Fourier Transforms and the Wave Equation. Key Mathematics: More Fourier transform theory, especially as applied to solving the wave equation.

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Estella Cornelia Hubbard
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1 Lur 7 Fourir Transforms and th Wav Euation Ovrviw and Motivation: W first discuss a fw faturs of th Fourir transform (FT), and thn w solv th initial-valu problm for th wav uation using th Fourir transform Ky Mathmatics: Mor Fourir transform thory, spcially as applid to solving th wav uation I Som Proprtis of th Fourir Transform In th last lur w introducd th FT h of a funion f through th two uations f h dk, (a) π h d π f (b) Hr w wish to point out a fw usful proprtis of th Fourir transform A Translation Th first proprty has to do with translation of th funion f Lt's say w ar intrstd in f ( + 0 ), which corrsponds to translation of f by 0 Thn, using E (a) w can writ f ( h k ) 0 π π [ h k ] 0 dk dk () 0 Thus, w s that th FT of f ( + 0 ) is h In othr words, translation of f 0 by 0 corrsponds to multiplying th FT h by B Diffrntiation Th scond proprty has to do with th FT of f, th drivativ of f Again, using E (a) w hav D M Riff -- /3/009

2 Lur 7 h ] [ f dk (3) π So w s that FT of f is h That is, diffrntiation of f corrsponds to multiplying h by C Intgration Lt's considr th dfinit intgral of f, d f d dk h π (4) Switching th ordr of intgration on th rhs producs d f dk h π π h dk d ( k ) ( ) (5) So if w dfin If to b th indfinit intgral of f, w can rwrit E (5) as If ( ) If ( ) ( k ) ( ) h dk π (6) This uation tll us that intgration of f ssntially corrsponds to dividing th Fourir transform h by You might think that E (6) could b simplifid to If D M Riff -- /3/009 π h dk, but this cannot b don bcaus indfinit intgration producs an undtrmind intgration constant Th constant dos not appar If in E (6) bcaus it is an uation for th diffrnc of If and

3 Lur 7 D Convolution Th last proprty concrning th a funion and its FT has to do with convolution Bcaus you may not b familiar with convolution, lt's first dfin it Simply put, th convolution of two funions f and g, which w dnot ( f * g), is dfind as ( f * g) f ( ) g( ) d (7) Prhaps th most common plac that convolution ariss is in sproscopy, whr g is som intrinsic sprum that is bing masurd, and f is th rsolution funion of th spromtr that is bing usd to masur th sprum Th convolution ( f * g) is thn th masurd sprum Not that ( f * g) is indd a funion of, and so w can calculat its FT, which w dnot ( f *ˆ g) Using E (b) w can writ ( f *ˆ g) d d f ( ) g( ) (8) π Now this may not look too simpl, but w can chang variabls of intgration on th intgral,, d d Thn E (8) bcoms ( + ( f *ˆ g) k d d f g ) (9) π which can b rarrangd as π π ( f *ˆ g) π d f ( ) d g( ) (0) Th rsolution funion is oftn uit clos to a Gaussian of a particular, fid width D M Riff -3- /3/009

4 Lur 7 Notic that E (0) is simply th produ of th FT of f and th FT of g (along with th faor of π ) Dnoting ths FT's as fˆ and ĝ, rspivly, w thus hav 3 fˆ gˆ () ( f *ˆ g) π So w s that th FT of th convolution is th produ of th FT's of th individual funions On way you may har this rsult prssd is that convolution in ral spac ( ) corrsponds to multiplication in k spac II Solution to th Wav Euation Initial Valu Problm Way back in Lur 8 w discussd th initial valu problm for th wav uation t (, (, c () on th intrval < < For th initial conditions (, 0) a, (3a) t (, 0) b, (3b) w found that th solution to E () can b writtn as + c (, a( + c + a( c + b( ) d (4) With th hlp of th Fourir transform w ar now going to rdriv this solution, and along th way w will larn somthing vry intrsting about th FT of (, W start by dfining th (spatial) FT of (, as ˆ d π ( k, (,, (5a) 3 Although w hav not don it up to this point, it is fairly common praic to dnot th FT of a funion such as f by fˆ You will vn find th praic of dnoting th FT of f as f won't b doing that hr! At last w D M Riff -4- /3/009

5 Lur 7 so that w also hav dk (5b) π (, ˆ ( k, W also dfin th FT of E (3), th initial conditions, ˆ ( k,0) aˆ, (6a) ˆ &( k,0) (6b) Now ach sid of th E () is a funion of and t, so w can calculat th FT of both sids of E (), t (, (, d c d (7) On th lhs of this uation w can pull th tim drivativ outsid th intgral Th lhs is thn just th scond tim drivativ of ˆ ( k, Th rhs can b simplifid by rmmbring that th FT of th ( ) drivativ of a funion is tims th FT of th original funion Thus th FT of (, is just k tims ˆ ( k,, th FT of (, Thus w can rwrit E (7) as ( k, ˆ t k c ˆ ( k, (8) This uation should look vry familiar to you What uation is it? Non othr than th harmonic oscillator uation! What dos this tll us about ˆ ( k,? It tlls us that ˆ ( k, (for a fid valu of k ) oscillats harmonically at th fruncy ω kc Thus w can intrprt th funion ˆ ( k, as th st of normal mods coordinats for this problm This furthr mans that th FT has dcoupld th uations of motion for this systm [as rprsntd by E (), th wav uation] Notic also that th disprsion rlation ω kc has also falln into our lap by considring th FT of E () As w should know by this point, th solution to E (8) can b writtn as ˆ ( k, A + B, (9) D M Riff -5- /3/009

6 Lur 7 whr A and B ar funions of k And as you should susp, ths two funions ar dtrmind by th initial conditions, as follows First, stting t 0 in E (9) and using E (6a) producs a ˆ A + B, (0a) and calculating th tim drivativ of E (9), stting t 0, and using E (6b) givs us A k B (0b) W can solv Es (0a) and (0b) for A and B by taking thir sum and diffrnc, which yilds aˆ A k B aˆ +, (a), (b) which givs us th solution for ˆ ( k, in trms of th initial conditions ˆ k, t aˆ k + + aˆ k () W ar ssntially don W hav now prssd th FT of (, in trms of th FT's of th initial conditions for th problm Th solution (, is just th invrs FT of E () [s E (5b)], π (, aˆ + + aˆ + ( ) dk (3) This is th initial-valu-problm solution 4 W can mak it look aly l E (4) with a littl bit mor manipulation To s this lt's first rwrit E (3) as 4 Notic that (, as prssd in E (3) is th sum of two funions, f ( + c and ( c g! D M Riff -6- /3/009

7 Lur 7 π ( + [ ) t a k a (, ˆ + ˆ )] + c ( ( + ) ( ) ) dk (4) Th first two trms w rcogniz as ( ) [ a( + c + a( c ], whil w can us E (5) to rcogniz th scond half of th rhs of E (4) as + c b( ) d Thus E (4) can b r-prssd as + c (, a( + c + a( c + b( ) d, (5) which is idntical to E (4) Erciss *7 FT Proprtis If th FT of f is (a) show that th FT of [ f ] 0 is h ( k k ) 0 ; (b) show that th FT of [ f ] is i h ; (c) show that th FT of f is h [ ] h, *7 Show that ( f * g) ( g * f ) by (a) dirly by manipulating E (7), th dfinition of convolution; (b) by using E (), th rsult for th FT of ( f * g) *73 Convolution and th Gaussian Th funion that has th sam form as its Fourir transform is th Gaussian Spcifically if σ f, its FT is givn by σ k ( σ 4 h k ) Using this fa, show that th convolution ( f * f ) of two σ σ Gaussian funions f and f is proportional to th Gaussian funion ( σ ) +σ [Hint: you nd not calculat any intgrals to do this problm] **74 Th Rangular Puls Considr th funion f which is a rangular puls of hight H and width L cntrd at 0 (a) Graph f (b) Find fˆ, th FT of f D M Riff -7- /3/009

8 Lur 7 (c) Graph fˆ (d) Th funion ( f * f ), th convolution of f with itslf, is a triangl funion of hight LH and bas 4 L cntrd at zro Graph this funion () Writ down th funional form of ( f * f ) that you graphd in (d) Thn dirly calculat ( f *ˆ f ) using your funional form for ( f * f ) (Do not st H and L to spcific valus) (f) Graph your calculatd transform ( f *ˆ f ) (g) Lastly, us fˆ and th convolution thorm to find ( f *ˆ f ) Show that this is ual to th rsult in part () **75 FT Solution to th D Wav Euation E (3), π (, aˆ + + aˆ + ( ) dk, (3) is th formal solution to th initial-valu problm (a) What kind of wav is dscribd by th funion p [ ( + c ] (travling wav, standing wav, tc) B as spcific as possibl What kind of wav is dscribd by th funion p [ ( c ] From a vor-spac point of viw, what ar ths funions? (b) From a vor-spac point of viw (whr funions ar viwd as lmnts of a i vor spac), what do th trms i aˆ k and aˆ k + rprsnt? kv kv (c) Givn your answrs in (a) and (b), how would you dscrib th solution (, as writtn abov? [Hint: th trm linar combination should appar in your answr] (d) How ar â and rlatd to th initial conditions (,0) and & (,0)? That is, writ down prssions for â and in trms of (,0) and & (,0) () Assum that th initial conditions (,0) and & (,0) ar ral Using your answr to (d), show that aˆ aˆ * ( k) and b ˆ b ˆ* ( k) (f) Using th rsults from () you can now show that (, is ral if th initial conditions (,0) and & (,0) ar both ral, as follows First, in E (3) rplac â * and by aˆ ( k) and *( k), rspivly Thn mak th chang of variabl k k, dk dk in th intgral (taking car with th limits of intgration) Thn compar E (3) with your nw prssion for (, and notic how thy ar rlatd From your comparison you should b abl to conclud that (, is ral D M Riff -8- /3/009

A Propagating Wave Packet Group Velocity Dispersion

A Propagating Wave Packet Group Velocity Dispersion Lctur 8 Phys 375 A Propagating Wav Packt Group Vlocity Disprsion Ovrviw and Motivation: In th last lctur w lookd at a localizd solution t) to th 1D fr-particl Schrödingr quation (SE) that corrsponds to