Homogeneous Constant Matrix Systems, Part I

Transcription

1 39 Homognous Constant Matrix Systms, Part I Finally, w can start discussing mthods for solving a vry important class of diffrntial quation systms of diffrntial quations: homognous constant matrix systms Ths systms ar important for at last thr rasons: W can dvlop straightforward procdurs for solving thm 2 Thy aris in applications 3 Thy provid approximations for mor gnral systms and, bcaus of this, will b instrumntal latr in analyzing th solutions to nonlinar autonomous systms Our goals for now ar twofold: To dvlop mthods for finding th solutions to ths systms 2 To carfully xamin th possibl pattrns for th trajctoris of 2 2 constant matrix systms (Th rsults of this analysis will latr b usd whn w study nonlinar systms) In this chaptr w will concntrat on finding th basic fundamntal solutions, and analyzing thos 2 2 systms for which ths solutions suffic Thos cass that cannot b dalt with using just thos basic solutions will b dalt with in th nxt chaptr 39 Basics, and Som Fundamntal Solutions W will now limit ourslvs to homognous linar systms of th form x = Px in which th componnts of th N N matrix P ar all ral-valud constants To signify this, w will us A instad of P for this matrix, and rfr to our systm x = Ax as a (basic) (ral) constant matrix systm So considr th systm x = Ax 3//2

2 Chaptr & Pag: 39 2 Constant Matrix Systms, Part I As alrady notd, th constant function x(t) = for all t is an quilibrium solution And if A is invrtibl (i, dt(a) = ), thn x = will b th only quilibrium solution What about th nonquilibrium solutions? On approach to finding thm is to gnraliz th solution to th abov whn N = and th matrix A is just on ral numbr A Thn our systm is simply th first-ordr diffrntial quation x = Ax, whos solution is At multiplid by som arbitrary constant But that approach rquirs figuring out just what w could man by At whn A is a gnral N N matrix It is an intrsting approach, but lt s sav that for latr Instad, lt us b inspird by th way w solvd a singl arbitrary N th -ordr homognous ordinary diffrntial quation a y (N) + a y (N ) + + a N y + a N y = back in chaptrs 6 and 8 Thr, th basic approach was to find a fundamntal st of N solutions by initially assuming that ach was of th form y(x) = r x whr r is a valu dtrmind by plugging this formula for y into th diffrntial quation and sing what workd If w wr lucky, w found N diffrnt valus for r r, r 2,, r N and could thn writ out our gnral solutions as a linar combination of ths r k x s, c r x + c 2 r 2x + + c N r N x Sinc w now know that a gnral solution to any homognous linar systm can b constructd by a similar linar combination c x (t) + c 2 x 2 (t) + + c N x N (t) whr { x (t), x 2 (t),, x N (t) } is a fundamntal st of solutions to our systm x = Ax, this sms to b a rasonabl approach Thr is on slight issu: th x k (t) s ar vctor-valud functions whil th xponntial rt is not So lt s try an xponntial multiplid by som constant vctor u, x(t) = u rt Plugging this into our systm x = Ax and simplifying: d [ u rt = A [ u rt dt u r rt = [Au rt ur = Au

3 A Short Rviw of Eignvalus and Eignvctors (with Applications) Chaptr & Pag: 39 3 So, for u rt to b a solution to our systm, th constant vctor u and scalar r must satisfy Au = ru If w can find N pairs (r, u ), (r 2, u 2 ), and (r N, u N ) satisfying this quation and such that th Wronskian of { u r t, u 2 r 2t,, u N r N t } is nonzro at on point (say, at t = ) (which will crtainly b th cas if {u, u 2,,u N } is a linarly indpndnt st of column vctors), thn w know this st is a fundamntal st of solutions for x = Ax, and that x(t) = c u r t + c 2 u 2 r 2t + + c N u N r N t is a gnral solution So, our approach to actually solving th systm of diffrntial quations x = Ax bgins with th attmpt to find th pairs (r, u) satisfying th algbraic quation Au = r u Looks lik it s tim to rviw ignvctors and ignvalus 392 A Short Rviw of Eignvalus and Eignvctors (with Applications) Eign-Things Lt r b som scalar (i, a numbr, ral or complx) and u som nonzro column vctor W say that r is an ignvalu and u is a corrsponding ignvctor for som givn N N matrix A if and only if Au = ru (39) Altrnativly, w can call r an ignvalu corrsponding to ignvctor u W can also simply say that (r, u) is an ignpair for th matrix A Not that, whil ignvalus can b zro, w insist that ignvctors b nonzro (bcaus quation (39) rducs to th trivial quation = if u = no mattr what r is)! Exampl 39: [ 2 A = 2 Lt, u = [, v = [ and w = [ 2 Obsrv that Au = [ [ 2 2 = [ = [ 3 3 = 3 [ = 3u and Av = [ [ 2 2 = [ = [ = [ = ( )v

4 Chaptr & Pag: 39 Constant Matrix Systms, Part I So u = [, T is an ignvctor for A with corrsponding ignvalu 3, and v = [, T is an ignvctor for A with corrsponding ignvalu That is, ( 3, [ ) and (, [ ) ar ignpairs for th matrix A On th othr hand, [ [ 2 2 Aw = = 2 [ = [ 5 = r [ 2 for any r So w is not an ignvctor for A As w discuss ign-things it is worthwhil to kp in mind that w wr ld to this discussion by th discovry that x(t) = u rt is a solution to th systm x = Ax whnvr (r, u) satisfy Au = ru ; that is, whnvr (r, u) is an ignpair for A So lt s actually solv our first systm of diffrntial quations:! Exampl 392: Considr th systm x = x + 2y y = 2x + y In matrix/vctor form, this is x = Ax whr A is th matrix from xampl 39 Thr w found two ignpairs (3, u) and (, v) with [ [ u = and v = So { [ 3t, [ } ( t i, {[ [ }) 3t t, is a st of 2 solutions to our 2 2 first-ordr systm of diffrntial quations By inspction, it is clar that nithr of ths two solutions is a constant multipl of th othr, so this st is clarly a linarly indpndnt pair of solutions (if it s not clar, comput th st s Wronskian at t = ) Hnc (invoking thorm 388), w know this st is a fundamntal st of solutions for our systm of diffrntial quations, and that a gnral solution is givn by x(t) = c [ 3t 3t [ + c 2 t, t or, quivalntly, by [ x(t) y(t) = [ c 3t c 2 t c 2 3t + c 2 t

5 A Short Rviw of Eignvalus and Eignvctors (with Applications) Chaptr & Pag: 39 5 It is important to not that ignvctors ar not uniqu If u and v ar both ignvctors corrsponding to th sam ignvalu r for A, and α and β ar any two scalars, thn A[αu + βv = αau + βav = α[ru + β[rv = r[αu + βv From this w gt th following: Lmma 39 Any nonzro linar combination of ignvctors corrsponding to a singl ignvalu r for a matrix A is also an ignvctor corrsponding to ignvalu r Th st of all linar combinations of ignvctors corrsponding to a singl ignvalu r for A is, itslf, a vctor subspac of th st of all vctors W will call this spac th ignspac corrsponding to ignvalu r for A Th dimnsion of this spac (i, th numbr of vctors in any basis for this ignspac) is calld th gomtric multiplicity of ignvalu r! Exampl 393: In an xampl just abov, w saw that ( 3,[, T) is an ignpair for th 2 2 matrix in that xampl Thus, for any nonzro constant c, th vctor [ [ c c = c is also an ignvctor corrsponding to ignvalu 3 It turns out that thr ar no othr ignvctors corrsponding to this ignvalu, so th ignspac corrsponding to 3 is th vctor spac of all constant multipls of [, T (including [, T = [, T ) Clarly this is a on-dimnsional vctor spac, so th gomtric multiplicity of ignvalu 3 is Finding (and Using) Eign-Things Th ky to finding ach ignpair (r, u) lis in rwriting th basic dfining quation Au = ru using th N N idntity matrix I = Rcall that Iu = u for any column vctor u So Au = ru Au = riu Au riu = [A ri u = (392)

6 Chaptr & Pag: 39 6 Constant Matrix Systms, Part I Sinc th ignvctor u is rquird to b nonzro, th last quation can hold only if A r I is not invrtibl Hnc, w must hav dt [A ri = (393) Computing this dtrminant yilds an N th dgr polynomial quation with r as th unknown This quation is calld th charactristic quation for A, and th polynomial is th calld th charactristic polynomial for A From what w alrady know about such polynomial quations, w know that this quation will hav N solutions r, r 2,, r N som of which may b rpatd Th numbr of tims a particular valu of r is rpatd in this st (which w normally call th multiplicity of that valu of r ) will b calld th algbraic multiplicity of that valu of r so w don t confus it with th gomtric multiplicity of r This lads to th following procdur for finding all ignpairs for an N N matrix A : By computing th dtrminant, rwrit th charactristic quation, as a polynomial quation dt [A ri =, 2 Find all valus of r r, r 2,, r M satisfying this charactristic quation Som of ths may b rpatd roots for th charactristic polynomial, and som may b complx 3 For ach diffrnt r k, st u k = u k u k 2 and solv ithr for all possibl u k j s Two practical nots on this: (a) Don t rally us th notation u k j u k sms convnint (b) u k N Au k = r k u k or [A ri u k = Us whatvr notation for th componnts of Sinc thr is a whol spac of ignvctors corrsponding to ach ignvalu, you will not obtain an ignvctor u k Instad, you will obtain a dscription of all th ignvctors in this ignspac in trms of a minimal st of arbitrary constants Thr is at last on mor stp w will want to carry out bfor using our rsults to solv our systm of diffrntial quations, x = Ax, but mayb w had bttr do an xampl, first! Exampl 39: Lt us find th ignvalus and corrsponding ignvctors for 5 A = 6 7 S sction 8 on pag 29 for a discussion of N th dgr polynomial quations

7 A Short Rviw of Eignvalus and Eignvctors (with Applications) Chaptr & Pag: 39 7 First, w rduc th charactristic quation to polynomial form: dt [A ri = 5 dt 6 r = 7 5 r dt 6 r = 7 r r 5 dt 6 r = 7 r ( r)(6 r)(7 r) + 5(6 r) = (6 r) [( r)(7 r) + 5 = (6 r) [ r 2 8r + 2 = Not that w did not compltly multiply out th trms Instad, w factord out th common factor 6 r to simplify th nxt stp, which is to find all solutions to th charactristic quation Fortunatly for us, w can asily factor th rst of th polynomial, obtaining (6 r) [ r 2 8r + 2 = (6 r) [(r 2)(r 6) = (r 2)(r 6) 2 as th compltly factord form for our charactristic polynomial Thus our charactristic quation dt[a r I rducs to (r 2)(r 6) 2 =, which w can solv by inspction W hav two ignvalus r = 2 and r = 6 (with r = 6 having algbraic multiplicity 2 ) To find th ignvctors corrsponding to ignvalu r = 2, w st r = 2 and solv [A riu = for th componnts of u For convninc, lt u = [α,β,γ T Thn [A ri u = 5 α 6 2 β = 7 γ

8 Chaptr & Pag: 39 8 Constant Matrix Systms, Part I 2 5 α 6 2 β = 7 2 γ 5 α β = 5 γ So th unknowns, α, β and γ must satisfy th algbraic systm α + β 5γ = α + β + γ = α + β + 5γ = Using whichvr mthod you lik, this is asily rducd to β = and α + 5γ = W can choos ithr α or γ to b an arbitrary constant Choosing γ, w thn must hav α = 5γ to satisfy th last quation abov Thus, th ignvctors corrsponding to ignvalu r = 2 ar givn by α 5γ 5 u = β = γ γ = γ whr γ can b any nonzro valu To find th ignvctors corrsponding to ignvalu r = 6, w st r = 6 and solv [A riu = for th componnts of u For convninc, lt us again us th notation u = [ α,β, γ T Thn which rducs to [A ri u = 5 α 6 6 β = 7 γ 5 5 α β =, γ α + γ = and β =

9 A Short Rviw of Eignvalus and Eignvctors (with Applications) Chaptr & Pag: 39 9 In this cas lt us tak α to b an arbitrary constant with γ = α so that th first quation is satisfid Th scond quation is satisfid no mattr what β is, so β is anothr arbitrary constant Thus, th ignvctors corrsponding to ignvalu r = 2 ar givn by u = α β = γ α β = α whr α and β ar arbitrary constants α + β = α α + β For our applications, w do not want a dscription of all possibl ignvctors corrsponding to ach ignvalu r Instad w want a st of spcific ignvctors { u, u 2,, u M } such that all th ignvctors in this st corrsponding to any on particular ignvalu form a basis for that ignvalu s ignspac So, th last stp in our procdur for finding th ignvalus and ignvctors for our matrix is: For ach distinct ignvalu, choos appropriat valus for th arbitrary constants dscribing th corrsponding ignvctors so as to obtain spcific vctors for a basis for ach ignspac Thn writ down th collction of all th ignvctors so found, { u, u 2,, u M }, kping track of thir corrsponding ignvalus By th way, rcall that th numbr of vctors in any basis for an ignspac corrsponding to ignvalu r (i, th dimnsion of that ignspac) is calld th gomtric multiplicity of that ignvalu With a littl thought, you will raliz that gomtric multiplicity of ignvalu r = numbr of arbitrary constants ndd to dscrib all corrsponding ignvctors Along ths lins, w might as wll mntion that, for any ignvalu r k, gomtric multiplicity of r k algbraic multiplicity of r k Lt us just accpt this fact as somthing usually vrifid in a cours on linar algbra! Exampl 395: In th last xampl, w saw that 5 u = γ dscribs all th ignvctors corrsponding to ignvalu r = 2 for th matrix A from xampl 39 Sinc w only hav on arbitrary constant, γ, th gomtric multiplicity for r = 2 is Stting γ = givs us th on ignvctor w nd corrsponding to r = 2, 5 u =

10 Chaptr & Pag: 39 Constant Matrix Systms, Part I (Any othr nonzro valu for z would also hav givn us a suitabl ignvctor) W also saw that th ignvctors corrsponding to ignvalu r = 6 for A ar all givn by α + β In this cas, w hav 2 arbitrary constants, so th gomtric multiplicity is 2 Taking α = and β = givs us th ignvctor u 2 =, and taking α = with β = givs us th ignvctor u 3 = Clarly, ths two ignvctors form a linarly indpndnt pair In summary, {u, u 2, u 3 } is our st of ignvctors with u corrsponding to r = 2, and u 2 and u 3 both corrsponding to r = 6 Lt s not forgt why w ar finding ths ignvalus and ignvctors W want to us thm to construct solutions to systms of diffrntial quations! Exampl 396: Considr th 3 3 systm of diffrntial quations x 5 x y = 6 y z 7 z From xampls 39 and 395, abov, w know that th matrix in this quation has ignvalus r = 2, r 2 = 6 and r 3 = 6, and corrsponding ignvctors 5 u =, u 2 = and u 3 =, rspctivly And from our discussion in th prvious sction, w know this givs th corrsponding st of 3 solutions to our 3 3 systm of diffrntial quations 5 2t, 6t, 6t Chcking th Wronskian at t =, w gt 5 W() = dt = 6 =,

11 A Short Rviw of Eignvalus and Eignvctors (with Applications) Chaptr & Pag: 39 assuring us (via thorm 389) that our st of solutions is a fundamntal st, and that a gnral solution to our systm is givn by or, quivalntly, by x(t) 5 y(t) = c 2t z(t) x(t) 5c 2t y(t) = z(t) + c 2 6t c 2 6t c 3 6t c 2t + c 2 6t + c 3 6t, Do not that, had w chosn diffrnt valus for th arbitrary constants in xrcis 395, thn th spcific vctors in our gnral solution would b diffrnt Howvr, th nd rsult would hav bn quivalnt to th abov Nots on Finding and Using Eignpairs Complt and Incomplt Sts of Eignvctors From what w (should) know about linar algbra, w ar assurd that th procdur just dscribd will gnrat a linarly indpndnt st of ignvctors { u, u 2,, u M } for our N N ral constant matrix A, with th siz M of this st bing no largr than N Morovr: M will not dpnd on how w choos th u k s, but only on th matrix A 2 Th componnts of u k can b chosn as ral numbrs providd its corrsponding ignvalu is ral If w ar lucky, M = N Thn w will hav a linarly indpndnt st of N ignvctors { u, u 2,, u N } corrsponding, rspctivly, to an ordrd st of N ignvalus { r, r 2,, r N } (with som of th ignvalus, possibly, bing rpatd) If this happns, w say that th matrix has a complt st of ignvctors, with any st of ignvctors bing calld a complt st of ignvctors if and only if it is a linarly indpndnt st of N ignvctors for A W lik for our st of ignvctors to b complt, bcaus thn { u r t, u 2 r 2t,, u N r N t } is a st of N solutions to our N N constant matrix systm x = Ax Morovr, it is a linarly indpndnt st of vctors whn t = That mans th Wronskian for this st is nonzro,

12 Chaptr & Pag: 39 2 Constant Matrix Systms, Part I which, in turn mans that this st is a fundamntal st of solutions for our systm of diffrntial quations, and, thus, x(t) = c u r t + c 2 u 2 r 2t + + c N u N r t is a gnral solution to x = Ax That was th situation in xampl 396, abov Thr ar two cass whr w ar assurd that a givn N N matrix A has a complt st of ignvctors bfor w bgin sking ths ignvctors: Whn th matrix has N diffrnt ignvalus 2 Whn th matrix is symmtric (i, A T = A ) Vrifying th first is asy: Sinc ach of th N ignvalus must hav at last on corrsponding ignvctor, th procss abov will gnrat a linarly indpndnt st of at last N ignvctors But, as notd abov, this st can contain at most N ignvctors Hnc, it must contain xactly N ignvctors Morovr, it follows with just a littl work that, for ach ignvctor r k, = gomtric multiplicity of r k = algbraic multiplicity of r k Vrifying that w hav a complt st of ignvctors whn th matrix is symmtric is a bit hardr to show Fortunatly for us, you almost crtainly discussd this in a cours on linar algbra sinc th following is a standard part of most introductory algbra courss: Thorm 392 Lt A b a symmtric N N matrix (with ral-valud componnts) Thn both of th following hold: All th ignvalus ar ral 2 Thr is an orthogonal basis for N consisting of ignvctors for A Shortcuts with Triangular Matrics Givn a matrix a a 2 a 3 a N a 2 a 22 a 23 a 2N A = a 3 a 32 a 33 a 3N, a N a N2 a N3 a N N th main diagonal is diagonal st of ntris from th uppr lft cornr to th lowr right cornr of th matrix That is, it is th diagonal consisting of a, a 22, and a N N W say that our matrix A is uppr triangular or lowr triangular if, rspctivly, all th ntris blow or all th ntris abov th main diagonal ar zro, a a 2 a 3 a N a a 22 a 23 a 2N a 2 a 22 a 33 a 3N or a 3 a 32 a 33 a N N a N a N2 a N3 a N N

13 Eignpairs and Corrsponding Solutions Chaptr & Pag: 39 3 And if A is both uppr and lowr triangular; that is, if all th ntris off of th main diagonal ar zro, a a 22 a 33, a N N thn w say that th matrix A is diagonal Whn A is any of ths forms, you can asily us th ruls for computing dtrminants to driv dt(a λi) = (a λ)(a 22 λ)(a 33 λ) (a N N λ), And so th charactristic quation dt(a λi) = immdiatly rducs (a λ)(a 22 λ)(a 33 λ) (a N N λ) = tlling us that th lmnts of this matrix s main diagonal ar th ignvalus of any uppr or lowr triangular matrix What s mor, if A is diagonal, thn, virtually by inspction, you can s that an ignvctor u k corrsponding to λ = a kk is th column vctor whos componnts ar all xcpt for th k th componnt, which w can tak to b So if you ar daling with a uppr or lowr triangular matrix, you can simply rad off th ignvalus from th main diagonal And if th matrix is diagonal, you can also immdiatly giv th corrsponding ignvctors 393 Eignpairs and Corrsponding Solutions Sts of Solutions from Sts of Eignpairs As just discussd, our mthod will gnrat a st of ignpairs { (r, u ), (r 2, u 2 ),, (r M, u M ) } for our constant N N matrix A, with th st { u, u 2,, u M } bing a linarly indpndnt st of column vctors Morovr, this mthod will find th largst possibl such st of ignvctors Any othr ignvctor will b a linar combination of vctors from th abov st Our intrst in ths sts cam from th obsrvations mad in sction 39 that can b summarizd, using trminology from th last sction, as th following lmma and thorm: Lmma 393 Lt { u, u 2,, u M }

14 Chaptr & Pag: 39 Constant Matrix Systms, Part I b a linarly indpndnt st of ignvctors for a constant N N matrix A, and, for ach u k, lt r k b th corrsponding ignvalu Thn { u r t, u 2 r 2t,, u M r M t } is a st of solutions for th systm x = Ax on th ntir ral lin Morovr, this is a linarly indpndnt st of vctors whn t = Thorm 39 Lt { u, u 2,, u N } b a complt st of ignvctors for a constant N N matrix A, and, for ach u k, lt r k th corrsponding ignvalu Thn { u rt, u 2 r2t,, u N } r N t b is a fundamntal st of solutions for th systm x = Ax on th ntir ral lin, and x(t) = c u r t is a gnral solution to x = Ax + c 2 u 2 r 2t + + c N u N r t Consquntly, if th matrix A has a complt st of ignvctors, th mthod discussd in th last sction, along with th obsrvations mad in th last thorm, will lad to a gnral solution to x = Ax Issus W ar still lft with th issu of just what to do if our matrix A dos not hav a complt st of ignvctors W can still find a linarly indpndnt st of ignvctors, { u, u 2,, u M }, and { u r t, u 2 r 2t,, u M r M t } is still a st of solutions to x = Ax, but it is not a fundamntal st W will nd to discovr how to find additional solutions to fill out this st to a complt fundamntal st of N solutions W will do this in th nxt chaptr Thr is anothr littl issu that w hav sidstppd: th fact that th ignvalus and th componnts of th ignvctors can b complx valus W will also dal with that issu in th nxt chaptr For now, lt us tak what w hav and xamin th possibl solutions and thir trajctoris, with particular mphasis on th trajctoris of solutions to 2 2 systms W will find that this will lad to som vry intrsting picturs and rsults that will vn b usful whn daling with mor gnral systms of diffrntial quations

15 Two-Dimnsional Phas Portraits: Prliminaris Chaptr & Pag: Two-Dimnsional Phas Portraits: Prliminaris It turns out that th ignvalus and ignvctors tll us much about th qualitativ bhavior of th solutions to a constant matrix systm x = Ax To s this most radily, w will carfully dtrmin th possibl trajctoris whn A is 2 2 A major goal is to dtrmin how th ignvalus and ignvctors dtrmin th possibl pattrns of th trajctoris in th phas portraits for x = Ax Th Critical Point at th Origin Rmmbr that th origin (, ) is a critical point for any systm x = Ax, and that this singl point is th ntir trajctory of th quilibrium solution [ [ x(t) = for all t y(t) If dt A = thn th origin is th only critical point, and x(t) = is th only quilibrium solution Much of our work in what follows will b to dtrmin th stability of this critical point, and to dtrmin gomtris of th trajctoris in rgions both clos to (, ) and in rgions far away from (, ) In th procss w will introduc trminology for furthr classifying th critical point (, ) according to ths gomtris Som of th trms w will introduc ar nods, saddl points, cntrs and spiral points (Howvr, w won t fully dfin ths trms until latr, in sction 3) Symmtris Th sktching of a phas portrait for x = Ax can b aidd by noting simpl symmtris inhrnt in th dirction fild of a basic constant matrix systm To s this, lt (x, y ) b any point in th plan othr that th origin, and lt x = [x, y T Rmmbr, th dirction arrow for our systm at (x, y ) is a short arrow drawn at this point in th sam dirction as v whr v = Ax Now considr th dirction arrow at any nonzro point (x, y ) on th straight half lin starting at th origin and going through (x, y ) Thn, for som c > [ [ (x, y ) = (cx, cy ), x x cx = = = cx y cy and th dirction arrow for our systm at (x, y ) is a short arrow drawn at this point in th sam dirction as Ax = Acx = cax = cv But multiplying a vctor by a positiv constant dos not chang its dirction So, as illustratd in figur 39a, Th dirction arrow at ach point on a half lin having th origin as an ndpoint all point in xactly th sam dirction (or ar all )

16 Chaptr & Pag: 39 6 Constant Matrix Systms, Part I y x (a) (b) Figur 39: Symmtris in dirction filds and phas portraits for a 2 2 constant matrix systm On th othr hand, th dirction arrow at ( x, y ) is in th sam dirction as [ x A = Ax = v y So, as also illustratd in figur 39a, Th dirction arrow at ( x, y ) is in th xact opposit dirction as is th dirction arrow at (x, y ) In othr words, th dirction fild is antisymmtric across th origin Using th abov obsrvations, w can sktch a usful dirction fild aftr computing th dirction arrows at only a fw wll-chosn points Morovr, ths pattrns in th dirction fild for must b carrid ovr in th corrsponding phas portrait, with, say th trajctoris in th lowr-half plan bing mirror imags rflctd across th origin of th trajctoris in th uppr half plan, as illustratd in figur 39b 395 Two-Dimnsional Phas Portraits from Ral, Complt Eign-Sts Throughout this sction, w will assum A is a 2 2 constant matrix with ral componnts and a complt st of ignvctors {u, u 2 } W will furthr assum that th corrsponding ignvalus r and r 2 ar ral and nonzro 2 That mans our solutions will b of th form x(t) = c u r t + u 2 r 2t whr ach componnt of ach u k is a ral numbr As t varis, ach solution tracs out a trajctory in th plan W want to discovr th possibl pattrns of th trajctoris, and how ths pattrns dpnd on th ignvalus and ignvctors 2 ou ll dal with zro ignvalus in xrcis 393

17 Two-Dimnsional Phas Portraits from Ral, Complt Eign-Sts Chaptr & Pag: 39 7 In fact, w will discovr that th ignpairs of A, (r, u ) and (r 2, u 2 ), alon, giv us nough information to sktch crud, yt nlightning, phas portraits for x = Ax If w thn wish, w can rfin our sktchs of th phas portraits by using minimal dirction filds and th symmtris discussd in th prvious sction 3 For convninc, lt us us th notation [ [ [ x(t) x(t) =, u x = and u 2 x2 =, y(t) y y 2 so that our solutions will look lik [ x(t) y(t) [ x = c r t y [ x2 + c 2 r 2t y 2, and our trajctoris will b curvs on th plan Lt us also assum w v labld our ignvalus so that r r 2 Trajctoris Corrsponding to On Eignpair W should first considr th trajctory of a singl solution of th form x(t) = cu rt whr c is any nonzro constant, and (r, u) is any ignpair with r bing ral and nonzro! Exampl 397: Suppos x(t) = cu rt with (r, u) = ( [ ) 3, and c is any ral numbr Thn x(t) = cu 3t = [ x(t) y(t) = c [ 3t, which mans x(t) is just som multipl of u for ach t, and that, in turn, mans that, for ach ral valu t, (x(t), y(t)) is a point on th lin through th origin and paralll to u, as illustratd in figurs 392a and 392b Furthrmor: If c =, thn our solution rducs to th alrady-known quilibrium solution x(t) =, whos ntir trajctory is just th critical point (, ) 2 If c > and t is any ral numbr, thn cu 3t is a positiv multipl of u, which mans (x(t), y(t)) is always on th half lin xtnding from (, ) in th dirction of u, as in figur 392a Morovr, sinc 3t is a continuous incrasing function on th ral lin, and lim (x(t), y(t)) = lim ( c 3 3t, c 2 3t) = (, ), t t 3 Of cours, you can also us a suitabl computr math packag to sktch phas portraits, as th author has don for this txt But w will find th knowldg and xprinc gaind by hand sktching phas portraits invaluabl latr on

18 Chaptr & Pag: 39 8 Constant Matrix Systms, Part I (x(t), y(t)) u cu 3t cu 3t u (x(t), y(t)) (a) (b) (c) Figur 392: Plotting th trajctory of cu 3t with u = [, T (a) whn c > and (b) whn c < (c) Trajctoris corrsponding to ignpairs (3, [, T ) and ( 3, [, T ) whil lim (x(t), y(t)) = lim t + ( c 3 3t, c 2 3t) = (+,+ ), t + it should b clar that, in fact, (x(t), y(t)) tracs out this ntir half lin with as (x(t), y(t)) going from (, ) to (+,+ ) t gos from to + So th trajctory is this ntir half lin paralll to u with ndpoint (, ), and with th dirction of travl bing th dirction away from th origin (Strictly spaking, th origin is not part of th trajctory sinc t can only approach, nvr qual ) 3 If c < and t is any ral numbr, thn cu 3t is a ngativ multipl of u, and, for rasons vry similar to thos givn whn c >, it should b clar that th trajctory is th half lin xtnding from (, ) in th dirction of u, and with th dirction of travl bing in th dirction away from th origin (as illustratd in figur 392b) (Again, whil (, ) is an ndpoint for this trajctory, it is not, strictly spaking, in th trajctory)! Exampl 398: Lt us now considr th trajctoris for [ [ x(t) x(t) = = c y(t) whr c is any ral numbr Th gnral idas givn in th prvious xampl still apply, and lad us to th trajctoris all bing parts of th straight lin through th origin and paralll to th vctor u Again, if c =, thn th trajctory is just th critical point (, ) ; if c >, thn th trajctory is th ntir half lin with ndpoint (, ) and xtnding in th dirction of u ; if c <, thn th trajctory is th ntir half lin with ndpoint (, ) and xtnding in th dirction of u This tim, howvr, lim (x(t), y(t)) = lim t + 3t ( c 3t, c 3t) = (, ) t +

19 Two-Dimnsional Phas Portraits from Ral, Complt Eign-Sts Chaptr & Pag: 39 9 (a) (b) Figur 393: Phas portraits xhibiting star nods with (a) th ignvalu bing ngativ and (b) th ignvalu bing positiv So (x(t), y(t)) approachs (, ) as t incrass That is, th dirction of travl on ths trajctoris ar towards th origin, as illustratd in figur 392c Clarly, th discussion carrid out for ths two xampls can b carrid out in gnral, and will lad to th obsrvation that, if r is any nonzro ral numbr, thn th trajctoris for any nonquilibrium solution of th form x(t) = cu rt ar th two straight half lins starting at th origin and paralll to th vctor u Th dirction of travl will b away from th origin if r > and towards th origin if r < For simplicity, lt us call ths th straight-lin trajctoris corrsponding to th ignpair (r, u) (assuming (r, u) is an ignpair for th matrix in our systm of diffrntial quations) Sktching ths will usually b on of th first things w will want to do whn sktching phas portraits (Rmmbr, vn though w ar calling ths straight-lin trajctoris, thy ar rally straight half-lin trajctoris Each has (, ) as an ndpoint, though, strictly spaking, (, ) is not a point in ths trajctoris) Trajctoris Whn r = r 2 = Suppos th matrix in our 2 2 systm x = Ax has a singl rpatd nonzro ignvalu and a complt st of ignvctors {u, u 2 } Thn r = r 2, and x(t) = c u r t + c 2 u 2 r 2t = [ c u + c 2 u 2 r t But sinc {u, u 2 } is a complt st of ignvctors, c u + c 2 u 2 dscribs all 2-dimnsional vctors So th abov gnral solution simplifis to x(t) = u r t whr u is an arbitrary column vctor Consquntly, th st of all trajctoris of this systm is th st of all half lins in th plan with ndpoints at (, ), with th dirction of travl bing away from th origin if r >, and towards th origin if r < (as illustratd in figur 393) In cass lik this, th critical point (th trajctory for th quilibrium solution) is said to b a nod for our systm x = Ax Somtims, such a point is also calld a propr nod or vn a star nod bcaus of th way th trajctoris sm to radiat from th origin If th ignvalus ar ngativ, thn this critical point and corrsponding quilibrium solution x(t) =

20 Chaptr & Pag: 39 2 Constant Matrix Systms, Part I ar asymptotically stabl (s figur 393a) If th ignvalus ar positiv, thn this critical point and quilibrium solution ar unstabl (s figur 393b) By th way, in xrcis 39, you will show that any 2 2 matrix having a complt st of ignvctors all corrsponding to th sam ignvalu r is a vry simpl diagonal matrix Trajctoris Whn < r < r 2 W start with an illustrativ xampl! Exampl 399: Considr a 2 2 systm x = Ax whos matrix has ignpairs ( [ ) ( [ ) (r, u ) = 2, and (r 2, u 2 ) = 5, Th gnral solution is thn x(t) = c u r t + c 2 u 2 r 2t [ [ = c 2t + c 2 5t (39) Obsrv that lim x(t) = t lim t c [ [ 2t + c 2 5t = This tlls us that th trajctoris all start at th origin Th fact that both 2t and 5t bcom infinit as t + tlls us that all nonquilibrium trajctoris xtnd infinitly out on th plan W start our sktch by plotting a dot at th origin for th quilibrium solution, and sktching th straight-lin trajctoris corrsponding to th ignpairs (r, u ) and (r 2, u 2 ) Not that th dirction of travl on ach of ths straight-lin trajctoris is away from th origin This taks car of th trajctoris of solution (39) whn ithr c = or c 2 = To sktch any othr trajctory, assum c and c 2 ar both nonzro, and considr som convnint tangnt vctors using th th drivativ of x(t), x (t) = 2c u 2t + 5c 2 u 2 5t, both as t and as t + Rmmbr, this vctor is tangnt to th trajctory at (x(t), y(t)) and points in th dirction of travl along th trajctory Th sam is tru with any positiv multipl of x (t) In particular, lt us b clvr and us [ and 2t x (t) = 2t [ 2c u 2t + 5c 2 u 2 5t = 2c u + 5c 2 u 2 3t 5t x (t) = 5t [ 2c u 2t + 5c 2 u 2 5t = 2c u 3t + 5c 2 u 2 Now lt t As alrady notd, w ll thn hav (x(t), y(t)) (, ) In addition, lim t 2t x (t) = lim t [ 2c u + 5c 2 u 2 3t = 2c u = 2c [ This tlls us that, as t, th tangnt to th trajctory at (x(t), y(t)) bcoms paralll to th vctor u = [, T, which, itslf, is paralll to th straight half-lin trajctory corrsponding to (r, u ) With a littl thought and som sktching on your own, you will

21 Two-Dimnsional Phas Portraits from Ral, Complt Eign-Sts Chaptr & Pag: 39 2 r2 r 2 r r (a) (b) Figur 39: Phas portraits for th systms (a) in xampl 399 (whr < r < r 2 ), and (b) in xampl 39 (whr r < r 2 < ) In ach, r and r 2 idntify th straight lin trajctoris corrsponding to ignvalus r and r 2 raliz that this mans (x(t), y(t)) approachs th origin along a path tangnt to th half-lin trajctory corrsponding to (r, u ) Now lt t + As alrady notd, w ll thn hav (x(t), y(t)) tracing a path furthr and furthr away from th origin In addition, lim t + 5t x (t) = [ lim 2c u 3t + 5c 2 u 2 [ = 5c 2 u 2 = 5c 2 t + This tlls us that, as t +, th tangnt to th trajctory at (x(t), y(t)) bcoms paralll to th vctor u 2 = [, T, which, itslf, is paralll to th straight half-lin trajctory corrsponding to (r 2, u 2 ) Thus, as t gts largr, th path bing tracd out by (x(t), y(t)) approachs a path paralll to th straight half-lin trajctory corrsponding to (r 2, u 2 ) Howvr, th trajctory of x(t) will not gt clos to this half-lin trajctory sinc th othr trm, c u 2t is also gtting largr, not smallr, as t incrass Som of ths trajctoris hav bn sktchd in figur 39a In gnral, if < r < r 2, thn a phas portrait for x = Ax can b roughly sktchd by doing th following: Plot th critical point (, ), and sktch th straight-lin trajctoris corrsponding to (r, u ) and (r 2, u 2 ) 2 Thn sktch a rasonabl collction of othr trajctoris with ach starting at th origin and tangnt thr to a straight-lin trajctory corrsponding to r (th smallr ignvalu), and thn curving around so that it bcoms somwhat paralll to th narst straightlin trajctory corrsponding to r 2 (th largr ignvalu) (A minimal dirction fild may hlp in rfining your sktching of th trajctoris) 3 B sur to sktch small arrows indicating that th dirction of travl along ach trajctory is away from th origin

22 Chaptr & Pag: Constant Matrix Systms, Part I Again, nod is th classical trm for th critical point (, ) in this cas, and it is clarly th trajctory of an unstabl quilibrium solution This tim (for rasons to givn dscribd latr) this nod is somtims considrd to b impropr By th way, instad of mmorizing which nd of th trajctoris bcom paralll to which ignvctors, you may just want to look at which trm in x (t) = 2c u 2t + 5c 2 u 2 5t bcoms dominant as t + and t using th fact that α β whn α β W ll illustrat this approach in th xampl for th nxt cas Trajctoris Whn r < r 2 < Th analysis hr is vry similar to that don whn < r < r 2 Th main diffrnc is that, bcaus of th ngativ ignvalus, th dirction of travl along ach trajctory will b towards th origin, instad of away ou should do this analysis yourslf ou will find that, in gnral th trajctoris gnratd by x(t) = c u r t + c 2 u 2 r 2t whn r < r 2 < can b roughly sktchd by doing th following: Plot th critical point (, ), and sktch th straight-lin trajctoris corrsponding to (r, u ) and (r 2, u 2 ) 2 Thn sktch a rasonabl collction of othr trajctoris with ach trajctory starting at th origin and tangnt thr to a straight-lin trajctory corrsponding to r 2 (th largr ignvalu), and thn curving around so that it bcoms somwhat paralll to th narst straight-lin trajctory corrsponding to r (th smallr ignvalu) (Again, a minimal dirction fild may hlp in rfining your sktching of th trajctoris) 3 B sur to sktch small arrows indicating that th dirction of travl along ach trajctory is towards from th origin Th critical point is, onc again, classifid as an impropr nod for our systm This tim, howvr, sinc r and r 2 ar ngativ, w hav [ lim x(t) = lim c u r t + c 2 u 2 r 2t = c u + c 2 u 2 = t t So th critical point (, ) is stabl In fact, it is asymptotically stabl! Exampl 39: Lt s considr a 2 2 systm x = Ax whos matrix has ignpairs ( [ ) ( [ ) (r, u ) = 5, and (r 2, u 2 ) = 2, In this cas, th gnral solution is x(t) = c u r t + c 2 u 2 r 2t [ [ = c 5t + c 2 2t, (395)

23 Two-Dimnsional Phas Portraits from Ral, Complt Eign-Sts Chaptr & Pag: and its drivativ is x (t) = 5c u 5t 2c 2 u 2 2t Again, w start our sktch by plotting a dot at th origin for th quilibrium solution, and sktching th straight-lin trajctoris corrsponding to th ignpairs (r, u ) and (r 2, u 2 ) Not that th dirction of travl on ach of ths straight-lin trajctoris is towards th origin This taks car of th trajctoris of solution (395) whn ithr c = or c 2 = Now assum c and c 2 ar both nonzro Bcaus th ignvalus ar both ngativ, w know (from th abov commnts) that, as t +, th trajctory for x(t) approachs (, ) along a path that closly parallls on of th ignvctors Rathr that try to rmmbr which ignvctor that is, lt us just obsrv that, for larg valus of t, 5t 2t So th trm with 5t valus of t w hav bcoms ngligibl as t + A littl mor xplicitly, for larg x(t) = c u 5t + c 2 u 2 2t = [ c u 3t }{{} and, by th sam approximation, + c 2 u 2 2t c 2 u 2 2t, x (t) = 5c u 5t 2c 2 u 2 2t 2c 2 u 2 2t d [ = c 2 u 2 2t dt Clarly, thn, as t +, x(t) approachs (, ) along a trajctory that bcoms mor and mor lik a straight-lin trajctory of a solution corrsponding to ignvalu r 2 = 2 Convrsly, this also tlls us that th part of th trajctory of x(t) corrsponding to larg ngativ valus of t must b far from (, ), bcoming mor and mor paralll to th th othr ignvctor, u as t Using ths facts, you should b abl to sktch a phas portrait similar to that in figur 39b (By th way, a similar analysis can b carrid out as t using th fact that 2t 5t whn t But thr is a proviso: Whil th 2t trm in x(t) is ngligibl compard to th 5t for larg ngativ valus of t, it is still a vry larg trm which incrass as t So, whil an analysis similar to th abov can justify th claim that th path of x(t) bcoms mor and mor paralll to u as t, it dos not show that x(t) is actually gtting closr to a straight-lin trajctory of an solution corrsponding to ignvalu r ) Trajctoris Whn r < < r 2 Again, w start with an xampl that, ssntially, xplains vrything! Exampl 39: Now considr a 2 2 systm x = Ax whos matrix has ignpairs ( [ ) ( [ ) (r, u ) = 2, and (r 2, u 2 ) = 5,

24 Chaptr & Pag: 39 2 Constant Matrix Systms, Part I Th gnral solution is thn x(t) = c u r t + c 2 u 2 r 2t [ [ = c 2t + c 2 5t (396) W again bgin our sktch by plotting a dot at th origin for th quilibrium solution, and sktching th th straight-lin trajctoris corrsponding to th ignpairs (r, u ) and (r 2, u 2 ) Not that th dirction of travl on th straight-lin trajctory corrsponding to (r, u ) is towards th origin, whil th dirction of travl on th straight-lin trajctory corrsponding to (r 2, u 2 ) is away from th origin This taks car of th trajctoris of solution (396) whn ithr c = or c 2 = To sktch any othr trajctory, assum c and c 2 ar both nonzro, and considr solution (396) both whn t is a larg ngativ valu, and whn t is a larg positiv valu If t is a larg ngativ valu, thn 2t is a larg positiv valu whil 5t So, for larg ngativ valus of t, [ [ [ x(t) = c 2t + c 2 5t c 2t In othr words, that part of th trajctory corrsponding to t bing a larg ngativ valu is clos to on of th straight-lin trajctoris corrsponding to r = 2 As t incrass, th trm with 2t dcrass in magnitud whil th trm with 5t incrass in magnitud Onc t bcoms sufficintly larg, 2t bcoms ngligibl and, [ [ [ x(t) = c 2t + c 2 5t c 2 5t Hnc, this part of th trajctory bcoms vry clos to on of th straight-lin trajctoris corrsponding to r = 5 So, to sktch a trajctory othr than that for th quilibrium solution or for on of th straight-lin trajctoris corrsponding to an ignpair, start your curv clos to and narly paralll to on of th straight lin trajctoris corrsponding to r = 2 and draw it moving towards th origin and curving away from that on straight-lin trajctory As you continu sktching, bnd th curv towards a straight-lin trajctory for th positiv ignvalu (without crossing any othr trajctoris) and try to nd your curv clos to and narly paralll to that straight-lin trajctory Th rsulting curv will look a littl lik a hyprbola with th straightlin trajctoris as asymptots B sur to add a littl arrow or two to indicat th dirction of travl A phas portrait for this systm has bn sktchd in figur 395 In gnral, if r < < r 2, thn a phas portrait for x = Ax can b roughly sktchd by doing th following: Plot th critical point (, ), and sktch th straight-lin trajctoris corrsponding to (r, u ) and (r 2, u 2 ) Not that th dirction of travl on thos corrsponding to th ngativ ignvalu is towards th origin, whil th dirction of travl on thos trajctoris corrsponding to th positiv ignvalu is away from th origin 2 Thn sktch a rasonabl collction of othr trajctoris with ach bing a roughly hyprbolic shapd curv with th straight-lin trajctoris as asymptots

25 Additional Exrciss Chaptr & Pag: r 2 r Figur 395: A phas portrait of th systm in xampl 39 (whr r < < r 2 ) Again, r and r 2 idntify th straight lin trajctoris corrsponding to ignvalus r and r 2 3 B sur to sktch small arrows indicating that th dirction of travl along ach trajctory, with ths dirctions of travl closly matching th dirctions of travl on th narby straight-lin trajctoris Finally, at long last, w hav an xampl in which th critical point is not calld a nod W call it a saddl point Also not that, whil two trajctoris do approach (, ) as t +, th rst go away from (, ) as t + So, th quilibrium solution for this systm is unstabl Additional Exrciss 39 Considr th systm x = x + 2y y = 3x y a If w writ this systm in matrix/vctor form x = Ax, what is th matrix A? b Vrify that th following two vctors ar ignvctors for th matrix A found in th abov xrcis, and find thir corrsponding ignvalus r and r 2, rspctivly [ [ 2 u = and u 2 = 3 c Using th ignpairs just found abov, writ out a pair of solutions to th abov systm of diffrntial quations, and vrify that th Wronskian of this pair is nonzro at t = d Writ out a gnral solution to th abov systm of diffrntial quations 392 Considr th systm x = x 3y y = 6x 7y

26 Chaptr & Pag: Constant Matrix Systms, Part I a If w writ this systm in matrix/vctor form x = Ax, what is th matrix A? b Vrify that th following two vctors ar ignvctors for th matrix A found in th abov xrcis, and find thir corrsponding ignvalus r and r 2, rspctivly [ [ 3 u = and u 2 = 2 3 c Using th ignpairs just found abov, writ out a pair of solutions to th abov systm of diffrntial quations, and vrify that th Wronskian of this pair is nonzro at t = d Writ out a gnral solution to th abov systm of diffrntial quations 393 Considr th systm x = x 3y + 3z y = 3x 5y + 3z z = 6x 6y + z a If w writ this systm in matrix/vctor form x = Ax, what is th matrix A? b Vrify that th following thr vctors ar ignvctors for th matrix A found in th last xrcis, and find thir corrsponding ignvalus r, r 2 and r 3, rspctivly u =, u 2 = and u 3 = 2 c Using th ignpairs found in th abov xrciss, writ out a st of thr solutions to th abov systm of diffrntial quations, and vrify that th Wronskian of this pair is nonzro at t = d Writ out a gnral solution to th abov systm of diffrntial quations 39 For ach of th following matrics, do th following: i Find th ignvalus and a corrsponding linarly indpndnt st of ignvctors for th matrix in th systm ii iii List th corrsponding ignpairs Stat whthr th matrix has a complt st of ignvctors (Not: Som of th ignvalus and ignvctors may b complx) a [ [ [ b c 3 2 d [ f

27 Additional Exrciss Chaptr & Pag: g j 2 h 8 2 i Find a gnral solution for ach of th following systms: a x = Ax whr A is th matrix in xrcis 39 a b x = Ax whr A is th matrix in xrcis 39 b c x = Ax whr A is th matrix in xrcis 39 d x = Ax whr A is th matrix in xrcis 39 f x = Ax whr A is th matrix in xrcis 39 i [ [ [ x 3 x x f y = g = x + 2y 6 y y = 5x 2y h x = 8x + 2y y = x + y i x = 6x y + 2z y = x z = 3y j x = 2y 3z y = 2x + 7z z = 3x + 7y k x = 2x y + 3z y = 2x + y z = 2x y + 3z l x = x + 2x 2 x 2 = 2x 2 + 3x 3 x 3 = 3x 3 + x x = x 396 Solv th initial-valu problm x = Ax with x() = x for ach of th following choics of A and x : a A is th matrix in xrcis 39 b and x = [, 6 T b A is th matrix in xrcis 39 b and x = [, 5 T c A is th matrix in xrcis 39 and x = [,, 6 T [ 2 d A = and x = [5, T For ach of th following, considr th 2 2 constant matrix systm x = Ax whr A has th givn two ignpairs Using just th ignpairs:

28 Chaptr & Pag: Constant Matrix Systms, Part I i Writ out th gnral solution to th systm ii Dscrib th stability of th nod at th origin a c g iii Sktch a phas portrait ( [ ) ( [ ) 2, and 2, ( [ ) ( [ ) 2, and, ( [ ) ( [ ), and 2, ( [ ) ( [ ) 5, and 3, b d f h ( [ 2, ( 2, ( 3, ( 3, ) and [ ) and [ ) and [ ) and ( 2, (, ( 5, [ ( 5, [ ) [ ) ) [ ) 398 For ach of th following, considr th 2 2 constant matrix systm x = Ax whr A has th givn two ignpairs, and, using just th ignpairs: i Writ out th gnral solution to th systm a c ii (Roughly) sktch (by hand) a phas portrait ( [ ) ( [ ) 2, and 2, b ( [ ) ( [ ) 3, and 3, d ( 2, ( 3, [ ) [ ) and and ( [ 2, ( 3, ) ) [ 399 For ach of th following 2 2 systms blow: i Find th gnral solution a c g ii iii [ x y [ x y [ x y [ x y Dscrib th stability of th quilibrium solution x =, and stat whthr th corrsponding critical point is a nod or saddl point Sktch a rough phas portrait, using th a minimal dirction fild to rfin your sktch basd on th matrix s ignpairs [ [ [ [ [ 5 x x 6 7 x = b 3 3 y y = 3 2 y [ [ [ [ [ 3 x x 3 3 x = d 3 y y = 3 3 y [ [ [ 5 3 x x = f 3 5 y y = [ [ 9 2 x y = [ [ [ 6 2 x x h y y = [ [ 3 2 x 9 3 y

29 Additional Exrciss Chaptr & Pag: Lt A b any constant 2 2 matrix with a complt st of ignvctors, all corrsponding to th sam ignvalu r Show that [ r A = r 39 In figurs 39 and 395 many of th trajctoris appar to b similar to ithr parabolas or hyprbolas In this xrcis, w will s that ths trajctoris ar not, in gnral, actually parabolas or hyprbolas by considring th trajctoris of x = Ax whr [ α A = β a Find th gnral solution to x = Ax whr A is as abov and α and β ar ral numbrs b Lt A b as abov with < α < β i Sktch a phas portrait for this systm ii Lt [x(t), y(t) T b a solution to this systm which is nithr an quilibrium solution nor a solution corrsponding to a singl ignpair Show that, for som constant c and vry ral valu t, y(t) = c[x(t) p whr p = β α iii For what choics of α and β ar th trajctoris of this systm parabolas (xcluding th critical point and straight lin trajctoris)? c Lt A b as abov with α < < β i Sktch a phas portrait for this systm ii Lt [x(t), y(t) T b a solution to this systm which is nithr an quilibrium solution nor a solution corrsponding to a singl ignpair Show that, for som constant c and vry ral valu t, y(t) = c[x(t) p whr p = β α iii For what choics of α and β ar th trajctoris of of this systm hyprbolas (xcluding th critical point and straight lin trajctoris)? 392 Lt A b a constant N N matrix Show that dt A = A has a zro ignvalu (Thr ar svral ways to prov this A particularly simpl approach is to us th charactristic quation dt[a ri = ) 393 In th txt, w did not discuss trajctoris whn an ignvalu is zro So: a Suppos (, u) is an ignpair for an N N matrix A Dmonstrat that vry point on th straight lin through th origin and paralll to u is a critical point for x = Ax

30 Chaptr & Pag: 39 3 Constant Matrix Systms, Part I b Suppos (, u ) and (r, u 2 ) ar ignpairs for a 2 2 matrix A, with r bing a nonzro ral numbr i What ar th trajctoris of th nonquilibrium solutions to x = Ax? And what is th ffct of th sign of r? ii What can b said about A and th phas portrait of x = Ax if r = and {u, u 2 } ar linarly indpndnt? c For ach of th following 2 2 systms blow: i Find th gnral solution ii iii Dscrib th stability of th quilibrium solutions Sktch a phas portrait, using th appropriat ignpairs i iii v x = 8x + 2y x y ii = x y = x + y y = x y [ [ [ x 3 x x y = iv = 3x + y 6 2 y y = 6x + 2y [ [ [ x a a x y = whr a is a nonzro ral numbr a a y

31 Additional Exrciss Chaptr & Pag: 39 3

32 Chaptr & Pag: Constant Matrix Systms, Part I Som Answrs to Som of th Exrciss WARNING! Most of th following answrs wr prpard hastily and lat at night Thy hav not bn proprly proofrad! Errors ar likly! a A = [ 2 3 b r = 2, [ r 2 = 5 d x(t) = c 3 [ 2t + c 2 2 5t 2a A = [ b r = 2, r [ 2 = 5 2d x(t) = c 32 [ 2t + c 3 2 5t [ 3 3 3a A = b r = 2, r 2 = 2, r 3 = 3d x(t) = c [ 2t + c 2 [ 2t + c 3 [ 2 { a Eignvalus: 3, 3 ; Eignvctor st: [, T,[, T} ; Eignpairs: ( ( ) 3,[, T) and 3,[, T ; Th matrix has a complt st of ignvctors b Eignpairs: (,[, T) and (,[2, T) ; Th matrix has a complt st of ignvctors c Eignpairs: (,[, T) ; Th matrix dos not hav a complt st of ignvctors d Eignpairs: ( 2i,[, i T) and ( 2i,[, i T) ; Th matrix has a complt st of ignvctors Eignpairs: (,[, 2, T), (,[,, T) and (2,[,, ) ; Th matrix has a complt st of ignvctors f Eignpairs: (,[,, T), (,[,, T) and (9,[,, ) ; Th matrix has a complt st of ignvctors g Eignpairs: (,[,, 8 T) and (,[,, T) ; Th matrix dos not hav a complt st of ignvctors h Eignpairs: (,[,, T), ( i,[, 2 + 2i, i T) and ( i,[, 2 2i, + i T) ; Th matrix has a complt st of ignvctors i Eignpairs: ( 2,[,,, T), (,[,,, T), (,[,,, T) and ( 2,[,,, T) ; Th matrix has a complt st of ignvctors j Eignpairs: (,[,,, T) and (,[,,, T) ; Th matrix dos not hav a complt st of ignvctors [ 5a c [ 3t + c 2 3t [ 5b c [ t + c 2 2 t [ [ 5c c 2 t [ + c 2 t + c 3 2t [ 5d c + c 2 [ + c 3 [ 5 c [ 2t + c 2 [ 5f c [ 3 3t + c 2 [ 7t [ 5g c 2 [ 5 t + c 2 3t [ 5h c [ + c2 2 9t 9t t t + c 3 [ t + c 3 [ t [ 2 3 [ 5i c 2t 2 [ + c 2 2t 32 + c 3 9t [ [ 7 [ 5j c + c t 7 + c 3 [ [ 3 [ 5k c 9 + c 2 2t + c 3 t [ 5 [ 2 [ 3 3 5l c t + c 2 2t + c t [ 3t + c 6 3 t

33 Additional Exrciss Chaptr & Pag: a [ t 2 [ 2 t 6b 5 [ 2 t 2 [ 6c 3[ t [ t + 2 2t [ 6d 5 7t + [ 2t 7a c [ 2x + c 2 [ 2x Unstabl 7b c [ 2x + c 2 [ 2x Asymptotically stabl 7c c [ 2x + c 2 [ x Unstabl 7d c [ 2x + c 2 [ x Asymptotically stabl 7 c [ x + c 2 [ 2x Unstabl

34 Chaptr & Pag: 39 3 Constant Matrix Systms, Part I 7f c [ 3x + c 2 [ 5x Unstabl 7g c [ 5x + c 2 [ 3x Unstabl 7h c [ 3x + c 2 [ 5x Asymptotically stabl Phas Portrait is that of f with th arrows rvrsd 8a c [ 2x + c 2 [ 2x 8b c [ 2x + c 2 [ 2x 8c c [ 3x + c 2 [ 3x

35 Additional Exrciss Chaptr & Pag: d c [ 3x + c 2 [ 3x 9a c [ 3 2t + c 2 [ 6t An unstabl nod 9b c [ t + c 2 [ 73 9t An unstabl saddl point 9c c [ 3 5t + c 2 [ 3 5t An unstabl saddl point 9d c [ 3 2t + c 2 [ 3 t An asymptotically stabl nod

36 Chaptr & Pag: Constant Matrix Systms, Part I 9 c [ 3 t + c 2 [ 3 t An unstabl saddl point 9f c [ 2 3 t + c 2 [ 3 t An asymptotically stabl nod 9g c [ 2 3 t + c 2 [ 3 t An unstabl nod 9h c [ 3 t + c 2 [ 2 3 t An unstabl saddl point b iii Only if β = 2α c iii Only if α = β 3b i Each is a straight half lin paralll to u 2 with ndpoint at a critical point on th lin through th origin and paralll to u, with th dirction of travl towards th critical point if r < and away from th critical point if < r 3b ii A = [ and vry point in th plan is a critical point 3c i c [ + c2 [ 2 9t Unstabl