Eigenfunction Expansion. For a given function on the internal a x b the eigenfunction expansion of f(x):
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1 Eigefuctio Epsio: For give fuctio o the iterl the eigefuctio epsio of f(): f ( ) cmm( ) m 1 Eigefuctio Epsio (Geerlized Fourier Series) To determie c s we multiply oth sides y Φ ()r() d itegrte: f ( ) ( ) r( ) d c ( ) ( ) r( ) d m m m 1 Becuse of orthogolity of eigefuctios: f ( ) ( ) r( ) d c ( ) r( ) d c f ( ) ( ) r( ) d ( ) r( ) d Dr. Bghli
2 For emple, cosider the previous prolem: y y y, y(), y( ) e si We wt to epd fuctio f() i terms of eigefuctios: f ( ) c( ) 1 ( ) ( ) ( si ) r d e e d si d c f ( ) c ( ) 1 f ( ) ( ) r( ) d 1 ( ( ) si ) f e d e si f ( ) e si d Dr. Bghli
3 For emple tke f()= ( is costt): c ( ) f e si d e c e si d e ( 1) (1 ) e ( 1).. e si e (1 ) 1 For =1 : e ( 1) 1 e si e (1 ) 1 Dr. Bghli
4 y Eigefuctio Epsio of y=1 Usig 5 terms Ect Eigefuctio Epsio Dr. Bghli
5 y Eigefuctio Epsio of y=1 Usig terms Ect Eigefuctio Epsio Dr. Bghli
6 y Eigefuctio Epsio of y=1 Usig 5 terms Ect Eigefuctio Epsio Dr. Bghli
7 Method of Eigefuctio Epsio for Solvig Ordiry Differetil Equtios Lu f To solve prolem : 1u u 3u 4u I which L is formlly self-djoit secod-order differetil opertor: d d L ( p ) q d d We cosider the ssocited S.L. eigevlue prolem (r=1 for simplicity): L fid ( ) Dr. Bghli
8 u ( ) 1 f c ( ) 1 c f d [ ] d Lu f L c ( ) 1 1 (1) : 1 1 L LHS L L (1) L 1 1 c ( c ) Sice eigefuctios re orthogol d hece idepedet: c c Dr. Bghli
9 u 1 c ( ) Whe r 1 we do ot epd f() d directly write: Lu f L f ( ) 1 1 L f ( ) L r r 1 f ( ) Usig orthogolity of eigefuctios with respect to r(): f d r( )[ ] d Dr. Bghli
10 Discussio: c u 1 c ( ) If oe of the eigevlues of S.L. system sy j is zero, u() fils to eist if c j c j c If s well, the is idetermite. j c Sice we hve: c j () j j j j j Ad j c e chose ritrry. j [ ] d f d j f d j c u k j j I which k is ritrry costt d hece there is ifiitely my solutios for the prolem. 1 Dr. Bghli
11 Emple 1- Use the method of eigefuctio epsio to solve: y y 1 y () y (1) Ly f ( ) f 1 1 for l ( ) l si l si f c ( ) 1 1 c si 1 c 1 si d 1 si d 1 si d 1 Dr. Bghli
12 c c 1 1 [ cos ] [1 1 ] eve si odd 1 odd y ( ) 1 1 si y y 1 si si si odd 4 4 ( )si si 1 1 odd Dr. Bghli
13 4 ( ) eve odd y 1 odd 4 si ( ) Ect Solutio: Use Lplce trsform method: Dr. Bghli
14 Dr. Bghli
15 Ect Solutio: Lplce trsform: y y 1 y () y (1) L( y ) L( y ) L(1) L( y ) sl( y ) y () sl( y ) y () L( y ) L(1) s[ sl( y ) y ()] y () L( y ) L(1) y () y () s L( y ) L( y ) s 1 L( y) s 1 s s s ( s ) s [ ] s s ( s ) L 1 1 ( y ) L ( si ) ( ) ( cos ) L L y 1 1 si cos y (1) cos 1 y ect cos ( )si si Dr. Bghli
16 y Solutio of y"+y=1 y()=y(1)= -. Ect Solutio Method of Eigefuctio Epsio (3 terms) Dr. Bghli
17 Emple - Use the method of eigefuctio epsio to solve: y y si y y For the system y 4y si with the sme B.C., wht goes wrog? Solutio: π r r '' Cse ( ) tke : ( ) c Dr. Bghli
18 Cse o eigevlue (verify t home) Cse r i ( ) Acos Bsi ( ) A si B cos B A si si 1,,... ( ) Acos ( ) cos Dr. Bghli
19 si f c f si c cos (It is ideed Fourier Cosie Series) 1 si d c d c c c 4 c f ( ) ( ) d [ ( )] d si cos d c cos d si cos d c [1 ( 1) 1 ] c 4 1 eve odd Dr. Bghli
20 si 4 1 eve cos We lso epd the solutio i series of eigefuctios: y cos 1 Sustitute i differetil equtio: y y si cos cos cos (1 ) eve 4 cos cos (1 ) 1 eve Dr. Bghli
21 4,4,6, 1 ( ) y 1 4 eve 1 ( ) cos For the system: y 4y si cos 4 cos eve 4 cos (1 ) eve 4 4 cos cos (1 ) Dr. Bghli
22 1 4,4, 1 (4 ) We hve troule to fid?? ( ) cos f si. cos d No Solutio! j Dr. Bghli
23 It is iterestig to see why the prolem hs o solutio whe we solve it i differet wy, usig for emple Lplce trsform. We will fid: 1 y C1cos C si si 3 1 y C 1si C cos cos 3 y y C C 3 Icosistet! No Solutio! Dr. Bghli
24 No-homogeeous Boudry Coditios To illustrte the procedure i this cse let us cosider: y y f ( ) y y y 1 y 1 Choose: y u v y u v y u v u v u v f u u v v f ( ) Tke u d v such tht: u u f ( ) v v Dr. Bghli
25 Boudry coditios: y y u u v v u v u v Tke oudry coditios such tht: u u v v y y u u v v u v u v Tke oudry coditios such tht: u 1 u 1 v 1 v 1 Dr. Bghli
26 u u f u () u() u (1) u(1) No-homogeeous prolem with homogeeous B.C. (S.L. prolem) Solve y the method of eigefuctio epsio. v v v () v() v (1) v(1) Homogeeous prolem with o-homogeeous B.C. Solve y direct methods, e.g. here we hve: r r i v A si B cos Ad we fid A d B usig B.C. s, filly we we will hve: y ( ) u( ) v ( ) Dr. Bghli
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