3. Differential Equations

Transcription

1 3. Differenial Equaions 3.. inear Differenial Equaions of Firs rder A firs order differenial equaion is an equaion of he form d() d ( ) = F ( (),) (3.) As noed above, here will in general be a whole la of funions (;) (parameerized b ) ha saisfies (3.). We need more informaion, like an iniial ondiion for, pin down he soluion eal. Resul Given ha F is oninuous and has oninuous firs derivaives, here is going o be a one funion () ha saisfies (3.) and he iniial ondiion The Simples Case If F is independen of, he soluion is rivial. Then sine ( ) = f() (3.) he la of funions saisfing his is a primiive funion of f plus an onsan, i.e., () = f( sds ) + = F () F ( ) +, (3.3) where df()/d f(). Noe ha F( ) is a onsan. We an hus merge he wo onsans and wrie () = F() +, (3.4) Cerainl, for all, (3.4) saisfies (3.) or, equivalenl, for an, (3.3) saisfies (3.). For eample, f () = e F () = a a e a (3.5) The arbirar onsan is pinned down wih some oher piee of informaion. So, if we wan o find () and we know he value of (), we ge () = F() + = F() F() + a a e e = + = + a a a (3.6) Where = ()+/a.this saisfies ( ) = e a and ()=().

2 A noe on noaion Above, we have used he proper noaion of an inegral, where boh he lower and upper limis and he dumm variable o inegrae over have separae names and are all wrien ou. fen, a more slopp noaion is used, where i is undersood ha f () sds f() d F () (3.7) d d z f () d = f (). (3.8) This noaion, alled an indefinie inegral, saves on variables, bu an be onfusing. Neverhele, I will use i someimes. Using his noaion in (3.4), we wrie () = f() d+, (3.9) 3... inear Firs rder Differenial Equaions Ver ofen, a soluion o a more ompliaed differenial equaion is derived b ransforming he original equaion ino somehing ha has he form of (3.). inear firs order differenial wih onsan oeffiiens equaions an be solved direl using suh a ransformaion. Consider ( ) + () = (3.) In his ase, we mulipl boh sides b e noe ha he HS beomes (ofen alled he inegraing faor). Afer doing ha, b g de () e ( ) + () = e ( ) + e () = d d i (3.) Thus, hinking of e () as simpl a funion of, as () in (3.), we ge a HS ha is he ime derivaive of a known funion of and he RHS is also a funion onl of. Then he soluion is found as in (3.3). John Haler, age 3:

3 d d de () i= e z s s e () e ds e = = N M + F HG () = e e + e F HG I KJ e = + e e ~ = + I KJ (3.) If we know () a some poin in ime, e.g. a =. e ( ) ~ = + = ~ = (3.3) Noe ha here is onl one degree of freedom in he onsans and. Choosing anoher simpl means ha he onsan has o hosen in anoher wa. Thus, one piee of informaion is suffiien o pin down he soluion eal Variable R.H.S. Aume he R.H.S. is a funion of. Use he same mehod as above. ( ) + () = () d e () = e () d z e () = e () d + d i (3.4) E.g., if ()= z z s e e se ds s e s s e ds s e s s () ~ ~ = + = N M + = N M N M + e e = +, ( ) = + e (3.5) where we had o use inegraion b pars Variable Coeffiiens If also he oeffiien on () is variable we have o use a more general inegraing faor o make he.h.s. ino he ime differenial of a known funion. Here he inegraing faor is e z ( s) ds o John Haler, age 3:3 (3.6)

4 wih Using he slopp noaion: d d e z o ( s) ds ( s) ds = () e z o (3.7) Chek ha ou an verif ha d d F HG z ( ) + () () = () d d F H z I K z = d () d () e () e () z z d () d () = z + e () e () d sds ( ) sds ( ) sds ( ) e () e () () e ( ) (3.8) I KJ z z = + (3.9) z sds Noe ha his is a generalizaion of he onsan oeffiien ase, sine e = e Eample; mone on a bank aoun wih variable ineres rae. +. ( ) = r() () d F () e z r d () I d H K = rd () ez () = r() d ( ) = ( ) ez ( ) e r (3.) Separaing Variables Someimes, we an wrie a differenial equaion suh ha he HS onl onains funions of and and he RHS onl a funion of. g ( ) = h ( ) (3.) eing G() be defined as a primiive of g(), i.e., from G'()= g(), we have John Haler, age 3:4

5 dg( ( )) = G'( ( )) = g ( ) = h( ) d G ( ( )) = hd ( ) + (3.) We an reover b invering G. An eample; () g ( ) = = 3 G ( ) = = d+ = /3+ ( ) ( ) 3 3 = G /3 + = /3 +. (3.3) 3.. inear Differenial Equaions of Higher rder 3... inear Seond rder Differenial Equaions A linear seond order differenial equaion has he form. () + ()( ) + () () = R() (3.4) This anno be solved direl b ransformaions in he simple wa we did wih firs order equaions. Insead we use a more general mehod (ha would have worked above also). Resul The general soluion (he omplee se of soluions) o a differenial equaion is he general soluion o he homogeneous par plus an pariular soluion o he omplee equaion. Resul The general soluion of he homogeneous par of a seond order linear differenial equaion an be epreed as () + () where () and () are wo linearl independen pariular soluions o he homogeneous equaions. Two funions () and () are linearl independen in a region Ω if here is no Ω., {, }, s.. ( ) = ( ), (3.5) 3... Homogeneous Equaions wih Consan Coeffiiens A homogeneous (par of a) differenial equaion has a zero HS when epreed as in (3.4); () + ( ) + () = (3.6) To solve his we sar b solving he haraerisi equaion, whih alwas is a polnomial, in his ase John Haler, age 3:5

6 As we know his polnomial has eal wo roos. r + r+ = (3.7) 4 r, = ± (3.8) Now we an hek ha boh e r i are soluions o (3.6) b noing ha. so using (3.9) in (3.6) ields r r h h i h r i i i i () = e, () = re, () = r e (3.9) r r r r i i d i i i (3.3) i i i i re + re + e = e r + r+ = So hen, he general soluion o he homogenous equaion is ()= e + e (3.3) h r r provided ha he wo pars are linearl independen, whih he are unle r =r Comple Roos The general soluion o he homogenous equaion is here found he same wa. r, r, 4 = ± 4 = ± a ± bi ( a+ bi) ( a bi) () = e + e a a = e osb + i sinb + e osb i sinb a b g b g b g = e ( + )os b+ i( )sin b. (3.3) where we used Resul 8 o ge he fourh equali. If we onl are abou real soluions, we resri he onsans in a wa o make sure he soluion is onl on he real line b a g (3.33) () = e osb+ sinb Repeaed Roos The general soluion o he homogenous equaion is in his ase ()= e + e (3.34) Chek ha he are boh soluions and onvine ourself ha he are linearl independen. John Haler, age 3:6 r r

7 3..5. Non-Homogeneous Equaions wih Consan Coeffiiens () + ( ) + () = R() (3.35) Reling on Resul, he onl added problem is ha we have o find one pariular soluion o he omplee equaion. Tpiall we gue a form of his soluion and hen use he mehod of undeermined oeffiiens. fen i works if we gue on a form similar o R(), e.g., if i is a polnomial of degree n we gue on anoher n degree polnomial wih unknown oeffiiens. Eample: Gue ha a pariular soluion is () ( ) + () = 3 + (3.36) A + B + C (3.37) for some onsans A,B,C. We hen have o solve for hese onsans b subsiuing ino he differenial equaion. () ( ) + () = A ( A + B) + A + B + C = A ( 4A B) + A B + C = 3 + (3.38) If his is going o hold for eah i is neear ha A = 3 ( 4A B) = B = 3 (3.39) A B+ C = C = So a pariular soluion is ()= (3.4) The haraerisi equaion is r r + = ( r ) = r =, (3.4) So he general soluion is ()= e + e (3.4) inear nh rder Differenial Equaions n n () + () () + + () = R() (3.43) n John Haler, age 3:7

8 The soluion ehnique is here eal analogous o he seond order ase. Firs we find he roos of he haraerisi equaion n n r + r + + = (3.44) The general soluion o he homogenous par is hen he sum of he n soluions orresponding o eah of n roos. The onl hing o noe is ha if one roo, r, is repeaed k imes, he soluions orresponding o his roo is given b r r n k r n e + e + + e (3.45) Repeaed omple roos are handled he same wa. Sa he roo a bi is repeaed in k pairs. Their orresponding soluion is given b a k k e dosb + sinb+ + k osb + k sinbi. (3.46) A pariular soluion o he omplee equaions an ofen be solved b he mehod of undeermined oeffiiens Sabili From he soluions o he differenial equaions we have seen we find ha he erms orresponding o roos ha have posiive real pars end o eplode as goes o infini. This means ha also he soluion eplodes unle he orresponding inegraion onsan, i, is zero. Terms wih roos ha have negaive real pars, on he oher hand, alwas onverge o zero. Global sabili, i.e., regardle of iniial ondiions, is hus equivalen o all roos being negaive. ook a he nonlinear differenial equaion oal Sabili ( ) = () 4 (3.47) Alhough we have no learned how o solve suh an equaion, we an sa somehing abou i. lo John Haler, age 3:8

9 ( ) = () 4 We see ha ()= and ()=- are saionar poins. We also see ha ()=- is loall sable. In he region [,) onverge o =-. is an unsable saionar poin and in he region (, ] eplodes over ime. In a plo loal sabili is equivalen o a negaive slope a he saionar poin Ssems of inear Firs rder Differenial Equaions Consider he following ssem of wo firs order differenial equaions () a a () () () = N M a a N M () + N M () = A + (3.48) As in he one equaions ase we sar b finding he general soluions o he homogeneous par. This plus some pariular soluion is he general soluion o he omplee ssem. If he off diagonal erms are zero he soluion o he homogeneous par is rivial, sine here is no inerdependen beween he equaions. () a () () = N M a N M () a () e a () e N M = N M N M (3.49) John Haler, age 3:9

10 The soluion an be showed in a graph, a phase diagram. Two Sable Roos ne Sable Roo If he roos are sable, i.e., negaive,. he homogeneous par alwas goes o zero. Wih onl one roo sable, here is jus one sable pah. (3.49) sugge a wa of finding he soluion o he homogeneous par of (3.48). We make a ransformaion of he variables so ha he ransformed ssem is diagonal. Sar b defining he new se of variables; () () () N M B () () () () B BA BAB () () () N M N M = N M = N M () (). (3.5) If we an find a B suh ha BAB - is diagonal we are half wa. The soluions for is hen () () r e r = N M e N M (3.5) where ri are he diagonal erms of he mari BAB -.The soluion for hen follows from he definiion of () () = N M r () e B B (3.5) r () e = John Haler, age 3: N M N M From linear algebra we know ha B - is he eigenveors of A and ha he diagonal erms of BAB - are he orresponding eigenvalues. The eigenvalues are given b he haraerisi equaion of A

11 a r a a a r = a r a r a a = b gb g r r( a + a ) + aa aa =. " "! "" ""! r( A) de( A) (3.53) The onl ru is ha we need he roos o be disin, oherwise B - is no alwas inverible. Disin roo implies ha B - is inverible. (If A is smmeri B - is also inverible.) e us draw a phase diagram wih he eigenveors of A. The dnami ssem behaves as he diagonal one in (3.49) bu he eigenveors have replaed he sandard, orhogonal aes. ne Sable and ne Unsable Roo Wha is remaining is o find a pariular soluion of he omplee ssem. ne wa is here o use he mehod of undeermined oeffiiens. Anoher is o look for a sead sae of he ssem, i.e., a poin where he ime derivaives are all zero. This is eas if he seond erm in (3.48) is onsan. We hen se he differenial equal o zero so = N M a a N M () + N M a a () N M = N M () a a () a a N M N M. (3.54) Given ha we know he value of () we an now give he general soluion of (3.48). The formula is given in mari form and is valid for an dimension of he ssem. Firs define John Haler, age 3:

12 and le bold leers define maries. Then we have r e r r e r e n = B + = B r C+ = B r C+ = B C+ C= B d = B r B + d i i (3.55) (3.56) The mehod oulined above works also in he ase of omple roos of he haraerisi equaion. If he roos are a bi we have Eample; So from using (3.57) we ge So if we know () we ge = () () N M ( a+ bi) e ( a bi) = B e a e (osb + isin b) M B a e (osb isin b) N () () = N M r, = ± i B M = i N M + N M + N M () () i + () i i e (os isin ) () = + Ne (os isin ) i ( )os ( + )sin = e ( + )os + i( )sin + N M ~ os ~ sin = e ~ os + ~ sin + N M (3.57) (3.58) + N M (3.59) (3.6) John Haler, age 3:

13 where Md i d i N () ( ) os ( ) sin e () = M ( ) os + ( ) sin + N M d i d i N M = A phase diagram of he ssem in (3.6) looks like (3.6) =. (3.6) If he roo has a real par loser o zero we ge more pronouned irles. Here is a phase diagram wih a real par of -.. John Haler, age 3:3

14 In he repeaed roo ase he mari of he eigenveors ma be singular, so ha we anno find B -. Then we use he mehod of equivalen ssems Equivalen Ssems A linear nh order differenial equaion is equivalen o a ssem of n firs order differenial equaions. So an be ransformed b he following subsiuion () + a () + a ( ) + a 3 () = () (3.63) ( ) = () () = () = () () = () ( ) () M () = M a a a N N 3 M + () () M () () (3.64) Sine he equaions are equivalen he onsequenl have he same soluions. Someimes one of he ransformaions is more onvenien o solve. e us also ransform a wo dimensional ssem ino a seond order equaion = a + a + = a + a + (3.65) John Haler, age 3:4

15 Firs use he firs equaion o epre and hen ake he ime derivaive of he firs. Then we an eliminae and is imederivaive. a = a a a = a + a = a + a a + a a + a = a + aa + a aa a + a ( a + a ) + ( a a a a ) = ~ 3.5. hase Diagrams and inearisaion hase diagrams are onvenien o analze he behavior of a dimensional ssem qualiaivel. E.g., b b g g ( ) = g (), k () + p k () = g (), k () + p (3.66) The firs sep here is o find he wo urves in he,k-spae where and k, respeivel are onsan. Seing (3.66) equal o zero defines wo relaions beween and k, whih we denoe b G and G () = = G ( k) k () = = G( k) (3.67) We hen draw hese urves in he,k-spae. For eample, ou will soon be able o show ha he dnami soluion o he Ramse onsumpion problem is given b he following ssem of differenial equaions bg bg bg u () = f k θ u () k = f k bg h (3.68) where is onsumpion, u is some uili funion k is a apial sok and f a ne produion funion. Seing he ime derivaives o zero we ge Draw hese urves in he,k spae u = () f bg k θh f bg θ = k u () = f k = f k b g b g (3.69) John Haler, age 3:5

16 () k () We hen have o find he signs of k, and above and below heir respeive zero moion urves. From (3.68) we see ha bg bg u () = k u () f k < k bg= k (3.7) This means ha is posiive o he lef of ( ) = and negaive o he righ. For k we find ha i is posiive below k () = and negaive above. Then draw hese moions as arrows in he phase diagram. Noe ha no pahs ever an ro. () k () We onlude ha his ssem has saddle poin haraerisis and hus have onl one sable rajeor owards he sead sae. The behavior lose o he sead sae should also be evaluaed b means of linearizaion around he sead sae. We do ha b approimaing in he following wa k John Haler, age 3:6

17 = N M ( ) g(, k ) g(, k ) k k () g (, k ) g (, k ) k wih an obvious generalizaion o higher dimensions. N M k k (3.7) We now evaluae he roos of he mari of derivaives. In he eample we find ha he haraerisi equaion is ( f θ ) ( u / u ) r f u / u f r =. (3.7) Sine f = θ a he sead sae, he, elemen of he oeffiien mari is zero and he roos of he ssem are given b r( f r) f u / u = r = so ha one roo is sable and he oher eplosive. f ( f ) + 4 f u / u (3.73) ± John Haler, age 3:7