Solving Matrix Differential Equations

Transcription

1 Solving Marix Differenial Equaions Seps for Solving a Marix Differenial Equaion Find he characerisic equaion ofa, de( A I) = Find he eigenvalues of A, which are he roos of he characerisic equaion 3 For each eigenvalues, find as many linearly independen eigenvecors as possible 4 The soluion of he marix differenial equaion akes he form: n x() c ve = + c ve + + c v e n n Example : Solving a sysem of wo linear differenial equaions using eigenvalues dx = x + 3y d Solve dy = x + y d 3 The sysem of equaions has he form x = Ax where A = In Example, we found he eigenvalues of his marix o be = and = 4 The corresponding eigenvecors were v = and 3 v =, respecively The soluion of he equaion is herefore 4 ce + ce 4 x() = cve + cve = c e c e 3 + = 4 3ce + ce Mahemaica: 8x@D,y@D<,D DSolve@8x'@DŠ x@d+ 3 y@d,y'@dš x@d+ y@d<, :x@d 5 ã- H + 3 ã 5LC@D+ 3 5 ã- H- +ã 5LC@D, 5LC@D> y@d 5 ã- H- +ã 5LC@D+ 5 ã- H3 + ã

2 Example 3: Solving a sysem of wo linear differenial equaions using eigenvalues dx = x + y d Solve dy = 4x + y d Soluion: The sysem of equaions has he formx = Ax where A = 4 The eigenvalues of his marix are given by he soluions of he equaion de( A I) = Since A = and, we have 4 I = A I = 4 Seing he deerminan of his marix equal o zero gives: de( A I) = ( )( ) 4 = Muliplying ou, we have or = and = 3 3 = So, he wo eigenvalues are =, 3 Nex, find he eigenvecors = : Wih his choice of, he marix A I = 4 becomes 4 v So ha he marix equaion we mus solve is( A I) v = 4 v = This marix equaion can be solved using he augmened marix 4 This can be reduced o 5 so he soluion is v + v = or v = v We choose v = so ha we answer involves inegers and we have v = = 3 : Wih his choice of : A I = 4 So he sysem o be solved v is( A I) v = 4 v = The associaed augmened marix is 4

3 5 This can be reduced o so he soluion is v v = or v = v We choose v = hen v 3 = The soluions are hen e x() = and y() = e We check he Wronskian o see if he soluions form a fundamenal se: 3 e e 3 3 W = = e ( e ) ( e ) e = 4e 3 e e Therefore he soluions are independen and he general soluion of he sysem is x() = c ve + c ve = c e c + e 3 Example 4: Solving a sysem of wo linear differenial equaions using eigenvalues dx = x y d Solve dy = x 3y d Soluion: The sysem of equaions has he form x = Ax where we have A I = 3 Seing he deerminan of his marix equal o zero gives The characerisic equaion is ( ) ( )( ) de A I = = = The wo eigenvalues are =, 4 or = 4 and = Nex, use he eigenvalues o find he corresponding eigenvecors A = 3 Then,

4 = : Wih his choice of, he marix A I = 3 becomes v So we mus solve he equaion( ) A I v = = v The corresponding augmened marix is This can be reduced o so he soluion is v + v = or v = v We choose so ha v = v = = 4 : Wih his choice of : A I = Then, v ( A I) v = = v The associaed augmened marix is This can be reduced o so ha v v = or v = v We choose hen v = v = 4 The soluions are hen () x = e and y () = e Noe: Check o see if he soluions form a fundamenal se Te general soluion of he sysem is x() = c ve + c ve = c e + c e 4

5 Analysis of Example 4 Using Mahemaica

6 Equilibrium Soluions of Linear Sysems dx = ax + by A linear sysem of wo differenial equaions of he form d has (, ) as an dy = cx + dy d equilibrium soluion The ype of equilibrium poin depends on he eigenvalues, of a b he marix A = For now we consider he cases where he sysem has nonzero, c d disinc, real eigenvalues Three Types of Equilibrium Poins If a linear sysem of wo differenial equaions has wo nonzero, disinc, real eigenvalues,, hen: If < <, hen he origin is a saddle There are wo lines in he phase plane ha correspond o sraigh-line soluions If < <, hen he origin is a sink (sable node) All soluions end o (, ) as and mos go o zero in he direcion of he -eigenvecors If < <, hen he origin is a source (unsable node) All soluions excep he equilibrium soluion end o infiniy as and mos soluions leave he origin in he direcion of he -eigenvecors The sink (sable node) of Example 4

7 Graphical Analysis of Example 3 y x 3 4 Saddle poin a (,) from Example 3 The wo lines correspond o he wo lines of eigenvecors The line y = x corresponds o he eigenvecor vecor and he line y = - x corresponds o he eigenvecor vecor 3 3 We can also see ha since, and, we have x () = ce y () = ce ( ) 3 yx () = ce = x 4 3 y x 3 4

8 Example 5: Solving a sysem of hree linear differenial equaions using eigenvalues dx = y + z d dy Solve he sysem of hree linear differenial equaions = x + z d dz = x + y d Soluion: Firs, wrie he sysem as a marix equaion: x = x

9 Soluions involving Complex Eigenvalues Example 6: Solving a sysem wih complex eigenvalues Solve he sysem dx = x 3y d dy = 3x y d Soluion: The characerisic equaion of his sysem is de( A I) = ( )( ) + 9 = which simplifies o = The eigenvalues are = + 3i and = 3i To find he eigenvecor corresponding o = + 3i we subsiue ino 3i 3v he equaion ( A I) v = o ge ( A I) v = 3 3i v = We solve he 3iv 3v = sysem of equaions by using he boom equaion o ge v = iv 3v 3iv = i (which also saisfies he firs equaion) We choose as he eigenvecor In order o finish, we need he following resul from precalculus: ib a ib a Euler s Formula: e = cosb + isinb e + = e ( cosb + isinb) ( + 3 i ) Coninuing wih our example, we see ha e = e [(cos 3 ) + isin(3 ) ] I follows ha he soluion has he form: i x() = e [(cos3) + isin(3) ] ie [(cos 3 ) i sin(3 ) ] + ie cos( 3 ) e sin(3 ) = = e [(cos 3 ) + isin(3 ) ] e cos( 3) + ie sin(3 ) e sin(3 ) e cos( 3) = + i e cos( 3) e sin(3 ) The wo pieces are he real and imaginary pars of he vecor funcion x()

10 a b Theorem: If x () is a complex-valued soluion o a linear sysem x () = x() c d where he coefficien marix has all real enries, and ha x() can be wrien as he sum of real and imaginary pars as x() = x() re + x() im where boh x() re, x() im are realvalued funcions of Then x() and x() are boh soluions of he original sysem re Example 6: Conclusion e sin(3 ) e cos( 3) The funcions xre() = and xim() = are boh soluions of e cos( 3) e sin(3 ) he sysem They are independen since heir iniial values x re () = and x im () = are independen Therefore, he general soluion of he sysem is given by im e sin(3 ) e cos( 3) x() = c + c e cos( 3) e sin(3 )

11 Example 7: Solving a sysem wih complex eigenvalues Solve he iniial value problem x () = Ax() where A = 3 and x() =

12 Example 8: Solving a sysem wih complex eigenvalues A harmonic oscillaor can be modeled by he second-order equaion my + by + ky = This is considered undamped if b = Choose b = and choose he mass o be m = and he spring consan o be k = Then, he equaion can x = x wrien as a sysem by leing x = y and x = x, so ha or x = x X () = () X Sorry bu I don have his one yped up ye!!

13 Deficien Eigenvalues Le he sysem of linear differenial equaions wih consan coefficiens be given by x () = Ax Suppose is he only eigenvalue We seek a soluion of he form: x() = ( v + v ) e Here v, v are consan column marices Differeniaing he proposed soluion gives = ( ) + = + ( + ) x () v e v e v e v v Subsiuing ino he differenial equaion gives ( ) ( ) v e e + v + v e = A v + v Dividing ou he exponenial, we have or v + ( v + v ) = A( v + v ) v + v + v = Av + A v Equaing coefficiens gives he sysem of equaions Av = v Av = v + v Then v is he eigenvecor of A associaed wih and v is he soluion of he second equaion The firs equaion is how we go he original eigenvalue and eigenvecor Solving he second equaion is equivalen o solving ( A I ) v = v

14 Example 9: Solving a sysem wih repeaed eigenvalues Solve he linear sysem x () = Ax() where A = Soluion: Firs noe ha he phase plo shows ha here is only one sraigh line of soluions which indicaes ha he sysem has a single eigenvalue The eigenvalue of A = is given by he soluion of de( A I ) = = ( ) = -3 So he sysem x () = () has as is only eigenvalue The associaed x = eigenvecor is found from Av or ( ) Since( ) = v A+ I v = A+ I =, we v mus solve he sysem v = I follows ha v = The eigenvecor can be chosen o be The firs soluion is given by x () = e To find a second eigenvecor and a second soluion, we mus solve he sysem Av = v + v which is equivalen o( A I) v = v or in his case: v v =

15 This resuls in v = So we can choose he second eigenvecor v = Then he second soluion will be of he form: () = e + e x The general soluion will hen be of he form: () c () c () c e c e x = x + x = + + e or c c x() c x () c x () e e = + = c + Theorem: Suppose x () = Ax() is a linear sysem in which he marix A has a repeaed real eigenvalue bu only one line of eigenvecors Then he general soluion has he form () e x = V + Ve x where V = y is an arbirary iniial condiion and V is deermined from by V V = ( A I) V If V =, hen V is an eigenvecor and x() is a sraigh-line soluion Example 9: Conclusion x The sysem x () = x() has as is only eigenvalues Le be = V = y an arbirary iniial condiion Then x y V = ( A+ I) V = y = So he general soluion x y is x() = y e + e

16 Example : Solving a sysem wih repeaed eigenvalues Solve he linear sysem x () = Ax() where A = 3 Soluion: The eigenvalues are soluions of de( A I) = = ( ) ( 3 ) + = 3 The equaion ( ) ( 3 ) + = = ( ) = has = as is only soluion So A has = as is only eigenvalue The associaed eigenvecor is found from Av = v or ( A I) v = Since( A I ) =, we mus solve he v sysem I follows ha The eigenvecor can be chosen o v = v = v be The firs soluion is given by x () = e To find a second eigenvecor and a second soluion, we mus solve he equaion v ( A I) v = v or in his case: v = This sysem has he augmened marix which reduces o I follows ha v = v We choose v = hen he eigenvecor is Then he second soluion will be of he form: () e x = + The general soluion will hen be of he form: e x() = cx() + cx () = c e + c e + e ce + ce or x() = ( c + c) e ce

17 Theorem: Suppose x () = Ax() is a linear sysem in which he marix A has a repeaed real eigenvalues bu only one line of eigenvecors Then he general soluion has he form () e x = V + Ve x where V = y is an arbirary iniial condiion and V is deermined from by V V = ( A I) V If V =, hen V is an eigenvecor and x() is a sraigh-line soluion Example : Conclusion x The sysem x () = x() has as is only eigenvalues Le be = V = y an arbirary iniial condiion Then x y V = ( A+ I) V = y = So he general soluion x y is x() = y e + e

18 Example : A Harmonic Oscillaor Consider he harmonic oscillaor modeled by he second-order equaion y + y + y =, wih mass m = and spring consan k = and damping coefficien b = The equaion can wrien as a sysem by leing x = y and x = x x = x, so ha or x = -x - x X () = () X