1. The vibrating string problem revisited.

Transcription

1 Weeks 7 8: S eprtion of Vribes In the pst few weeks we hve expored the possibiity of soving first nd second order PDEs by trnsforming them into simper forms ( method of chrcteristics. Unfortuntey, this process often does not hep much. If we strt with n rbitrry second order PDE, nd reduce it to cnonic form, most ikey we sti do not know how to find the gener soutions. In the foowing few weeks, we wi introduce method tht re much more powerfu t finding soutions thn the method of chrcteristics. The min ide is seprtion of vribes.. The vibrting string probem revisited. We consider the system u t t c u x x =, < x <, t > ( u( x, = f( x, x, ( u t ( x, = g( x, x, ( 3 u(, t =, t, ( 4 u(, t =, t. ( 5 In the pst week we hve tried to sove this system using the gener soution formus of the wve equtions. We hve seen tht it is not possibe to write cen formu for the soution nd therefore very hrd to extrct much informtion from it. Now we try nother pproch. Insted of trying to get the gener soution, we sk, wht is the simpest nontrivi soution of the wve eqution ( without considering the initi nd boundry conditions? One possibiity is the speci form u( x, t = X( x T( t. ( 6 We pug this into the eqution to see when we wi be so ucky to hve such soutions. We obtin which eds to X T = c X T ( 7 X X = c T T. ( 8 This seems to be compicted eqution unti we reized tht X /X is function of x ony nd T /T is function of t ony. The equity cn hod ony if both X /X nd T /T re constnts. From this we see tht u( x, t = X( x T( t is soution to the wve eqution if there is constnt λ such tht From ODE theory we know tht the soutions re X λ X =, ( 9 T λ c T =. ( X( x = A e λ / x + B e λ / x, T( t = A e λ / c t + B e λ / c t. ( To fix the rbitrry constnts, we now consider the boundry conditions u(, t = u(, t = ( but sti negect the initi conditions, which gives X( = X( =. ( Discussing the cses λ >, =, < ( see pp of the textbook we see tht this cn be stisfied by nonzero X ony if. A =, nd λ = ( 3 Therefore. To see this, differentite the eqution by, we hve x of x ony. X( x = X n ( x for some n N, ( 4 x ( X = which gives X X X = constnt becuse it is function

2 where X n ( x = B n sin x for some rbitrry constnt B n. It foows tht c c T( t = T n ( t = C n cos t + D n sin t for rbitrry constnts C n nd D n. We hve shown tht soutions of the form X( t T( t re ( c c α cos t + β sin t sin x where α, β re constnts. Now it s time to tke into ccount the initi conditions. If we tke t =, we hve f( x = u( x, = α sin x g( x = u t ( x, = β n c sin ( 5 ( 6 ( 7, ( 8 x. ( 9 which re cery not true for most f nd g! Does this men the method fis? Not so fst. Rec tht for iner equtions, finite or even infinite sums of soutions re sti soutions ( certin conditions ppy. Therefore, if we cn find finitey mny constnts α n, β n : n =,, m or infinitey mny constnts α n, β n such tht or m f( x = f( x = α n sin x, g( x = m α n sin x, g( x = β n n c β n n c sin x, ( sin x, ( then the soution u cn be written s either u( x, t = m ( c c α n cos t + β n sin t sin x ( or u( x, t = ( c c α n cos t + β n sin t sin x. ( 3 It turns out tht finite sum representtion is in gener not possibe, we hve to rey on the infinite sums. To indeed crry this out, we need to nswer the foowing questions:. Cn we represent rbitrry f, g?. If we cn, how to compute the coefficients α n, β n? 3. Does the infinite sum represent function? Tht is, does the sum converge? 4. If the infinite sum converges to function, does this function sove the eqution nd stisfy the initi nd boundry conditions? The nswers to these questions form the bsics of the theory of Fourier series. We wi study Fourier series in the foowing one to two ectures before returning to the method of seprtion of vribes. Remrk. Wht is the retion between our new formu u( x, t = ( c c α n cos t + β n sin t sin x ( 4 nd our previous one using gener soutions? Using bsic trignometric formus: cos( x ± y = cos x cos y sin x sin y, sin( x ± y = sin x cos y ± cos x sin y ( 5

3 we esiy obtin u( x, t = or equiventy with { α n } sin ( x + c t sin ( x c t + β n cos ( x c t cos ( x + c t φ( x = u( x, t = φ( x + c t + ψ( x c t ( 6 α n sin x β n cos x, ( 7 ψ( x = β n cos x α n sin x. ( 8 We see tht ψ( x = φ( x nd φ( x = ψ( x s we know. Remrk. As we wi soon see, oftentimes we cnnot obtin cosed form formu for u nd hve to ive with the infinite sum. Recing our motivtion to do better thn the method of chrcteristics one my wonder wht the difference is. The difference is the foowing. When we cnnot sove the probem by method of chrcteristics, we re toty stuck; On the other hnd, the coefficients in the infinite sum te us much informtion of the soution.. Fourier Series. Rec the issues we woud ike to sette.. Cn we represent rbitrry f, g?. If we cn, how to compute the coefficients α n, β n? 3. Does the infinite sum represent function? Tht is, does the sum converge? 4. If the infinite sum converges to function, does this function sove the eqution nd stisfy the initi nd boundry conditions?.. Representtion of functions by Fourier series. We first try to sette the first two questions. We study whether ny function f cn be represented in the form f( x = α n sin x. ( 9 First we notice tht, for ny f representbe in the bove fshion, formy ( tht is negecting ny convergence issues reted to infinite sums f( k = f( x + = α n sin k = α n f( x = α n sin ( x + = In other words, ny function tht is possibe to be represented must stisfy i. f( k =, ii. f( x dx =, α n sin( k n =, ( 3 sin x dx =, ( 3 α n sin x + n. ( 3

4 iii. f( x + = f( x tht is f is periodic function with period ; Or equiventy, we cn ony represent functions defined over interv of ength. As we woud ike to represent s mny functions s possibe, we woud try to fix the bove restrictions. i. f( k = : The reson for this restriction is tht sin k =, it is esiy fixed by introducing cos x into the series. ii. f( x dx = : Even with cos s introduced, we sti hve f( x dx =. Therefore we need to introduce one more term constnt reting to f( x dx into the series. iii. f( x + = f( x : There is currenty no wy to fix this. We wi mention bit bout this in few ectures. Now the representtion becomes ( we hve chnged the nottions bit to be consistent with the textbook f( x = + ( x x n cos + b k sin. ( 33 for f with period. In the foowing we wi show how to find the coefficients.. We strt with. As we hve seen, the introduction of is to ow us to represent function with nonzero men. Therefore we integrte the representtion over (, : f( x dx = dx + ( ( k x ( k x k cos dx + b k sin dx =. ( 34 k = Therefore = f( x dx. ( 35 To obtin n nd b n, we notice x ( m x { n = m cos cos dx = ( 36 n m x ( m x cos sin dx = ( 37 x ( m x { n = m sin sin dx = ( 38 n m Therefore n = x f( x cos dx; b k = Remrk 3. The interv (, cn be repced by ny interv of ength. From the bove formus, we cery see tht x f( x sin dx. ( 39 f( x is odd n = n =,,, ( 4 f( x is even b n = n =,, ( 4 Tht is, when f is odd, ony sin s re invoved in the Fourier series nd when f is even ony cos s re invoved. In these cses, we ony need hf of f to determine the coefficients: f even: n = x f( x cos dx, b n =, n =,,, ( 4 f odd: b n = x f( x sin dx, n =, n =,, ( 43

5 Exmpe 4. ( 6. 4 Find the Fourier series of the foowing function { x < x < f( x = h is constnt. ( 44 h < x < Soution. As f is defined over (,, =. We compute = f( x dx = x dx + h dx = + h = h ; ( 45 n = f( x cos( n x dx = x cos( n x dx + h cos( n x dx = x dsin( n x + h n n sin( n x = n x sin( n x sin( n x dx + h n n sin( n x = n ( cos( n x = ( n n. ( 46 b k = f( x sin( n x dx = x sin( n x dx + h sin( n x dx = x dcos( n x + h sin( n x dx n = x cos( n x cos( n x dx + h cos( n x n = ( ( n n n sin( n x + h ( ( n = ( n + h ( ( n n = h ( h + ( n. ( 47 n Therefore the representtion is f( x = h 4 + ( n ( n cos( n x + n h ( h + ( sin( n x. ( 48 n Exmpe 5. ( c Obtin the Fourier cosine series representtion for the foowing functions: f( x = x, < x <. ( 49 Soution. Obtin the Fourier cosine series effectivey mens extending f eveny nd obtin its Fourier series. Or equiventy, use the formus n = x f( x cos dx, b n =, n =,,, ( 5 Thus =. We compute = x dx = 3. ( 5

6 nd n = x cos( n x dx = x dsin( n x n = x sin( n x n 4 = n x dcos( n x 4 = n x cos( n x The Fourier cosine series representtion is then x sin( n x dx cos( n x dx = 4 ( n n. ( ( n n cos( n x. ( 53 Remrk 6. Mthemticy speking, wht we hve done is the foowing. We hve shown tht, if function f( x cn be represented by f( x = + ( x x n cos + b n sin. ( 54 Then the coefficients must be given by n = x f( x cos dx, n =,,, ( 55 b n = x f( x sin dx, n =,, ( 56 To estbish sound mthemtic theory, we hve to study whether the sequence + ( x x n cos + b n sin, ( 57 with coefficients given by the bove formus, converges to the function f... C onvergence nd other issues. We mention quicky theories reting the 3rd nd 4th questions. In fct, due to imited time, we wi ony mention quicky the questions tht wi be nswered by the theory of Fourier series. First we review wht we hve so fr. We re concerned with soving the initi-boundry probem u t t c u x x =, < x <, t > ( 58 u( x, = f( x, x, ( 59 u t ( x, = g( x, x, ( 6 u(, t =, t, ( 6 u(, t =, t. ( 6 We re be to construct the foowing infinite sum: ( c c α n cos t + β n sin t sin x ( 63

7 nd hope tht this is our soution. Mthemticy the foowing re required:. The sum gives function. Tht is we cn define u( x, t = ( c c α n cos t + β n sin. The eqution is stisfies by this function. Tht is i. t t u( x, t nd x x u( x, t re we-defined for < x <, t >, ii. u t t c u x x = for < x <, t >. t sin x. ( 64 We note tht if we cn differentite the sum termwise, then the eqution wi be stisfied. 3. Correct initi vues re tken. f( x = g( x = in pproprite senses, nd furthermore im u( x, t = u( x,, im t α n sin x β n n c sin x ( 65 ( 66 t u t( x, t = u t ( x,. ( 67 Tht is, the t imit is the sme s the resut of setting t = in ech term of the infinite sum. 4. Correct boundry vues re tken. im u( x, t = u(, t, im x u( x, t = u(, t ( 68 x where u(, t, u(, t re the vues of the infinite sum when we repce x by nd in the infinite sum. From the bove we see tht the foowing questions need to be nswered for Fourier series:. Given function, is the corresponding Fourier series converging to this function?. Given n infinite sum of sin s nd cos s, when is it convergent to resonbe function? 3. If the series converges to function, when is this function differentibe? Does the derivtive coincide with the sum of termwise derivtive of ech term? Tht is if f( x = + ( n cos x + b n sin x ( 69 when do we hve the existence of f, f, etc. nd furthermore when do we hve ( d p dx p f( x = 4. Cn we tke termwise imits. Tht is ( x x im n cos + b n sin =? = x x ( d p ( d p n dx pcos x + b n dx psin x. ( 7 ( x x n cos + b n sin. ( 7 Of course in fct the bove questions need to be nswered for doube Fourier series. The nswers to the bove questions re highy invoved. We just give the shortest ( reds: crudest nswers here. Answer to. If f is piecewise continuous, then its Fourier series converges to f t of its continuous points, nd converges to f ( x + + f ( x t discontinuous points. In prticur, if f is uniformy continuous, the convergence is uniform.

8 Answer to nd 4. Yes s ong s n + b n <. Answer to 3. Yes s ong s p ( n + b n <. Remrk 7. Finy we mention few resons why Fourier series is widey ppied in modern mthemtics nd science. Fourier series cn be obtined for wide css of functions. b The prti sum f N = + N ( x x n cos + b n sin is the best pproximnt of f in the foowing sense f N = rgmin g VN ( f g dx ( 73 where V N is the spce of functions of the form α + N ( x x α n cos + β n sin. ( 74 ( 7 Furthermore we hve the Prsev s retion + ( n + b n = f( x dx. ( 75 c Most importnty, the sin s nd cos s re eigenfunctions of derivtive opertors. In prticur, differentition of one Fourier series resuts in nother Fourier series. This is why Fourier series is ubiquitous in modern theory of PDEs.. 3. C ompex Fourier series. Recing we hve cos x = ei x + e i x + ( n cos x + b n sin x where, sin x = eix e ix, ( 76 ( = c + k = c n e inx / + c n e inx / = c n e i nx/. ( 77 c =, c n = n i b n, c n = n + i b n. ( 78 This series is ced the compex Fourier series. Note tht, insted of computing c k from k nd b k, we cn obtin the coefficients directy: c = f( x dx, ( 79 c n = f( x e inx / dx, n = ±, ±, ( 8 Remrk 8. We emphsize gin tht the interv (, cn be repced by ny interv of ength. In this cse the Prsev s retion tkes more estheticy stisfctory form: c n = f( x dx. ( 8

9 Exmpe 9. ( 6. 4, 5 b Expnd the foowing function to compex Fourier series f( x = cosh x, < x <. ( 8 Soution. We hve =. Rec tht cosh x = e x + e x. We compute c = e x + e x dx = ( e e ; ( 83 c k = e x + e x e ik x dx = e ( ik x dx + e ( + ik x dx 4 = 4 i k e( ik x + i k e ( + ik x = ( e ( k e ( k ( e ( k e ( k 4 i k + i k = 4 ( i k + + i k ( e e ( k = ( k ( e + k e. ( 84 Noticing + =, the compex Fourier series cn be written in the foowing compct form: f( x = k = ( k ( e + k e e ik x. ( Doube Fourier series. We just mention tht the theory of Fourier series cn be extended to functions with more thn one vribes. For exmpe, in D, insted of cos x nd sin x, we hve four combintions cos x cos y, cos x sin y, sin x cos y, sin x sin y. ( 86 See 6. of the textbook for detis. Remrk. In higher dimensions, the compex representtion becomes more convenient to use, s insted of d combintions of sin s nd cos s ( d is the dimension, we cn simpy write the gener form of the bsis function s where k = k k d, x = x x d. e i k x ( Non-periodic functions nd Fourier trnsform. We hve presented stisfctory theory for periodic functions. Now how bout non-periodic functions? Here we give hint of wht hppens in this gener cse. We tke the interv to be (,. Rec the compex Fourier series formus for f with period : with c n = Furthermore we hve the Prsev s retion f( x = c n = c n e i n x, ( 88 f( x e inx / dx. ( 89 f( x dx. ( 9

10 Now setting we hve λ = n, c λ = c n, ( 9 dλ = dn =, ( 9 nd the bove formus cn be formy re-written s ( tking λ f( x = c n e i n = c n e i λ x dn = c λ e iλ x dn = c λ e iλ x dλ, ( 93 c λ = c n = f( x e iλ x dx f( x e iλ x dx, ( 94 c ( λ dλ = c n dn = cn = f( x dx f( x dx. ( 95 The formus c λ = f( x e iλ x dx, f( x = c λ e iλ x dλ ( 96 re ced the Fourier trnsform. It turns out tht these formus, incuding the Prsev s retion c λ dλ = f( x dx ( 97 cn be rigorousy derived for resonby rge css of functions. For trignometric version of the Fourier trnsform theory, see 6. 3 of the textbook. 3. Appictions of the Fourier series theory. The theory of Fourier series cn be ppied to mny PDE probems. 3.. The wve eqution. We return to the probem u t t c u x x =, < x <, t > ( 98 We hve seen tht the soution cn be represented by u( x, t = ( ( n cos n c c t + b n sin with the coefficients determined from u( x, = f( x, x, ( 99 u t ( x, = g( x, x, ( u(, t =, t, ( u(, t =, t. ( f( x = u( x, = n sin b n c x t sin x ( 3 ( 4 sin x. ( 5 g( x = u t ( x, = Now we try to compute n, b n. First we shoud note probem here. The formu determining the coefficients for the Fourier series reds n = f( x sin x dx, ( 6. In fct, the Fourier trnsform theory is the foundtion of compete gener theory of iner prti differenti equtions, the deveoper L. Hörmnder ws wrded the Fieds Med due to this contribution.

11 but our f( x is ony defined on (,. Therefore we need to extend f( x to (, in n pproprite wy. This extension must not produce ny cos terms. Therefore we shoud extend f oddy, { f( x < x < f ( x = ( 7 f( x < x < which eds to n = f ( x sin x dx = f( x sin x dx. ( 8 Simiry we hve b n = g( x sin x dx. ( 9 n c Exmpe. ( 7. 9 Sove u t t = c u x x, < x <, t > ( u( x, = x ( x, x, ( u t ( x, =, x, ( Soution. We hve =. Compute n = x ( x sin x dx = x ( x dcos x n = x ( x cos x n = ( x dsin x = ( x sin x 4 = sin x dx 4 = 3 cos x It is cer tht b n = s u t ( x, =. Therefore the soution is u(, t =, t, ( 3 u(, t =, t. ( 4 cos x ( x dx sin x ( dx = 4 ( ( n 3. ( 5 u( x, t = 4 ( ( n 3.. The het eqution. We consider the initi-boundry probem of the her eqution 3 cos c t sin x. ( 6 u t k u x x =, < x <, t > ( 7 u(, t =, t ( 8 u(, t =, t ( 9 u( x, = f( x, x. (

12 We consider soution of the form Substituting into the eqution we obtin u( x, t = X( x T( t. ( X T = k X T Simir to wht we hve done for the wve eqution, we concude for some constnt λ. Now considering the boundry condition, we hve nd consequenty ( up to constnt fctor X = X n sin nd ( up to constnt fctor T T = k X X. ( T = k λ T, X = λ X ( 3 X = λ X, X( = X( = ( 4 x ( 5 T = k n T T( t = e k t. ( 6 Thus we expect the gener soution to tke the form u( x, t = n e ( k t n sin x. ( 7 where the coefficients n = f( x sin x dx The Lpce eqution. Consider the probem We sove it using seprtion of vribes. Consider Substituting into the eqution we obtin which eds to u x x + u y y =, < x <, < x < b, ( 8 u( x, = f( x, x ( 9 u( x, b =, x, ( 3 u x (, y =, y b, ( 3 u x (, y =, y b. ( 3 u( x, y = X( x Y( y. ( 33 X Y + X Y = X X = Y Y ( 34 X λ X =, Y + λ Y = ( 35 for some constnt λ. Tking into ccount the boundry conditions for X, we hve, X λ X =, X ( = X ( = λ = X = An cos x ( 36 nd therefore, when n, Y Y =, Y(, Y( b = Y = B n e n y e n b e n y n = B n sinh( ( y b. ( 37

13 When n =, Thus the soution shoud be represented by Y =, Y( =, Y( b = Y = B ( y/b. ( 38 u( x, y = ( b y b To determine the coefficients, we use + f( x = u( x, = + n cos x sinh ( y b. ( 39 ( n sinh n b cos x which gives ( note tht we re expnding into cosine series here = x f( x dx, n = ( n b f( x cos. ( 4 sinh Nonhomogeneous probems. Now we show the power of the method of seprtion of vribes ( more precisey, the power of Fourier series by studying probem tht is beyond the power of the method of chrcteristics we hve erned. ( 4 u t t c u x x = h( x, t, < x <, t > ( 4 u( x, = f( x, x ( 43 u t ( x, = g( x, x ( 44 u(, t =, t ( 45 u(, t =, t. ( 46 As h( x, t is not seprted, it is not possibe to find ny soution of form u( x, t = X( x T( t for the eqution u t t c u x x = h( x, t. ( 47 However, it is sti possibe to sove this probem using our knowedge of Fourier series. We ssume the soution tkes the form 3 u( x, t = u n ( t sin x. ( 48 Reders fmiir with ODE theory my recognize this s PDE version of the vrition of constnts method. Aso note tht We need to represent h( x, t by sin x too. h( x, t = h n ( t sin x, h n ( t = h( x, t sin x dx. ( 49 Now the eqution becomes Now it is cer tht with initi conditions u n ( t + c un ( t sin x = h n ( t sin x. ( 5 u n ( t + c un ( t = h n ( t ( 5 c u n ( = n, u n ( = b n ( 5 ( n 3. Note tht, s we hve u(, t = u(, t =, Fourier sine series is reevnt. Therefore we ony keep the sin x terms.

14 where n nd b n re from the sine series of f nd g: f( x = n sin x n = g( x = c b n sin x To sove this eqution, we notice tht b n = n c f( x sin x dx, ( 53 g( x sin x dx. ( 54 u n ( t = v n ( t + w n ( t ( 55 where v n nd w n sove c c v n ( t + vn ( t =, v n ( = n, v n ( = b n, ( 56 c w n ( t + vn ( t = h n ( t, w n ( =, w n ( =. ( 57 The former eqution yieds c c v n ( t = n cos t + b n sin t ; ( 58 To sove the second eqution, we need the foowing Duhme s principe: The soution to the eqution is given by where w( t; s soves v ( t + λ v( t = h( t, v( =, v ( = ( 59 v( t = Using the bove principe we hve c w n ( t = Therefore the soution is { n u( x, t = } s ds sin x. t w( t; s ds ( 6 w ( t + λ w( t =, w( s =, w ( s = h( s. ( 6 c cos t Exmpe. ( 7. 9 Sove the probem t c h n ( s sin ( t s ds. ( 6 c + b n sin Soution. Rec tht we shoud use the formu u( x, t = { c c n cos x + b n sin } s ds sin x. t + c t h n ( s c sin ( t ( 63 u t t c u x x = A x, < x <, t > ( 64 u( x, =, x ( 65 u t ( x, =, x ( 66 u(, t =, t > ( 67 u(, t =, t > ( 68 x + c t h n ( s c sin ( t ( 69

15 with n = f( x sin x dx, b n = n c g( x sin x dx, h n ( t = Now tht f = g =, h( x, t = A x, =, we hve n = b n = nd h n ( t = A x sin x dx = A x dcos x n = A x cos x cos x dx n Therefore We cn check h( x, t sin x dx. ( 7 = A ( n+. ( 7 n u( x, t = { } t A ( n+ sin c ( t s ds sin x n c n = { ( } A ( n+ cos c ( t s t sin x n c n = n+ A ( n 3 3 c cos c t sin x. ( 7 u t t c u x x = { } A ( n+ cos c t sin x n c A ( n+ n c cos c t sin x = n+ A ( sin x = A x. ( 73 n