Sturm-Liouville Eigenvalue Problem

Transcription

1 Sturm-Liouville Eigevalue Problem d dx d p q r 0 a x b dx i which p, q ad r are fuctios of x, is called Sturm-Liouville (S.L.) eigevalue problem. d d L p q dx dx L r 0

2 Examples 1- Bucklig of a colum: d u dx u 0 p 1, q 0, r 1 - Heat flow i a ouiform rod: d dx dt K T c T 0 dx p K, q, r c 3- Vibratios of a ouiform strig: d u 0 0 T u u 0 p T 0 ( cte ), q, r 0 dx

3 Regular Sturm-Liouville Problem d d p q r 0 a x b dx dx 1( a) ( a) 0 3( b ) 4( b ) 0 i which all i are real, the coefficiets p, q, r must be real ad cotiuous everywhere i a x b, ad p 0, r 0 everywhere i, the the S.L. problem is called regular. a x b

4 Importat Properties of Regular S.L. Problem 1- All eigevalues are real. - There exist a ifiite umber of eigevalues: a- there is a smallest eigevalue usually deoted by 1 b- there is ot a largest eigevalue ad as 3- Correspodig to each eigevalue, there is a eigefuctio deoted by (which is uique to withi a arbitrary multiplicative costat). has exactly -1 zeros for a x b (ot icludig the ed poits)

5 4- The eigefuctios ( x ) form a complete set, so ay piecewise smooth fuctio f(x) ca be represeted by a geeralized Fourier series of eigefuctios: f ( x ) a( x ) 1 Furthermore, this ifiite series coverges to: 1 If the coefficiets are properly chose. f ( x ) f ( x ) for a x b

6 5- Eigefuctios belogig to differet eigevalues are orthogoal relative to the weight b fuctio r(x): ( x ) ( x ) r( x ) dx 0 if a m m 6- Ay eigevalue ca be related to its eigefuctio by the Rayleigh quotiet : b d b d ( ) ( ) ) dx a a dx b p p q dx a rdx Note: For S.L problems that are ot regular, theorems maybe still valid.

7 Illustrative Example Cosider: d 0 (0) ( ) 0 dx It is a S.L. problem with: p(x)=1 q(x)=0 r(x)=1 For 0 characteristic polyomial is: r 0 r We explore three possible cases for the solutio:

8 Case I 0 d 0 ( ) 1 dx x C C x (0) 0 ( ) 0 C 0 ( x ) C x 1 0C C 0 0 Trivial solutio So λ=0 is ot a eigevalue

9 Case II: 0 r let s 0 s 0 ( x ) C cosh s x C sih s x ( ) 0 0 C4 sih s r ( x ) C e C e s x 1 e e e e Recall that: cosh x sih x (0) 0 0 C 3 s 3 4 s x x x x x ( x ) C sih s x 4 Two real roots sl 0 Ad Sih is ever zero for a positive argumet C 4 0 ( x ) 0 Trivial solutio

10 Case III: 0 r Has imagiary roots r i (0) 0 C cos x C si x 1 ( x ) C si x C 1 0 ( ) 0 0 C si For otrivial solutio: si l 0 l The eigevalues are: ( ) 1,,3,... l ( ) si x C x Take arbitrary costat : C 1 (=0 is ot allowed ) ( ) ( x ) si x

11 1- We ote that all eigevalues are real. - We see that eigevalues are ifiite. There is a smallest eigevalue: but o largest. 1 ( ) 3- Cosider eigefuctios: ( x ) si x ( x ) has exactly -1 zeros for 0 x (ot icludig the ed poits):

12 1: ( x) si x 1 10 zeros : ( x) si x 11 zeros 3: ( x) si 3 3 x 1 zeros

13 4- We see that ay piecewise smooth fuctio f(x) ca be represeted by series of eigefuctios: f ( x ) a( x ) Which is Fourier Sie Series ad it coverges to f(x) if it is cotiuous i 0 x ad it coverges to 1 1 ( ) ( ) f x f x f ( x ) a( x ) 1 asi 1 x for piecewise smooth fuctio. is called a eigefuctio expasio of f(x) or a Geeralized Fourier Series of f(x).

14 5- Orthogoality of eigefuctios: b a ( x ) ( x ) r( x ) dx 0 if m m Here we have: 0 x mx Si Si dx 0 for m Sice: 0 x mx 0 m Si Si dx / m

15 Some useful itegrals: x mx 0 m Si Si dx m 0 m x mx Cos Cos dx m 0 m 0 x mx Si Cos dx 0

16 6- Rayleigh quotiet: b d b d ( ) ( ) ) dx a a dx b p p q dx Here we have: p=1, q=0, r=1 a rdx x x x [ Si ( ) Cos ] ( ) Cos ) dx 0 0 x Si dx 0 ( )

17 Physical Meaig of Eigevalues ad Eigefuctios X X=0 P u(x) p u u 0 EI u(0) u( l) 0 (1) Compare with the eigevalue problem: X=l P u 0, (0) ( l) 0 l x l ( ) si( )

18 u(0)=u(l)=0 بنابراین جواب معادله دیفرانسیل )1( با شرایط مرزی 0=u خواهد بود مگر اینکه P/EI یک eigevalue باشد که در این صورت جواب si(пx/l) u=a خواهد شد یعنی: P EI P Bucklig loads EI l l x si l Bucklig modes For desig purposes the lowest bucklig load P 1 is of most iterest ad is called critical bucklig load P cr : The fact that l l 0 l P cr P cr P cr 0 EI Is i agreemet with our experiece ad ituitio