Non-Homogeneous Systems, Euler s Method, and Exponential Matrix

Transcription

1 Non-Homognous Systms, Eulr s Mthod, and Exponntial Matrix W carry on nonhomognous first-ordr linar systm of diffrntial quations. W will show how Eulr s mthod gnralizs to systms, giving us a numrical approach to solving systms. W thn continu on th xponntial matrix.. Variation of Paramtrs 2. Eulr s Mthod 3. Exponntial Matrix 4. Diagonalization Non-Homognous Systms A nonhomognous first-ordr linar systm of diffrntial quations has th form y = P(ty + g(t, whr g(t is a nonzro matrix. P(t is a squar matrix, but th othr symbols y, y, and g(t dnot column matrics. (This is in kping with th notation introducd at th bginning of th chaptr that column matrics would appar in bold lowrcas lttrs, but squar matrics would appar in ITALIC CAPITAL lttrs. In our study of nonhomognous quations from prvious chaptrs, w notd that th solution to a nonhomognous quation took th form y = y C + y P, whr y C rprsntd th complmntary solution th gnral solution to th rlatd homognous quation, and y P rprsntd a particular solution to th nonhomognous part of th quation.

2 Exampl: Considr th systm of diffrntial quations: 5 3t + y = y + t 3 2t t 2 Show that y = 3t 5 t t 3 + t is a solution to th systm. Find th gnral solution to th abov systm: 2 Solving Nonhomognous Equations In sctions 3.8 and 3., w discussd two tchniqus for solving nonhomognous highr-ordrd quations: th Mthod of Undtrmind Cofficints, and Variation of Paramtrs. Both tchniqus xtnd to nonhomognous systms, although th Mthod of Undtrmind Cofficints is a much wakr mthod of solution in th cas of homognous systms. (Exampl, p. 278, and th discussion immdiatly following this xampl point out th waknsss in th mthod. Thrfor, w ll focus our attntion on th mthod of Variation of Paramtrs. 2

3 2. Varation of Paramtrs W know that th complmntary solution has th form y C = Ψ(t c, whr Ψ(t dnots th solution matrix a squar matrix whos columns form a fundamntal st and c dnots a column of constants In th Variation of Paramtrs mthod, w rplac th constants c,..., c n, with functions u,...,u n, and solv th rsulting quation. Whn w usd this mthod in sction $3., w discovrd that it was asir to solv first for thir drivativs, and thn intgrat; this continus to b tru. Howvr, solving first for th drivativs is complicatd by th fact that w ll hav to find th invrs of th squar matrix Ψ(t in th procss. c. c n. Exampl 3 Using variation of paramtrs, solv th initial valu problm: Solution: y = 2 2 2t y + 2t, y( = First, w find th complmntary solution to th homognous quation:. Find th ignvalus: λ = and λ = Find an ignvctor for ach ignvalu. Whn λ =, th ignvctor is, and whn λ = 3, th ignvctor is. 3. Writ th complmntary solution: y C (t = t t 3t 3t c c 2 To find th particular solution to th nonhomognous quation:. Comput th invrs of th solution matrix: 3

4 t 3t t 3t 2. Multiply th invrs of th solution matrix by th matrix g(t: 2 t 2 t 2t = 2t 2 3t 2 3t 3. Find th antidrivativ of Stp 2: (W usd intgration by parts hr. 2 3t + t t 2 t t 3t dt = 3t 6 t + t t t 2 + 3t + t 3t 3 4. Multiply th complmntary solution matrix Ψ(t by th rsult of stp 3: 3t y P (t = t 3t t 3t 6 t + t t = 2t 3 + 4t 3 8 t 2 + 3t + t 3t 22t 3 3 2t Combin th complmntary solution with th particular solution: y(t = y C (t + y P (t = t 3t c 2t t 3t t 3 8 c 2 22t 3 2t Us th initial condition to solv for c and c 2 : 4

5 At tim t =, y( = c c = Solving this systm of quations, w gt c = 5 and c 6 2 = 7, so: 8 5 y(t = t 3t 6 2t t 3t t t 8 3 2t 3 + 5

6 Exampl 3: A shortcut If w did not hav an initial condition, stps 5 abov would find th gnral solution. Howvr, w can us th initial condition as a limit of intgration in Stp 3 to sav us som work: 3. Intgrat th product ovr th intrval t, t. Notic that sinc t is on of our ndpoints, w ll hav to us a diffrnt variabl of intgration: t 2 3s + s s 3t ds = 6 t + t t s s 3s t 2 + 3t + t 3t Multiply th invrs of th solution matrix Ψ(t (w r valuating th invrs at t by th initial condition. Add this rsult to Stp = 2 2 3t 6 t + t t t 2 + 3t + t 3t = 3t 6 t + t t t 2 + 3t + t 3t Multiply th complmntary solution matrix Ψ(t by th rsult of stp 4. This complts th procss w don t hav to solv for c and c 2. 3t y(t = t 3t t 3t 6 t + t t t 2 + 3t + t 3t + 7 = 8 3t 2t t + 4t t 2 3 2t 5 6 t 2t 3 + (Chck that this is th sam solution w obtaind on th prvious pag. 6

7 p. 286, #7 Solv th initial valu problm: y = y + y( = 3 Eulr s Mthod Rcall that Eulr s mthod givs an approximat solution to an initial valu problm givn in th form y = f(t, y, y(t = y. W choos a stp siz h, and w will approximat th solution y at a sris of qually spacd points t = t + h, t 2 = t + 2h,..., t n = t + nh. W will us y i to dnot th approximation at t i, so y(t i y i. W gt from th approximation at t i to th approximation at t i+ using th formula y i+ = y i + hf(t i, y i which uss th linarization of y to approximat th chang in going from t i to t i+. Exampl: Suppos w wish to approximat th solution to th first ordr non-linar quation y = cos(y + t, y( =, using a stp siz h =.. W know that thr xists a uniqu solution for all t, sinc both f and f y ar continuous vrywhr. Thus w start with our initial condition, y( =. Thn w can approximat y(t = y(. y = y( + hf(, W can thn us this to approximat y(.2: = +.(cos( +.46 y(.2 y 2 =.46 + (.(cos( W can continu for as many stps as w nd, approximating y(t i+ using t i and y(t i. Eulr s mthod can b gnralizd asily to systms of first ordr quations. Eulr s mthod uss th linarization for y at ach point t i to prdict th rsult at t i+. If 7

8 w now considr a function x (which is a vctor valud function, w can still us a linarization at th point t i to prdict th valu of x at t i+. W will still us th sam formula, but it will now b basd on th vctor valud function x: x(t i+ x(t i + hx (t i Exampl: Suppos w wish to solv th following systm: x = 3 2 x x 2 x 2 = 3 + x 5 x 2 with initial conditions x ( = 25 and x 2 ( = 5. (Although w will larn how to solv homognous systms shortly, w will not b studying how to find xact solutions to non-homognous systms such as this. W will approximat th solution numrically using Eulr s mthod. In this cas, our initial condition for our vctor valud function x is ( 25 x( =, 5 and w know that x = ( 3 2 x x x 5 x 2 Thrfor, if w choos h =., w gt th following approximations: ( ( 3 25 x(. + ( ( = ( ( x(.2 + ( ( If w continu this procss, w gt th following tabl of approximations for th valus of x and x 2 : t x x

9 W can of cours us this tchniqu with a non-linar systm as wll: Exampl: Suppos w hav th initial valu problm givn by th systm x = x x 2 + t x 2 = 2x x 2 with initial conditions x ( =, x 2 ( =. Lt s us Eulr s mthod with stp siz h =.2. W will dnot th approximation of x i at stp j by x ij. So our initial conditions ar x = and x 2 =, and x will b th approximation of x (t at t =.2. Thus, for our first stp w gt th following: and x = x + (.2( x x 2 + t = + (.2( + =.2 x 2 = x 2 + (.2(2x x 2 = + (.2(2 =.2 Not that w ndd both x and x 2 to calculat th approximation at th nxt stp for ithr function. W can thn procd to x 2 x (.4 and x 22 x 2 (.4: whil x 2 = x + (.2( x x 2 + t =.2 + (.2( x 22 = x 2 + (.2(2x x 2 =.2 + (.2 (2(.2(.2.46 W can continu as w nd to, using our prvious approximations for x and x 2 to gnrat nw approximations. W can now of cours also solv highr ordr diffrntial quations using Eulr s mthod by first rwriting th quation as a systm of first ordr quations. Exampl: Approximat th solution to y + y = y + t with initial conditions y( = and y ( =, using Eulr s mthod with a stp siz of h =.25. W bgin by rwriting th quation as a systm. W lt u = y, and u 2 = y. Thn w hav first th quation u = u 2

10 and scond th quation u 2 + u 2 = u + t or u 2 = u u 2 + t In matrix form our quation is ( u = u 2 ( ( u u 2 + ( t Our initial conditions ar thn u ( = and u 2 ( =, or ( u =. Eulr s mthod bcoms u k+ = u k +.25 So w st out to find u and u 2 : ( ( u k + tk u = ( ( +.25 ( + ( = ( ( +.25 = ( u 2 = ( ( +.25 ( (.25 = ( ( = ( u = u 2 = (initial conditions W tak th first stp: u = u + (.25(u 2 = +.25 =.25 u 2 = u 2 + (.25(u u 2 + t = + (.25( + =.75 W thn tak a scond stp: u 2 = u + (.25(u 2 =.25 + (.25(.75 =.4375 u 22 = u 2 + (.25(u u 2 + t =.75 + (.25( =.6875

11 Th actual solution to y + y = y + t with initial conditions y( = and y ( = can b found by finding th complmntary solution y c (t = c xp ( + 5t/2 + c 2 xp ( 5t/2, and thn using th mthod of undtrmind cofficints to find th non-homognous trm. Th solution is 5 2 xp ( 5 t xp ( + 5 t t 2 Evaluatd at t =.5, w gt.42863, so our approximation of u 2 =.4375 is off by about.88. W also not that w happn to hav an approximation u 22 =.6875 for y (.5. Th actual valu of y (.5 is about.8738, which is not quit as good. Of cours, thr ar mor sophisticatd tchniqus than Eulr s mthod for stimating solutions (although thy ar basd loosly on Eulr s mthod, but w will not study ths mor complicatd tchniqus. 4 Th Exponntial Matrix Th xponntial function y = at has turnd up ovr and ovr in our study of diffrntial quations. On of th rasons this function has bn ubiquitous is that th function rtains th sam form whn diffrntiating th only thing that changs is th constant cofficint. Is it possibl to dfin a function on matrics that bhavs th sam way? In othr words, is thr som matrix-valud function f(a whos drivativ f (A = Af(A is a multipl of itslf? Th answr is ys, and w ll us th xact sam notation for this function: Lt A b a squar matrix. W dfin th xponntial matrix At to b th matrix whos drivativ is A tims itslf. 4.. What th xponntial matrix is NOT: Lt A = 3 2. Find At. Unfortunatly, th asy and obvious thing to try 3t t 2t

12 dos not work. 3 3t W can chck that it dosn t work by taking th drivativ: t 2 2t and multiplying A tims th matrix: 3t 3 3 3t + 2 t 2 2t. Ths arn t th sam matrix, which mans that w hav to look for anothr mthod of finding th xponntial matrix. 4. Th Maclaurin sris for at Instad, w mak us of th Maclaurin sris for at : at = n= a n t n. n! If w rplac th scalar a with th matrix A, w gt: At = n= t n n! An. In gnral, this summation is difficult to comput by hand, bcaus of th tdiousnss of calculating rpatd powrs of A n, howvr, thr ar som shortcuts that w can us: Exampl A matrix is in diagonal form (or simply a diagonal matrix if vry ntry not on th main diagonal of th matrix is zro. 3 For xampl, is a diagonal matrix. 5 2

13 λ Lt A = λ 2 b a diagonal matrix. Find At. 5 Diagonalization In th last sction, as on of th stps in prforming th Mthod of Variations on a systm, w multiplid by a matrix, and thn latr multiplid by th invrs of that matrix. A similar procss, calld diagonalization can b usd to find th xponntial of a matrix. Exampl 2 Lt A = λ λ. Find At. 3

14 Exrcis 7 Lt A = 2. Find At. If matrix A has a full st of ignvctors x, x 2,, x n, thn A is diagonalizabl: T AT = D, whr λ T = λ 2 x x 2 x n and D =... λ n Not th positions of T and T : th ignmatrix T appars aftr A and T appars bfor. λ t Also not that: At = T Dt T, Dt λ 2t =.... λnt 5. Dcoupling transformation Considr th linar systm y = Ay + g(t. 4

15 If A is diagonalizabl (i.. T AT = D, thn th abov systm can b rwrittn as T y = T AT(T y + T g(t. Us th substitution z = T y, w hav a dcoupld linar systm whr ĝ = T g. z = Dz + ĝ(t, In th dcoupld systm, ach row only concrns on unknown scalar and thrfor can b solvd as a first ordr linar diffrntial quation: Finally, w can transform z back to y by z = λ z + ĝ (t z 2 = λ 2 z 2 + ĝ 2 (t y = Tz Exampl: Find th gnral solution of th following linar systm by dcoupling transformation. y = 3y + 2y y 2 = y + 4y 2 + 5