Fourier transforms. Up to now, we ve been expressing functions on finite intervals (usually the interval 0 x L or L x L) as Fourier series: a n cos L

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1 ourier trnsforms Motivtion nd definition Up to now, we ve been expressing functions on finite intervls (usully the intervl 0 x or x ) s ourier series: fpxq 0 8 nπx n cos n b n sin nπx where 0» fpxq dx 2 nd n» fpxq cos nπx dx, b n» fpxq sin nπx dx. We lso occsionlly thought bout the complex exponentil version of ourier series: Since e iθ cos θ i sin θ nd e iθ cos θ i sin θ, or equivlently cos θ eiθ e iθ 2 nd sin θ eiθ e iθ, 2i we cn rewrite the bove series s: fpxq 0 e 0ix 0 e 0ix 8 n 8 e nπix{ n n 8 n c n e nπix{ n ib n 2 e nπix{ 2 b n e nπix{ e nπix{ 2i e nπix{ n ib n e nπix{ 2 where $ '& c n '% 2 n ib n q for n 0 0 for n 0 2 n ib n q for n 0

2 2 fourier trnsforms Using the formuls for n nd b n given bove, we see tht, for n 0. If n 0 we hve c n 2 p n 2 2 2»»» ib n q nπx i fpxq cos dx 2 nπx fpxq cos i sin fpxqe nπix{ dx.» nπx fpxq sin dx nπx dx c n 2 p n ib n q» fpxq cos nπx dx i» 2 2» nπx nπx fpxq cos i sin 2 2» fpxqe nπix{ dx fpxq sin dx nπx dx becuse cosine is n even function nd sine is odd. So the sme formul works for ll the coefficients (even c 0 ) in this cse nd we hve 8 fpxq c n e nπix{ where c n» fpxqe nπix{ dx. 2 n Wht we wnt to do here is let tend to infinity, so we cn consider problems on the whole rel line. To see wht hppens to our ourier series formuls when we do this, we introduce two new vribles: ω nπ{ nd ω π{. Then our complex ourier series formuls become fpxq 8 n c n e iωx where c n ω» fpxqe iωx dx nd the n in the formul for c n is hiding in the vrible ω. Now, if we let c ω c n { ω, we cn rewrite these s 8 fpxq c ω e iωx ω where c ω» fpxqe iωx dx. n The vrible ω nπ{ tkes on more nd more vlues which re closer nd closer together s Ñ 8, so c ω begins to feel like function of the vrible ω defined

3 mth for ll rel ω. ikewise, the sum on the left looks n wful lot like Riemnn sum pproximting n integrl. Wht hppens in the limit s Ñ 8 is: fpxq cpωqe iωx dω where cpωq fpxqe ixω dx. The formul on the right defines the function cpωq s the ourier trnsform of fpxq, nd the formul on the left defines fpxq s the inverse ourier trnsform of cpωq. These formuls hold true (nd the inverse ourier trnsform of the ourier trnsform of f pxq is f pxq the so-clled ourier inversion formul) for resonble functions fpxq tht decy to zero s x Ñ 8 in such wy so tht fpxq nd/or fpxq 2 hs finite integrl over the whole rel line. There re mny stndrd nottions for ourier trnsforms (nd lterntive definitions with the minus sign in the ourier trnsform rther thn in the inverse, nd with the fctor in different plces, so wtch out if you re looking in books other thn our textbook!), including nd Properties nd exmples. fpωq pωq rfpxqspωq q pxq fpxq r pωqspxq fpxqe ixω dx pωqe ixω dω. The ourier trnsform is n opertion tht mps function of x, sy fpxq to function of ω, nmely rfspωq q fpωq. It is clerly liner opertor, so for functions fpxq nd gpxq nd constnts α nd β we hve rαfpxq βgpxqs α rfpxqs β rgpxqs. Some other properties of the ourier trnsform re. Trnsltion (or shifting): rfpx qs pωq e iω rfpxqs pωq. And in the other direction, re ix fpxqs pωq rfpxqs pω q. 2. Scling: f x pωq rfpxqs pωq, nd likewise rfpxqs pωq rfpxqs ω. 3. Opertionl property (derivtives): rf pxqs pωq iω rfpxqs pωq, nd rxfpxqs pωq i d p rfpxqs pωqq. dω The opertionl property is of essentil importnce for the study of differentil equtions, since it shows tht the ourier trnsform converts derivtives to multipliction

4 4 fourier trnsforms so it converts clculus to lgebr (or might reduce prtil differentil eqution to n ordinry one). Here re the proofs of the first of ech of the three pirs of formuls to give sense of how to work with ourier trnsforms, nd leve the other three s exercises. or the first shifting rule, we mke the substitution y x (so dy dx nd x y ) to clculte rfpx qs pωq fpx qe iωx dx fpyqe iωy e iω dx e iω rfpxqs pωq or the first scling rule, we mke the substitution y x{ (so dx dy) nd get x f pωq x f e iωx dx fpyqe iωy dy rfpxqs pωq or the opertionl property we first point out tht since the ourier trnsforms of both f pxq nd fpxq exist, we must hve tht fpxq Ñ 0 nd f pxq Ñ 0 s x Ñ 8. Therefore the endpoint terms will vnish when we integrte by prts (with u e iωx nd dv f pxq dx, so du iωe iωx nd v fpxq): rf pxqs pωq f pxqe iωx dx x8 eiωx fpxq 0 iω x iω rfpxqs pωq fpxqe iωx dx iωfpxqe iωx dx et s clculte few bsic exmples of ourier trnsforms: Exmple. et S pxq be the function defined by " if x S pxq 0 otherwise

5 mth Then rs pxqs pωq» e iωx dx eiω e iω iω sin ω πω. Exmple 2. et upxq e x2 {2, so the grph of upxq is Gussin or bell-shped curve. Then upxq stisfies the differentil eqution u xu 0. We cn use this fct nd the properties of the ourier trnsform to clculte û s follows: Tke the ourier trnsform of the differentil eqution nd use linerity nd both prts of property (3) bove to get 0 ru xus pωq ru s rxus d rus iω rus i dω Therefore rus stisfies the differentil eqution the solution of which is rus ω rus 0 rus Ce ω2 {p2q. The constnt C is the vlue of rus p0q, i.e., C c e x2{2 dx 2 e y2 dy?, using the substitution y x nd the fmilir (or t lest ccessible) fct tht ³ 2 8 e y2 dy? π. Therefore e x2 {2 pωq? e ω2{p2q, so the originl Gussin is trnsformed into different one. An interesting observtion is wht hppens for : Then we hve e x2 {2? e ω2{2, x2{2 so the specific Gussin e is eigenvlue {?. n eigenfunction of the ourier trnsform with Exmple 3. et fpxq e x, so fpxq " e x if x 0 e x if x 0.

6 6 fourier trnsforms Then e x pωq πp 2» 0» 0 e x e iωx dx e x e iωx dx e p iωqx dx p iωqx x0 e iω x iω ω 2 q iω 0 e ω e iωx dx e p iωqx dx 0 p iωqx x8 e iω x0 (the limiting vlues of the exponentils t 8 re zero becuse e x goes to zero s x goes to 8 nd e iωx stys bounded). An observtion. Becuse the formuls for the ourier trnsform nd the inverse ourier trnsform re so similr, we cn get inverse trnsform formuls from the direct ones nd vice vers. In prticulr, note tht if we let y x then rfpxqs pωq ikewise r pωqs pxq fpxqe iωx dx pωqe iωx dω fp yqe iωy dy rfp yqs pωq p αqe iαx dα r p αqs pxq So if we know ourier trnsform formul or n inverse ourier trnsform formul, we cn get nother one for free by reversing the inverse. or exmple, since we immeditely hve tht rs pxqs pωq sin ω πω, sin ω pxq S pxq. πω rom either of the formuls bove nd the fct tht sin x{x is n even function, we hve sin x pωq πx S pωq,

7 mth or Similrly, since we know tht sin x x e x 2 S pωq. πp 2 ω 2 q for 0, we cn immeditely write πp 2 e x. ω 2 q And since is n even function of ω, we hve πp 2 or 2 πp 2 x 2 q x 2 ω 2 q e ω, 2 e ω. Convolutions. We need one more ourier trnsform formul, nd it involves n opertion on functions tht might seem new to you. It is clled convolution nd it strts with two functions, fpxq nd gpxq nd produces new one, denoted f g or pf gqpxq (or sometimes just f gpxq), defined by pf gqpxq fpyqgpx yq dy. If you think bout it, convolution is like multipliction of polynomils or series, wherein n bn c n where c n m b n m. This motivtes the definition of convolution s n opertion tht might hve hs its ourier trnsform the product of the trnsforms of the individul functions, nd it does (up to nuisnce fctor of ), but more on tht in moment. irst some bsic properties of convolutions:

8 8 fourier trnsforms. Convolution is liner in ech of the two functions. In other words, if f, g nd h re functions nd α nd β re constnts, then pαf βgq h αf h βg h nd f pαg βhq αf g βf h. 2. Convolution is commuttive: f g g f. To prove this, we mke the substitution z x y in the integrl (so we think of x s being constnt while we re doing the integrl, nd y x z nd dy dz) nd get pf gqpxq 3. Now for the ourier trnsform:» 8 pg fqpxq fpyqgpx yq dy fpx zqgpzq dz gpzqfpx zq dz rf gs pωq rfs pωq rgs pωq or pf gqpωq fpωqĝpωq. To see this we use the substitution z x y gin, nd brek up e iωx e iωpx y yq e iωy e iωpx yq to get: pf gqpωq fpyqgpx yq dy e iωx dx fpyqgpx yqe iωx dx dy fpyqe iωy gpx yqe iωpx yq dx dy gpx yqe iωpx yq dx fpyqe iωy dy fpωqĝpωq gpzqe iωz dz gpzqe iωz dz fpyqe iωy dy fpyqe iωy dy s ikewise, the trnsform of the product of two functions is the convolution of the trnsforms, except without the fctor of, nmely r Gs pxq q pxq q Gpxq.

9 mth One lst thing: the convolution product of two functions hs n interesting property reltive to derivtives: d df pf gq dx dx g f dg dx so when you tke derivtives of convolution of two functions, you get to stick the derivtive on whichever of the two functions is more convenient (or differentible) this is n esy consequence of the commuttivity of convolution, nd it hs the powerful consequence tht the convolution of two functions hs the better of the differentibility properties of the two individul functions. So if f is discontinuous but g is smooth, then f g will be smooth. The het eqution on the whole line. Now we re redy to use ll of this for something! We seek to solve the het eqution u t ku xx for t 0 nd x 8, with initil conditions upx, 0q fpxq nd ssuming u decys to zero t x nd x 8. We ll strt by tking the ourier trnsform of both sides of the differentil eqution in the x-vrible. By this we men so û stisfies nd ûpω, tq Bû Bt kω2 û ûpω, 0q fpωq. upx, tqe ixω dx, This is n ordinry differentil eqution where the independent vrible is t nd ω should be treted s constnt. The generl solution of the differentil eqution is ûpω, tq cpωqe kω2t. Putting t 0 shows tht cpωq fqωq. Therefore ourier trnsform of the solution of our problem is ûpω, tq fpωqe kω2t. We cn now recover upx, tq by tking the inverse ourier trnsform of both sides, using the rule tht the ourier trnsform of convolution is times the product of the individul ourier trnsforms, so upx, tq fpxq e kω2 t

10 0 fourier trnsforms nd we hve to clculte the inverse ourier trnsform of e kω2t. But we hve the rule for Gussins: e x2 {2? e ω2{p2q, or e ω2 {p2q? e x2{2. Since we wnt to clculte the inverse ourier trnsform of e kω2t, we should set {p2q kt, or {p2ktq. Then we get c e kω2 t π {p4ktq kt e x2, nd so upx, tq c π fpxq {p4ktq kt e x2? 4πkt fpyqe px yq2 {p4ktq dy. The function Gpx, y, tq? 4πkt e px yq2 {p4ktq is clled the fundmentl solution of the het eqution. It is the solution to the initil vlue for the het eqution for the sitution where the initil conditions re such tht single unit of het energy is introduced t the point y t time t 0. So the bove formul for upx, tq sys tht we cn solve the het eqution for rbitrry initil conditions upx, tq by integrting together ll the contributions to the temperture t time t 0 t ll points, s described by the differentil fpyq dy. The wve eqution on the whole line. Next, let s look t the initil-vlue problem for the wve eqution on the whole line. We ll solve the wve eqution together with initil conditions u tt c 2 u xx upx, 0q fpxq nd u t px, 0q gpxq. As we did with the het eqution, we ll tke the ourier trnsform of both sides of the differentil eqution in the x-vrible. So once gin, let ûpω, tq upx, tqe iωx dx

11 so û stisfies nd mth 425 B 2 û Bt 2 c2 ω 2 û ûpω, 0q fpωq nd û t pω, 0q ĝpωq. We will gin tret this s n ordinry differentil eqution in t with ω treted s constnt. The generl solution of this eqution is nd the initil conditions imply u p ω, tq c pωq cos ωct c 2 pωq sin ωct, upω, 0q c pωq fpωq nd u t pω, 0q cωc 2 pωq ĝpωq. Therefore, c pωq fpωq nd c 2 pωq ĝpωq{pcωq nd we hve obtined the ourier trnsform of the solution: ûpω, tq fpωq ĝpωq cos ωct sin ωct. cω Therefore ĝpωq upx, tq fpωq cos ωct cω sin ωct. We ll tke the two terms one t time. or the first, we write cos ωct in complex exponentil form, so we re trying to compute fpωq eiωct e iωct fpωqre iωct e iωct se iωx dω 2 2 fpωqe iωpx ctq dx fpωqe iωpx ctq dx 2 rfpx ctq fpx ctqs 2 using the definition of the inverse ourier trnsform. or the second term, we proceed differently. Reclling tht Ŝpωq sin ω{pπωq nd tht the inverse ourier trnsform of product of two functions is their convolution (divided by q, we hve ĝpωq sin ωct cω π gpxq c S ctpxq 2c 2c 2c» ct ct» x ct x ct gpx yqs ct pyq dy gpx yq dy gpuq du

12 2 fourier trnsforms where in the lst step we mde the substitution u x y (so y x u nd dy du). We put both terms together to get the solution to our initil-vlue problem for the wve eqution: upx, tq 2 rfpx ctq fpx ctqs 2c» x ct x ct gpuq du. This is clled d Alembert s solution of the wve eqution, nd clerly shows tht signls propgte with speed c, since the vlue of the solution t point x nd time t depends only on the initil position nd velocity vlues in the intervl rx ct, x cts, nd conversely tht the initil vlues t point x influence the solution t time t only within the intervl rx ct, x cts. iner lgebric properties of the ourier trnsform: Prsevl s theorem nd Hermite functions We remrked erlier tht the ourier trnsform is liner trnsformtion from functions of x to functions of ω defined on the whole line (which cn be integrted etc.). We lso found tht the function e x2 {2 is n eigenfunction of the ourier trnsform with eigenvlue {?. We explore some further consequences of these observtions in this section. irst, we cn compre the inner product of the ourier trnsforms of two functions with the inner product of the functions themselves (remember, in the complex (or Hermitin) inner product, we hve to tke the complex conjugte of the second fctor, nd we write z for the complex conjugte of z): xf, gy fpxqpgpxqq dx fpxq ĝpωqe dω iωx dx fpxqe iωx pĝpωqq dx dω fpxqe iωx dx pĝpωqq dω fpωqpĝpωqq dω A E f, ĝ. A E In prticulr, if f g we hve tht }f} 2 xf, fy f, f } f} 2 or } f} 2 }f}2.

13 mth This is clled Prsevl s equlity or Prsevl s theorem it sys tht the ourier trnsforms shrinks the norms of ll functions by fctor of {? (so it wsn t n ccident tht we found n eigenfunction with tht eigenvlue!), nd it hs number of interesting consequences. As simple exmple. since we know tht (reclling tht S pxq is the step function for the intervl r, s, sin ω rs pxqs pωq πω we cn conclude tht 2 sin ω πω }S pxq} 2. The right side is esy to compute, it s simply» dx π. But we lern something interesting by compring this to the left side: sin 2 ω π 2 ω 2 dω π or sin 2 ω ω 2 dω π, which is certinly something we didn t know before. Now, bck to the liner lgebr. We sk curious question, relted to the observtion we mde on pge 6: Wht hppens if you tke the ourier trnsform of the ourier trnsform of function? et s see: r rfpxqss pωq fpzqe iωz dz fpzqe ip ωqz dz fp ωq, since the lst integrl is the inverse ourier trnsform evluted t ω. But this shows tht doing the ourier trnsform twice to function gives bck its reverse multiplied by {pq. So if we do the ourier trnsform four times to function we should get {p4π 2 q times the reverse of the reverse, which is the originl function. In other words, 4 rfpxqs 4π fpxq. 2 We could write this s n eqution bout the ourier trnsform opertor: 4 4π 2 I 0,

14 4 fourier trnsforms where I is the identity opertor. Knowing this, we cn see tht if f is ny eigenfunction of with eigenvlue λ, then 4 rfs 4π Irfs λ 4 f 0, 2 4π 2 which mens tht ll of the eigenvlues of the ourier trnsform must stisfy the eqution λ 4 4π 2 0, so the only possible eigenvlues of re?, i?,? nd i?. We lredy hve tht the Gussin e x2 {2 is n eigenfunction of with eigenvlue {?. Now we ll find eigenfunctions for the other eigenvlues. We hve two ourier trnsform rules tht look quite similr: df dx pωq iω fpωq nd xfpxqpωq i d f dω. So the ourier trnsforms of ech of the opertions multiply by the vrible nd tke the derivtive on the x-side is the other opertion on the ω-side (well, there re fctors of i to keep trck of, nd we will). Becuse of this, the sum nd difference of these two opertors hve specil reltionship with the ourier trnsform opertor: df d rfs d rfs xfpxq iω rfs pωq i dx dω i ω rfs pωq dω nd df dx xfpxq iω rfs pωq d rfs i dω i d rfs dω ω rfs pωq. These equtions show tht if fpxq is n eigenfunction of the ourier trnsform opertor with eigenvlue λ then f pxq xfpxq will be n eigenfunction of the ourier trnsform with eigenvlue iλ (unless f pxq xfpxq 0), nd f pxq xfpxq will be n eigenfunction of the ourier trnsform with eigenvlue iλ (unless f pxq xfpxq 0). et s try this with the eigenfunction we know, nmely fpxq e x2{2. Unfortuntely, f xf 0, but f pxq xfpxq xe x2 {2 xe x2 {2 2xe x2 {2, so 2xe x2{2 is n eigenfunction of the ourier trnsform with eigenvlue i{? : 2xe x2 {2 pωq? i p 2ωe ω2{2 q.

15 mth And we cn keep going: If we set f pxq 2xe x2 {2, then f xf 2e x2 {2 2fpxq, so we get nothing new here. But f xf p4x 2 2qe x2 {2, so we cll f 2 pxq p4x 2 2qe x2{2 nd we hve p4x 2 2qe x2 {2? p4ω 2 2qe ω2{2. We cn continue in this wy nd obtin n infinite sequence of eigenfunctions of the ourier trnsform, strting from f 0 pxq e x2 {2, where f n pxq f n pxq xf npxq, nd the eigenvlue of f n will be i n {?. There will be more to sy bout this in the homework!