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1 Geeral Liear Spaes (Vetor Spaes) ad Solutios o ODEs Deiitio: A vetor spae V is a set, with additio ad salig o elemet deied or all elemets o the set, that is losed uder additio ad salig, otais a zero elemet (), ad satisies the ollowig axioms: For all, gh, Vad salars,, () ( + g) + h= + ( g+ h) (5) ( + g) = + g () + g = g+ (6) ( + ) = + (3) + = (7) ( ) = ( ) (4) + ( ) = (8) = We ll deal iitially with the ase where the salars are real umbers. Suh a vetor spae is alled a real vetor spae. We will also use omplex salars i whih ase we ll all this a omplex vetor spae. Though we a give deiitios ad prove theorems about vetor spaes i geeral, it s helpul to develop a library o examples to whih we a reer.. R is a vetor spae. Ideed, the motivatio or our deiitio ad axioms is to deie vetor spaes to be spaes whih a udametally lie R. All o the required axioms are amiliar ats about vetors i R.. Ay subspae o R is a vetor spae. All o the axioms are iherited ad every subspae otais the zero vetor, ad the deiitio o subspae esures that a subspae is losed uder additio ad salig. 3. The omplex umbers = { a + bi : a, b are real umbers, i =} a be viewed as a real vetor spae where additio is deied by ( a+ bi ) + ( a + bi ) = ( a+ a) + ( b+ b) i ad salig by a real umber is deied by ( a + bi) = a + bi. Note that we do ot deie multipliatio o omplex umbers i this otext. The omplex umbers otai the zero elemet = + i ad all o the axioms ollow rom orrespodig ats about real umbers. m 4. = M( m, ) = { m matries with real etries} R is a real vetor spae with additio ad salig o matries deied ompoet-wise. The m zero matrix is the zero elemet ad the axioms are all ow properties o matrix algebra. Note that i this otext we do ot deie the produt o matries. 5. F(, ) = F( ) = { utios : with domai } RR R R R R is a real vetor spae where additio o utios ad salig o utios are deied by poitwise by ( + g)( x) = ( x) + gx ( ) ad ( )( x) = ( x). The zero elemet i this ase is the utio that is idetially zero or all x. (This is quite dieret tha just the real umber.) Oe agai, the axioms all ollow rom amiliar ats about real umbers. 6. P = { real polyomials o degree } = { a + ax + + ax : a, a,, a R} is a real vetor spae. Note that we must allow all polyomials less tha or equal to beause we might add two polyomials (or sale by ) ad get a polyomial o lesser degree. 7. C (, ) = C ( ) = { otiuous utios : } RR R R R is a real vetor spae. Closure ollows rom theorems o Calulus that the sum o otiuous utios is otiuous ad a salar multiple o a otiuous utio is also otiuous. The zero utio is learly otiuous, ad the axioms are all easily veriied. Deiitio: A subspae W o a vetor spae V is a subset that is losed uder additio ad salig o elemets. That is, or ay vetors v, v W ad salars,, it must be the ase that v+ v W. We write W V. 8. C (, ) = C ( ) = { dieretiable utios : } RR R R R is a real vetor spae. Closure ollows rom the theorems o Calulus that ( + g) = + g ad ( ) =. The zero utio is learly dieretiable.

2 Note: All polyomials are dieretiable, ad a (hopeully) amiliar theorem o Calulus tells us that dieretiable utios must be otiuous, so or ay, P C ( R) C ( R) F( R ). 9. C (, ) = C ( ) = { utios : that are at least times dieretiable} RR R R R is a real vetor spae. This ollows similarly rom Calulus theorems. The zero utio is dieretiable to all orders.. C (, ) = C ( ) = { utios : that are dieretiable to all orders} RR R R R is a real vetor spae. This also ollows rom the Calulus theorems above. The zero utio is dieretiable to all orders. Note: For ay, P P P P P C ( R) C ( R) C ( R) C ( R) F( R). It s also importat to ote that whe dealig with spaes o utios, these spaes are muh larger that R. To better uderstad this, we ll eed some more deiitios, but some o the importat details will have to wait util you tae ourse i aalysis ad topology. May o the deiitios whe worig with vetors spaes are essetially the same as those i R. Deiitio: Give a olletio o elemets {,,, } V spa {,, } = { + + where, are salars} Deiitio: A set o elemets { }, we deie the spa o these elemets as:.,,, V is alled liearly idepedet i give ay liear ombiatio o the orm + + =, this implies that = = =. That is, there is o otrivial way to ombie these vetors to yield the zero elemet. Deiitio: Give a subspae W V, a olletio o elemets {,,, } W is alled a basis o W i Spa {,,, } = W ad {,,, } are liearly idepedet. A basis is a miimal spaig set ad, as was the ase i R, i a basis or W osists o iitely may elemets the ay other basis will have the same umber o elemets, the dimesio o W. It is importat to ote, however, that it will ote be the ase, espeially i the ase o utio spaes, that a subspae might ot be spaed by iitely may elemets. Coordiates relative to a basis Deiitio: I B = {,, } is a basis or a iite dimesioal vetor spae V (or a subspae o V), ad i V, the a be expressed uiquely as = + + or salars {,, }. These uiquely determied salars are alled the oordiates o relative to this basis. I we express these oordiates as a olum vetor (eetively a vetor i. B R ), we deote this oordiate vetor by [ ] Deiitio: Give two vetors spaes V ad W, a utio T : V W is alled a liear trasormatio i or all elemets, V ad or all salars,, T satisies T( + ) = T ( ) + T( ). This a also be expressed by sayig that T preserves additio ad salar multipliatio. We all the iput spae V the domai o T ad we all the output spae W the odomai. Deiitio: Suppose T : V W is a liear trasormatio. We deie: image( T ) im( T ) T ( ): V W erel( T) = er( T) = V : T( ) = V = = { } ad { } These are both subspaes. The argumet is the same as we ve see beore.

3 Deiitio: Give a liear trasormatio T : V W, we deie: ra( T) = dim(im T) ad ullity( T) = dim(er T) whe these subspaes have iite dimesio. We a also state (without proo) the orrespodig at regardig the relatioship betwee ra ad ullity. Ra-Nullity Theorem: I T : V ra( T) + ullity( T) = dim( V). W is a liear trasormatio ad V has iite dimesio, the Deiitio: A liear trasormatio T : V W is alled a isomorphism i it is oe-to-oe ad oto its odomai. That is, or every g W there is a uique V suh that T( ) = g. Appliatio to Liear Ordiary Dieretial Equatios I ay spae osistig o dieretiable utios, the dieretiatio operator D( ) = is a liear trasormatio. This ollows rom the Calulus ats that D( + g) = ( + g) = + g = D( ) + D( g) ad D( ) = ( ) = = D( ). The erel o this liear trasormatio osists o the ostat utios. This liear trasormatio is thereore NOT a isomorphism. I ay spae o utios, aother importat liear trasormatio is multipliatio by a ixed utio g. That is, i we deie Mg ( ) = g, the or ay salars, ad ay utios, we have: M ( + ) = g( + ) = g + g = M ( ) + M ( ) g g g I we ombie the dieretiatio operator ad various multipliatio operators via additio ad ompositio, we a build more ompliated liear dieretial operators. For example, D = D D is a liear operator orrespodig to taig the d derivative. Similarly, D = D D D orrespods with taig the th derivative. I pt () is a utio, the T = D + pi is a irst order liear dieretial operator alulated by: T( xt ()) = ( D+ pi)( xt ()) = + ptxt () () More geerally, give utios p (), t, p(), t p() t, we a orm the liear dieretial operator T= D + p D + + pd+ pi. Whe applied to a suiietly dieretial utio xt (), this gives: d x d = + x = T ( xt ()) ( D p D pd pi )( xt ()) p () t p () t p () txt () Solvig a th order liear ODE o the orm + () () () () () p t p t p txt qt = a thus be see as solvig T( xt ()) = qt (). I the ase o a homogeeous liear ODE, the beomes simply T( xt ( )) =, i.e. we are tryig to id the erel o the liear dieretial operator. 3 d x d x Perhaps the greatest beeit o this ormulatio is that i we a id several (liearly idepedet) homogeeous solutios x (), t x () t [that is, T( x( t)) = T( x( t)) = = T( x ( t)) = ], the or ay salars,, we have (by liearity) T( x() t + + x()) t = T ( x()) t + + T( x()) t =. That is, ay liear ombiatio o solutios will also be a solutio [Priiple o Superpositio]. Liear ODEs with ostat oeiiets A importat (ad easily solvable) ase is where all o the oeiiet utios are ostat, i.e. a operator o the orm T= D + a D + + ad + ai where a,, a, a are ostats. We ll irst osider the homogeeous ase, i.e. solvig a homogeeous liear ODE with ostat oeiiets o the orm d x a d + x + + a + ax =. We a geerate solutios to suh a ODE by (iitially) osiderig rt expoetial utios o the orm xt () = e where r is ostat. Substitutig this ito the ODE gives:

4 d x d x rt rt rt rt rt + a + + a + a x = r e + a r e + + are + ae = ( r + a r + + ar + a ) e = The polyomial P() r = r + a r + + ar + a is alled the harateristi polyomial o this ODE. Ay root o this polyomial will yield a expoetial solutio to the ODE, ad the Fudametal Theorem o Algebra guaratees roots, though some may be omplex ad some may be repeated. The roots o the harateristi polyomial are alled the harateristi roots or harateristi values o the ODE. Our strategy will be to try to reate a basis out o suh expoetial solutios rom whih ALL homogeeous solutios a be ostruted. First Order Example: I we solve x =, we have the harateristi polyomial Pr () = r = ad t t harateristi value r =. This yields the expoetial solutio e. By liearity, x() t = e will also be a t solutio ad, i at, this yields ALL homogeeous solutios. That is { e } is a basis or all suh solutios. Seod Order Example: Suppose we wat to solve the ODE x =. The orrespodig liear dieretial operator is T = D + 3D+ I ad the harateristi polyomial is Pr ( ) = r + 3r+ = ( r+ )( r+ ). t This yields the harateristi values r = ad r = ad expoetial solutios e t ad e. By liearity, ay t t utio o the orm x() t = e + e will also be a homogeeous solutio or ay salars,. Does this t t give ALL solutios? That is, does { e, e } ostitute a basis or all solutios? We a show that this is atually the ase by thiig o T = D + 3D+ I = ( D+ ) I ( D+ I), i.e. as a ompositio o irst-order liear operators. We are seeig a solutio xt () suh that ( D+ I)[( D+ I)( xt ( ))] =. I we let yt ( ) = ( D+ I)( xt ( )), the we a irst solve ( D+ I)[ yt ( ))] = y + y = to get all solutios o the orm t y() t e t =. Next, or ay suh solutio we ll have ( D+ I)( x( t)) = x + x= e. This is a ihomogeeous t irst-order ODE. The homogeeous solutios are all o the orm xh () t = e ad we a use the Method o t Udetermied Coeiiets to produe a partiular solutio to the ihomogeeous equatio, i.e. xp () t = Ae t t t or some salar A. Substitutio gives x + x = ( A + A) e = Ae = e, so A=. We the add the t t homogeeous ad partiular solutios to get that all solutios must be o the orm x() t = e e. I we t t t t ow simply reame the oeiiets we see that all solutios are o the orm x() t = e + e, i.e. { e, e } does ostitute a basis or all solutios. (It s straightorward to show that these utios are liearly idepedet.) The above argumet wors the same i the ase o a th order liear homogeeous ODE with ostat oeiiets as alog as all the roots o the harateristi polyomial are distit. It s eve valid i the ase o omplex roots (though it s geerally preerable to use Euler s Formula to re-express the expoetial solutios i terms o sie ad osie utios). Repeated roots have to be hadled dieretly. A example will illustrate what should be doe i that ase. Repeated root example: Suppose we wat to solve the ODE dieretial operator is T = D + 4D+ 4I ad the harateristi polyomial is yields oly the harateristi value r = with orrespodig expoetial solutios x =. The orrespodig liear Pr ( ) r 4r 4 ( r ) = + + = +. This t e t. Does x() t = e yield ALL homogeeous solutios, i.e. does { e t } ostitute a basis or all solutios? The aswer is NO. I we proeed i the same maer as the previous example, we write ( D+ I)[( D+ I)( x( t))] = ad let t yt ( ) = ( D+ I)( xt ( )). We solve ( D+ I)[ yt ( ))] = y + y = to get y() t = e ad the see solutios o t ( D+ I)( x( t)) = x + x= e. I this ase, it may ot be so lear how to use Udetermied Coeiiets to

5 guess a partiular solutio to this ihomogeeous equatio. We a use Variatio o Parameters to id t t e solutios. I we see a solutio o the orm xp () t = vte (), the method gives that v () t = = t, so e v() t = t + (we atually do t eed to ilude the + i we re just seeig a partiular solutio). So we a t hoose xp () t = te t. The homogeeous solutios are all o the orm xh () t = e. Thereore, ALL solutios t t t t are o the orm x() t = te + e. That is, { e, te } orms a basis or ALL solutios. This a be geeralized or ay repeated root o ay multipliity. For example, a similar alulatio shows that 3 3t 3t 3t or the homogeeous liear ODE ( D+ 3 I) ( xt ( )) =, { e, te, t e } gives a basis or all solutios. That is, 3t 3t 3t all solutios are expressible (uiquely) i the orm x() t = e + te + t e } 3 As log as we a easily ator the harateristi polyomial, the above example illustrate how we a simply geerate a basis or all homogeeous solutios or i the ase o ostat oeiiets. Example: Fid all solutios to the ODE: d4x d3x x =. 4 3 Solutio: The harateristi polyomial i the ase is Pr ( ) = r + r + 5r + 8r+ 4 = ( r+ ) ( r + 4) =. The harateristi values are r = (with multipliity ), r = i, ad r = i. From these we get a basis or all t t it it t t it it solutios { e, te, e, e }. We ould express all solutios i the orm x() t = e + te + 3e + 4e where we eessarily have to allow or omplex oeiiets. However, i we ote (usig Euler s Formula) that it it e = os( t) + isi( t) ad e = os( t) isi( t) we a easily see that we ould also (perhaps more simply) use { t t e, te,os t,si t} as a basis or all solutios, i.e. all solutios may be expressed i the orm ( ) t t x t = e + te + os t+ si t where all the oeiiets are real salars. 3 4 Our ext several tass will be to osider ihomogeeous equatios ad iitial value problems. This will eessitate a slightly more uaed view o what we mea by liearly idepedet utios or the purpose o idig uique solutios to well-posed iitial value problems. This will ivolve what is ow as the Wrosia assoiated with a give basis o homogeeous solutios. Notes by Robert Witers 5