that the right hand side is merely the left hand side expanded and integrated term by term However it is the fact that the result is an asymptotic expansion that is nontrivial This is because the summation need not con verge uniformly in t for all t in the range of integration Thus it is not clear that we can interchange the order of integration and summation Under more restrictive circumstances we could just integrate by parts to show this However Watson7s lemma works in more general situations and a more subtle proof is required I wont give the proof here although if we have time I may come back and supply it later Laplaee s method is a way to calculate asymptotic expansions for functions of the form FmfWm am a as x a 00 x real The rough argument is that the largest contribution comes from the biggest value of hu say hu0 which is exponentially larger than any other contribution We7ll see how this works in 2 distinct situations In both these Watson7s lemma is crucial to obtaining the nal result 1 h u0 0 g a calculus type maximum Begin by taking Taylor series of h and 9 about uo b 1 exp r hu0 7u 7 u02h u0 2 0 1 emw exp yer110 gltu0gtidn lt76 9 u i u0g u0 du PHY581 Methods of Theoretical Physics I Fall 2005 20 where 739 u iuo and we can extend the range of integration to 7007 00 since any extra contributions are negligible the dominant contribution comes from 739 0 Now integrate term by term using Watson7s lemma7 to obtain 27139 mh u 73 2 N 5 O 9W0 W OW 77 2 h u0 7 0 In this case we have uo b or 1 Now a Taylor expansion about uo yields 0 F96 e h 09uo GEM 0 d7 1 N W 0 2 78 e gltuogthuo z gt lt gt Lets see immediately how this works by applying what we7ve just learned to an example that is well known the result at least to some of you Consider expanding Px 1 as x a 007 for z real If you know what the P function is7 you7ll know that the answer we hope to get is known as Stirling s formula7 and is very useful in all types of situations in physics If you havent heard of the l function7 then this will still be a good example of how to use Laplace7s method The P function has an integral expression given by m 1 e ttmdt 79 0 Although this appears to already be in the correct form to apply Laplace7s method to7 we must transform it because the largest value of the exponential occurs at t 07 where tw vanishes Therefore7 we7ll write P1 0 eit logmdt xx 0 ewltu10gltugtgt du 80 0 where we have made the change of variables t zu not because it is essential7 but because it makes things neater because the position of the maximum stays at a xed point and doesnt go to in nity as we take the asymptotic limit PHY581 Methods of Theoretical Physics I Fall 2005 21 Now hu 7u logu has a maximum at u 1 We will only be interested in this example in getting the leading term of the expansion Therefore we Taylor expand everything about u 1 as far as the rst non constant term This gives 1 N95 1 zw exp x h1 u 7 12h 1 du 0 N z167m 00 67ms22 d8 7 81 71 with s u 7 1 The N here comes from Watson7s lemma We could have expanded to higher order but have chosen not to We can extend the limit of integration since any contribution from the range 700 71 is subdominant Thus to leading order we obtain 2 12 W 1 N Weir 82 which you may recognize as the leading term in Stirling7s formula 22 RiemannLebesgue Lemma and Method of Stationary Phase Lemma 22 RiemannLebesgue Let qt be piecewise continuous on the compact interval ab Then for real z b IQ E emqt dt 01 as x 7 00 83 Proof 23 Assume wlog that qt is continuous on ab so that for any given 6 gt 0 the interval ab can be divided into n 7 1 subintervals in each of which qt varies by less than 26 Then 3tn such that a to lt t1 lt lt tn b with qt 7 qtk lt 6 fort E tk1tk Also qt is bounded in ab so 3Q such that qt lt Q V t 6 ab Then n ti I ti Iltzgt qm t dt Z t W 7 qmem dt lt84 1 23971 1 23971 Now ti I imti 7 iztiil emdt L g 3 85 tH ix a and ti t W 7qtemdt ea 7t1 86 23971 Putting these together we obtain 2 W901 S an 6b i a 7 87 which can be made as small as you like by choosing 6 small enough andor choosing z large enough PHY581 Methods of Theoretical Physics I Fall 2005 22 The method of stationary phase is a way to calculate asymptotic expansions for functions of the form Ix b airmail du 88 a with h twice differentiable and 9 once differentiable as x a 00 x real The rough argument is that the largest contribution comes from the place where the integrand oscillates least7 since where rapid oscillations occur7 one expects cancellations to occur More formally7 making the substitution hu t7 and using the Riemann Lebesgue lemma7 we see that the above expression is 01 unless there7s a place where h 0 3 Solution of Ordinary Differential Equations Let I be an interval of the real line C I is the set of functions f de ned on I such that d f 7 E D E r 89 M f f lt gt exists and is continuous lf g 6 C I7 then so are f g and af Thus7 C I is a vector space De nition 31 Suppose add 0 S i S n are de ned and bounded on I an 31 0 Then L ow a 001 f a Lf 96 ZaMXDifX 90 i0 is a linear di erential operator LDO of order ii If anx 31 0 on I then L is normal De nition 32 Let L be a LDO of ordern on I and let f be n times di erentiable on I An equation Ly f 91 is a linear di erential equation LDE of order n on I If f E 0 on I the LDE is homogeneous We refer to solutions of the homogeneous equation as complementary functions CFs7 and the set of CFs as the Kernel of the operator L Speci c solutions of the non homogeneous equation we then refer to as particular integrals Pls PHY581 Methods of Theoretical Physics I Fall 2005 23 31 Normal LDES of Order 1 The general form is Camix loanz f 96 92 on I Since a1x 31 0 on 17 we divide by a1x and rewrite as Mr Myth96 Wt 93 We rst consider the homogeneous equation 11W Myth96 0 94 De ne the integrating factor as 6P7 where Pltzgt zpltugt du 7 lt95 which exists7 since we assume that p is bounded Then7 yo py y pyeP Thus7 if satis es the homogeneous equation7 then yo py 07 which implies Ce Pm 96 Now consider the non homogeneous equation Clearly yo py rep Therefore the general solution is Ce Pm e Pm rueP du 97 Example 31 Show that 962 Uzi96 1 zz96 966 98 has solution 99 PHY581 Methods of Theoretical Physics I Fall 2005 24 32 Normal Second Order LDES with Constant Coef cients The general form is My y p96y 1001 r06 100 Once again here are two sub problems to solving this equation determining the Kernel and the particular integrals In general both are dif cult Example 32 Lly Ey 3y 2y pzew 101 Consider rst the CFs Try y e 02302e 0 e 71or72 102 m 21 Therefore independent solutions are yl e and y2 e and the CF is a linear combination of these The fastest way to nd a PI is to guess one Guess y of bx dew Then routine algebra gives a 16 b 7518 d 19108 The general solution is therefore 1 ele m Cgeizm EOSH 7 30x 19ew 103 Example 33 Lly E y 2y y cosx 104 Consider rst the CFs Try y e Yields 0 71 repeated So there are not independent roots here To nd the other linearly independent CF try y uxe m This then gives Lly u e m 0 105 So u apb Therefore two linearly independent solutions are yl e m andyg ze w Now guess a PI y ecosx dsinp This gives 0 0 and d 12 Therefore the general solution is 01 02m sinw 106 PHY581 Methods of Theoretical Physics I Fall 2005 25 Why did this trick work in the latter example This is a particular case of reduction of order If one solution of a nth order LDE is known7 the equation can be converted to an order n 7 1 one Let7s verify this explicitly when n 2 L191 y M96 qy 0 107 Suppose y Mr is a solution Then try Then Luv 11 211 1 uLM puv 108 But LM 07 so by writing w u we obtain the rst order equation i w lt2Zpgtw0 109 for w 33 Green s Functions This primarily concerns nding PIS for second order equations7 although the concept can be generalized to higher order systems The technique assumes that one can nd the general CF There are two standard cases in which to do this First7 consider y Ay By07 110 where A and B are constants Let 711 and 712 be the roots of n2 An 1 B 0 Then 7 016mm 1 0267mm 711 7 712 01 026n1z n1 n2 7 as we have seen Second7 consider A B yn iy 7y 039 r r Now 711 and 712 are roots of nn 71 An B 0 Then sz0m n n yltzgt 1 2 1 2 113 01 021Hn1 n1 712 where the second solution here can be found by reduction of order PHY581 Methods of Theoretical Physics I Fall 2005 26 331 Initial value Problems In this section wlog we will use I 0 00 and for appropriateness of notation will use It time instead of x as our variable The general problem is MMU f6 7 Y0 yo 7 114 If this solution exists it can be shown to be unique By assumption we can solve the homogeneous problem with the same initial condition We therefore consider the standard problem MMU f6 7 Y0 0 7 115 since a solution of the homogeneous problem with the given boundary condition added to a solution of this equation is the general solution to the equation Let us write the standard problem as Llyl E W 1000116 10096 1 00 7 116 with y0 0 A heuristic approach is as follows Suppose we can solve 116 when ft 6t 7 s with 5 xed Let the associated solution be Gts ie MG 6t i s 117 with G0s Gt0 s 0 Now consider gm Gma aw no 0 Clearly y0 0 and yt0 0 Also mm fummwww f Amatis sds ft 119 Thus 118 is the solution of 116 PHY581 Methods of Theoretical Physics I Fall 2005 27 Now7 what does 117 mean Clearly7 if t 31 5 we can assume that G is a smooth function of t In addition7 assume that G and G are bounded as t a 5 Now integrate 117 fromtsietose7egt0 5 95 5 GttthqGdt 6tisdt1 120 Thus 95 otijtg th qG dt 1 121 By assumption the integrand is bounded7 and so the integral is 06 as e a 07 so in this limit we get Gt 1 122 Thus7 G is not continuous7 but has a jump of 1 at t s Lets be a little more formal about all this De nition 33 The Green s function for the initial value problem posed earlier is a function Gt7 s satisfying 1 fort 2 0 s 2 0 t7 s G is smooth and LG 0 for cced s 2 G07s Gt07s 0 for s gt 0 3 G is CO at t s but Gt f 1 De nition 34 If y1x and y2x are linearly independent solutions to a second order LDE then the wronshian is Wiyh 121 E lawn2amp6 291 7 123 and can be shown to be nonzero on I Lemma 31 G epists and is unique Proof 31 By epplicit construction Let y1 y2 be two independent solutions of Lly 0 so that the wronshian is nonzero Let 0 tltsltoo 0 G028 0191t 021209 0 lt 5 S t lt 00 124 PHY581 Methods of Theoretical Physics I Fall 2005 28 Clearly the rst two conditions are satis ed We need to impose continuity at t s and ajump of1 in Gt These conditions read 0191502925 0 Cly18 ng28 1 By the de nition of the wronshian there epists a unique solution Cl 71125 108 915 7 126 CZ ws So given these de nitions the solution of the initial value problem is W Gt7sf8d8 127 0 Example 34 y wa 7 128 with t gt 0 y0 0 The Green s function is sinwt 7 sw show this Therefore the solution is yt OtGtse 5ds 1 t is 7 50 e s1nwtis ds 1 i t 1 wz equot snafu 7 coswt 129 Example 35 W 7 t2 2m t 2n W 7 130 with y0 0 We have 2 1 2 A LlylE ilt1Egtyltgt gtyfEti2 131 By inspection one solution of My 0 is y t Using reduction of order the second solution is tet Set 0 t lt s Gltt7sgt gt S 7 132 Continuity at t 5 implies c1 c2 0 A jump of1 in Gt implies c2 15 7C1 Therefore 8 tOt 15H 1 fsds 133 yt Atltet sitfsds PHY581 Methods of Theoretical Physics I Fall 2005 29 332 TwoPoint Boundary Value Problems New set I a7b Consider a n th order system The kernel has dimension n In a 2 point pr7 we impose m gt 0 conditions at z a7 and n 7 m gt 0 conditions at z b7 to x a complementary function Such a problem may have 07 1 or in nitely many solutions Example 36 Consider y y 0 for which the candidate functions are sinx and cos x Consider the following possibilities for boundary conditions bcs Z y0 1 y 7r 0 This has one solution cos x 2 y0 y7r 0 This has an in nite number of solutions Asinz for arbitrary A 3 y0 0 y 7r 0 This has no non trivial solutions Example 37 Consider y 7y 0 and consider the following possibilities for boundary conditions bcs Z y0 1 y bounded as x a 00 This has one solution e w 2 y0 0 y bounded as x a 00 This has no non trivial solutions De nition 35 Suppose the problem 0 with boundary values at z a and z b has no non trivial solutions Then a b are conjugate points De nition 36 A boundary condition Cy7 a is homogeneous if whenever it is satis ed by y it is also satis ed by Ay with A an arbitrary constant Homogeneous bcs usually come in the form c1ya egya 0 7 134 for example De nition 37 Consider a 2 point bup My E y p96y WM fz 135 z E a7b with bcs Clyy a C2yy b The Green s function Cz satis es PHY581 Methods of Theoretical Physics I Fall 2005 30 Z Gz7 is smooth LG 0fora S x and S b x 31 g 2 Considered as afunetion of x G satis es the has 3 G is CO at z but G1 has ajump 1 I will state7 but not prove7 that if the bcs are homogeneous7 and a7 b are conjugate7 then G exists and is unique It can then be shown that the solution of the problem 135 is b we Goa om d5 136 Example 38 1 96 Wt f96 7 137 on 077139 with y0 0 y 7r 0 It is easy to see that the homogeneous equation has solutions y1x sinx satisfying y0 0 y2x cosr satisfying y 7r 0 138 The wronshian is then it 71 so that the Green s function is Gltz7 gt 132 8 7 lt13 and the nal solution to the problem is 7 cosxOm sin f d 7 sinsm7r cos f d 140 4 Transform Calculus 41 The Fourier Transform I711 assume that you know something about Fourier series Suppose g is continuous on 77139 to 7T7 and that giir 0 Then 9a 2 One 7 141 where 1 w on 7 myd 142 2W 49106 1 PHY581 Methods of Theoretical Physics I Fall 2005 31 Now consider changing the interval to 7L27 L2 Set y Lox w 27rL7 gy x on LC Then we have f 96 2007mm 143 with L2 en fpe wm dp 144 iLZ In the sum7 h E Lon changes by Ah 27rL at each term 1 fp genelkmAh 145 We now take the limit as L a 00 The sum now samples points increasingly close together and in the limit becomes an integral Denoting on by we obtain the relationships fa l fltkgtemdk k fze dz 146 De nition 41 Suppose fid x dz lt 00 Then de ned by the above is the Fourier Transform and the eppression for fx is the inversion formula Note that I will use these de nitions consistently7 however7 physicists often switch the signs in the exponents7 and make this more symmetric by having a 1 27139 in front of each integral These are just issues of convention 411 Fundamental Relations The so called shifting relations are extremely useful Lemma 41 Shifting Relations Suppose epists Let 9p fz 7 0 and hp eiMfx with A and p0 constant Then k G ikmo k Mk k 4 A 147 Proof 41 Do it in class Easyl PHY581 Methods of Theoretical Physics I Fall 2005 32 Lemma 42 Suppose exists Let 9a ax with a 31 0 real Theh gltkgt 215 148 al a Pr00f42 gltkgt Oofcw6 md 0 dy i ih ai 7 829namey6 y a 1 h Wf a 149 Lemma 43 Suppose 9a f a hh d dh f h Theh assuming all the integrals converge 7 W00 fizfx 150 D z A A 82 P V V H Proof 43 Easy Simple extensions of these results show a general trend that the faster f falls off as x a ioo7 the smoother is7 and Vice versa An important point is that Fourier transforms can be used to solve differential equa tions with constant coef cients 21 gay96 qu Wt 151 Because of linearity7 270217 mm qzj f 152 and so 1 00 k m W Lmdk 153 This is clearly related to the Greens function approach7 and we shall return to it later PHY581 Methods of Theoretical Physics I Fall 2005 33 412 A Digression on Distributions The delta function7 5z7 is an example of a generalized function7 or distribution some thing which may fail to satisfy either smoothness7 boundedness7 or asymptotic properties required of a given class of functions7 but which can still be manipulated like a function For example7 neither the delta function nor the function f 1 comes in the class of functions for which Fourier transforms are normally de ned7 but the results 5k 1 i 2m5k 7 154 are familiar For these purposes7 both these functions must be regarded as distributions De nition 42 Let f be a class of good functions on 700700 for emample C with edponential decay at ioo Then g is a distribution with respect to f if ltf7g de ned by ltmgt E lfltzgtgltzgtdz 155 is nite V f E f ie for all test functions Note that a different de nition of good77 would lead to a different class of distribu tions but 0 is usually required because this implies that the derivative of a distribution de ned below is also a distribution With the de nition above7 the space of distributions which is dual to the space of test functions has many of the nice properties of a space of functions eg linearity Each distribution is de ned by its action on test functions For example7 6W is the distribution de ned by lt6 fgt f0 w e r 156 The derivative of a distribution 9 is de ned by lt9Cfgt 7lt97f gt W E f 157 ie by integration by parts The Fourier Transform of a distribution 9 is de ned similarly lt 7fgt lt97fgt W E f 158 It is straightforward to show that most of the properties of Fourier transforms hold also for the Fourier transforms of distributions PHY581 Methods of Theoretical Physics I Fall 2005 34 1 The Fourier transform ofthe delta function By the above de nition lt5 fgt lt6 The RHS is L 6ltkgtfltkgt dk fltogt L we dz 2 L M dk 159 and the LHS is 5kfk dk 160 E0 00 The Fourier transform of the Heaviside function A Comparing these which must be equal for all test functions fx gives the result 5001 The Fourier transform of a constant By the above de nition lt1 fgt lt1 fgt The RHS is 001fkdk L k dk 2M0 161 and the LHS is L00 1fkdk 162 These must be equal for all test functions so 1 2W6k This result is consistent with the Fourier inversion theorem but the conditions of the theorem do not hold here naive approach gives the wrong answer One could argue that since H z 6x and for any f f k z k k then since 1 it follows that 1 which is correct However it does not follow that 1z39k because when distributions are allowed the full solution of the equation 1 should be Hk A6k 163 where A is a constant which is not determined by this method Since H7z 1 the real part of the Fourier transform of H must be the Fourier transform of 12 Therefore A 7T PHY581 Methods of Theoretical Physics I Fall 2005 35 413 Convolution Integrals Unfortunately there exists no simple formula relating E to f and g Instead ng need not exist There is however another kind of multiplication which is physically very important and for which Fourier transforms are easy to evaluate De nition 43 Suppose fowl zdx lt 00 and 30 lglzdz lt 00 The convolution of fp and 9p is mm l fy9zydy 9ufudu 9f96 164 Note that fgtk6 f for all f Thus 6 is the identity for considered as multiplication Theorem 44 The Convolution Theorem Suppose f g and f gtk g epist Then fg epists and m fi 165 Proof 45 See homework A useful result obtained from the inversion theorem to be seen soon is so emdk 166 Theorem 46 Rayleigh 1899 Plancherel 1910 Suppose complep t is such that and 3 00 l zdx both epist Then l fx2d96 7 mwdk 167 Proof 47 RHS l N 8 m A R V m A R V R R sz 168 PHY581 Methods of Theoretical Physics I Fall 2005 36 Theorem 48 Parseval7s Theorem If f g are real and f g and ff fgdp all erist then 00 1 00 fltzgtgltzgtdz gl fk9kdk 169 Theorem 49 Suppose u is continuous and f eccists Then 1 00 ihm fx 7 m mag dk 170 Proof 410 Let x be ped and let gt ft Then taking Fourier transforms with respect to t we have 1 27139 Lfkeikmdk if kdk fa 171 Note that if f is discontinuous at7 say7 z zo then is continuous and the inversion integral is also continuous It can be rigorously shown that 1 27139 l were dk gum o mo 7 0 172 42 The Laplace Transform In this section I will use t as the independent variable and p as the transform variable De ne the Laplace Transform by mp Ooe ptftdt3 cm 173 0 The Laplace transform traditionally treats only t 2 0 It is therefore conventional to regard ft 0 for t lt 0 In the Laplace transform7 p may be complex7 de ned at rst for Ep gt y where y is as required for convergence of the transform However7 it is important to be aware that no such 7 may exist For example7 this is true for ft etZ PHY581 Methods of Theoretical Physics I Fall 2005 37 Here are some examples7 for which the Laplace integrals are easy to compute 71 1 7 171 174 71 7 171 p i 0 171 gt 171 175 71 cosw1 7 171 1 176 71 111111 7 171 1 177 71 smhw1 7 171 132 178 71 coshw1 7 171 132 7 179 71 617 a 7 1711 7 180 71 917 a 7 1711 7 181 71 17 7 171 7 p111 182 421 Properties of Laplace Transforms The Change of scale property f7 F 183 The shift theorems e tft Fp e A 184 1197 77 7 a for 7 gt a and is zero otherwise then 191 e777Fp 185 A very important result is that the Laplace transform of a derivative is given by df 5 p ft 7 f0 7 186 and7 similarly7 we obtain dzf 2 by 1 AW pf0 7 f 0 187 The Laplace transform of an integral 5 0t 71 du rm 188 PHY581 Methods of Theoretical Physics I Fall 2005 38 Theorem 411 Initial Value Theorem m f t ggrgop p 189 provided both limits exist Proof 412 d pFltpgt Adi f0 pti f0 1 e dt dt 190 As p a 00 the right hand side becomes f0 as required provided ft is bounded near t 0 There exists a similar Final Value Theorem tlim ft lirpFp 191 Aux 1H Convolutions are also important for Laplace transforms Recall that for Laplace transforms we assume that functions vanish for t lt 0 Therefore7 in a convolution integral ht ff ft 7 du the integrand is nonzero only for t gt u gt 0 Thus we have t ht ft 7 ugu do 7 tgt 0 7 192 0 ht0 7 tlt0 193 Theorem 413 Ifh f gtk 9 then Hp Fpgp Proof 414 FltpgtGltpgt 0 WW8 ds 0 7ng do l e l sfewe ds 1 WWW du Ld 9ltugt9ltugtldse Plt gtfltsgteltsgt dugut9u dteiptm 7 um 7 u m7 du9uft u6t 7 mom 1 dt 57mm 0t dufltt umu 000 dte pt 0 duft 7 A dte pth t HQ 194 PHY581 Methods of Theoretical Physics I Fall 2005 39 To illustrate the usefulness of the Laplace transform7 we7ll tackle an example of a differential equation with non constant coef cients The Bessel function7 J0t obeys d dJO amp 1 we 7 0 7 195 with the boundary conditions J00 17 J60 0 First rewrite the equation as mg J we 0 196 Now Laplace transform and use the result t ft iddp Fp to get d 5J0 0 197 d 77 J6 513 7 dp Next 1711 write Kp E LJO and use our results on the Laplace transforms of derivatives to get 192 1K p pK 0 198 This is now a simple rst order equation that we can solve to give Kp a C 1 1 gt712 199 p p2 7 where c is a constant We can now use the initial value theorem to x 0 via 1 J00 limZHOO pKp 0 So nally7 Kp 171 1 i 712 200 Kltpgt 0 7 201 where V 1 n 271 an W 202 Finally7 we can use that the Laplace transform of t is 71lpn 7 to invert and get 01 th i0 39 203 PHY581 Methods of Theoretical Physics I Fall 2005 40 In this example we used a Laplace transform that we know to invert another Laplace transform However for general problems we7ll need a general inversion formula anal ogous to the one that we used for Fourier transforms Remember that for Fourier transforms the transform variable was real therefore the inversion formula was a real integral However as I mentioned with Laplace transforms the transform variable is complex in general Therefore we7ll end up needing a complex contour integral to invert and recover f To get to the appropriate inversion formula we7ll postulate a form for the integral and then show how it can be an inversion Remember that I said we7d need to require 3X10 gt y for some 7 in order for the Laplace transform to converge With this is mind let7s examine integrals of the form I m dp eptFp 204 77100 where Fp is the Laplace transform of a function ft We would like this integral to yield ft Substituting in for the actual Laplace transform we get 7in0 dp em 00 du eimfu0u 205 77100 700 Set p y M We then get m L dkem L due ik e l fu6u 206 But by the Fourier transform and inversion formula this is 27r elte ltft6t 207 So we can nally rearrange things to get the Bromwz39ch Inversion Formula for the Laplace transform NW 1 WW dpeptFltpgt 208 H00 Now notice that in deriving this we have not speci ed what 7 is For a given Fp y is not known a priori In fact 7 must be chosen so that the right hand side of the inversion integral is zero for t lt 0 to match the left hand side To do this start with y gt 0 and close the contour using a semicircle in 3X10 gt y gt 0 to form a closed contour C Now for t lt 0 the factor of 6 quot ensures that the contribution from the integral over the semicircle at in nity vanishes Thus for the integral around PHY581 Methods of Theoretical Physics I Fall 2005 41 C to yield zero 7 Fp must have no singularities inside 0 Therefore7 all singularities of Fp must lie to the left of the line 3X10 y This xes 7 For t gt 07 we close the contour in 3X10 lt y This gives ft fcemFpdp 7 tgt 0 209 If the only singularities of Fp are isolated poles7 the inversion integral can be performed by the calculus of residues N 7 ZRemlemFWN poles 7 Zepj mj 7 210 for poles at p pj Suppose the pole of Fp with largest real part is p 10 Then ft em as t a 007 and therefore we require 7 gt gap Let me give some examples of how to use Laplace transforms to solve ordinary differential equations7 in particular initial value problems Consider i2ixe t7 211 with initial values 0 17 0 Write zt E Xp Laplace transforming the equation7 and using our results about the Laplace transforms of derivatives7 gives l 1 p2X10 7 19960 7 960l 21pXp7 960l X0 7 m 7 7 212 which7 after a little rearranging7 implies that 1 1 1 X p 213 7 7 7 p1 19 1 19 1 But now7 using our example from earlier7 we can invert each of these term by term to obtain t2 t 6 to 56 214 Here7s another example yye2y07 215 subject to y0 1 and y 7 0 as t a 00 Write yt E Yp Transforming the equation we get p2p72Yp py01 7 216 PHY581 Methods of Theoretical Physics I Fall 2005 42 This is now easily inverted to give OHQ t 17110 2 yt 7 Te T6 7 217 Now7 requiring y a 0 as t a 00 implies that 2 07 and therefore the solution is m Ht 218 What7s interesting about this example is that we use the rst boundary condition just after transforming7 to dispose of one of the terms generated by Laplace transforming a derivative7 but use the second boundary term only after inverting the transform The next example deals with a pair of coupled rst order differential equations Consider i z 2y 621 7 2zyiz 0 219 with 20 y0 0 Laplace transforming gives p 1Xp 2Yp 2p 1Xp pYp V H D 220 V These can be trivially solved to give X p WP 221 We could do this by partial fractions and using earlier results However7 instead we7ll use this as our rst Bromwich inversion formula example Clearly we will need 7 gt 2 The right hand side of the equation for Xp has poles at p 1 and p 27 with Resp 1 MUN 1Xp6 t 6t 7 pa d M Resp 2 ppi 1 7 222 which yields zt 2t7152t at 223 PHY581 Methods of Theoretical Physics I Fall 2005 43 Similarly we obtain m 1 i 3052 i at 224 Sometimes7 a convolution trick is useful Consider 239 wzx ft 7 225 with 20 0 Laplace transforming we get Fp p2w239 Xp 226 Now7 as we learned earlier7 Gp E 1p2 LUZ is the Laplace transform of gt sinwtw0t Thus7 we can write Xp GpFp 7 227 and use the convolution theorem to tell us that t g gtk ft This reads t i t 7 t m dt who 7 228 0 w for t gt 0 Therefore7 gt is the Greens function of the problem As a nal example for Laplace transforms7 consider the diffusion equation 3 1 3U 7 229 in z 2 07 t 2 07 subject to u07t ft7 given7 u7 0 07 and u7t a 07 as x a 00 This problem could describe7 for example7 a prescribed heating ft applied to the z 0 end of a semi in nite rod7 initially unheated Perform the Laplace transform with respect to t Uz7p dte ptuz7t 230 0 Note that7 evaluating this at z 0 gives U07p ft E Now7 the diffusion equation7 using u70 07 gives 62U p w 7 EU 7 231 which is easily solved to give Ultz7pgt Altpgt expemz 32 expwz z 232 PHY581 Methods of Theoretical Physics I Fall 2005 44 Now7 since the x dependence of U and u will be the same7 we require Ux7p a 07 as x a 007 which gives Bp 0 So7 we have Uz7p Ap exp7pkz At z 07 we have U07p Ap7 and so we write UWJ Ummwmmevzia F pGp 7 233 with Gp exp7 pkz Therefore7 the convolution theorem tells us that u7t 9 i an 1 remains 0 I1 g u I OHI an ear 181 resu we HOW IS 7t d t t d 7t Btf 1 1t k th gz t th SZ exp 7L2 234 7 47Tk 4kt Therefore7 the complete solution to the diffusion equation problem is Mm gOtdt t 7 rsZap ft 235 I7ve given you a lot of examples using the Laplace transform During this time you7ve had some time to get used to the Fourier transform 17d now like to go back to the Fourier transform for one example that is of particular physical signi cance Consider the problem of nding the Greens function that satis es lt 62 i 12gt G7x 62 7 2 7 236 7 where q is a xed7 real7 positive number7 and 700 lt Lz lt 00 This Green7s function describes one dimensional scattering in quantum mechanics Set x 0 wlog7 and then Fourier transform We obtain that Wi amh em with the function we7re looking for given by the inversion formula 62 L dke ika Ug 233 Now7 we would like to solve for However7 1k2 7 qz will not do7 because it puts poles on the real k axis7 and this gives problems for the inversion integral To proceed7 we will apply Feynman s Rule This technique is extremely important in quantum mechanics and quantum eld theory Replace 12 by 12 i 2396 This enables PHY581 Methods of Theoretical Physics I Fall 2005 45 one to de ne two independent Green7s functions Gix by taking the limit 6 a 0 at an appropriate later stage Consider 1 G k kgi 7 1 k 12 12llk 12 2012 7 i 1 239 T kiqiz39e kq 7 where 6 E 62q2 and limeno and limeH0 are equivalent So we have c 2 dxe ik 240 700 kiqiz39e kq equot Now for z gt 0 we can evaluate this integral by closing the contour in the upper half k plane gt 0 and for z lt 0 we can evaluate this integral by closing the contour in the lower half k plane lt 0 We use the residue theorem and then take the limit limeno at the very end The result is 6mm 67m Gx 7W6ltgt q m 07zT 241 One can calculate G similarly Given our technique you should check that these Green7s functions obey the differential equation 5 Sturm Liouville Theory Let7s begin with an example to get the feel of the kind of problems we7ll tackle with these techniques Consider a uniform string with xed ends The displacement of this string obeys the wave equation 1 32y i 32y 02 3t 7 3x2 7 with boundary conditions y 0 at z 0 and at z l for all time To start we separate variables making the ansatz yx t XTt This yields 242 1 T X 7 77 const A 243 say Vibrations correspond to T 6 so A LUZCZ We seek solutions of X 4X 244 PHY581 Methods of Theoretical Physics I Fall 2005 46 subject to X 0 at z 0 and at z l The solutions are of the form Xz anm 245 with 71 123 Note that there exist solutions only for A E S a discretely distributed set of real eigenvalues S is referred to as the spectrum of eigenvalues The corresponding eigen functions Xn are orthogonal on 0 l The equation of motion for Xx is typical of a wide class of eigenvalue problems which arise from some partial differential equations of physics 51 General Remarks We will study at rst second order linear differential equations for z E I ab Restrict attention to linear differential operators of the form L IZDZ 7 a1D 10 for given functions Mr with 12x gt 0 on I Actually it will prove suf cient to restrict our attention to L of the self adjoin form L iDpD q 246 with p gt 0 qx given functions on I the Sturm Lz39oum39lle Problem is speci ed by the differential equation LyW MUWMW 7 247 to be solved for for z E I subject to the boundary conditions to be speci ed with wz gt 0 on I and A an eigenvalue parameter The solution has some general features 1 3 nontrivial solutions which obey the boundary conditions in use iff A E S the spectrum of the problem S is a monotonic set of discretely distributed real eigen values A so that A0 lt A1 lt A2 lt with An a 00 like 712 as n a 00 E0 The eigenfunctions yn corresponding to the An 6 S are unique to within nor malization Also yn and ym are orthogonal in a sense that we shall see soon if 9 The yn provide a basis in the in nite dimensional vector space of functions on I which obey the boundary conditions in use and suitable smoothness properties PHY581 Methods of Theoretical Physics I Fall 2005 47 52 Orthogonality and Boundary Conditions Suppose yl and y2 obey the Sturm Liouville equation Then pyi qy1 Alwyh 248 py z qy2 Azwy27 249 and suitable boundary conditions7 for distinct values A1 and A2 of the eigenvalue param eter Form the object 17 dz 42 x 248 iyl x 249 250 This yields 17 b 4142 a dzwltzgtylltzgtyzltzgt a dzieltpyagtwzltpygiylm pylwz py wlli 251 Now7 appropriate boundary conditions are those that make this vanish For then7 since A1 31 A2 we have b dxwxy1xy2x 0 252 1 This is the sense in which yl and y2 are orthogonal with respect to the weight function A good example is given by Legendre7s equation and polynomials This arises from the Laplace equation in cylindrical coordinates 7quot7 072 Writing z E cos 07 the Legendre equation is d d if 1 7 2 7 dx dx with I 711 Note that this is a Sturm Liouville problem with w 17 and 135 APz 7 253 p 1 7 2 The suitable boundary conditions are automatically imposed if Pz is nite at z i17 since p a 0 at the endpoints The solutions are a set of polynomials7 the rst few of which are P0 41 A0 4 0 254 P1 i 7 A1 7 27 1 P2 4 z 4 g A2 6 256 More generally7 there exists a unique Pn x with An nn 1 It can also be checked that the Pn are orthogonal on I PHY581 Methods of Theoretical Physics I Fall 2005 48 53 Real Eigenvalues Let us allow the possibility the An and yn are complex Then7 we may write two Sturm Liouville equations as py Y qy My 7 257 py if wy 258 We now form the object 17 dz y gtlt 257 7y gtlt 258 259 This yields A e x Abdzwltzgtyltzgtyltzr jazz py y pggt211 7 py y Mpg31111 7 07 260 for the suitable boundary conditions Now7 since wz gt 0 on I7 this implies that f drwlylz is strictly positive on I Therefore7 x A 261 ie7 A is real 54 Formal Vector Space View Let us regard suitably behaved functions x7 9z7 which obey the boundary conditions of our Sturm Liouville problem as elements of an in nite dimensional vector space V spanned by the Thus7 we can write me cam 262 De ne a scalar inner product on V by 1 79E Abd cwltgtfltgt9ltgt 7 263 and note that suitably well behaved77 requires that f7f12 exists Then7 ymym 0 for n 31 m Further7 choose the scale of yn to achieve orthonormality ymyn 67ml 264 PHY581 Methods of Theoretical Physics I Fall 2005 49 Also ymif chymymi Zcm nmi em 265 This is to be compared with the formulae for Fourier series Thus7 if we assume that the yn are known7 then for a given f we can obtain the em via an Abd w yn f 266 An important and useful result follows if we substitute this into the expansion for We obtain we Abdtwemom W 7 Abd wlttgtrtlttgtrtltzgtl f 267 Since this is true v f 6 127 we may therefore infer moimm 6ltz e 6 268 This is a formal completeness relation for the Sturm Liouville problem Note that we can Check this if we assume that 6W 7 g obeys the boundary conditions for a lt g lt b We can then expand the delta function as 6ltz e 6 i enema 269 0 Then7 our expression for en gives 17 one a dzwltzgtrtltzgt6ltzetgti matte 270 55 Inhomogeneous Equations and Green s Functions Suppose we have solved the Sturm Liouville problem L974 AmeMnW 7 271 PHY581 Methods of Theoretical Physics I Fall 2005 50 where the yn obey suitable boundary conditions on I ab and wz gt 0 on I By solvedl mean that the An and yn are determined and gmym on has been arranged We would like to solve the problem L Awly f96 7 272 for yx for z E I subject to the same boundary conditions Here A is a xed real number and f is a given function naturally assumed to obey the same boundary conditions To proceed write 95 wh9 7 h ahnynw 273 where we can calculate the coef cients hn when f and hence h is given and the yn are known We now posit the expansion WC ZuniAx 7 274 n0 and seek the unknowns an to complete the speci cation of the solution Substituting in to the problem 272 the left hand side becomes 2 anL 7 Awyn Z annw 7 Awyn 0 n0 114 anO n 7 39 7 n0 Now the right hand side is 1W Z hnydz 275 n0 Finally equating these rnultiplying both sides by ymz and integrating over a lt z lt b we obtain amm i A hm 276 Now if A S we have h V L an 7 277 PHY581 Methods of Theoretical Physics I Fall 2005 51 We now have the solution in terms of quantities calculated for the homogeneous equation It is instructive to write this in another way The solution is z e i x M y i 01 A 7 A 7 mam l emotional 7 n0 7 a 75616 In this form it is easy to identify the Green s Function in the form m 7 278 Gag J y 279 We can check that this Green7s function behaves as it should7 by acting on it with the left hand side of 272 L7 Awltzgticltz7 gt aw 7 Awgty77ltzgt 7 iL uwwzmz 7 y77lt gtwltzgty77ltzgt 7 which7 using our earlier results7 is L AwlGx7 596 i 6 7 280 as expected 56 SelfAdjointness It has probably not escaped your notice that the Sturm Liouville problem and its solu tions are related to things that you7ve seen in quantum mechanics Here7 we shall see precisely what this relation is In particular7 we shall compare the terms self adjoin in Sturm Liouville theory and Hermitz39tm in quantum mechanics 1 will begin by considering the case of wz 1 Consider a typical Sturm Liouville problem Ly A117 L iDpDq7 281 PHY581 Methods of Theoretical Physics I Fall 2005 52 with boundary conditions ya yb 0 In the vector space V of possibly complex functions f g obeying the boundary conditions we will use the inner product g Abdzmrgm 282 We are interested in operators A such that Af E V Vf E V De ne the Hermitian conjugate or Herrnitian adjoint operator A by A1129 fig 7 283 V f g E V We say that the operator A is herrnitian if A A or Af 9 f A9 7 284 With this de nition note that our self adjoint operator L is herrnitian for the special case w 1 You can check this trivially using the de nition of the inner product Do this as a brief exercise Now turn to the case of w 31 1 In this course and in Sturrn Liouville theory in general we refer to the operator L as self adjoint even when w 31 1 we also use Ly Amy and f g dx wfg from which real An and orthogonal yn follow However with this de nition Lfg 31 f Lg because of w To understand the relationship to herrniticity we de ne M E iw lZDprlZ q 285 which is herrnitian To see this 17 Mf7g drww 12Dpr12fg 7 b dxpr12fDw129 7 f7 My 286 Therefore care is needed with terrninology even though either de nition leads to real eigenvalues and orthogonal eigenfunctions