Concourse Math Spring 2012 Worked Examples: Matrix Methods for Solving Systems of 1st Order Linear Differential Equations

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1 Concourse Mah 80 Spring 0 Worked Examples: Marix Mehods for Solving Sysems of s Order Linear Differenial Equaions The Main Idea: Given a sysem of s order linear differenial equaions d x d Ax wih iniial condiions x (0), we use eigenvalue-eigenvecor analysis o find an appropriae basis { v,, v n } for R n and a change of basis marix S v vn such ha in coordinaes relaive o his basis ( u S x) he sysem is in a sandard form wih a known soluion Specifically, we find a sandard marix [ A] S AS, ransform he sysem ino d u d u, A solve i as u( ) [ e ] u (0), hen ransform back o he original coordinaes o ge x( ) [ e ] x (0) where A [ e ] S[ e ] S Tha is () [ A x e ] S[ e ] S x (0) This is acually easier o do han i is o explain, so here are a few illusraive examples: The diagonalizable case dx 5x 6y Problem: Solve he sysem d wih iniial condiions x(0), y(0) x y d Soluion: In marix form, we have d x d 5 6 Ax where A and x (0) 5 6 We sar by finding he eigenvalues of he marix: IA, and he characerisic polynomial is pa ( ) ( )( ) This gives he eigenvalues and The firs of hese gives he eigenvecor v, and he second gives he eigenvecor v The change of basis marix is S 0 and wih he new basis (of eigenvecors) { v, v} we have [ A] S AS 0 D, a diagonal marix [There is no need o carry ou he muliplicaion of he hree marices if you know ha you have a basis of eigenvecors I will always yield a diagonal marix wih he respecive eigenvalues on he diagonal] D e 0 The evoluion for his diagonal marix is [ e ] 0 e, and he soluion of he sysem is A D e 0 e e e e x() [ e ] S[ e ] S x(0) 0 e e e e e I s worh noing ha his may also be expressed as e e The complex eigenvalue case dx x 5y Problem: Solve he sysem d x y d wih iniial condiions x(0) 0, y(0)

2 Soluion: In marix form, we have d x d 5 0 Ax where A and x (0) 5 We again sar by finding he eigenvalues of he marix: IA, and he characerisic polynomial is pa ( ) ( ) This gives he complex eigenvalue pair i and i 5 0 i We seek a complex eigenvecor for he firs of hese: i 0 gives he (redundan) equaions ( i) 5 0 and ( i) 0 The firs of hese can be wrien as 5 ( i) and an easy soluion o his is where 5, i This gives he complex eigenvecor w i i i u v We have shown ha wih he specially chosen basis { vu, }, he new a b sysem will have sandard marix [ A] S AS b a where a is he real par of he complex cos sin eigenvalue and b is is imaginary par We also showed ha [ ] b b a e e sin b cosb In his example, 0 5 a and b, Sv u, 5 cos sin S 5 0,, and [ ] e e sin cos The A 0 5 cos sin 5 0 soluion o he sysem is herefore () [ ] [ ] (0) e x e S e S x 5 sin cos 0 e 5sin 5cos 5 5sin 5 e cossin sin cos 0 cossin Tha is, x () 5sin y() cossin Repeaed eigenvalue case (wih geomeric mulipliciy less han he algebraic mulipliciy) dx y Problem: Solve he sysem d wih iniial condiions x(0), y(0) xy d Soluion: In marix form, we have d x d 0 Ax where A and x (0) We again sar by finding he eigenvalues of he marix: IA, and he characerisic polynomial is pa ( ) ( ) This gives he repeaed eigenvalue wih (algebraic) mulipliciy We 0 seek eigenvecors: 0 gives he (redundan) equaions 0 and 0 Therefore, so we can choose v or any scalar muliple of his as an eigenvecor, bu we are unable o find a second linearly independen eigenvecor (We say ha he geomeric mulipliciy of he eigenvalue is ) The sandard procedure in his case is o seek a generalized eigenvecor for his repeaed eigenvalue, ie a vecor v such ha ( I A) v is no zero, bu raher a muliple of he eigenvecor v Specifically, we seek a vecor such ha Av vv This ranslaes ino seeking v such ha ( I A) v v Tha is, This gives redundan equaions he firs of which is or If we

3 0 Av v (arbirarily) choose 0, hen, so v The fac ha ells us ha wih he Av v v 0 change of basis marix S, we will have [ A] S AS 0 We have shown ha e e [ e ] 0 e for a marix in his sandard form The soluion o he sysem is herefore A 0e e 0 e e x() [ e ] S[ e ] S x (0) 0 e e e e e e e e x() e () e 6e 8e e e 8e 8 Tha is, y() e (8) I s worh noing ha his can also be expressed as x () e e In general, you should expec o encouner sysems more complicaed han hese by examples To illusrae he line of reasoning in a significanly more complicaed case, here a ig Problem ig Problem: a) Find he general soluion for he following sysem of differenial equaions: dx xx x5 d dx x xx 5 d dx x x b) Find he soluion in he case where x (0) d dx x d dx 5 x x5 d 0 0 Soluion: This is a coninuous namical sysem of he form d x d 0 0 Ax where A We sar by seeking he eigenvalues We have IA The characerisic polynomial is pa ( ) ( ) ( )( ) which yields he repeaed eigenvalue (wih algebraic mulipliciy ), he disinc eigenvalue, and he complex pair i and 5 i

4 00 The repeaed eigenvalue yields jus one eigenvecor v, so is geomeric mulipliciy if jus 0 0 We hen seek a generalized eigenvecor v such ha Av v v where Tha is, we seek a vecor v such ha v Av ( IA) v v This is jus an inhomogeneous sysem which yields soluions of he form v 0 For simpliciy, ake he soluion wih 0, ie v The eigenvalue yields he eigenvecor v A sraighforward calculaion wih he complex i eigenvalue i yields he complex eigenvecor v i0 v5 iv in accordance wih he mehod derived in class Using he basis v 0, v 0, v, v 0, v5 and change of basis marix S 0 0 0, we compue he inverse marix S We know ha Av v Av vv Av v, and he marix of A relaive o he basis is S AS Av v v Av5 v v Since A SS, i will be he case ha he evoluion marices are relaed via e e e e 0 0 e e cos e sin e sin e cos A e Se S where

5 The soluion is hen e e e x( ) e (0) e (0) e x S S x x (0) A e cos e sin e sin e cos c c If we muliply he lefmos marices and wrie S x (0) c, his yields he general soluion: c c 5 e e e 0 0 c e (cossin ) e (cossin ) c () A e (0) x e (0) 0 0 e e sin e cos x S S x c 0 0 e 0 0 c 0 e e 0 0 c5 x() ce ce ce x( ) ce (cossin ) c5e (cossin ) or x() ce ce sinc5e cos x() ce x5() ce ce 5 If, on he oher hand, we use he iniial condiion x (0), we ge he specific soluion: e e e e (cossin ) e (cossin ) x () 0 0 e e sin e cos 0 0 e e e 0 0 x () e e e x() e (cos sin ) or x() e e (sincos) x() e x5 () e e Moral of he Sory: I s always possible o find a special basis relaive o which a given linear sysem is in is simples possible form The new basis provides a way o decompose he given problem ino several simple, sandard problems which can be easily solved Any complicaion in he algebraic expressions for he soluion is he resul of changing back o he original coordinaes 5