DE Dr. M. Sakalli

Transcription

1 DE-0 Dr. M. Sakalli DE 55 M. Sakalli a n n 0 a Lh.: an Linar g Equaions Hr if g 0 homognous non-homognous ohrwis driving b a forc. You know h quaions blow alrad. A linar firs ordr ODE has h gnral form whr p g can b consans and/or variabls. onsan officin as: sraighforward soluion is d / a b / a Variabl officin as: Mhod of Ingraing Facors. Using h produc rul duvvdu udv. Mulipling h quaion b a funcion µ so ha h nir quaion mus b asil ingrad. DE 55 M. Sakalli d b / a a d p g a a b ln b / a a b / a k k ±

2 Variabl officin as: Mhod of Ingraing Facors. From h produc rul mulipling h s ordr linar DE b a funcion µ so ha h rsuling quaion mus b asil ingrad. This is h Gnral as. Proof is an am qusion. ' p g µ ' µ p µ g d [ µ ] d dµ µ µ g d [ g ] µ µ [ µ ] [ µ g ] DE 55 M. Sakalli dµ µ p dµ µ lnµ µ p p k p Mhod of Ingraing Facors: Variabl Righ Sid g a g dµ aµ > µ d µ aµ µ g a d a a a g > a a g DE 55 M. Sakalli a a d a [ ] a g

3 Eampl : / Obsrv ha quilibrium soluion of slops is shifing du o h dpndnc.. / ' 0 / / Wih µ w solv h original quaion as follows: / d / µ µ µ d 5 / >> 5 / >> 5 d 5 >> µ [ ] / 5 / DE 55 M. Sakalli Eampl : Gnral Soluion of a a a g Ingraing b pars udvduv-vdu 5 Thus [ 5 5 ] / DE 55 M. Sakalli

4 5 5 Equilibrium poins '0-5 0 and 5 0 Nds ingraing b pars a a a g [ 5 5 ] / 5 [ 5 ] 5 DE 55 M. Sakalli Eampl for gnral cas of s ordr DE IVP probl. EXAM WARNING linar!!? 5 Firs pu ino sandard form: Ingraing Facor 5 for 0 and hnc h gnral and paricular soluion for rspcivl. µ g 5 µ µ 5 p ln ln Ingral curvs for h diffrnial quaion and a paricular soluion in rd for h iniial poin. ln ln / 5 5 DE 55 M. Sakalli

5 5 DE 55 M. Sakalli Sparabl DEs: gd fd or d/ ' f/g. Two Eampls and implici soluions and isoclins. Linari? d d d d d d d d d d d d plici implici ± DE 55 M. Sakalli In nd Eampl domain of h soluion Thus h soluions o h iniial valu problm ar givn b From plici rprsnaion of i follows ha and hnc domain of is -. Smallr han - ngas insid sqr and - ilds which maks dnominaor of d/d zro vrical angn. onvrsl domain of can b simad b locaing vrical angns on graph usful for implicil dfind soluions. plici implici 0 d d

6 cos 0 ln sin DE 55 M. Sakalli h.: Diffrncs Bwn Linar and Nonlinar Equaions Rcall ha a firs ordr ODE has h form ' f and is linar if f is linar in and nonlinar if f is nonlinar in rgardlss of. Eampls: ' - '. Firs ordr linar and nonlinar quaions diffr in a numbr of was: Th hor dscribing isnc and uniqunss of soluions and corrsponding domains ar diffrn. Soluions o linar quaions can b prssd in rms of a gnral soluion which is no usuall h cas for nonlinar quaions. Linar quaions hav plicil dfind soluions whil nonlinar quaions picall do no and nonlinar quaions ma or ma no hav implicil dfind soluions. For boh ps of quaions numrical and graphical consrucion of soluions ar imporan. DE 55 M. Sakalli 6

7 Linari muliplici scalabili and addiivi suprposiion. Linari Dfinion: wih rspc o dpndn variabl hrfor h dgr of h indpndn variabls as cofficins of h drivaions is nor a concrn. Scalabili affa; Suprposiion uv fufv? fuvfufv fau bv fau fbv afu bfv Eampl: Lz z ''' z k z a bz ''' a bz k a bz az ''' bz ''' az bz k az k bz afzbfz So is s linar. W can find i o look dgr of funcions f z oo. Dgr of z is and no an rig combinaions is involvd. L ' a b'- a b a b a' b' a b a b So i s no linar.and h dgr of is indicaing nonlinari. DE 55 M. Sakalli Thorm.. onsidr h linar firs ordr iniial valu problm: d p g 0 0 If h funcions p and g ar coninuous on an opn inrval α β conaining h poin 0 hn hr iss a uniqu soluion φ ha saisfis h IVP for ach in α β. Proof: µ g 0 p s ds 0 0 whr µ µ DE 55 M. Sakalli 7

8 Thorm.. onsidr h nonlinar firs ordr iniial valu problm: Suppos f and f/ ar coninuous on som opn rcangl α β γ δ conaining h poin 0 0. Thn in som inrval 0 - h 0 h α βhr iss a uniqu soluion φ ha saisfis h IVP. Sinc hr is no gnral formula for h soluion of arbirar nonlinar firs ordr IVPs his proof is difficul and bond h scop of his cours. I urns ou ha condiions sad in Thm.. ar sufficin bu no ncssar o guaran isnc of a soluion and coninui of f nsurs isnc bu no uniqunss of φ. DE 55 M. Sakalli d f 0 0 Eampl : Linar IVP Rcall h iniial valu problm from hapr. slids: 5 5 ln Th soluion o his iniial valu problm is dfind for > 0 h inrval on which p -/ is coninuous. If h iniial condiion is - hn h soluion is givn b sam prssion as abov bu is dfind on < 0. In ihr cas Thorm.. guarans ha soluion is uniqu on corrsponding inrval. Qusion wha is h inrval hr for hr.. DE 55 M. Sakalli 8

9 Eampl : Nonlinar IVP onsidr nonlinar iniial valu problm from h.: d 0 d Th funcions f and f/ ar givn b f f and ar coninuous cp on lin. Thus possibl o draw an opn rcangl abou 0 - on which f and f/ ar coninuous as long as i dosn covr. How wid is rcangl? Rcall soluion dfind for > - wih DE 55 M. Sakalli Eampl : hang Iniial ondiion SKIP Our nonlinar iniial valu problm is d 0 d wih f f which ar coninuous cp on lin. If w chang iniial condiion o 0 hn Thorm.. is no saisfid. Solving his nw IVP w obain ± > 0 Thus a soluion iss bu is no uniqu. DE 55 M. Sakalli 9

10 Eampl :!!!linar IVP Vr simpl o draw angns onsidr iniial valu problm / Th funcions f and f/ ar givn b / f / f Thus f coninuous vrwhr bu f/ dosn is a 0 and hnc Thorm.. is no saisfid. Soluions is bu ar no uniqu. Sparaing variabls and solving w obain / / d c ± 0 Posiiv sinc canno b ngaiv du o sqr If iniial condiion is no on -ais whr 0 hn Thorm.. dos guaran isnc and uniqunss. / DE 55 M. Sakalli SKIP o acnss. Eampl :!!!linar IVP onsidr iniial valu problm 0 Th funcions f and f/ ar givn b f f Thus f and f/ ar coninuous a 0 so Thm.. guarans ha soluions is and ar uniqu. Sparaing variabls and solving w obain d c c Th soluion is dfind on -. No ha h singulari a is no obvious from original IVP samn. DE 55 M. Sakalli 0

11 Inrval of Dfiniion: Linar and Nonlinar ass B Thorm.. h soluion of a linar iniial valu problm iss hroughou an inrval abou 0 on which p and g ar coninuous. Vrical asmpos or ohr disconinuiis of soluion can onl occur a poins of disconinui of p or g. Howvr soluion ma b diffrniabl a poins of disconinui of p or g. In h nonlinar cas h inrval on which a soluion iss ma b difficul o drmin. Th soluion φ iss as long as φ rmains wihin rcangular rgion indicad in Thorm... This is wha drmins h valu of h in ha horm. Sinc φ is usuall no known i ma b impossibl o drmin his rgion. Furhrmor an singulariis in h soluion ma dpnd on h iniial condiion as wll as h quaion. DE 55 M. Sakalli Gnral Soluions For a firs ordr linar quaion i is possibl o obain a soluion conaining on arbirar consan from which all soluions follow b spcifing valus for his consan. For nonlinar quaions such gnral soluions ma no is. Tha is vn hough a soluion conaining an arbirar consan ma b found hr ma b ohr soluions ha canno b obaind b spcifing valus for his consan. onsidr Eampl : Th funcion 0 is a soluion of h diffrnial quaion bu i canno b obaind b spcifing a valu for c in soluion using sparaion of variabls: DE 55 M. Sakalli d c

12 Eplici Soluions: Linar Equaions B Thorm.. a soluion of a linar iniial valu problm p g 0 0 iss hroughou an inrval abou 0 on which p and g ar coninuous and his soluion is uniqu. Th soluion has an plici rprsnaion µ g whr µ µ p s ds and can b valuad a an appropria valu of as long as h ncssar ingrals can b compud. DE 55 M. Sakalli Eplici Soluion Approimaion For linar firs ordr quaions an plici rprsnaion for h soluion can b found as long as ncssar ingrals can b solvd. If ingrals can b solvd hn numrical mhods ar ofn usd o approima h ingrals. 0 µ g 0 µ µ g n k k whr µ µ g k k p s ds 0 DE 55 M. Sakalli

13 Implici Soluions: Nonlinar Equaions For nonlinar quaions plici rprsnaions of soluions ma no is. As w hav sn i ma b possibl o obain an quaion which implicil dfins h soluion. If quaion is simpl nough an plici rprsnaion can somims b found. Ohrwis numrical calculaions ar ncssar in ordr o drmin valus of for givn valus of. Ths valus can hn b plod in a skch of h ingral curv. Rcall h following ampl from h. slids: cos 0 ln sin DE 55 M. Sakalli Dircion Filds In addiion o using numrical mhods o skch h ingral curv h nonlinar quaion islf can provid nough informaion o skch a dircion fild. Th dircion fild can ofn show h qualiaiv form of soluions and can hlp idnif rgions in h -plan whr soluions hibi inrsing faurs ha mri mor daild analical or numrical invsigaions. hapr.7 and hapr 8 focus on numrical mhods. DE 55 M. Sakalli

14 h.6: Eac Equaions chain rul!!!. onsidr a firs ordr ODE of h form Suppos hr is a funcion ψ such ha and such ha ψ c dfins φ implicil. Thn ψ ψ d d ψ [ φ ] and hnc h original ODE bcoms d d d ψ [ φ ] 0 d Thus ψ c dfins a soluion implicil. In his cas h ODE is said o b ac. DE 55 M. Sakalli M N 0 ψ M ψ N Thorm.6.- oninui and Eisnc of ψ and h condiion of Eacnss. Suppos an ODE can b wrin in h form M N 0 whr h funcions M N M and N ar all coninuous in h rcangular rgion R: α β γ δ. Thn Eq. is an ac diffrnial quaion iff M Tha is hr iss a funcion ψ saisfing h condiions iff M and N saisf Equaion. Think hr.. How o solv i. N R ψ M ψ N DE 55 M. Sakalli

15 5 DE 55 M. Sakalli Eampl : Eac Equaion of onsidr h following diffrnial quaion. Thn and hnc From Thorm.6. Thus B Thorm.6. h soluion is givn implicil b 0 d d N M ODEis ac N M ψ ψ d d ψ ψ k ψ c k ψ c 8 DE 55 M. Sakalli Eampl : From Thorm.6. 0 sin cos sin cos N M ODEis ac cos N M sin cos N M ψ ψ c d d sin cos ψ ψ k sin sin ψ c k sin ψ

16 Eampl : Non-Eac Equaion Trad b Ingraing Facors. Inrsing hrfor ponial Eam Qusion I is somims possibl o convr a inac DE ino an ac quaion b raing wih a suiabl ingraing facor µ: M N 0 µ M µ N 0 For his quaion o b ac w nd M µ N Mµ Nµ M N µ 0 µ This parial diffrnial quaion ma b difficul o solv. If µ is a funcion of alon hn µ 0 and hnc w solv dµ M N µ d N providd righ sid is a funcion of onl. Similarl if µ is a funcion of alon. S for mor dails. DE 55 M. Sakalli Non-Eac Equaion Eampl rad. onsidr h following non-ac diffrnial quaion. 0 Sking an ingraing facor w solv h linar quaion dµ M N dµ µ µ µ d N d Mulipling our diffrnial quaion b µ w obain h ac quaion 0 which has is soluions givn implicil b c DE 55 M. Sakalli 6

17 Eam qusion and HW 7.b a pag 7 and 7 solv brnoulli problms a las wo o prov ha q rducs o a rs ordr linar DE. DE 55 M. Sakalli dsolv'd^' an >> dsolv'd^''0 an/*pi >> diff '' ans an/*pi^ DE 55 M. Sakalli 7