ODEs II, Lecture 1: Homogeneous Linear Systems - I. Mike Raugh 1. March 8, 2004

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1 ODEs II, Lecure : Homogeneous Linear Sysems - I Mike Raugh March 8, 4 Inroducion. In he firs lecure we discussed a sysem of linear ODEs for modeling he excreion of lead from he human body, saw how o ransform a linear ODE of degree n ino a linear sysem of n firs-order ODEs, and noed he form of he general sysem of n firs-order equaions, for boh linear and non-linear sysems. We also solved a simple linear sysem of wo firs-order equaions. Here we examine linear sysems in greaer deail. General Linear Sysem. Given a se of funcions real-valued funcion ij gi defined on some inerval I of he real line, where he indices i and j run from o n, we can wrie a sysem of firs-order linear equaions, a and x a a n g x =, A =, g = x a a g n n nn x = Ax + g, I (.) The erm g is called he driving erm or forcing erm. If g ( ), he sysem is called homogeneous, oherwise i is called non-homogeneous. In he former case, he sysem may be called a driven sysem, and in he laer case he sysem may be called undriven. We begin our sudy wih a homogeneous linear sysem wih consan coefficiens. When he value of he soluion x ( ) is specified a an iniial ime, he sysem is referred o as an iniial value problem (IVP): IVP: x x a a n =, A = xn an a nn ( ) x = Ax, x = x (.) Based on a course given joinly wih Michael Moody. If boh A and g are consan funcions of, hen he sysem is said o be auonomous. We will sudy nonlinear auonomous sysems laer in he course. mrr/hmc/mah 64B of 6 3//4, :5 AM

2 Noe ha, if he wo vecor funcions ( ), ( ) x x saisfy hen, defining d x = x = Ax, x( ) = x d d x = x = Ax, x ( ) = x d (.3) i follows ha ( ) = c ( ) + c ( ) x x x (.4) d d x = cx + c x = cx + c x d d = c Ax + c Ax = A cx + c x = Ax (.5) This is why he sysem is called linear. Noe ha he iniial condiions could also correspond: Then i can be seen by inspecion ha a soluion o he IVP x = Ax, x = x = cx + c x (.6) is obained by summing he soluions of he wo IVPs (.3). This sum is in fac he unique soluion o IVP (.6). 3 In case he marix A= A n nhas consan enries and n linearly independen eigenvecors, here is an amazingly simple and direc way o solve he homogeneous T sysem Eq. (.). Try x = e v for some consan vecor v = ( v v n ) and scalar r, in oher words, le x( ) v = e (.7) xn v n So v x = re = re v v n And, 3 We will discuss he relevan Fundamenal Exisence and Uniqueness heorem in he nex lecure. mrr/hmc/mah 64B of 6 3//4, :5 AM

3 Therefore, plugging hese resuls ino v v Ax= Ae = e A = e Av (.8) v v n n = x Ax, we ge: Bu e, herefore, e Av= re v (.9) Av = rv (.) This shows ha v is an eigenvecor of A corresponding o he eigenvalue r, and, working backward wih hese values, we see ha v e v x = e v= e = (.) v n e v n is a soluion of he sysem. A remarkable feaure of his derivaion is ha i conveed a calculus problem ino a linear algebra problem, in fac, an eigenvalue/eigenvecor problem. This is grea evidence for he usefulness of linear algebra, bu here s more o come. For example, le s see wha happens when we have a complee se of eigenvecors n for A, ha is a se of eigenvecors ha spans R. To keep i simple, we ll consider, bu he approach is applicable o higher dimensions. Example. Find linearly independen funcions x ( ) and x ha saisfy ( ) x = x+ x x x = x = 4x+ x x x (.) We ll solve his sysem in general hen come back o he specific case: x 3 = x (.3) Rewrie Eqs. (.) in vecor/marix noaion, x x x x = = = Ax, x = (.4) x 4 x x mrr/hmc/mah 64B 3 of 6 3//4, :5 AM

4 Our approach is o find eigenvalues and corresponding eigenvecors of he marix A. In order o do so, we firs find he eigenvalues of A: λ = λ λ 3= ( λ 3)( λ+ ) = 4 λ ( λ, λ ) ( 3, ) = 3 (.5) Corresponding eigenvecors v, where Av = λv, can be found using a familiar rowreducion procedure symbolized as follows: 3 λ = 3: v = λ = : v = (.6) I is obvious on inspecion ha he eigenvecors a he righ-hand side of Eq. (.6) are linearly independen, bu even wihou looking we know hey mus be because hey belong o disinc eigenvalues. 4 Now, in accord wih our consrucion, x = e v, where now we see ha v and r = λ, as found in Eq. (.6). So, le x λ e x = = e v = e = (.7) x e In he represenaion x ij above, he second index indicaes he vecor idenificaion number and he firs index gives he componen number for ha vecor. For example, below when referencing he vecor x, we will use x and x o indicae x he componens of x =. The reason for choosing his order of he index will x become clear below when we place boh vecors in a common marix, so ha he second index will indicae he column in which he vecor appears. For economy of expression, we ofen omi explici saemen of he dependency of variables like x and y on he independen variable. We have explained why Eq. (.7) mus be a soluion of Eq. (.); bu since ODE work is ofen error prone, i s a good idea o doublecheck he resul by direc calculaion. From Eq. (.7), x 3 3e x = = (.8) x 6e 4 Noe ha we could pick any scalar muliples of he eigenvecors shown; o simplify subsequen arihmeic, we picked muliples o ensure ha all componens of he vecors are inegers. mrr/hmc/mah 64B 4 of 6 3//4, :5 AM

5 Therefore, x = 3e = e + e = x + x x = 6e = 4e + e = 4x + x (.9) Thus we confirm ha = ( e e ) T x solves he sysem, as we knew i should. Now le x λ e x = = e v = e = (.) x e We could easily confirm ha x is also a soluion of Eq. (.), so now we have wo disinc soluions of our homogeneous equaion: e e =, 3 = x x (.) e e Neiher of hese soluions saisfies he iniial condiions of Eq. (.3). Bu hey are linearly independen funcions of for, as is easily verified in his case by noing simply ha x canno be a fixed linear muliple of x valid for all. 5 Because he soluions are linearly independen, hey can be combined o produce a general soluion of Eq. (.), as demonsraed nex. As noed above, any linear combinaion of soluions of a homogeneous linear sysem of ODEs mus also be a soluion. Therefore, any linear combinaion of he soluions in Eq. (.) mus saisfy he homogeneous Eq. (.4). So, all we have o do is see wheher we can mach he given iniial condiions o some linear combinaion of x( ), x ( ). This ranslaes ino he quesion of wheher we can find consans c and c such ha cx( ) + cx( ) = ce x (.) ce + ce = = ce x Generally, any n-by-n marix comprised of n columns of eigenvecors belonging o disinc eigenvalues mus be non-singular. So, he wo-by-wo marix in Eq. (.) consising of wo such eigenvecors mus be non-singular. In our presen case, his is obvious, bu i will usually no be obvious in higher dimensions, so i is good o keep he general condiion in mind. In any case, he linear independence of he columns in he 5 Laer we will discuss a Wronskian es, similar o he one encounered in Mah 3 bu ailored for linear sysems, which will permi us o deermine linear dependence or independence of a se of n soluions of a homogeneous linear sysem of n equaions. mrr/hmc/mah 64B 5 of 6 3//4, :5 AM

6 square marix of Eq. (.) implies ha he equaion has a unique soluion for ce and ce, hence a unique soluion for c and c because e, e. In oher words, Eq. (.) can be solved for arbirary iniial condiions. This proves ha ( ) = c ( ) + c ( ) x x x (.3) is a general soluion of he sysem of Equaion (.4). The soluion we have found is also complee, i.e., here aren any oher soluions han he ones we have found. 6 This follows from a Fundamenal Exisence and Uniqueness Theorem for sysems of ODEs alluded o in a previous foonoe. Le s proceed o he paicular case anicipaed above. Using he iniial values 3 x = in Eq. (.) we obain ce x = ce x c 3 = c (.4) 3 3 c = R+ R 4 4 c Therefore, he soluion of he IVP wih iniial condiions defined by Eq. (.3) is and e + e x = e + e = (.5) 4e e x e as (.6) We see in his example he impoan role ha linear algebra can play in he sudy of differenial equaions. 6 Noice ha we have inroduced some discriminaing vocabulary here. A general soluion of an ODE or sysem of ODEs is one ha provides a soluion for any arbirary se of iniial condiions. A complee soluion is one ha provides all possible soluions. The disincion arises because i is possible in some circumsances o have more han one soluion for a given iniial value problem, so ha a general soluion (providing only one soluion for any given se of iniial values) isn necessarily complee. For he problems encounered in his course, he general soluion will also be he complee soluion. mrr/hmc/mah 64B 6 of 6 3//4, :5 AM