MATH 39, WEEK 5: Th Fundamntal Matrix, Non-Homognous Systms of Diffrntial Equations Fundamntal Matrics Considr th problm of dtrmining th particular solution for an nsmbl of initial conditions For instanc, suppos w ar considring th diffrntial quation x + 3y dy 3x y which w know has th gnral solution x(t) C y(t) 4t + C t Suppos, howvr, that instad of dtrmining th particular solution for a singl initial condition x() x, y() y, w wish to dtrmin an nsmbl of particular solutions for a varity conditions As things stand now, w ar rquird to solv for C and C individually ach tim w wish to apply an initial condition This could b a lot of work! Fortunatly, this will not b rquird, but w will hav to mak us of a littl linar algbra in ordr to s how to gt around th problm First of all, w should rcogniz that w can writ th solution in th quivalnt form x(t) 4t t C y(t) 4t t In othr words, w can writ th quation in th matrix form C x(t) Ψ(t)c () whr Ψ(t) is th matrix with th fundamntal solutions x (t) and x (t) along th columns
Our goal is to rlat th undtrmind cofficints c R to th initial conditions x() in gnral This quation tlls us xactly how to do that! W hav x() Ψ()c c Ψ() x() Ψ ()x () whr Ψ() R is th matrix Ψ(t) valuatd at zro and Ψ () is th invrs of this matrix W can now plug () into () to gt x(t) Φ(t)x (3) whr Φ(t) Ψ(t)Ψ () R This is xactly what was wantd! Onc w hav dtrmind th matrix Φ(t), w hav a rlationship which immdiatly givs th solution x(t) for an arbitrary initial vctor x For th xampl, w hav that 4t Ψ(t) t Ψ() 4t t which implis that It follows that Ψ () 4t t Φ(t) Ψ(t)Ψ () 4t t 4t + t 4t + t 4t + t 4t + t and thrfor th solution can b writtn in th form x(t) Φ(t)x as x(t) 4t + t 4t + t x y(t) 4t + t 4t + t As xpctd, th rsult rlats th solutions dirctly to th initial conditions W no longr hav to comput th constants C and C at vry itration! W stop to mak th following nots: Th matrics Ψ(t) and Φ(t) ar calld fundamntal matrics Th matrix Φ(t) will b of particular concrn to us as it has many xcptionally usful proprtis (Mor on this in th nxt sction) y
It follows from (3) that th fundamntal matrix Φ(t) has th proprty Φ() I (Othrwis, w would hav x() Φ()x ) This can b vrifid in th abov xampl sinc w hav Φ() 4() + () 4() + () 4() + () 4() + () 3 Th matrix Φ(t) also hav th proprty that it satisfis th diffrntial quation Φ (t) AΦ(t), sinc ach column of Φ(t) is a solution of x (t) Ax(t) Exampl : Dtrmin th fundamntal matrix Φ(t) R for th systm of diffrntial quations x 4y dy x 3y Solution: W dtrmind last wk that this systm of diffrntial quations has th gnral solution ( ( )) x(t) C y(t) + C t + W now want to dtrmin th fundamntal matrix Φ(t) Ψ(t)Ψ () whr Ψ(t) is th matrix with th fundamntal solutions abov along th columns W will hav to b a littl carful whn stablishing our fundamntal solutions thy ar componnt-wis cofficints of th undtrmind constants C and C In this cas, w hav that Ψ(t) + t It follows that Ψ() t Ψ () Consquntly, w hav Φ(t) Ψ(t)Ψ () + t t + t 4t t t 3
Matrix Exponntials W will now tak a stp backward to tak a stp forward Rconsidr th gnral linar and homognous systm x (t) Ax(t) (4) If w carry through with our arlir analogy with a singl first ordr quation of th form x (t) ax(t), w would rcogniz th gnral form of th solution as x(t) x at So prhaps thr is som sns in which w could writ x(t) At x as th solution for our systm (4) (Not that this is vry clos to th solution x(t) Φ(t)x w just discovrd!) Th qustion thn bcoms: how might w dfin At? It should b clar that w cannot simply tak th xponntial of ach trm in th matrix as this dos not gnrally lad a function x(t) which is a solution of (4) Considr th following two possibl approachs: Rcall that th xponntial at can b xpandd as th Taylor sris at (at) n n! n + at + (at)! + (at)3 3! + This suggsts that, instad of dfining th matrix xponnt At dirctly, w dfin it as th infinit sris At n n! An t n I + At +! A t + 3! A3 t 3 + (5) (Notic hr that A A A, A 3 A A A, tc) W should hav som halthy skpticism about this formula Aftr all, it is an infinit sum, which w hav no hop of computing xplicitly in our liftims Th bst w might hop for is som rsult rgarding convrgnc But bfor w giv up hop on this intrprtation ntirly, considr th following obsrvation: It is a solution of th systm of diffrntial quation! To chck this, w can simply valuat W clarly hav x() A() x (I + A() +! A () + )x x 4
On th lft-hand sid of th systm, w hav ( d ( At ) d x I + At +! A t + ) 3! A3 t 3 + x ( A + A t + )! A3 t + x and on th right-hand sid w hav A ( At ) x A (I + At +! A t + 3! A3 t 3 + ( A + A t + )! A3 t + x ) x Howvr uncomfortabl this dfinition may sn, w cannot scap th implication that it is maningful! (Whthr it is usful is anothr discussion ntirly) Rcall our arlir discussion, whr w dfind th fundamntal matrix Φ(t) Ψ(t)Ψ () and notd that th function x(t) Φ(t)x was a solution of our diffrntial quation and that it satisfis th initial conditions (sinc Φ() I) In othr words, w rally hav two functions which w ar claiming ar solutions of th initial valu problm x (t) Ax(t), x() x, (6) spcifically, x(t) At x and x(t) Φ(t)x But how can this b? How can an initial valu problm has two solutions? Th answr is that it can t It is a wll-known fact (Thorm 7 of txt) that th initial valu problm (6) always has a uniqu solution W ar inscapably drawn to conclud that At Φ(t) (7) This quation should b surprising! What it will allow us to do is intrprt th fundamntal matrix Φ(t) in two ways: a solution of a linar systm of diffrntial quations (right-hand sid of (7)), or as a matrix xponntial with many of th proprtis and intuitions which com from th standard xponntial function (lft-hand sid of (7)) Exampl: Dtrmin th matrix xponntial At for th matrix 5 A 5
Solution: Without th prcding motivation, w might b a littl lost By dfinition, w hav At I + At +! A t + 3! A3 t 3 + which, byond a fw trms, w hav no hop of computing xplicitly byond a fw trms in th sris Nvrthlss, this is th dfinition w ar now allowd to tak th xponntial of A componnt-wis as this will produc an incorrct rsult What w raliz givn th prvious discussion is that At Φ(t) whr Φ(t) is th particular fundamntal solution of th diffrntial quation Ax with th proprty Φ() I In othr words, all w nd to do is find Φ(t) for th first-ordr systm x + 5y dy x + y W found last wk that this systm had th gnral solution ( ) 3 x(t) C cos(3t) sin(3t) ( ) 3 + C sin(3t) + cos(3t) Th fundamntal matrix Ψ(t) is thrfor givn by cos(3t) + 3 sin(3t) 3 cos(3t) + sin(3t) Ψ(t) cos(3t) sin(3t) so that It follows that Ψ() 3 Ψ() 6 3 cos(3t) + 3 sin(3t) 3 cos(3t) + sin(3t) 3 At Φ(t) 6 cos(3t) sin(3t) 6 cos(3t) sin(3t) sin(3t) 6 4 sin(3t) 6 cos(3t) + sin(3t) cos(3t) 3 sin(3t) 5 3 sin(3t) 3 sin(3t) cos(3t) + 3 sin(3t) 6
Notic that, as xpctd, w hav A() Φ() I W hav accomplishd our task, without having to vr dirctly considr proprtis of th matrix xponntial At! 3 Non-Homognous Linar Systms of Diffrntial Equations with Constant Cofficints Now considr bing askd to solv th diffrntial quation x dy + x y by using th matrix algbra mthods w hav bn mploying ovr th past two wks W would notic quickly that w could rwrit th lft-hand sid as a vctor drivativ x (t) (x (t), y (t)) and collct most of th right-hand sid as Ax x(t) y(t) But what do w do with th xtra trms in th quation (i th s)? Thy do not fit into ithr of ths forms! Th answr is that th bst w can do is writn thm as a sparat vctor, so that w hav x (t) y (t) x(t) y(t) + In gnral, w may considr diffrntial quations of th form Ax + g(t) (8) Such systms ar calld non-homognous for th sam rason as w calld scond-ordr systm homognous or non-homognous thr ar trms which affct th systm which ar indpndnt of solutions and thir drivativs (i x, y, x, and y) How might w go about solving such a systm? Lt s considr on of th first intgration tchniqus w larnd for firstordr diffrntial quations If w had a gnral first-ordr linar diffrntial quation ax(t) + g(t) 7
w would r-arrang th xprssion as and thn obtain th intgrating factor so that w could writ ax(t) g(t) µ(t) a at d at x(t) at g(t) Dpnding on whthr w wr intrstd in th gnral solution or a particular solution (i if w hav initial conditions), w could ithr comput th indfinit intgral to gt x(t) C at + at as g(s) ds or intgrat xplicitly from s t to s t to obtain x(t) x at + at t t as g(s)ds whr x(t ) x Thr is nothing stopping us from attmpting this tchniqu with our linar systm (9)! W may writ th xprssion as Ax(t) g(t) and thn obtain th intgrating factor µ(t) At (W will suspnd discussion of th matrix xponntial until latr, just pausing now to acknowldg that w hav sn this objct bfor and know how to handl it!) It follows that w hav d At x(t) At g(t) If w ar intrstd in th gnral solution, w can comput th indfinit intgral and rarrang to gt x(t) At c + At As g(s) ds (9) 8
If th initial conditions ar spcifid, w can intgrat from s t to s t to gt x(t) At x + At t t As g(s) ds () If w had ncountrd this solution bfor our prvious discussion of matrix xponntials, w would b vry distrssd! W would lik to maningfully dfin th matrix xponntial At in som closd form, but would only hav th infinit sris dfinition availabl to us W now know, howvr, that At Φ(t) whr Φ(t) is th fundamntal matrix of th corrsponding homognous systm Ax which satisfis th condition Φ() I W can rwrit (9) and () to tak advantag of this simplification Allowing th proprty At At Φ (t) (tru, but byond th scop of this cours!), for problms without initial conditions w hav th formulas x(t) Φ(t)c + Φ(t) Φ (s)g(s) ds () x(t) Ψ(t) c + Ψ(t) Ψ (s)g(s) ds whr c (C, C ) and c ( C, C ) ar vctors of undtrmind constants, whil for problms with initial conditions w hav x(t) Φ(t)x + Φ(t) t t Φ (s)g(s) ds () In othr words, now that w undrstand th matrix xponntial At, w can solv first-ordr linar systms of diffrntial quations by xactly th sam mthod intgrating factor w solvd first-ordr linar diffrntial quations with during th scond wk of th cours! W stop now to mak a fw nots: Th two quations in () ar quivalnt (dtails blow) but it is gnrally asir to us th Ψ(t) drivd from th standard solution from than th modifid Φ(t) That is to say, th scond formula is typically asir to solv It will b vry important to rmmbr whn to intgrat th dfinit intgral from s t (usually t ) to s t whn initial conditions ar spcifid! Othrwis th formula will fail 9
3 (Tchnical dtails ahad!) Th quivalnc of th two formulas in () and () can b sn by noting that Φ(t) Ψ(t)Ψ () and Φ (t) Ψ(t)Ψ () Ψ()Ψ (t) (proprtis of matrics, but byond th scop of th cours) W consquntly hav Φ(t) Φ (s)g(s) Ψ(t)Ψ () Ψ()Ψ (s)g(s) ds Ψ(t) Ψ (s)g(s) ds sinc Ψ ()Ψ() I W hav th othr quivalnc of Φ(t)c Ψ(t)Ψ ()c Ψ(t) c whr c Ψ ()c Th ky thing to not is that, sinc th vctor of constants c is arbitrary, c is just as arbitrary Exampl: Lt s s how this applis to our prvious xampl Sinc w hav initial conditions, w will want to find th fundamntal matrix Φ(t) W thrfor solv th linar systm of diffrntial quations x(t) Ax(t) with th matrix A W can quickly comput that th charactristic polynomial is ( λ)( λ) so that λ and λ ar th ignvalus For λ w hav th ignvctor v (, ) and for λ w hav th ignvctor v (, ) It follows that th gnral solution is x(t) C + C This givs th fundamntal matrix Ψ(t) In ordr to comput Φ(t), w valuat Ψ() Ψ () so that Φ(t) Ψ(t)Ψ ()
To find th gnral solution x(t) by th formula () w nd to comput th matrix Φ (t) First of all, w hav It follows that dt(φ(t)) ( )() 3t Φ (t) 3t + t t t t W now valuat th intgral Sinc t, w hav t t Φ (s)g(s) ds t t t t t s s ds t t ds W can now comput t Φ(t) Φ (s)g(s) ds t t It follows that th gnral solution is t x(t) Φ(t)x + Φ(t) Φ (s)g(s) ds x + y In particular, if w us th initial conditions x() and y() thn w obtain th solution x(t) y(t) + as w had bfor Dspit th amount of work w hav don, w should b ncouragd that non of th individual stps wr particularly challnging, and that th rsult obtaind is consistnt with our arlir work!
Not: On thing worth noting is that w computd Φ(t) and Φ (t) t t t t This should b striking th scond matrix is simply th first with th argumnt instad of t That is to say, w hav Φ (t) Φ() W might wondr if this is th gnral cas Th answr in gnral is a dfinitiv NO For a gnral fundamntal matrix Ψ(t) it is not tru that Ψ (t) Ψ() That said, it is tru in th vry spcific cass w will b considr that of th fundamntal matrix Φ(t) This is du, in analogy, to th idntity Φ(t) At and th proprty that Φ() At At Φ(t) In othr words, it is bcaus w can rlat Φ(t) to a matrix xponntial that w ar justifid in bliving Φ(t) Φ() Exampl : Find th gnral solution of th systm of diffrntial quations 4x 3y + dy 3x + y + Solution: Sinc w ar looking for a gnral solution, any fundamntal solution Ψ(t) will work W first solv th homognous quation x (t) Ax(t) whr A 4 3 3 W hav th charactristic polynomial ( 4 λ)( λ) + 9 λ + λ + (λ + ) It follows that λ So w hav a rpatd ignvalu W comput th corrsponding ignvctor quickly by solving (A + I)v This simplifis to so that v + v Stting v, w hav v (, ) Sinc w hav only obtaind a singl ignvctor, w must solv th quation (A + I)w v for w This givs 3 3 3 3 3 It follows that w hav w + w 3 Stting w w hav w 3 so w hav w ( 3, ) W thrfor hav th gnral solution to th homognous
systm x(t) ( C ( + C t + 3 )) It follows that th fundamntal matrix Ψ(t) is givn by Ψ(t) 3 t t so that Ψ (t) 3 t 3 + t 3t t t 3t t 3 t 3 t It follows that w hav Ψ 3s s (s)g(s) s 3s s 3 s 3 s 6s + 6 s s W thrfor hav Ψ(s) 6s + g(s) ds 6 3t ds + t 6t and so Ψ(t) Ψ(s) g(s) ds 3 t t 3t + t 3t + t 3t + t 6t It follows that th gnral solution is givn by x(t) Ψ(t)c + Ψ(t) Ψ (s)g(s) ds x(t) 3 t C 3t y(t) t + + t 3t + t C 3