After the completion of this section the student. Theory of Linear Systems of ODEs. Autonomous Systems. Review Questions and Exercises

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1 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 45 V.5 SYSTEMS OF FIRST ORDER LINEAR ODEs Objecives: Afer he compleion of his secion he suden - should recall he definiion of a sysem of linear s order ODEs - should be able o find he general soluion of a sysem wih consan coefficiens - should be able o invesigae for sabiliy he equilibrium poin of a plane sysem Conens: V.5.. Definiions and Noaions V.5.. V.5.3. V.5.4. V.5.5. V.5.6. V.5.7. Theory of Linear Sysems of ODEs The Fundamenal Se of a Linear Sysem wih Consan Coefficiens Auonomous Sysems Examples Review Quesions and Exercises Sysems of ODEs wih Maple

2 46 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions V.5 SYSTEMS OF s ORDER LINEAR ODEs. DEFINITIONS AND NOTATIONS In his secion we will sudy he heory of he sysems of linear s order ODEs. I can be shown ha such sysems are equivalen o a single linear differenial equaion of a higher order; and for boh of hem he mos heoreical resuls have a similar descripion. Alhough, we will ry o avoid duplicaion of he heoreical jusificaion, in a pracical approach, he mehods of soluion for a single equaion and for he sysems are differen. In many cases, he descripion of he physical model is more naural o perform wih he sysems of ODEs, and invesigaion of he physical models such as dynamic, sabiliy ec. is more visual when i is made wih he help of sysems. Normal form Consider a sysem of n linear firs order ODEs wrien in he normal form (solved for he derivaives of unknown funcions): ( ) x + a ( ) x + + an ( ) xn f( ) ( ) x + a ( ) x + + a ( ) x f ( ) x = a x = a + n n + () x = a x + a x + + a x f where x ( ), x ( ),, xn ( ) sysem (), and coefficiens ( ) n n n nn n + are unknown funcions o be deermined from he a ij,, j =,,... f i are coninuous funcions in D R. i and funcions n Marix form Inroduce he following column vecors and a marix wih enries which are he elemens of he sysem (): x = x x x n ( ) ( ) ( ) x x x = x n ( ) ( ) ( ) f f f = f n ( ) ( ) ( ) a a an a a an A = an ( ) an ( ) ann ( ) Then he sysem () can be wrien in he compac marix form: x = Ax + f () This is a non-homogeneous sysem. Wihou a free erm f, he sysem is homogeneous: x = Ax (3) Soluion vecor The soluion vecor (paricular soluion) is any column vecor x ( ), D which saisfies he sysem (). Iniial value problem Find he soluion of a sysem of ODEs subjec o iniial condiions: x x x = Ax + f subjec o x( ) = x where x =, x i (4) x n

3 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 47 Exisence Theorem Theorem (exisence and uniqueness heorem) Le he sysem of linear s order ODEs x = Ax + f be normal for D (means ha he funcions ( ) and f i ( ) are coninuous on D ) and le xi. Then here exis exacly one soluion x ( ) such ha x ( ) x x = x n D a ij The paricular soluion x ( ) of he sysem () or (3) is a vecor valued funcion n ( ) : x n which can be reaed as a parameric definiion of he curve in he space : x = x( ) x = x( ) x( ) = D (5) xn = xn( ) wih he coordinaes x,x,...,x n. phase porrai rajecory phase space This space is called a phase space, and he soluion curve defined paramerically by equaion (5) is called he rajecory or he orbi in he phase space. For -D and 3-D cases, he radiional noaions for he coordinae sysem and, correspondingly, for unknown funcions in he sysem are used: x = a x+ a y+ a z+ f 3 y = a x+ a y+ a z+ f (b) 3 z = a x+ a y+ a z+ f wih he paricular soluions wrien as x= x x ( ) = y = y D (5b) z = z The graph of he equaion (5b) defines a rajecory in he phase space (phase plane, for -D case). The independen variable D can be reaed as he ime (can also be negaive), and for any momen of ime equaion (5b) defines he posiion of a poin on he rajecory herefore, he parameric equaion (5b) can be inerpreed as a moion of a maerial poin along he rajecories defined by he linear sysem of ODEs. The arrows on he rajecories indicae he direcion of moion wih he increase of ime. The family of all rajecories of he linear sysem is called he phase porrai. An iniial value problem defines he rajecory which goes hrough he prescribed poin. According o he Exisence Theorem, he soluion of any iniial value problem of he linear sysem is unique i means ha here is only one rajecory which goes hrough any poin of he phase space, and ha he rajecories of he linear sysem do no inersec.

4 48 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions. THEORY OF LINEAR SYSTEMS OF ODEs I can be shown ha a linear n h order ODE can be ransformed o a sysem of n linear s order ODEs, and a sysem of n linear s order ODEs can be reduced o a linear n h order ODE. Therefore, he descripion and he properies of he general soluion of he sysem will be similar o hose of he general soluion of a linear n h order ODE. Here, we will lis he major resuls of he heory of he sysems of linear s order ODEs. I) Homogeneous Sysem: x = Ax (3) linear independen soluions Vecors (vecor funcions) ( ), ( ),..., ( ) x x x are linearly independen if n heir linear combinaion equals zero for all D c x + c x c x = (6) n n only if all coefficiens are zero c = c =... = cn =. If vecors ( ), ( ),..., ( ) x x x are no linearly independen, hen hey are n linearly dependen. I means ha in he linear combinaion a leas one coefficien ck can be non-zero. Wronskian The Wronskian of he se of soluion vecors of he homogeneous sysem x, x,..., x is defined as a deerminan of he marix whose columns n are he vecors x k ( ) : W ( ) = de ( ) ( )... ( ) x x x n (7) There is a relaionship of he Wronskian (7) o he Wronskian defined in 5.3. If W( ) a leas a one poin D, hen he soluion vecors x ( ), x ( ),..., x ( ) are linearly independen. n There always exis n linear independen soluions ( ), ( ),..., ( ) x x x of he n homogeneous sysem (3). Bu any n+ soluions of he homogeneous sysem (3) are linearly dependen. Fundamenal se Any se of n linearly independen soluions of he sysem (3) x, x,..., x is called a fundamenal se. n I is obvious ha homogeneous sysems always possess a zero soluion x( ) (rivial soluion). Bu any se which includes he zero vecor is linearly dependen. Therefore, he fundamenal se canno include he rivial soluion. Fundamenal marix A marix wih columns which are consruced from he fundamenal se is said o be he fundamenal marix: ( ) = ( ) ( )... ( ) X x x x n (8) General Soluion Any soluion of he homogeneous sysem (3) can be wrien as a linear combinaion of he vecors from he fundamenal se. Therefore, he general soluion (complee soluion, complimenary soluion) of he homogeneous sysem is a se of all is soluions and i is given by all linear combinaions of he

5 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 49 vecors from he fundamenal se (span of he fundamenal se) and i can be defined as: ( ) c ( ) c ( )... c ( ) x = x + x + + x = Xc (9) c n n where c is a vecor of arbirary consans. Equaion (9) defines a family of curves in he phase space which represens he soluions of he homogeneous sysem. Soluion of IVP The soluion of he iniial value problem for a homogeneous sysem: = x = x () is given by x Ax subjec o = x X X x () where X ( ) is he fundamenal marix and X is he inverse of he fundamenal marix a = II) Non-Homogeneous Sysem: x = Ax + f () Denoe by ( ) x any paricular soluion of he sysem (). A paricular soluion p can be found by he mehod of undeermined coefficiens (similar o he case of linear ODE) or by he mehod of variaion of parameer: xp = X X f d () The general soluion of he non-homogeneous sysem () is given by a sum of he general soluion of he homogeneous sysem (complemenary soluion) and a paricular soluion: x ( ) = x ( ) + x ( ) c p General Soluion Using equaions (9) and (), he general soluion of he non-homogeneous sysem can be wrien as: x ( ) = X( ) c + X X f d (3a) Soluion of IVP The soluion of he IVP for he non-homogeneous equaion (4) can be given by: x ( ) = X X x + X X s f s ds (3b) In a conclusion, he general soluion and he soluion of IVP for homogeneous and non-homogeneous sysems (9), (), () and (3) can be deermined if he X is known. In he nex fundamenal marix of he homogeneous sysem secion we will consider he case of linear sysems wih consan coefficiens (auonomous sysems) for which here exis he developed mehods of finding he fundamenal marix.

6 4 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 3. FUNDAMENTAL SET OF LINEAR SYSTEMS WITH CONSTANT COEFFICIENTS Consider he homogeneous sysem of linear s order ODEs x = ax + ax + + an xn x = a x + a x + + a x n n (4) x = a x + a x + + a x n n n nn n where all coefficiens aij, i, j =,,... are consans. In marix form: a a an a a a n x = Ax A = (5) an an ann The Eigenvalue Problem, marix diagonalizaion, marix exponenial are among he echniques used for consrucion of he fundamenal se for a sysem wih consan coefficiens. Eigenvalue Problem: Because he linear sysem is of he firs order, we look for he non-rivial soluion of he exponenial form k x( ) = k e λ, k = (6) k n k is a non-zero (non rivial) vecor of consans, where k i and λ can be he real or he complex numbers which have o be found from saisfying equaion (5). Subsiue he rial form (6) ino marix equaion (5): ( k λ e ) = A λ ( k e ) λ λ e = k Ak e λ λ λ λ e e = k Ak Ike Ake I is he ideniy marix λ λ λ = e λ ( λi A) k = ( λ ) = λi A k = can be divided by e λ > A I k (7) This is he homogeneous sysem of algebraic equaions, which according o he Theorem has a non-rivial soluion if he deerminan of he marix of coefficiens is equal o zero. Therefore, he following condiion should be saisfied: de A λi = (8) Equaion (8) is he de h n order algebraic equaion for consan λ : ( A λi ) a a an a a a n de λ = an an ann a λ a an a a λ a n de = an an ann λ = c λ + c λ c λ+ c n n n n

7 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 4 Expansion of he deerminan yields an algebraic equaion wih real coefficiens which is called he characerisic equaion. According o he Fundamenal Theorem of Algebra i has n roos λ, λ,..., λn which can be real or complex, disinc or repeaed. These roos are called he eigenvalues. Afer he eigenvalues are deermined, hey can be subsiued ino equaion (7) and he corresponding non-zero soluions k, k,..., k n of he vecor equaion can be found. These soluions k, k,..., k n are called he eigenvecors corresponding o eigenvalues λ, λ,..., λ n. The soluion of he eigenvalue problem (7) is no unique; hey can be chosen in such a way ha he desired soluions (6) have only he real-valued componens. Then he consruced fundamenal marix also will have only real-valued enries. Le us show how i can be done: Fundamenal Marix: Case : All eigenvalues λ, λ,..., λ n are real and disinc. Then he corresponding eigenvecors vecors k, k,..., k n are also real-valued and linearly independen. Therefore, he fundamenal marix can be defined as n ( ) e λ e λ... e λ X k k k (9) = n Exercise: show ha he Wronskian is no equal o zero for any. In general, for real disinc eigenvalues λ, λ,..., λ m he corresponding enries of he fundamenal marix are λ λ m k k k () e e... e λ m Case : Le eigenvalue λ be he repeaed real roo of he characerisic equaion (8) of mulipliciy m. Then if: a) here are m linearly independen eigenvecors k, k,..., k m corresponding o he eigenvalue λ. Then he fundamenal marix includes m ke λ ke λ... k me λ b) here is only one linearly independen eigenvecor k corresponding o he eigenvalue λ. Then he oher linearly independen soluions can be consruced in he following way: find he vecors kpq,,,... which are a soluion of he vecor equaions A λi k = ( λ ) ( λ ) A I p = k A I q = p Then he fundamenal marix will include he vecors ke λ, ke λ + pe λ, k e λ + pe λ + q e λ,... () I can be verified wih he help of he Wronskian ha hese vecors are linearly indeneden. Case 3: The eigenvalue λ is complex. We know ha he complex roos of he algebraic equaion wih real coefficiens appear in conjugae pairs: λ = α ± β, i The corresponding eigenvecors are also complex conjugaes k and k. Therefore for disinc eigenvalues λ, = α ± βi here are wo linearly independen soluions e ( α+ βi ) k and k e ( α βi ). Bu

8 4 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions hey are complex-valued funcions which is no convenien for represenaion of he physical problems. I can be shown ha he linear combinaions of hese wo soluions and applicaion of he a+ bi a e = cos b + i sinb e ) yield he wo Euler formula ( independen real-valued soluions which can be included in he fundamenal marix: ( cos β sin β) e α x = b b ( cos β sin β) e α x = b + b () where vecors are b = Re( k ) and = Im b k. Conclusion: The soluion of he eigenvalue problem for he homogeneous linear sysem of s order ODEs wih consan coefficiens yields he fundamenal marix X. Marix exponenial: The sysem of s order ODEs in marix form x = Ax resembles a s order ODE for which i is very emping o wrie he soluion in he radiional exponenial form e A. Bu how can an exponenial funcion wih he marix be calculaed? Again, we can use he analogy wih he calculus of funcions of a single variable and define he marix-valued exponenial funcion in he form of he Taylor series: k 3 A A 3 e = = k= k!! 3! I A A A (3) k imes in which we know how o calculae he powers of he marix A k = AA A (i can be shown ha he infinie series (3) is always convergen for any ). Then he fundamenal marix of he linear sysem (5) can be wrien as X (3) = e A Then he general soluion in marix exponenial form is A A A x = e c+ e e f d (4) The soluion of he IVP can be defined by ( ) A A sa x = e x + e e f s ds (5) The marix exponenial form of he fundamenal marix is no used very ofen for he acual soluion of a linear sysem of ODEs. Bu i is very convenien for derivaion and proofs of he heoreical resuls such as exisence heorems ec.

9 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions AUTONOMOUS SYSTEMS: A sysem of s order ODEs is called auonomous if i can be wrien in he form: x = f x,x,...x n x = f x,x,...x n x = f x,x,...x n n n (6a) and in a marix form: x = f x (6b The unknown funcions x( ),x( ),...xn( ) are funcions of, bu he independen variable does no appear explicily in he righ hand side of he sysem (6). Auonomous sysems are no necessarily linear. plane linear auonomous sysems Here, we will consider only plane linear auonomous sysems, which for simpliciy can be wrien as x = ax + by y = cx + dy a,b,c,d < < (7a) and in marix form x = Ax (7b) The paricular soluion of he plane sysem is a -dimensional vecor which paramerically describes a rajecory (orbi) on he phase plane: x ( ) x = y x= x y = y < < (8) The general soluion also includes an arbirary consan vecor ( ) x,c x(, c ) = (9) y,c I defines he family of rajecories in he phase plane (phase porrai) and describes he moion of he poins along he soluion curves wih he change of ime. The arrows on he rajecories indicae he direcion of he moion of he poin wih he increase of ime. This mapping defined by equaion (9) is called a dynamical sysem. x x (, x ) The iniial value problem = x Ax, x(,x ) (, ) y(,y ) x = x has a unique soluion x = x x = he rajecory which goes hrough he prescribed poin x. The righ hand side of he vecor equaion (7b) defines he vecor field in he phase plane. A any poin on he plane x we can draw a vecor Ax and obain a geomerical represenaion of he direcional field. These direcional vecors are angen o he rajecories defined by he sysem (7b). The direcional vecor field can be drawn even wihou solving sysem (7), bu i provides a qualiaive picure of he dynamical sysem.

10 44 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions Criical Poins: Sabiliy of criical poins: sable The imporan characerisics of he auonomous sysems are he criical (equilibrium, saionary) poins, which can be defined as he soluions no changing in ime (consan soluions): if a poin is placed a he equilibrium poin i will remain here forever. The criical poins can be defined as he soluions of he equaion: f x = For he plane sysem, criical poins are he soluions of he sysem of equaions: ax + by = cx + dy = If he deerminan a b de = ad bc c d here is only one criical poin x cr = (,) (isolaed criical poin). If he deerminan a b de = ad bc = c d ad = bc hen here are infiniely many criical poins which are locaed on he line a y = x b These criical poins are no isolaed. For a non-linear dynamical sysem, he siuaion wih he criical poins is more complicaed. Le cr f x x be he isolaed criical poin of he auonomous sysem (7):. x cr is said o be sable if for any neighborhood ( cr ) smaller neighborhood V ( x cr ) such ha for any V ( cr ) x(,x ) = (, ) = U ( cr ) y(,y ) cr = U x here exiss a x x he rajecory x x x x for all I means ha he rajecory which sars in V remains compleely in U. unsable. x cr is said o be unsable if i is no sable. I means ha i does no maer how close o he criical poin he saring poin x will be, some rajecory will go ouside of any neighborhood U ( xcr ) of he criical poin. asympoically sable 3. x cr is said o be asympoically sable if a) x cr is sable; b) here exiss a neighborhood W ( x cr ) such ha lim x(, x ) = x cr Here, we will invesigae he sabiliy of he plane dynamical sysems (7) which can have only one isolaed criical poin x cr = (,) depending on he marix of coefficiens A.

11 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 45 Phase Porrai of he Plane Sysem: x = ax + by y = cx + dy a b A = c d a b de A = de = ad bc c d Eigenvalue Problem: The characerisic equaion: a λ b A λi = = λ ( + ) λ+ = λ λ+ = c d λ A A de a d ad bc Tr de ( a + d ) ± ( a + d ) 4 ( ad bc) Eigenvalues: λ, = The form of eigenvalues depends on he expression under he square roo which is called he discriminan: Discriminan: ( A) = a + d 4 ad bc = Tr 4 de A I) If > hen he eigenvalues are real and disinc λ λ II) If = hen he eigenvalues are real and repeaed λ = λ = λ III) If < hen he eigenvalues are complex conjugaes λ, = α ± βi Consider he possible configuraions of he plane phase porrai (for simpliciy of presenaion, he deails of soluion will be skipped; derivaion of some of he resuls will be conduced in he examples and he exercises) : λ λ > General soluion: I) x = c k e + c k e a) λ λ >, > Boh eigenvalues are posiive Unsable node b) λ λ <, < Boh eigenvalues are negaive λ λ λ λ lim x = lim c k e + c k e = c k lim e + c k lim e = Sable node c) λ λ >, < The eigenvalues are of he opposie sign Saddle poin (unsable)

12 46 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions II) = λ = λ = λ λ > a) There are wo linearly independen eigenvecors k, k c e c e c c e General soluion: x = k + k = ( k + k ) λ λ λ i) λ > degenerae (proper) unsable node ii) λ < degenerae (proper) sable node λ > b) There is one linearly independen eigenvecor k (find p, q,... ). General soluion: λ λ c c λ x( ) = cke + c( k + p) e = k+ p+ ck e i) λ > degenerae (improper) unsable node ii) λ < degenerae (improper) sable node III) < λ, = α ± βi eigenvecors k = b + i b, k = b ib α > a) α General Soluion: x = c ( bcos β b sin β) + c( bcos β + b sin β) e α i) α > unsable focus (spiral poin) α < ii) α < sable focus (spiral poin) asympoically sable b) α =, λ, = ± βi (pure imaginary, when a = d ) α = General Soluion: ( b β b β ) ( b β b β ) x = c cos sin + c cos + sin sable cener (no asympoically sable)

13 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 47 Classificaion of he criical poins of he plane linear sysem sable degenerae node sable focus de A sable cener unsable focus = de A = ( TrA) 4 unsable degenerae node sable node unsable node TrA saddle poin Procedure for Soluion of he Linear Sysem of s Order ODEs wih Consan Coefficiens:. Wrie he sysem in he normal marix form (): x = Ax + f. Solve he Eigenvalue Problem (7): A λi k = o find eigenvalues λ i and eigenvecors 3. Consruc he fundamenal marix (8) X in correspondence wih equaions (9-). k i. 4. Calculae he general soluion according o he variaion of parameer formula (3a): x ( ) = X( ) c X X f + d 5. For soluion of he IVP (4) wih x = x, use he variaion of parameer formula (3b): x ( ) = X X x X X f + s s ds

14 48 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 5. EXAMPLES: ) (reducion of he sysem of s order ODEs o a higher order ODE) Consider he sysem of wo s order ODEs: x = 3x y y = x y Reducion is performed by differeniaion of he equaions and consecuive replacemen of he unknown funcions unil a differenial equaion for a single unknown funcion is obained. Consider he second equaion: y = x y x = y + y ( ) Differeniae he second equaion wih respec o y = x y x = y + y Subsiue expressions for x and x ino he firs equaion x = 3x y y + y = 3 y + y y Rearrange i ino he equaion for he funcion y y y + y = This is a single nd order ODE, linear homogeneous wih consan coefficiens, which can be solved by he sandard mehod: auxiliary equaion general soluion m m + = m, = y c e c e The funcion x can be found from equaion ( ): x y y = ce + ce + ce + ce = c + c + e = + = ( ce + ce ) + ( ce + ce ) Therefore, he general soluion of he sysem of equaions is: = + y c e c e x c c e = + + = + ( ) Tha can be verified by direc subsiuion ino he original sysem of equaions.

15 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 49 ) (reducion of a higher order ODE o a sysem of s order ODEs) h Consider a normal n order linear ODE = ( n) ( n ) n a x y a x y... a x y f x Divide he equaion by a ( x) : ( n) ( n ) n y y... y a x for all x ( ) a x a x f x = + ( ) a x a x a x Inroduce he se of new funcions: x = y x = y x = x x3 = y x = x3 xn Differeniae x n ( n ) y = x n = xn ( n) x n = y Wih hese noaions, he equaion ( ) can be rewrien as a( x) an( x) f ( x) x n = x n... x + a x a x a x Collec hese in he normal linear sysem of s order ODEs: x = x x = x3 x = x n n n n n a x a x f x x = x... x + ( ) a x a x a x 3) Solve he nd order ODE y y + y = by reducion o a sysem of s order ODEs. Applying ( ) for he nd order equaion, we obain x = x a x a x f x x = x x + = x x a x a x a x (Surprisingly, his sysem is no idenical o he sysem of Example ) In marix form: x x x = x Find he fundamenal se for a sysem wih consan coefficiens. Characerisic equaion (8): de λ = λ λ + = λ There is only one eigenvalue λ = of mulipliciy. Find eigenvecors by plugging λ = in vecor equaion (7): k = k

16 4 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions k = k I is a singular linear sysem of algebraic equaions, i has only one independen soluion: k = Find he vecor p by soluion of he equaion ( λ ) A I p = k p = p Then p,p can be found from he equaion p + p = One of he soluions can be p = Then he fundamenal marix is: λ λ λ X= ke ke + p e e [ ] X= k k + p X = e + Then he general soluion of he sysem is given by x = Xc or in he componen form = + = + ( + ) x c e c e x c e c e This is he soluion of he sysem of ODEs o which he ODE was reduced. Recall now ha x ( ) was defined as x y ( ) = x ( ) = c e + c e = y, herefore, he general soluion is Which coincides wih he previously obained general soluion ( ). The second soluion can be reaed as y = x = c e + c + e Inegraion of his equaion will duplicae he previous resul.

17 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 4 4a) (Linear Sysem of equaions General Soluion) Find he general soluion of he sysem of ODEs: Soluion: x = x + x + e 3 x = 4x + 3x + 8e ) Rewrie he given sysem in he marix form: 3 x e = x 8e x x ) Solve he eigenvalue problem: k λ 4 3 = k Characerisic equaion: λ 4 3 = = = 4 3 λ Eigenvalues: λ =, λ = 5 (real disinc) de λ λ 4λ 5 Eigenvecors: λ = k λ 4 3 = k k = k k 4 4 = k k 5 k λ = = 5 k 4 k = 4 k 3) Fundamenal marix: 5 e e X ( ) = 5 e e e e k = = X ( ) = de = e e = 3e 5 e e 4) Variaion of parameer formula (3a): Inverse of he Fundamenal marix: e e 3 3 X ( ) = 5 5 e e 3 3 e e e 8e + 6e X ( ) f ( ) = = e 4e + 6e e e 3 3

18 4 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions ( ) X f Paricular soluion: e d + 6 e d 4 e + 3 e 3 e d = = 3 e 4 e d + 6 e d e e e + e 4e xp = X( ) X ( ) f ( ) d = 5 3 = 3 e e e e 6e e Complimenary Soluion x c : Complimenary Soluion (soluion of he homogeneous sysem): k = k = 5 5 e e c ce + ce xc = X( ) c = 5 5 e e c = ce + ce General soluion: saddle poin (eigenvalues of opposie sign) x ( ) = x ( ) + x ( ) c p 5 ce + ce 4e = 3 3 ce + ce 6e e 4b) (Linear Sysem of equaions Iniial Value Problem) Find he soluion of he sysem of ODEs: x = x + x + e 3 x = 4x + 3x + 8e subjec o he iniial condiion: x ( ) 3 = Soluion: Use he fundamenal se of he previous example. 3) Fundamenal marix: 5 e e X ( ) = 5 e e 4) Variaion of parameer formula (3b): X X ( ) ( ) e e 3 3 = 5 5 e e = 3 3

19 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 43 X X x 5 5 e e 3 33 e + e = 5 5 e e = e + e 3 3 ( ) ( ) X f e e e 8e + 6e = = e 4e 6e e e = = = 4s 3s e ds + 6 e ds e e 4 3 e e ( s) ( s) ds + X f 3 3 s 3s e e e e e ds + 6 e ds e e e + e 4e + 4e X( ) X ( s) f ( s) ds = 5 3 = 3 5 e e e e + 4 6e e + 8e Soluion of IVP: x ( ) e + e 4e + 4e e 4e + 5e = = 3 5 e + e 6e e + 8e e 6e e + e x( )

20 44 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 5) (sabiliy of auonomous sysem) Invesigae for sabiliy he equilibrium poin and skech he phase porrai of he following auonomous sysem: x = x+ 3y y = 3x y Soluion: ) Rewrie he given sysem in marix form: x 3 x y = 3 y ) Solve he eigenvalue problem: 3 k λ = 3 k Characerisic equaion: 3 λ λ de λ = = λ + 4λ+ 3 = Eigenvalues: λ = + 3i, λ = 3i (complex) Eigenvecors: λ = + 3i 3 k λ = 3 k + 3i k 4 3 = + 3i k 3i 3 k = 3 3i k k = i i k = = + i 3) Fundamenal marix (use equaion ()): α x = b cos β b sin β e = cos 3 sin 3 e b =, b = x = b + b = + X ( ) α ( cos β sin β ) e cos 3 sin 3 e cos 3 e sin 3 e = sin 3 e cos 3 e 4) General soluion: = ( ) x X c cos 3 e sin 3 e c sin 3 e cos 3 e c = c cos 3 e + c sin 3 e = csin 3 e + c cos 3 e

21 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 45 In parameric form: = + = + x c cos 3 e c sin 3 e y c sin 3 e c cos 3 e According o case III) a) ii) his is an asympoically sable focus. Skech he phase porrai: For he paricular curve, choose c =, c =, hen = = x y cos 3 e sin 3 e A graphing calculaor can be used for skeching he graph of his curve, bu i is imporan o know how o skech he graph jus from he parameric equaion we can perform i qualiaively in he following way: For = x = y = The saring poin is defined. Now le us see where he curve will go under a small increase of ime x = ε = ε y = ε Then coninue he curve as a shrinking spiral in he deermined direcion: All oher rajecories will be of he same shape, covering he enire plane wihou inersecions. Here is he graph generaed by Maple: Conclusion: he equilibrium poin is he asympoically sable focus (spiral poin).

22 46 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 6. REVIEW QUESTIONS AND EXERCISES: ) Wha is a sysem of differenial equaions? ) Wha ype of sysems did we sudy in his secion? 3) How many soluions of a normal sysem of s order ODEs can go hrough an arbirary poin of he plane? 4) How many soluions of a homogeneous sysem of s order ODEs can go hrough he poin (, )? 5) Why is uniqueness no violaed for a saddle equilibrium poin? 6) Why is he sysem of wo s order ODEs called a dynamical sysem? 7) Wha is he sabiliy of an equilibrium poin? Wha does i mean? EXERCISES: ) Reduce he following ODEs o a sysem of s order ODEs: a) b) y + 5y + 3y 6 y = e ( iv) y 6y + y y 3y = cos Reduce he sysem of s order ODEs o a higher order ODE: c) x = 4x + x + x = x + 3x ) Marix exponenial: a) Using he definiion of marix exponenial, verify he differeniaion rule: d e d A = A e A b) Show ha if a a A = ann hen e A a e a e = ann e c) Consider IVP: solve = x = x. x Ax subjec o A Show ha e = X( ) X ( ), where ( ) X is he fundamenal marix. 3) Find he general soluion of he following sysems and skech he soluion curves of he homogeneous par of he sysems: a) x x = b) x = 5x + 3x + e x = x + sec x = x x + c) x = 4x + x + d) x = x + x + sin x = x + 3x x = x + cos

23 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 47 e) x = x + x f) x = 3x x x3 x = x + 4x3 x = x + x x3 + x 3 = x 4x3 x = x x + x + e 3 3 4) Find he soluion of he following Iniial Value Problems and skech he graph of he soluion: a) = + b) x 3x x 4e x = 3x x + 4 = + + x x 3x 4e x = 5x 3x + 3 subjec o x ( ) =, x ( ) = ( π ) ( π ) x =, x = c) x = 3x x x3 c) x = 3x x x3 x = x + x x3 + x = x + x x3 + x = x x + x + e 3 3 x = x x + x + e 3 3 = = = x, x, x 3 x =, x =, x = 3 5) Invesigae for sabiliy he equilibrium poin and skech he phase porrai of he following auonomous sysems: a) x = x+ 3y b) x = x+ 4y y = 3x+ y y = x+ y c) x = x+ 5y d) x = x 4y y = x y y = x+ 5y e) x = x+ y f) x = x+ y y = x+ y y = x+ y g) x = x y h) x = x α y y = x+ 4y y = x y

24 48 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 7. LINEAR SYSTEMS OF ODEs WITH MAPLE. a) Find he general soluion of he homogeneous sysem and skech he phase porrai: x = x x x = 3x x + 4 > wih(linalg): > k:=marix(,,[[],[]]); k := > f:=marix(,,[[],[4*]]); f := 4 > C:=marix(,,[[c[]],[c[]]]); C := Eigenvalue Problem: > A:=marix(,,[[,-],[3,-]]); c c A := > eigenvecs(a); [,, {[, ]}], [-,, {[, 3 ]}] Fundamenal marix: > X:=marix(,,[[exp(),exp(-)],[exp(),3*exp(-)]]); X := e e ( ) e 3 e ( ) Complimenary Soluion - General Soluion of Homogeneous Sysem: > Xc:=evalm(X&*C); e c Xc := + e ( ) c e c + 3 e ( ) c -Phase Porrai: > x():=exp()*c[]+exp(-)*c[]; x( ) := e c + e ( ) c > y():=exp()*c[]+3*exp(-)*c[]; y( ) := e c + 3 e ( ) c > p:={seq(seq(subs({c[]=i/*,c[]=j/*}, [x(),y(),=-4..4]),i=-4..4),j=-4..4)}: > plo(p,x=-5..5,y=-..,color=black);

25 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 49 b) Find he general soluion of he non-homogeneous sysem: Paricular Soluion - Variaion of Parameer: > Xinv:=simplify(inverse(X)); Xinv := > simplify(evalm(xinv&*f)); > map(in,%,); 3 ) e( ) e( e e e ( ) e e ( ) + e ( ) e e > Xp:=simplify(evalm(X&*%)); General Soluion: > GS:=evalm(Xc+Xp); Xp := e c GS := + e ( ) c + 4 e c e ( ) c 8 4 Soluion Curves: > xn():=exp()*c[]+exp(-)*c[]+4*; xn( ) := e c + e ( ) c + 4 > yn():=exp()*c[]+3*exp(-)*c[]+8*-4; yn( ) := e c + 3 e ( ) c > pn:={seq(seq(subs({c[]=i/,c[]=j/}, [xn(),yn(),=-..]),i=-3..3),j=-..)}: > plo(pn,x=-8..6,y=-..,color=black, numpoins=5);

26 43 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions c) Find he soluion of he Iniial Value Problem: Soluion of IVP - Variaion of parameer formula (3b): > Xinv:=simplify(subs(=,evalm(Xinv))); Xinv := > X:=evalm(evalm(X&*Xinv)&*k); X := 3 e e ( ) 3 e 3 e ( ) > X:=simplify(evalm(Xinv&*f)); X := e ( ) e > X3:=subs(=s,evalm(X)); X3 := e ( s ) s e s s > X4:=simplify(map(in,X3,s=..)); X4 := e ( ) + e e e + > X5:=simplify(evalm(X&*X4)); > XS:=evalm(X+X5); Graph of he soluion of IVP: X5 := 4 e + e ( ) 8 4 e + 6 e ( ) XS := e + e ( ) + 4 e + 3 e ( ) > u:=exp()+exp(-)+4*; u := e + e ( ) + 4 > v:=exp()+3*exp(-)+8*-4; v := e + 3 e ( ) > plo([u,v,=-3..]);

27 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions 43 Applicaion of he sandard Maple procedure for soluion of he sysem of ODEs.. Invesigae for sabiliy he equilibrium poin and skech he phase porrai of he following auonomous sysem: x = x+ 3y y = 3x+ y > eq:=diff(x(),)=-4*x()+3*y(); d eq := x( ) = 4 x( ) + 3 y( ) d > eq:=diff(y(),)=-*x()+*y(); d eq := y( ) = x( ) + y( ) d > Soluion:=dsolve({eq,eq},{x(),y()}); > assign(soluion): Soluion := { x( ) = _C e ( ) + _C e ( ), y( ) = _C ) e( + _C e ( ) } 3 > p:={seq(seq(subs({_c=i/,_c=j/}, [x(),y(),=-..]),i=-5..5),j=-5..5)}: > plo(p,x=-3..3,y=-3..3,color=black,scaling=consrained); The marix of coefficiens has wo real disinc negaive eigenvalues λ =, λ =. The equilibrium poin is a sable node.

28 43 Chaper V ODE V.5 Sysems of Ordinary Differenial Equaions