2.3 Matrix Formulation

Transcription

1 23 Matrix Formulation 43 A mor complicatd xampl ariss for a nonlinar systm of diffrntial quations Considr th following xampl Exampl 23 x y + x( x 2 y 2 y x + y( x 2 y 2 (233 Transforming to polar coordinats, on can show that In ordr to convrt this systm into polar form, w comput r r( r 2, θ This uncoupld systm can b solvd and such nonlinar systms will b studid in th nxt chaptr 23 Matrix Formulation W hav invstigatd svral linar systms in th plan and in th nxt chaptr w will us som of ths idas to invstigat nonlinar systms W nd a dpr insight into th solutions of planar systms So, in this sction w will rcast th first ordr linar systms into matrix form This will lad to a bttr undrstanding of first ordr systms and allow for xtnsions to highr dimnsions and th solution of nonhomognous quations latr in this chaptr W start with th usual homognous systm in Equation (25 Lt th unknowns b rprsntd by th vctor x(t x(t y(t Thn w hav that x ( x y ( ax + by a b x Ax cx + dy c d y Hr w hav introducd th cofficint matrix A This is a first ordr vctor diffrntial quation, x Ax Formrly, w can writ th solution as x x 0 At Th xponntial of a matrix is dfind using th Maclaurin sris xpansion

2 44 2 Systms of Diffrntial Equations W would lik to invstigat th solution of our systm Our invstigations will lad to nw tchniqus for solving linar systms using matrix mthods W bgin by rcalling th solution to th spcific problm (22 W obtaind th solution to this systm as x(t c t + c 2 4t, y(t 3 c t 2 c 2 4t (235 This can b rwrittn using matrix oprations Namly, w first writ th solution in vctor form x(t x y(t c t + c 2 4t 3 c t 2 c 2 4t c t c2 3 c t + 4t 2 c 2 4t c t + c 2 3 4t (236 2 W s that our solution is in th form of a linar combination of vctors of th form x v λt with v a constant vctor and λ a constant numbr This is similar to how w bgan to find solutions to scond ordr constant cofficint quations So, for th gnral problm (23 w insrt this guss Thus, x Ax For this to b tru for all t, w hav that λv λt Av λt (237 Av λv (238 This is an ignvalu problm A is a 2 2 matrix for our problm, but could asily b gnralizd to a systm of n first ordr diffrntial quations W will confin our rmarks for now to planar systms Howvr, w nd to rcall how to solv ignvalu problms and thn s how solutions of ignvalu problms can b usd to obtain solutions to our systms of diffrntial quations So, w dfin x A k0 k0 + x + x2 2! + x3 3! + I + A + A2 2! + A3 3! In gnral, it is difficult computing A unlss A is diagonal + (234

3 24 Eignvalu Problms 25 Solving Constant Cofficint Systms in 2D 45 W sk nontrivial solutions to th ignvalu problm Av λv (239 W not that v 0 is an obvious solution Furthrmor, it dos not lad to anything usful So, it is calld a trivial solution Typically, w ar givn th matrix A and hav to dtrmin th ignvalus, λ, and th associatd ignvctors, v, satisfying th abov ignvalu problm Latr in th cours w will xplor othr typs of ignvalu problms v For now w bgin to solv th ignvalu problm for v Insrting v 2 this into Equation (239, w obtain th homognous algbraic systm (a λv + bv 2 0, cv + (d λv 2 0 (240 Th solution of such a systm would b uniqu if th dtrminant of th systm is not zro Howvr, this would giv th trivial solution v 0, v 2 0 To gt a nontrivial solution, w nd to forc th dtrminant to b zro This yilds th ignvalu quation 0 a λ b c d λ (a λ(d λ bc This is a quadratic quation for th ignvalus that would lad to nontrivial solutions If w xpand th right sid of th quation, w find that λ 2 (a + dλ + ad bc 0 This is th sam quation as th charactristic quation (28 for th gnral constant cofficint diffrntial quation considrd in th first chaptr Thus, th ignvalus corrspond to th solutions of th charactristic polynomial for th systm Onc w find th ignvalus, thn thr ar possibly an infinit numbr solutions to th algbraic systm W will s this in th xampls So, th procss is to a Writ th cofficint matrix; b Find th ignvalus from th quation dt(a λi 0; and, c Find th ignvctors by solving th linar systm (A λiv 0 for ach λ 25 Solving Constant Cofficint Systms in 2D Bfor procding to xampls, w first indicat th typs of solutions that could rsult from th solution of a homognous, constant cofficint systm of first ordr diffrntial quations

4 46 2 Systms of Diffrntial Equations W bgin with th linar systm of diffrntial quations in matrix form dx a b dt x Ax (24 c d Th typ of bhavior dpnds upon th ignvalus of matrix A Th procdur is to dtrmin th ignvalus and ignvctors and us thm to construct th gnral solution If w hav an initial condition, x(t 0 x 0, w can dtrmin th two arbitrary constants in th gnral solution in ordr to obtain th particular solution Thus, if x (t and x 2 (t ar two linarly indpndnt solutions 2, thn th gnral solution is givn as x(t c x (t + c 2 x 2 (t Thn, stting t 0, w gt two linar quations for c and c 2 : c x (0 + c 2 x 2 (0 x 0 Th major work is in finding th linarly indpndnt solutions This dpnds upon th diffrnt typs of ignvalus that on obtains from solving th ignvalu quation, dt(a λi 0 Th natur of ths roots indicat th form of th gnral solution On th nxt pag w summariz th classification of solutions in trms of th ignvalus of th cofficint matrix W first mak som gnral rmarks about th plausibility of ths solutions and thn provid xampls in th following sction to clarify th matrix mthods for our two dimnsional systms Th construction of th gnral solution in Cas I is straight forward Howvr, th othr two cass nd a littl xplanation 2 Rcall that linar indpndnc mans c x (t+c 2x 2(t 0 if and only if c, c 2 0 Th radr should driv th condition on th x i for linar indpndnc

5 25 Solving Constant Cofficint Systms in 2D 47 Classification of th Solutions for Two Linar First Ordr Diffrntial Equations Cas I: Two ral, distinct roots Solv th ignvalu problm Av λv for ach ignvalu obtaining two ignvctors v,v 2 Thn writ th gnral solution as a linar combination x(t c λt v + c 2 λ2t v 2 2 Cas II: On Rpatd Root Solv th ignvalu problm Av λv for on ignvalu λ, obtaining th first ignvctor v On thn nds a scond linarly indpndnt solution This is obtaind by solving th nonhomognous problm Av 2 λv 2 v for v 2 Th gnral solution is thn givn by x(t c λt v + c 2 λt (v 2 + tv 3 Cas III: Two complx conjugat roots Solv th ignvalu problm Ax λx for on ignvalu, λ α + iβ, obtaining on ignvctor v Not that this ignvctor may hav complx ntris Thus, on can writ th vctor y(t λt v αt (cosβt + i sin βtv Now, construct two linarly indpndnt solutions to th problm using th ral and imaginary parts of y(t : y (t R(y(t and y 2 (t Im(y(t Thn th gnral solution can b writtn as x(t c y (t + c 2 y 2 (t Lt s considr Cas III Not that sinc th original systm of quations dos not hav any i s, thn w would xpct ral solutions So, w look at th ral and imaginary parts of th complx solution W hav that th complx solution satisfis th quation d [R(y(t + iim(y(t] A[R(y(t + iim(y(t] dt Diffrntiating th sum and splitting th ral and imaginary parts of th quation, givs d dt R(y(t + i d Im(y(t A[R(y(t] + ia[im(y(t] dt Stting th ral and imaginary parts qual, w hav d R(y(t A[R(y(t], dt and d Im(y(t A[Im(y(t] dt Thrfor, th ral and imaginary parts ach ar linarly indpndnt solutions of th systm and th gnral solution can b writtn as a linar combination of ths xprssions

6 48 2 Systms of Diffrntial Equations W now turn to Cas II Writing th systm of first ordr quations as a scond ordr quation for x(t with th sol solution of th charactristic quation, λ 2 (a + d, w hav that th gnral solution taks th form x(t (c + c 2 t λt This suggsts that th scond linarly indpndnt solution involvs a trm of th form vt λt It turns out that th guss that works is x t λt v + λt v 2 Insrting this guss into th systm x Ax yilds (t λt v + λt v 2 A [ t λt v + λt v 2 ] λt v + λt λt v + λ λt v 2 λt λt v + λt Av 2 Noting this is tru for all t, w find that Thrfor, λt (v + λv 2 λt Av 2 (242 v + λv 2 Av 2 (243 (A λiv 2 v W know vrything xcpt for v 2 So, w just solv for it and obtain th scond linarly indpndnt solution 26 Exampls of th Matrix Mthod Hr w will giv som xampls for typical systms for th thr cass mntiond in th last sction 4 2 Exampl 24 A 3 3 Eignvalus: W first dtrmin th ignvalus 0 4 λ λ (244 Thrfor, 0 (4 λ(3 λ 6 0 λ 2 7λ (λ (λ 6 (245 Th ignvalus ar thn λ, 6 This is an xampl of Cas I Eignvctors: Nxt w dtrmin th ignvctors associatd with ach of ths ignvalus W hav to solv th systm Av λv in ach cas

7 26 Exampls of th Matrix Mthod 49 Cas λ 4 2 v v 3 3 v 2 v 2 ( 3 2 v v 2 0 (246 (247 This givs 3v + 2v 2 0 On possibl solution yilds an ignvctor of ( v 2 v 2 3 Cas λ v v v 2 v 2 ( 2 2 v v 2 0 (248 (249 For this cas w nd to solv 2v + 2v 2 0 This yilds ( v v 2 Gnral Solution: W can now construct th gnral solution x(t c λt v + c 2 λ2t v 2 2 c t + c 3 2 6t 2c t + c 2 6t 3c t + c 2 6t ( Exampl 25 A Eignvalus: Again, on solvs th ignvalu quation 0 3 λ 5 λ (25 Thrfor, 0 (3 λ( λ λ 2 2λ + 2 λ ( 2 ± 4 4((2 2 ± i (252

8 50 2 Systms of Diffrntial Equations Th ignvalus ar thn λ + i, i This is an xampl of Cas III Eignvctors: In ordr to find th gnral solution, w nd only find th ignvctor associatd with + i 3 5 v v ( + i ( 2 i 5 2 i v 2 v v 2 v 2 ( 0 0 (253 W nd to solv (2 iv 5v 2 0 Thus, v 2 + i (254 v 2 Complx Solution: In ordr to gt th two ral linarly indpndnt solutions, w nd to comput th ral and imaginary parts of v λt λt 2 + i (+it 2 + i 2 + i t (cost + i sint t ( (2 + i(cost + i sint cost + i sint (2 cost sin t + i(cost + 2 sint t cost + i sint ( t 2 cost sin t + i cost t cost + 2 sint sint Gnral Solution: Now w can construct th gnral solution 2 cost sint x(t c t cost cost + 2 sint + c 2 t sint t ( c (2 cost sin t + c 2 (cos t + 2 sint c cost + c 2 sin t Not: This can b rwrittn as ( x(t t 2c + c cost 2 Exampl 26 A Eignvalus: 7 9 c + t sin t 2c2 c c 2 ( λ 9 λ (256

9 26 Exampls of th Matrix Mthod 5 Thrfor, 0 (7 λ( λ λ 2 8λ (λ 4 2 (257 Thr is only on ral ignvalu, λ 4 This is an xampl of Cas II Eignvctors: In this cas w first solv for v and thn gt th scond linarly indpndnt vctor ( ( 7 v v 4 9 ( v 2 v v 2 v 2 ( 0 (258 0 Thrfor, w hav 3v v 2 0, ( v v 2 3 Scond Linarly Indpndnt Solution: Now w nd to solv Av 2 λv 2 v ( 7 u 9 u 2 ( u 4 u 2 ( u ( 3 u 2 ( 3 (259 Expanding th matrix product, w obtain th systm of quations 3u u 2 9u 3u 2 3 (260 ( u Th solution of this systm is u 2 2 Gnral Solution: W construct th gnral solution as y(t c λt v + c 2 λt (v 2 + tv [( c 4t + c 3 2 4t 2 4t ( c + c 2 ( + t 3c + c 2 (2 + 3t + t ( 3] (26

10 52 2 Systms of Diffrntial Equations 26 Planar Systms - Summary Th radr should hav notd by now that thr is a connction btwn th bhavior of th solutions obtaind in Sction 22 and th ignvalus found from th cofficint matrics in th prvious xampls Hr w summariz som of ths cass Typ Figur Eignvalus Stability Nod Ral λ, sam signs λ > 0, stabl Saddl Ral λ opposit signs Mostly Unstabl Cntr λ pur imaginary Focus/Spiral Complx λ, R(λ 0 R(λ > 0, stabl Dgnrat Nod Rpatd roots, λ > 0, stabl Lin of Equilibria On zro ignvalu λ > 0, stabl Tabl 2 List of typical bhaviors in planar systms Th connction, as w hav sn, is that th charactristic quation for th associatd scond ordr diffrntial quation is th sam as th ignvalu quation of th cofficint matrix for th linar systm Howvr, on should b a littl carful in cass in which th cofficint matrix in not diagonalizabl In Tabl 22 ar thr xampls of systms with rpatd roots Th radr should look at ths systms and look at th commonalitis and diffrncs in ths systms and thir solutions In ths cass on has unstabl nods, though thy ar dgnrat in that thr is only on accssibl ignvctor 27 Thory of Homognous Constant Cofficint Systms Thr is a gnral thory for solving homognous, constant cofficint systms of first ordr diffrntial quations W bgin by onc again rcalling th spcific problm (22 W obtaind th solution to this systm as x(t c t + c 2 4t, y(t 3 c t 2 c 2 4t (262

11 27 Thory of Homognous Constant Cofficint Systms 53 Systm Systm 2 Systm 3 a 2, b 0, c 0, d 2 3 a 0, b, c -4, d 4 3 a 2, b, c 0, d 2 3 y 2 y 2 y 2 K3 K2 K x K K2 K3 K2 K x K K2 K3 K2 K x K K2 K3 2 0 x x x 0 2 K3 ( x x x 0 2 Tabl 22 Thr xampls of systms with a rpatd root of λ 2 K3 This tim w rwrit th solution as c x t + c 2 4t 3 c t 2 c 2 4t t 4t c c 2 3 t 2 4t Φ(tC (263 Thus, w can writ th gnral solution as a 2 2 matrix Φ tims an arbitrary constant vctor Th matrix Φ consists of two columns that ar linarly indpndnt solutions of th original systm This matrix is an xampl of what w will dfin as th Fundamntal Matrix of solutions of th systm So, dtrmining th Fundamntal Matrix will allow us to find th gnral solution of th systm upon multiplication by a constant matrix In fact, w will s that it will also lad to a simpl rprsntation of th solution of th initial valu problm for our systm W will outlin th gnral thory Considr th homognous, constant cofficint systm of first ordr diffrntial quations dx a x + a 2 x a n x n, dt dx 2 a 2 x + a 22 x a 2n x n, dt dx n dt a n x + a n2 x a nn x n (264 As w hav sn, this can b writtn in th matrix form x Ax, whr

12 54 2 Systms of Diffrntial Equations x x 2 x x n and a a 2 a n a 2 a 22 a 2n A a n a n2 a nn Now, considr m vctor solutions of this systm: φ (t, φ 2 (t, φ m (t Ths solutions ar said to b linarly indpndnt on som domain if c φ (t + c 2 φ 2 (t + + c m φ m (t 0 for all t in th domain implis that c c 2 c m 0 Lt φ (t, φ 2 (t, φ n (t b a st of n linarly indpndnt st of solutions of our systm, calld a fundamntal st of solutions W construct a matrix from ths solutions using ths solutions as th column of that matrix W dfin this matrix to b th fundamntal matrix solution This matrix taks th form φ φ 2 φ n Φ φ 2 φ 22 φ 2n φ φ n φ n φ n2 φ nn What do w man by a matrix solution? W hav assumd that ach φ k is a solution of our systm Thrfor, w hav that φ k Aφ k, for k,,n W say that Φ is a matrix solution bcaus w can show that Φ also satisfis th matrix formulation of th systm of diffrntial quations W can show this using th proprtis of matrics d dt Φ ( φ φ n Aφ Aφ n A ( φ φ n AΦ (265 Givn a st of vctor solutions of th systm, whn ar thy linarly indpndnt? W considr a matrix solution Ω(t of th systm in which w hav n vctor solutions Thn, w dfin th Wronskian of Ω(t to b W dtω(t If W(t 0, thn Ω(t is a fundamntal matrix solution

13 27 Thory of Homognous Constant Cofficint Systms 55 Bfor continuing, w list th fundamntal matrix solutions for th st of xampls in th last sction (Rfr to th solutions from thos xampls Furthrmor, not that th fundamntal matrix solutions ar not uniqu as on can multiply any column by a nonzro constant and still hav a fundamntal matrix solution Exampl 24 A ( 2 t Φ(t 6t 3 t 6t W should not in this cas that th Wronskian is found as W dtφ(t 2t 6t 3 t 6t 5 7t 0 ( Exampl 25 A Φ(t t (2 cost sin t t (cost + 2 sint t cost t sint 7 Exampl 26 A 9 Φ(t ( 4t 4t ( + t 3 4t 4t (2 + 3t So far w hav only dtrmind th gnral solution This is don by th following stps: Procdur for Dtrmining th Gnral Solution Solv th ignvalu problm (A λiv 0 2 Construct vctor solutions from v λt Th mthod dpnds if on has ral or complx conjugat ignvalus 3 Form th fundamntal solution matrix Φ(t from th vctor solution 4 Th gnral solution is givn by x(t Φ(tC for C an arbitrary constant vctor W ar now rady to solv th initial valu problm: x Ax, x(t 0 x 0 Starting with th gnral solution, w hav that

14 56 2 Systms of Diffrntial Equations x 0 x(t 0 Φ(t 0 C As usual, w nd to solv for th c k s Using matrix mthods, this is now asy Sinc th Wronskian is not zro, thn w can invrt Φ at any valu of t So, w hav C Φ (t 0 x 0 Putting C back into th gnral solution, w obtain th solution to th initial valu problm: x(t Φ(tΦ (t 0 x 0 You can asily vrify that this is a solution of th systm and satisfis th initial condition at t t 0 Th matrix combination Φ(tΦ (t 0 is usful So, w will dfin th rsulting product to b th principal matrix solution, dnoting it by Ψ(t Φ(tΦ (t 0 Thus, th solution of th initial valu problm is x(t Ψ(tx 0 Furthrmor, w not that Ψ(t is a solution to th matrix initial valu problm whr I is th n n idntity matrix x Ax, x(t 0 I, Matrix Solution of th Homognous Problm In summary, th matrix solution of is givn by dx dt Ax, x(t 0 x 0 x(t Ψ(tx 0 Φ(tΦ (t 0 x 0, whr Φ(t is th fundamntal matrix solution and Ψ(t is th principal matrix solution Exampl 27 Lt s considr th matrix initial valu problm x 5x + 3y y 6x 4y, (267 satisfying x(0, y(0 2 Find th solution of this problm W first not that th cofficint matrix is 5 3 A 6 4 Th ignvalu quation is asily found from

15 0 (5 λ(4 + λ + 8 λ 2 λ 2 28 Nonhomognous Systms 57 (λ 2(λ + (268 So, th ignvalus ar λ, 2 Th corrsponding ignvctors ar found to b ( v, v 2 2 Now w construct th fundamntal matrix solution Th columns ar obtaind using th ignvctors and th xponntials, λt : ( ( φ (t 2 t, φ (t 2t So, th fundamntal matrix solution is t Φ(t 2t 2 t 2t Th gnral solution to our problm is thn t x(t 2t 2 t 2t C for C is an arbitrary constant vctor In ordr to find th particular solution of th initial valu problm, w nd th principal matrix solution W first valuat Φ(0, thn w invrt it: Φ(0 ( 2 Φ (0 ( 2 Th particular solution is thn ( t x(t 2t 2 t 2t 2 2 t 2t 3 2 t 2t 4 3 t + 4 2t 6 t 4 2t (269 Thus, x(t 3 t + 4 2t and y(t 6 t 4 2t 28 Nonhomognous Systms Bfor laving th thory of systms of linar, constant cofficint systms, w will discuss nonhomognous systms W would lik to solv systms of th form

16 58 2 Systms of Diffrntial Equations x A(tx + f(t (270 W will assum that w hav found th fundamntal matrix solution of th homognous quation Furthrmor, w will assum that A(t and f(t ar continuous on som common domain As with scond ordr quations, w can look for solutions that ar a sum of th gnral solution to th homognous problm plus a particular solution of th nonhomognous problm Namly, w can writ th gnral solution as x(t Φ(tC + x p (t, whr C is an arbitrary constant vctor, Φ(t is th fundamntal matrix solution of x A(tx, and x p A(tx p + f(t Such a rprsntation is asily vrifid W nd to find th particular solution, x p (t W can do this by applying Th Mthod of Variation of Paramtrs for Systms W considr a solution in th form of th solution of th homognous problm, but rplac th constant vctor by unknown paramtr functions Namly, w assum that Diffrntiating, w hav that or x p (t Φ(tc(t x p Φ c + Φc AΦc + Φc, x p Ax p Φc But th lft sid is f So, w hav that, or, sinc Φ is invrtibl (why?, Φc f, c Φ f In principl, this can b intgratd to giv c Thrfor, th particular solution can b writtn as x p (t Φ(t t Φ (sf(sds (27 This is th variation of paramtrs formula Th gnral solution of Equation (270 has bn found as x(t Φ(tC + Φ(t t Φ (sf(sds (272

17 28 Nonhomognous Systms 59 W can us th gnral solution to find th particular solution of an initial valu problm consisting of Equation (270 and th initial condition x(t 0 x 0 This condition is satisfid for a solution of th form t x(t Φ(tC + Φ(t Φ (sf(sds (273 t 0 providd x 0 x(t 0 Φ(t 0 C This can b solvd for C as in th last sction Insrting th solution back into th gnral solution (273, w hav t x(t Φ(tΦ (t 0 x 0 + Φ(t Φ (sf(sds t 0 (274 This solution can b writtn a littl natr in trms of th principal matrix solution, Ψ(t Φ(tΦ (t 0 : t x(t Ψ(tx 0 + Ψ(t Ψ (sf(sds t 0 (275 Finally, on furthr simplification occurs whn A is a constant matrix, which ar th only typs of problms w hav solvd in this chaptr In this cas, w hav that Ψ (t Ψ( t So, computing Ψ (t is rlativly asy Exampl 28 x + x 2 cost, x(0 4, x (0 0 This xampl can b solvd using th Mthod of Undtrmind Cofficints Howvr, w will us th matrix mthod dscribd in this sction First, w writ th problm in matrix form Th systm can b writtn as x y y x + 2 cost (276 Thus, w hav a nonhomognous systm of th form ( x 0 x Ax + f + 0 y ( 0 2 cost Nxt w nd th fundamntal matrix of solutions of th homognous problm W hav that 0 A 0 Th ignvalus of this( matrix ar λ ±i An ignvctor associatd with λ i is asily found as This lads to a complx solution i

18 60 2 Systms of Diffrntial Equations it i ( cost + i sint i cost sin t From this solution w can construct th fundamntal solution matrix cost sin t Φ(t sint cost So, th gnral solution to th homognous problm is c cost + c x h Φ(tC 2 sint c sin t + c 2 cost Nxt w sk a particular solution to th nonhomognous problm From Equation (273 w s that w nd Φ (sf(s Thus, w hav Φ coss sins 0 (sf(s sin s coss 2 coss 2 sinscoss 2 cos 2 (277 s W now comput t cost sin t t 2 sinscoss Φ(t Φ (sf(sds t 0 sint cost t 0 2 cos 2 ds s ( cost sin t sin 2 t sint cost t + 2 sin(2t t sin t (278 sin t + t cost thrfor, th gnral solution is ( c cost + c x 2 sin t c sin t + c 2 cost ( + Th solution to th initial valu problm is ( cost sin t 4 x + sin t cost( 0 or x 4 cost + t sin t 3 sint + t cost t sint sint + t cost t sint sin t + t cost, 29 Applications In this sction w will dscrib svral applications lading to systms of diffrntial quations In kping with common practic in aras lik physics, w will dnot diffrntiation with rspct to tim as