1 Continuity of solutions with respect to initial conditions and parameters.

Transcription

1 Mah 275 Honors ODE I Spring, 23 Nos 7 Coninuiy of soluions wih rspc o iniial condiions and paramrs. Considr h iniial valu problm W know ha h uniqu soluion is y ky y () (; ; k) k Th funcion is coninuous wih rspc o all hr variabl s. Hr is anohr xampl y ky 2 y () Using sparaion of variabls w can show ha h uniqu soluion is (; ; k) k If hn on < < If k > hn h soluion xiss on ; k If k < hn h soluion xiss on ; So w can s ha h k domain of h soluion is f(; ; k) j k, < < g [ (; ; k) j k >, < < [ k (; ; k) j k <, k < < Wih a lil hough w can s ha his is an opn subs of R 3 Th funcion is coninuous in his s, maning ha if ( ; ; k ) 2 ; hn hr is a nighborhood of ( ; ; k ) in which is d nd, and is coninuous a ( ; ; k )

2 This rsul holds mor gnrally for a sysm of rs ordr quaions of h form y f (; y; k) y ( ) whr y (y ; ; y n ) ; ( ; ; n ) ; f (f ; ; f n ), and k (k ; ; k m ) (Th las of hs mans ha hr ar m paramrs in h quaion.) If f and all of is parial j ar coninuous in an opn subs of D of R n+m+ and ( ; ; k ) is in his subs, hn h soluion (; ; k) xiss on som nighborhood of his poin; and a any poin whr h soluion xiss, i is a coninuous in all m+n+ variabls. I will no prov his rsul, nor is i provd, in h x. I is provd in gradua xs in od s, such as Thory of Ordinary Di rnial Equaions by Coddingon and Lvinson. I is ofn usd, howvr, sinc quaions wih paramrs ar imporan in applicaions. 2 homognous sysms of rs ordr od s Ths ar sysms which can b wrin in h vcor-marix form whr x x x 2 x n C A and P () x () P () x; () p () p 2 () p n () p 2 () p 22 () p 2n () p n () p n2 () p nn () C A (2) I is assumd ha all of h funcions p ij () ar coninuous on h sam opn inrval I. (Th inrval I migh b ( ; ).) From his i can b shown ha vry soluion xiss on h nir inrval I. This is vry di rn from h nonlinar cas, as for xampl wih h quaion x x 2 + ; whr vry soluion blows up in ni im. Th proof ha vry soluion of a linar sysm xiss on h nir inrval whr h co cins of h od ar coninuous is a bi subl. In h homwork you wr askd o prov a spcial cas of his rsul. 2

3 In scion 7.4 of h x, svral horms ar sad abou hs sysms, which you should rad. Hr I will sa and prov quivaln rsuls, bu using linar algbra rminology. Thorm Th s V of all soluions of () form a vcor spac of dimnsion n, using h usual d niions of addiion and scalar muliplicaion on R n o d n addiion and scalar muliplicaion on V. Proof. To clarify h d niion of addiion and scalar muliplicaion, suppos ha x () and y () ar soluions of (). If c and c 2 ar scalars, hn w s (c x + c 2 y) () c x () + c 2 y () Wih his d niion, V is closd undr addiion and scalar muliplicaion, bcaus for ach in I, (c x + c 2 y) () c x () + c 2 y () c P () x () + c 2 P () y () P () (c x () + c 2 y ()) P () (c x + c 2 y) () To show ha V is of dimnsion n; w nd a basis for V consising of n linarly indpndn soluions. Following h noaion in h book, l () ; ; (n) dno h sandard basis vcors in R n. For xampl, () W apply h xisnc and uniqunss horm of h prvious s of nos, #6, wih f (; x) P () x Noic ha Hnc, f C j p ij () and all h parial j ar coninuous on I 3

4 Suppos ha is som poin in I. Thn for ach i, h iniial valu problm x P () x x ( ) (i) has a uniqu soluion on som inrval conaining As sad abov, ach of hs soluions xiss on h nir inrval I (S homwork problm #5 from las wk.) Hnc, w hav found n disinc soluions, say x () () ; ; x (n) (), all d nd on I. I claim ha hs soluions ar linarly indpndn. If no, hn hr ar consans c ; ; c n ; no all zro, such ha c x () () + + c n x (n) () for ach in I. Bu valuaing his a h poin ; w g c () + + c n (n) ; and his is impossibl, unlss c c n ; bcaus h basis vcors () ; ; (n) ar linarly indpndn in R n Hnc, h soluions x () () ; ; x (n) () form a linarly indpndn s in V. To show ha hy ar a basis for V, w mus show ha vry soluion is a linar combinaion of hs soluions. Bu if x () is a soluion such ha x ( ) c c 2 c n C A c () + + c n (n) hn using h xisnc and uniqunss horm, x () c x () () + + c n x (n) () ; sinc ach sid of his quaion is a soluion of () ; and ach sid quals h sam vcor x ( ) a Lik almos all vcor spacs, h vcor spac V has many di rn bass. Any s of n linarly indpndn soluions of () is a basis. You will s in Thorm ha hr is a Wronskian s for linar indpndnc. On xampl of wo di rn bass for v coms from h scond ordr quaion Th corrsponding sysm is y y x x 2 x 2 x 4

5 On basis for V in his cas is ; x (i) () (i) To g his sandard basis for V; w us 3 Fundamnal soluions This basis dos no saisfy cosh sinh ;. sinh cosh Suppos ha P () is an n n marix funcion, wih nris p ij () which ar coninuous on an inrval I. W considr h sysm of n rs ordr od s givn by x P () x (3) D niion 2 A fundamnal soluion for () is a marix () whos columns form a s of n linarly indpndn soluions of (). Exampl 3 Considr h sysm x x Hr ar wo fundamnal soluions () and cosh () sinh sinh cosh No ha h scond of hs has h propry ha () I This urns ou o b a convnin propry, as w will s blow.. Noic ha a fundamnal soluion is a marix funcion of ; no a vcor. i is calld a soluion bcaus i sais s h marix di rnial quaion Bu () P () () (4) This is ru bcaus ach column of () is a (vcor) soluion of (3). Suppos ha c is a vcor in R n. Thn x () () c 5

6 is a soluion of (3), bcaus x () () c P () () c P () x () Suppos ha is som poin in I. W can always nd a fundamnal soluion () ; wih ( ) I; by ling h i h column of () b h soluion of (3) such ha x ( ) (i) Thn i is asy o solv (3) wih h iniial condiion x ( ) x. Jus s x () () x Noic ha x ( ) dos qual x. Now assum ha is in IIn h following w will ak iniial condiions a If w alrady hav a fundamnal marix ; bu () 6 I; w could prhaps solv iniial valu problms mor asily by compuing () () () Howvr, for larg marics, h invrs is im consuming o compu, and gnrally i is br o us Gaussian liminaion o solv h quaion () c x for c Th sandard fundamnal marix is ofn convnin for horical purposs howvr. Indd, h x only uss h word fundamnal for his marix. 3. Wronskian If x () () ; ; x (n) () ar all soluions of (), linarly indpndn or no, hn h marix X () wih columns x () () ; ; x (n) () sais s (??) Th Wronskian of x () () ; ; x (n) () is d nd as W () d X () Thorm in h x sas ha W () is ihr zro vrywhr in I, or nvr zro. X () is a fundamnal soluion if and only if W () 6 Thr is a vrsion of Abl s horm, as follows W () c R rp ()d (5) Thus, X is a fundamnal soluion if and only if c 6 in his quaion. This formula follow from h following rsul Thorm 4 If () is an n n marix funcion saisfying () P () () 6

7 for ach in som inrval I; hn in his inrval, d d () rp () d () d Proof. I will only do h cas n 2 Thn i is no oo hard o compu dircly, assuming ha x () x () 2 () x 2 () x 22 () W hav so () p p 2 p 2 p 22 d x x 22 x 2 x 2 x x 2 x 2 x 22 (d ) x x 22 + x 22x x 2x 2 x 2x 2 ; (p x + p 2 x 2 ) x 22 + (p 2 x 2 + p 22 x 22 ) x (p x 2 + p 2 x 22 ) x 2 (p 2 x + p 22 x 2 ) x 2 Wih a lil concnraion, w can s ha all rms cancl xcp (p + p 22 ) (x x 22 x 2 x 2 ) ; which givs h rsul. As an xampl of (5), considr h sysm In marix form his is x x + x 2 x 2 x + x 2 x x (6) To solv his sysm w noic ha x 2 dos no dpnd on x ; so w can solv x 2 c 2 2 W can hn solv h x quaion, x x c 2 2 If c w g x 2 2 If c w g, using an ingraing facor, x 2 2. Thrfor, () 2!

8 Thn W () d () 2 W 2W rp () W 4 Marix xponnial Suppos ha M is an nn consan marix. Thn w d n h marix xponnial, M as follows X M I + n! M n No ha M 2 ; M 3, and so forh ar all d nd, by h usual ruls of marix muliplicaion. This would no b ru if M wr no squar. To show ha his d niion of M maks sns, w should prov ha h in ni sris convrgs. This mans ha h in ni sris ha appars in ach nry of h marix convrgs. Bu w will no ry o do his hr. I is don in som linar algbra books. I will do hr xampls 2 M 3 Thn and from his i follows ha Scond xampl Thn M M n M 2 n 3 n 2 3 ; A A ; M 8 A

9 Hnc, Third xampl Thn Also, and so If A ! M ! M ; hn M A; so 2 M A ; M ! 2 + ; A No ha if () A ; hn () + () (You should chck his marix muliplicaion.) Also, () I Hnc () is a fundamnal soluion of x x (7) This is always ru A is h fundamnal soluion of x Ax such ha () I. 9

10 5 Basic mhod for solving x Ax whr A is consan This marial is discussd in h x saring wih scion 7.5. I will no wri nos xcp on h cas of rpad ignvalus, scion 7.8. I will only considr a spcial cas of rpad ignvalus, rahr han h gnral cas discussd in h book so rad h marial blow insad of ha scion. In class I will also considr phas plans, which wr inroducd bri y in h las s of nos, for di rn xampls. 6 Rpad roos cas for sysms of od s. Th dscripion in h x is mor gnral han I will covr, or rquir for h xam. Undrsanding h following wo xampls should b nough. Exampl. y + 2y + y (8) Now w inrpr hs in h phas plan. As usual, l x y; x 2 y. W g h sysm In marix form his is whr x x 2 x 2 x 2x 2 A Th ignvalu quaion is h sam x Ax; 2 d (A ri) r 2 + 2r + ; so hr is a rpad ignvalu r r 2. To nd ignvalus, w hav v (A r I) v ; v 2

11 from which w asily s ha hr is only on linarly indpndn ignvcor, which w can s qual o This givs a vcor soluion x () () I is ricky o nd h scond soluion using linar algbra. scond scalar soluion y 2 Sinc x x 2 ; h scond soluion is x (2) () Insad, w us our Th phas plan for his sysm will b drawn in class. This xampl is ypical; s Figur for anohr xampl. This mhod for solving a sysm works in h n 2 cas whnvr P () is consan and is rs row is (; ), sinc his maks h rs quaion x x 2 ; and such a sysm is always quivaln o a scalar scond ordr linar quaion wih consan co cins. For n 3 w would wan h rs wo rows o b, in ordr o g a scalar hird ordr quaion, and so forh for highr n. 7 nonhomognous linar sysms W now urn o scion 7.9. Onc again I suggs ha you rad hs nos, as I will no b covring all of his scion. Th goal is o solv sysms of h form x () P () x + g () (9) As for scalar quaions, w can wri h gnral soluion as x () x h () + x p () ; whr x h is h gnral soluion of h homognous sysm x P () x; ()

12 and x p is a paricular soluion of h nonhomognous sysm (9). w will discuss in 7.9 is h following formula x p () () Z Th main im (s) g (s) ds () In his formula, () can b any fundamnal soluion of h homognous sysm (6) Proof of () Di rnia, using h produc rul. 7. Exampl Rcall ha in nos # w considrd h simpl sysm x x A fundamnal soluion for his sysm was found o b () Now considr h sysm x W rs nd o nd (s) s s s 2 Hnc, s s 2 s 2 s 2 s (s) g (s) s 2 x+ s s 2 s 2 s 2 s s s s ( 2) s s 2 2 ss W now hav o ingra ach rm. This rquirs ingraion by pars. I won giv h dails, bu Z Z s s ds s s ds + 2 (2)

13 Hnc, x p () 2 + ( ) + ( + ) 2 ( ) + ( + ) 2 ( ) + 2 ( + ) Obviously his can b qui laborious. Howvr, i didn hav o b ha hard. A hin can b sn in h answr noic sinh ha i is cosh sinh Also, w hav sn bfor ha is a soluion o h homognous sysm cosh x x Hnc ha dos no hav o b par of h paricular soluion and w can s x p () W could hav don his mor asily by saring wih a di rn fundamnal soluion, namly cosh sinh () sinh cosh And w could hav don i vn mor asily by using h quivaln scond ordr quaion. I is asy o s ha his is y y ; from which w asily spo h paricular soluion y Hnc w s x ; x 2 x 7.2 Exampl 2 L s rurn o h scalar cas. W saw how o solv y + y f () 3

14 whn f () had crain spci c forms, such as ; or was a polynomial. Bu w hav no mhod for solving i in h gnral cas. W now rwri his as a sysm or x x x 2 x 2 x + f () ; x+ f () A fundamnal soluion of h corrsponding homognous sysm is cos sin () sin cos No ha () I. Hnc so W us formula () ; wih () (). Sinc d (), h invrs is asily found o b cos sin sin cos Thrfor, (s) cos s sin s g (s) sin s cos s f (s) sin s f (s) cos s Z ; f (s) R (s) f (s) sin s ds g (s) ds f (s) cos s ds R cos sin f (s) sin s ds x p () sin cos f (s) cos s ds Our original y () is h rs componn, which is y () cos Z R R f (s) sin s ds + sin Z f (s) cos s ds Sinc funcions of ar consans as far as h ingral ds gos, w can wri his as y () This can b a usful formula. Z (sin cos s 4 cos sin s) f (s) ds

15 Midrm xam Th midrm xam will b on Wdnsday, Jam 27. Wha i covrs will b discussd lar. 8 Homwork, Du Fb. 2. Considr h iniial valu problm y y 2 y () Dno h soluion by (; ) W hav shown ha lim j (; )j! if 6 if Dos his conradic h coninuiy of h wih rspc o? Explain. 2. Scion 7.4, #5. ("Show ha h gnral soluion...") In h following problms allow plny of spac for your phas diagrams. Plo hm clarly. Don skch a lo of vagu curvs. Jus draw nough clar curvs o mak h picur obvious o a radr familiar wih phas plans, as I v rid o do in class scion 7.5 # 8 (a 3 3 sysm and h iniial 5 A ) ( ps.) 5 4. scion 7.5, # 24 ( ps.) B sur o add arrows o your phas porrais (no od s, jus ignvalus and ignvcors.) 5. scion 7.5, # 26 (sam dircions.) ( ps.) 5