Boyce/DiPrima/Meade 11 th ed, Ch 2.1: Linear Equations; Method of Integrating Factors. A linear first order ODE has the general form

Transcription

1 Boce/DiPrima/Meade h ed, Ch.: Linear Equaions; Mehod of Inegraing Facors Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce and Richard C. DiPrima, 7 b John Wile & Sons, Inc. A linear firs order ODE has he general form d d f (, ) where f is linear in. Eamples include equaions wih consan coefficiens, such as hose in Chaper, a b or equaions wih variable coefficiens: d d p( ) g( )

2 Consan Coefficien Case For a firs order linear equaion wih consan coefficiens, d a b, d recall ha we can use mehods of calculus o solve: d / d a b / a d b / a ln b / a a C b / a ke a, a d k e C

3 Variable Coefficien Case: Mehod of Inegraing Facors We ne consider linear firs order ODEs wih variable coefficiens: d d p( ) g( ) The mehod of inegraing facors involves mulipling his equaion b a funcion m(), chosen so ha he resuling equaion is easil inegraed.

4 Eample : Inegraing Facor ( of ) Consider he following equaion: d /3 e d Mulipling boh sides b, we obain We will choose so ha lef side is derivaive of known quani. Consider he following, and recall produc rule: m() ( ) d d d d Choose so ha ( m()) = m() d d + dm() d ( ) m() / ( ) m() ( ) ( ) ( ) e /3 e

5 Eample : General Soluion ( of ) Wih = e /, we solve he original equaion as follows: d d + = e/3 e / d d + e/ = e5/6 d d m() ( e / ) = e5/6 3 Sample Soluions : 3 5 e /3 Ce / e / = 3 5 e5/6 + c general soluion: = 3 5 e/3 + ce -/

6 Mehod of Inegraing Facors: Variable Righ Side In general, for variable righ side g(), he soluion can be found b choosing m() = e a : d + a = g() d m() d d + am() = m()g() e a d d + aea = e a g() d d ( e a ) = e a g() e a = = e -a ò ò e a g()d e a g()d + c + ce -a

7 Eample 3: General Soluion ( of ) We can solve he following equaion b mulipling b he inegraing facor m() = e - : giving us boh sides. Inegraing b pars, d d 4 d d (e- ) = 4e - - e - e - = ò which we can inegrae on 4e - - e - d e - = -e - + e- + 4 e- + c e - = e- + e- + c Thus = ce

8 Eample 3: Graphs of Soluions ( of ) d d 4 The graph shows he direcion field along wih several inegral curves. If we se c =, he eponenial erm drops ou and ou should noice how he soluion in ha case, hrough he poin (, -7/4), separaes he soluions ino hose ha grow eponeniall in he posiive direcion from hose ha grow eponeniall in he negaive direcion.. = ce

9 Mehod of Inegraing Facors for General Firs Order Linear Equaion Ne, we consider he general firs order linear equaion d p( ) g( ) d Mulipling boh sides b m(), we obain d ( ) p( ) ( ) ( ) g( ) d dm() Ne, we wan m() such ha = p()m(), from which i will follow ha d d d ( m()) = m() d d + p()m()

10 Inegraing Facor for General Firs Order Linear Equaion Assuming m() >, i follows ha d( ) p d p d ( ) ln ( ) ( ) ( ) k Choosing k =, we hen have ( ) e p( ) d, and noe m() > as desired.

11 Soluion for General Firs Order Linear Equaion Thus we have he following: d p( ) g( ) d d ( ) p( ) ( ) ( ) g( ), d Then where ( ) e p( ) d d d ( m()) = m()g() m() = ò m()g()d + c ( ) = m() ò m(s)g(s)ds + c where is some convenien lower limi of inegraion.

12 Eample 4: General Soluion ( of ) To solve he iniial value problem firs pu ino sandard form: Then and hence Giving us he soluion 4 4,,, for p( ) d d ln ln ( ) e e e e '+ = ( )' = 4 3 Þ = 4 + c Þ = + c = +

13 + = 4, ( ) =, Eample 4: Paricular Soluion ( of ) Using he iniial condiion () = and general soluion i follows ha = + c, = + c Þ c = = +, > The graphs below show soluion curves for he differenial equaion, including a paricular soluion whose graph conains he iniial poin (,). Noice ha when c=, we ge he parabolic soluion and ha soluion separaes he soluions ino hose ha are asmpoic o he posiive versus negaive -ais (,) = + c

14 Eample 5: A Soluion in Inegral Form ( of ) To solve he iniial value problem firs pu ino sandard form: Then and hence,, = e - /4 ( ) e p( ) d e d ( e s /4 ò ) ds + c = ( e- /4 e s /4 ò ) ds + /4 ce- e 4

15 ,, Eample 5: A Soluion in Inegral Form ( of ) Noice ha his soluion mus be lef in he form of an inegral, since here is no closed form for he inegral. ( = e - /4 ò ) e s /4 ds + ce- /4 Using sofware such as Mahemaica or Maple, we can approimae he soluion for he given iniial condiions as well as for oher iniial 3 condiions. Several soluion curves are shown ( = e - /4 ò ) e s /4 ds + ce- /4

16 Boce/DiPrima/Meade h ed, Ch.: Separable Equaions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. In his secion we eamine a subclass of linear and nonlinear firs order equaions. Consider he firs order equaion d d f (, ) We can rewrie his in he form d M (, ) N(, ) d For eample, le M(,) = f (,) and N (,) =. There ma be oher was as well. In differenial form, M(, ) d N(, ) d If M is a funcion of onl and N is a funcion of onl, hen M( ) d N( ) d In his case, he equaion is called separable.

17 Eample : Solving a Separable Equaion Solve he following firs order nonlinear equaion: d d Separaing variables, and using calculus, we obain ( - )d = ( )d ò ( - )d = = c 3-3 = 3 + c ò ( )d 4 4 The equaion above defines he soluion implicil. A graph showing he direcion field and implici plos of several soluion curves for he differenial equaion is given above. 4 4

18 Eample : Implici and Eplici Soluions ( of 4) Solve he following firs order nonlinear equaion: d d 3 4 Separaing variables and using calculus, we obain ( -)d = ( )d ò ( -)d = ò ( )d - = c The equaion above defines he soluion implicil. An eplici epression for he soluion can be found in his case: ( ) = Þ = ± ( c) c = ± C

19 Eample : Iniial Value Problem ( of 4) Suppose we seek a soluion saisfing () = -. Using he implici epression of, we obain Thus he implici equaion defining is Using an eplici epression of, I follows ha 4 3 C C C 3 ) ( ) ( 3 C C C d d

20 Eample : Iniial Condiion () = 3 (3 of 4) Noe ha if iniial condiion is () = 3, hen we choose he posiive sign, insead of negaive sign, on he square roo erm: d d This is indicaed on he graph in green.

21 Eample : Domain (4 of 4) Thus he soluions o he iniial value problem are given b From eplici represenaion of, i follows ha and hence he domain of is (, ). Noe = ields =, which makes he denominaor of d/d zero (verical angen). Conversel, he domain of can be esimaed b locaing verical angens on he graph (useful for implicil defined soluions). (eplici) 4 (implici) (), 4 3 d d

22 Eample 3: Implici Soluion of an Iniial Value Problem ( of ) Consider he following iniial value problem: d d 4-3 = 4 +, () = 3 Separaing variables and using calculus, we obain ( ) d 3 4 d 6 (4 4 (4 8 C c where C Using he iniial condiion, ()=, i follows ha C = ) d 4 3 ) d 4c

23 Eample 3: Graph of Soluions ( of ) Thus he general soluion is and he soluion hrough (,) is The graph of his paricular soluion hrough (, ) is shown in red along wih he graphs of he direcion field and several oher soluion curves for his differenial equaion, are shown: The poins idenified wih blue dos correspond o he poins on he red curve where he angen 3 line is verical: = -4» on hered curve, bu a all poins where he line connecing he blue poins inersecs soluion curves he angen line is verical , 3 4 C 8 7 () 4 4 3

24 Parameric Equaions The differenial equaion: d d F(, G(, ) ) is someimes easier o solve if and are hough of as dependen variables of he independen variable and rewriing he single differenial equaion as he ssem of differenial equaions: d d d F(, ) and G(, ) d Chaper 9 is devoed o he soluion of ssems such as hese.

25 Boce/DiPrima/Meade h ed, Ch.3: Modeling wih Firs Order Equaions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Mahemaical models characerize phsical ssems, ofen using differenial equaions. Model Consrucion: Translaing phsical siuaion ino mahemaical erms. Clearl sae phsical principles believed o govern process. Differenial equaion is a mahemaical model of process, picall an approimaion. Analsis of Model: Solving equaions or obaining qualiaive undersanding of soluion. Ma simplif model, as long as phsical essenials are preserved. Comparison wih Eperimen or Observaion: Verifies soluion or suggess refinemen of model.

26 Eample : Sal Soluion ( of 7) A ime =, a ank conains Q lb of sal dissolved in gal of waer. Assume ha waer conaining ¼ lb of sal/gal is enering ank a rae of r gal/min, and leaves a same rae. (a) Se up IVP ha describes his sal soluion flow process. (b) Find amoun of sal Q() in ank a an given ime. (c) Find limiing amoun Q L of sal Q() in ank afer a ver long ime. (d) If r = 3 & Q = Q L, find ime T afer which sal is wihin % of Q L. (e) Find flow rae r required if T is no o eceed 45 min.

27 Eample : (a) Iniial Value Problem ( of 7) A ime =, a ank conains Q lb of sal dissolved in gal of waer. Assume waer conaining ¼ lb of sal/gal eners ank a rae of r gal/min, and leaves a same rae. Assume sal is neiher creaed or desroed in ank, and disribuion of sal in ank is uniform (sirred). Then Rae in: (/4 lb sal/gal)(r gal/min) = (r/4) lb/min Rae ou: If here is Q() lbs sal in ank a ime, hen concenraion of sal is Q() lb/ gal, and i flows ou a rae of [Q()r/] lb/min. Thus our IVP is dq / d raein - raeou dq d r rq, 4 Q() Q

28 Eample : (b) Find Soluion Q() (3 of 7) To find amoun of sal Q() in ank a an given ime, we need o solve he iniial value problem To solve, we use he mehod of inegraing facors: or dq d m() = e a = e r/ Q() = e -r/ é ê ë rq ò re r/ 4 Q() = 5 + Q - 5 Q( ) 5 r, 4 [ ] e -r/ Q() ù d ú û = e-r/ r/ r/ e Q e Q éë 5e r/ + cù û = 5 + ce -r/

29 Eample : (c) Find Limiing Amoun Q L (4 of 7) Ne, we find he limiing amoun Q L of sal Q() in ank afer a ver long ime: Q lim Q( ) L This resul makes sense, since over ime he incoming sal soluion will replace original sal soluion in ank. Since incoming soluion conains.5 lb sal / gal, and ank is gal, evenuall ank will conain 5 lb sal. The graph shows inegral curves for r = 3 and differen values of Q. Q( ) 5 r/ Q 5 e 5 lb lim 5 r/ r/ e Q e

30 Eample : (d) Find Time T (5 of 7) Suppose r = 3 and Q = Q L. To find ime T afer which Q() is wihin % of Q L, firs noe Q = Q L = 5 lb, hence Q() = 5 + (Q - 5)e -r/ = 5 + 5e -.3 Ne, % of 5 lb is.5 lb, and hus we solve e.3T. e.3t ln(.).3t ln(.) T 3.4 min.3

31 Eample : (e) Find Flow Rae (6 of 7) To find flow rae r required if T is no o eceed 45 minues, recall from par (d) ha Q = Q L = 5 lb, wih r/ Q( ) 5 5e and soluion curves decrease from 5 o 5.5. Thus we solve r ln(.).45r r 5 5e e ln(.) r 8.69 gal/min

32 Eample : Discussion (7 of 7) Since his siuaion is hpoheical, he model is valid. As long as flow raes are accurae, and concenraion of sal in ank is uniform, hen differenial equaion is accurae descripion of he flow process. Models of his kind are ofen used for polluion in lake, drug concenraion in organ, ec. Flow raes ma be harder o deermine, or ma be variable, and concenraion ma no be uniform. Also, raes of inflow and ouflow ma no be same, so variaion in amoun of liquid mus be aken ino accoun.

33 Eample : Compound Ineres ( of 3) If a sum of mone is deposied in a bank ha pas ineres a an annual rae, r, compounded coninuousl, he amoun of mone (S) a an ime in he fund will sais he differenial equaion: ds d rs, S() S where S represenshe iniial invesmen. The soluion o his differenial equaion, found b separaing he variables and solving for S, becomes: r S( ) Se, where is measured in ears Thus, wih coninuous compounding, he amoun in he accoun grows eponeniall over ime.

34 S( ) Se Eample : Compound Ineres ( of 3) r In general, if ineres in an accoun is o be compounded m imes a ear, raher han coninuousl, he equaion describing he amoun in he accoun for an ime, measured in ears, becomes: S() = S (+ r m )m The relaionship beween hese wo resuls is clarified if we recall from calculus ha S (+ r m )m = S e r Growh of Capial a a Reurn Rae of r = 8% For Several Modes of Compounding: S()/S() m = 4 m = 365 ep(r) Years Compounded Compounded Compounded Quarerl Dail Coninuousl lim m A comparison of he accumulaion of funds for quarerl, dail, and coninuous compounding is shown for shor-erm and long-erm periods.

35 Eample : Deposis and Wihdrawals (3 of 3) Reurning now o he case of coninuous compounding, le us suppose ha here ma be deposis or wihdrawals in addiion o he accrual of ineres, dividends, or capial gains. If we assume ha he deposis or wihdrawals ake place a a consan rae k, his is described b he differenial equaion: ds ds rs k or in sandard form rs k and S() S d d where k is posiive for deposis and negaive for wihdrawals. We can solve his as a general linear equaion o arrive a he soluion: r r S( ) Se ( k / r)( e ) To appl his equaion, suppose ha one opens an IRA a age 5 and makes annual invesmens of $ hereafer wih r = 8%..8*4.8*4 A age 65, S(4) * e (/.8)( e ) $588,33

36 Eample 3: Pond Polluion ( of 7) Consider a pond ha iniiall conains million gallons of fresh waer. Waer conaining oic wase flows ino he pond a he rae of 5 million gal/ear, and eis a same rae. The concenraion c() of oic wase in he incoming waer varies periodicall wih ime: c() = + sin() g/gal (a) Consruc a mahemaical model of his flow process and deermine amoun Q() of oic wase in pond a ime. (b) Plo soluion and describe in words he effec of he variaion in he incoming concenraion.

37 Eample 3: (a) Iniial Value Problem ( of 7) Pond iniiall conains million gallons of fresh waer. Waer conaining oic wase flows ino pond a rae of 5 million gal/ear, and eis pond a same rae. Concenraion is c() = + sin g/gal of oic wase in incoming waer. Assume oic wase is neiher creaed or desroed in pond, and disribuion of oic wase in pond is uniform (sirred). Then dq / d raein - raeou Rae in: ( + sin())g/gal( )gal/ear If here is Q() g of oic wase in pond a ime, hen concenraion of sal is Q() lb/ 7 gal, hus Rae ou: 5 6 (5 6 )gal/ear(q() / 7 )g/gal = Q() / g/r

38 Eample 3: (a) Iniial Value Problem, Scaling (3 of 7) Recall from previous slide ha Rae in: ( + sin g/gal)(5 6 gal/ear) Rae ou: (Q() g/ 7 gal)(5 6 gal/ear) = Q()/ g/r. Then iniial value problem is dq d Q( ) 6 sin 5, Q() Change of variable (scaling): Le q() = Q()/ 6. Then dq d + q = + 5 sin, q() =

39 Eample 3: (a) Solve Iniial Value Problem (4 of 7) To solve he iniial value problem q q / 5sin, q() we use he mehod of inegraing facors: ( ) q( ) e e a / e e / / 5sin Using inegraion b pars (see ne slide for deails) and he iniial condiion, we obain afer simplifing, q() = e -/ é e / e/ cos() + ù 7 e/ sin() + c ë ê û ú q() = cos() + sin() e-/ d

40 Eample 3: (a) Inegraion b Pars (5 of 7) e / é ò sin()d = - e/ cos() + ëê 4 ( ) ò e/ cos()d é = - e/ cos() + æ 4 e/ sin() - ö ù ê 4 ò e/ sin()d è ç ø ú ë û é = - e/ cos() + 8 e/ sin() - ù 6 ò e/ sin()d ëê ûú 7 6 ò e/ sin()d = - e/ cos() + 8 e/ sin() + c ò e / sin()d = e/ cos() + 7 e/ sin() + c 5 e / sin()d ò = e/ cos() + 7 e/ sin() + c ù ûú

41 Eample 3: (b) Analsis of soluion (6 of 7) Thus our iniial value problem and soluion is dq d + q = + 5sin(), q() = q() = A graph of soluion along wih direcion field for differenial equaion is given below. Noe ha eponenial erm is imporan for small, bu decas awa for large. Also, = would be equilibrium soluion if no for sin() erm. 3 cos() + sin() e-/

42 Eample 3: (b) Analsis of Assumpions (7 of 7) Amoun of waer in pond conrolled enirel b raes of flow, and none is los b evaporaion or seepage ino ground, or gained b rainfall, ec. Amoun of polluion in pond conrolled enirel b raes of flow, and none is los b evaporaion, seepage ino ground, dilued b rainfall, absorbed b fish, plans or oher organisms, ec. Disribuion of polluion hroughou pond is uniform.

43 Eample 4: Escape Veloci ( of ) A bod of mass m is projeced awa from he earh in a direcion perpendicular o he earh s surface wih iniial veloci v and no air resisance. Taking ino accoun he variaion of he earh s graviaional field wih disance, he graviaional force acing on he mass is w() = - mgr where is he disance above he earh's surface (R + ) R is he radius of he earh and g is he acceleraion due o gravi a he earh s surface. Using Newon s law F = ma, dv dv d dv m dv d = - mgr (R + ), v() = v Since v d d d d and cancelling he m s, he differenial equaion becomes v dv d = - gr (R + ), since = when =, v() = v

44 Eample 4: Escape Veloci ( of ) We can solve he differenial equaion b separaing he variables and inegraing o arrive a: v The maimum heigh (aliude) will be reached when he veloci is zero. Calling ha maimum heigh A ma, we have We can now find he iniial veloci required o lif a bod o a heigh A ma : and, aking he limi as A ma, we ge he escape veloci. dv v d = gr R + + c = gr R + + v - gr A ma = v = gr A ma R + A ma v R gr - v gr ( R ) Noice ha his does no depend on he mass of he bod. v gr, v( ) v

45 Boce/DiPrima/Meade h ed, Ch.4: Differences Beween Linear and Nonlinear Equaions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Recall ha a firs order ODE has he form ' = f (, ), and is linear if f is linear in, and nonlinear if f is nonlinear in. Eamples: ' = - e, ' =. In his secion, we will see ha firs order linear and nonlinear equaions differ in a number of was, including: The heor describing eisence and uniqueness of soluions, and corresponding domains, are differen. Soluions o linear equaions can be epressed in erms of a general soluion, which is no usuall he case for nonlinear equaions. Linear equaions have eplicil defined soluions while nonlinear equaions picall do no, and nonlinear equaions ma or ma no have implicil defined soluions. For boh pes of equaions, numerical and graphical consrucion of soluions are imporan.

46 Theorem.4. Consider he linear firs order iniial value problem: ' p( ) g( ), () If he funcions p and g are coninuous on an open inerval I :a < < b conaining he poin =, hen here eiss a unique funcion = f() ha saisfies he IVP for each in I. Proof ouline: Use Ch. discussion and resuls: ( ) g( ) d, where ( ) e ( ) p( s) ds

47 Theorem.4. Consider he nonlinear firs order iniial value problem: ' f (, ), () Le he funcions f and f / be coninuous in some recangle a < < b, g < < d conaining he poin (, ). Then in some inerval h < < + h in he recangle here is a unique soluion = f() of he iniial value problem. Proof discussion: Since here is no general formula for he soluion of arbirar nonlinear firs order IVPs, his proof is difficul, and is beond he scope of his course. I urns ou ha condiions saed in Thm.4. are sufficien bu no necessar o guaranee eisence of a soluion, and coninui of f ensures eisence bu no uniqueness of. = f()

48 Eample : Linear IVP Recall he iniial value problem from Chaper. slides: 4, The soluion o his iniial value problem is defined for >, he inerval on which p() = / is coninuous. If he iniial condiion is ( ) =, hen he soluion is given b same epression as above, bu is defined on <. In eiher case, Theorem.4. guaranees ha soluion is unique on corresponding inerval. (-,) 4 (,)

49 Eample : Nonlinear IVP ( of ) Consider nonlinear iniial value problem from Ch.: The funcions f and are given b and are coninuous ecep on line =. Thus we can draw an open recangle abou (, ) in which f and are coninuous, as long as i doesn cover =. How wide is he recangle? Recall soluion defined for >, wih, 4 3 ), (, 4 3 ), ( f f (), 4 3 d d 4 3 f / f /

50 Eample : Change Iniial Condiion ( of ) Our nonlinear iniial value problem is wih which are coninuous ecep on line =. If we change iniial condiion o () =, hen Theorem.4. is no saisfied. Solving his new IVP, we obain Thus a soluion eiss bu is no unique., 4 3 ), (, 4 3 ), ( f f (), 4 3 d d, 3

51 Eample 3: Nonlinear IVP Consider nonlinear iniial value problem ' = /3, () = ( ³ ) The funcions f and f / are given b /3 f /3 f (, ), (, ) 3 Thus f coninuous everwhere, bu f / doesn eis a =, and hence Theorem.4. does no appl. Soluions eis bu are no unique. Separaing variables and solving, we obain /3 d d 3 /3 c If iniial condiion is no on -ais, hen Theorem.4. does guaranee eisence and uniqueness. 3 3/,

52 Eample 4: Nonlinear IVP Consider nonlinear iniial value problem ' =, () = The funcions f and f / are given b f f (, ), (, ) Thus f and f / are coninuous a =, so Theorem.4. guaranees ha soluions eis and are unique. Separaing variables and solving, we obain - d = d Þ - - = + c Þ = - + c Þ = - The soluion () is defined on ( -, ). Noe ha he singulari a = is no obvious from original IVP saemen.

53 Inerval of Eisence: Linear Equaions B Theorem.4., he soluion of a linear iniial value problem p( ) g( ), () eiss hroughou an inerval abou = on which p and g are coninuous. Verical asmpoes or oher disconinuiies of soluion can onl occur a poins of disconinui of p or g. However, soluion ma be differeniable a poins of disconinui of p or g. See Chaper.: Eample 3 of e. Compare hese commens wih Eample and wih previous linear equaions in Chaper and Chaper.

54 Inerval of Eisence: Nonlinear Equaions In he nonlinear case, he inerval on which a soluion eiss ma be difficul o deermine. The soluion = f() eiss as long as [, f()] remains wihin a recangular region indicaed in Theorem.4.. This is wha deermines he value of h in ha heorem. Since f() is usuall no known, i ma be impossible o deermine his region. In an case, he inerval on which a soluion eiss ma have no simple relaionship o he funcion f in he differenial equaion ' = f (, ), in conras wih linear equaions. Furhermore, an singulariies in he soluion ma depend on he iniial condiion as well as he equaion. Compare hese commens o he preceding eamples.

55 General Soluions For a firs order linear equaion, i is possible o obain a soluion conaining one arbirar consan, from which all soluions follow b specifing values for his consan. For nonlinear equaions, such general soluions ma no eis. Tha is, even hough a soluion conaining an arbirar consan ma be found, here ma be oher soluions ha canno be obained b specifing values for his consan. Consider Eample 4: The funcion = is a soluion of he differenial equaion, bu i canno be obained b specifing a value for c in soluion found using separaion of variables: d d = Þ = - + c

56 Eplici Soluions: Linear Equaions B Theorem.4., a soluion of a linear iniial value problem p( ) g( ), () eiss hroughou an inerval abou = on which p and g are coninuous, and his soluion is unique. The soluion has an eplici represenaion, ( ) g( ) d ( ), where ( ) e p( s) ds, and can be evaluaed a an appropriae value of, as long as he necessar inegrals can be compued.

57 Eplici Soluion Approimaion For linear firs order equaions, an eplici represenaion for he soluion can be found, as long as necessar inegrals can be solved. If inegrals can be solved, hen numerical mehods are ofen used o approimae he inegrals. n k k k k ds s p g d g e C d g ) ( ) ( ) ( ) ( ) ( ) ( where, ) ( ) ( ) (

58 Implici Soluions: Nonlinear Equaions For nonlinear equaions, eplici represenaions of soluions ma no eis. As we have seen, i ma be possible o obain an equaion which implicil defines he soluion. If equaion is simple enough, an eplici represenaion can someimes be found. Oherwise, numerical calculaions are necessar in order o deermine values of for given values of. These values can hen be ploed in a skech of he inegral curve. Recall he eamples from earlier in he chaper and consider he following eample cos 3, () ln sin 3 3

59 Direcion Fields In addiion o using numerical mehods o skech he inegral curve, he nonlinear equaion iself can provide enough informaion o skech a direcion field. The direcion field can ofen show he qualiaive form of soluions, and can help idenif regions in he -plane where soluions ehibi ineresing feaures ha meri more deailed analical or numerical invesigaions. Chaper.7 and Chaper 8 focus on numerical mehods.

60 Boce/DiPrima/Meade h ed, Ch.5: Auonomous Equaions and Populaion Dnamics Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. In his secion we eamine equaions of he form d/d = f (), called auonomous equaions, where he independen variable does no appear eplicil. The main purpose of his secion is o learn how geomeric mehods can be used o obain qualiaive informaion direcl from a differenial equaion wihou solving i. Eample (Eponenial Growh): Soluion: d d r, r e r

61 Logisic Growh An eponenial model ' = r, wih soluion = e r, predics unlimied growh, wih rae r > independen of populaion. Assuming insead ha growh rae depends on populaion d size, replace r b a funcion h() o obain. d = h() We wan o choose growh rae h() so ha r > when is small, h() decreases as grows larger, and h() < when is sufficienl large. The simples such funcion is h() = r a, where a >. Our differenial equaion hen becomes d r a, r, a d This equaion is known as he Verhuls, or logisic, equaion.

62 Logisic Equaion The logisic equaion from he previous slide is d r a, r, a d This equaion is ofen rewrien in he equivalen form d d where K = r/a. The consan r is called he inrinsic growh rae, and as we will see, K represens he carring capaci of he populaion. A direcion field for he logisic equaion wih r = and K = is given here. r, K

63 Logisic Equaion: Equilibrium Soluions Our logisic equaion is d d Two equilibrium soluions are clearl presen: ), ( ) K In direcion field below, wih r =, K =, noe behavior of soluions near equilibrium soluions: = is unsable, r, r, K K ( = is asmpoicall sable.

64 Auonomous Equaions: Equilibrium Solns Equilibrium soluions of a general firs order auonomous equaion ' = f () can be found b locaing roos of f () =. These roos of f () are called criical poins. For eample, he criical poins of he logisic equaion d d are = and = K. Thus criical poins are consan funcions (equilibrium soluions) in his seing. r K

65 Logisic Equaion: Qualiaive Analsis and Curve Skeching ( of 7) To beer undersand he naure of soluions o auonomous equaions, we sar b graphing f () vs.. In he case of logisic growh, ha means graphing he following funcion and analzing is graph using calculus. f ( ) r K

66 Logisic Equaion: Criical Poins ( of 7) The inerceps of f occur a = and = K, corresponding o he criical poins of logisic equaion. The vere of he parabola is (K/, rk/4), as shown below. f æ f () = r - ö è ç K ø éæ f '() = r - ö è ç K ø + æ - ö ù ê è ç K ø ú ë û æ è ç K = - r [ K - K ] se = Þ = K ö æ ø = r - K è ç K ö ø æ è ç K ö ø = rk 4

67 Logisic Soluion: Increasing, Decreasing (3 of 7) Noe d/d > for < < K, so is an increasing funcion of here (indicae wih righ arrows along -ais on < < K). Similarl, is a decreasing funcion of for > K (indicae wih lef arrows along -ais on > K). In his cone he -ais is ofen called he phase line. d d r, r K

68 Logisic Soluion: Seepness, Flaness (4 of 7) Noe d/d when or K, so is relaivel fla here, and ges seep as moves awa from or K. d d r K

69 Logisic Soluion: Concavi (5 of 7) Ne, o eamine concavi of (), we find '': d d d f ( ) d f ( ) f ( ) f ( ) Thus he graph of is concave up when f and f ' have same sign, which occurs when < < K/ and > K. The graph of is concave down when f and f ' have opposie signs, which occurs when K/ < < K. Inflecion poin occurs a inersecion of and line = K/. d d

70 Logisic Soluion: Curve Skeching (6 of 7) Combining he informaion on he previous slides, we have: Graph of increasing when < < K. Graph of decreasing when > K. Slope of approimael zero when or K. Graph of concave up when < < K/ and > K. Graph of concave down when K/ < < K. Inflecion poin when = K/. Using his informaion, we can skech soluion curves for differen

71 Logisic Soluion: Discussion (7 of 7) Using onl he informaion presen in he differenial equaion and wihou solving i, we obained qualiaive informaion abou he soluion. For eample, we know where he graph of is he seepes, and hence where changes mos rapidl. Also, ends asmpoicall o he line = K, for large. The value of K is known as he environmenal carring capaci, or sauraion level, for he species. Noe how soluion behavior differs from ha of eponenial equaion, and hus he decisive effec of nonlinear erm in logisic equaion.

72 Solving he Logisic Equaion ( of 3) Provided and K, we can rewrie he logisic ODE: d K rd Epanding he lef side using parial fracions, K Thus he logisic equaion can be rewrien as / K d rd / K Inegraing he above resul, we obain ln A ln K K B r C A B K B, A K

73 We have: ln Solving he Logisic Equaion ( of 3) ln If < < K, hen < < K and hence ln ln K K r C r C Rewriing, using properies of logs: ln or K r C K, where () r K e K e rc K ce r

74 Soluion of he Logisic Equaion (3 of 3) We have: K r K e for < < K. I can be shown ha soluion is also valid for > K. Also, his soluion conains equilibrium soluions = and = K. Hence soluion o logisic equaion is K r K e

75 Logisic Soluion: Asmpoic Behavior The soluion o logisic ODE is K r K e We use limis o confirm asmpoic behavior of soluion: lim lim K lim r K e K Thus we can conclude ha he equilibrium soluion () = K is asmpoicall sable, while equilibrium soluion () = is unsable. The onl wa o guaranee ha he soluion remains near zero is o make =. K

76 Eample : Pacific Halibu ( of ) Le be biomass (in kg) of halibu populaion a ime, wih r =.7/ear and K = kg. If =.5K, find (a) biomass ears laer (b) he ime (a) For convenience, scale equaion: Then such ha ( ) =.75K. K = K K +[- / K]e -r () K -(.7)() e and hence ().5797K kg K r K e

77 Eample : Pacific Halibu, Par (b) ( of ) (b) Find ime for which ( ) =.75K. K.75 = K K + ( - K )e -r é.75 K + æ - ö ù ê è ç K ø e -r ú ë û = K.75 K +.75( - K )e -r = K e -r = K.5 K.75 - K = -.7 ln æ.5 è ç 3.75 r K e ( ) = K 3( - K ) ( ) K ö ø» 3.95 ears

78 Criical Threshold Equaion ( of ) Consider he following modificaion of he logisic ODE: d d r, r T The graph of he righ hand side f () is given below.

79 Criical Threshold Equaion: Qualiaive Analsis and Soluion ( of ) Performing an analsis similar o ha of he logisic case, we obain a graph of soluion curves shown below. T is a hreshold level for, in ha populaion dies off or grows unbounded, depending on which side of T he iniial value is. See also laminar flow discussion in e. I can be shown ha he soluion o he hreshold equaion is d d r, r T T T e r

80 Logisic Growh wih a Threshold ( of ) In order o avoid unbounded growh for > T as in previous seing, consider he following modificaion of he logisic equaion: d d = -r æ - è ç T ö ø æ - è ç K ö ø, r > and < T < K The graph of he righ hand side f () is given below.

81 Logisic Growh wih a Threshold ( of ) Performing an analsis similar o ha of he logisic case, we obain a graph of soluion curves shown below righ. T is hreshold value for, in ha populaion dies off or grows owards K, depending on which side of T is. K is he carring capaci level. Noe: = and = K are sable equilibrium soluions, and = T is an unsable equilibrium soluion.

82 Boce/DiPrima/Meade h ed, Ch.6: Eac Equaions and Inegraing Facors Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Consider a firs order ODE of he form M(, ) N(, ) Suppose here is a funcion (, ) such ha (, ) M(, ), (, ) N(, ) and such ha (, ) = c defines = f() implicil. Then M(, ) + N(, )' = + d d = d d (,f()) and hence he original ODE becomes d d (,f()) = Thus (, ) = c defines a soluion implicil. In his case, he ODE is said o be an eac differenial equaion.

83 Eample : Eac Equaion Consider he equaion: I is neiher linear nor separable, bu here is a funcion φ such ha + = and = The funcion ha works is (, ) = + Thinking of as a funcion of and calling upon he chain rule, he differenial equaion and is soluion become d d = d d ( + ) = Þ (, ) = + = c

84 Theorem.6. Suppose an ODE can be wrien in he form M(, ) N(, ) () where he funcions M, N, M and N are all coninuous in he recangular region R: a < < b, g < < d. Then Eq. () is an eac differenial equaion if and onl if M (, ) N (, ), (, ) R Tha is, here eiss a funcion saisfing he condiions if and onl if M and N saisf Equaion (). () (, ) M(, ), (, ) N(, ) (3)

85 Eample : Eac Equaion ( of 3) ) (sin ) cos ( e e sin ), (, cos ), ( e N e M eac ODEis ), ( cos ), ( N e M sin ), (, cos ), ( e N e M Consider he following differenial equaion. Then and hence From Theorem.6., Thus (, ) = (, )d ò = cos + e ( )d ò = sin + e + h()

86 Eample : Soluion ( of 3) We have (, ) M and (, ) = (, )d cos e, (, ) N I follows ha (, ) = sin + e - = sin + e + h'() sin Þ h'() = - Þ h() = - + k Thus (, ) sin e k B Theorem.6., he soluion is given implicil b sin e e ò = ò ( cos + e )d = sin + e + h() c

87 Eample : Direcion Field and Soluion Curves (3 of 3) Our differenial equaion and soluions are given b ( cos e ) (sin sin e c A graph of he direcion field for his differenial equaion, along wih several soluion curves, is given below. e ),

88 Eample 3: Non-Eac Equaion ( of ) Consider he following differenial equaion. Then and hence To show ha our differenial equaion canno be solved b his mehod, le us seek a funcion such ha Thus ) ( ) (3 N M ), (, 3 ), ( no eac ODEis ), ( 3 ), ( N M N M ), (, 3 ), ( (, ) = (, )d ò = 3 + ( )d ò = h()

89 Eample 3: Non-Eac Equaion ( of ) We seek such ha and Then (, ) M 3, (, Because h () depends on as well as, here is no such funcion (, ) such ha ) N 3 d 3 / C( ) (, ) (, ) d (, ) = + = h'() Þ h'()=? - - d d = (3 + ) + ( + )'

90 Inegraing Facors I is someimes possible o conver a differenial equaion ha is no eac ino an eac equaion b mulipling he equaion b a suiable inegraing facor (, ): m M (, ) N(, ) (, ) M (, ) (, ) N(, ) For his equaion o be eac, we need ( m M ) = ( m N ) Û Mm - Nm + ( M - N )m = This parial differenial equaion ma be difficul o solve. If m is a funcion of alone, hen m = and hence we solve d M N, d N provided righ side is a funcion of onl. Similarl if m is a funcion of alone. See e for more deails.

91 Eample 4: Non-Eac Equaion Consider he following non-eac differenial equaion. Seeking an inegraing facor, we solve he linear equaion Mulipling our differenial equaion b, we obain he eac equaion which has is soluions given implicil b ) ( ) (3 d d N N M d d ) (, ) ( ) (3 3 c 3 m

92 Boce/DiPrima/Meade h ed, Ch.7: Numerical Approimaions: Euler s Mehod Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Recall ha a firs order iniial value problem has he form d f (, ), ( ) d If f and f / are coninuous, hen his IVP has a unique soluion = f() in some inerval abou. When he differenial equaion is linear, separable or eac, we can find he soluion b smbolic manipulaions. However, he soluions for mos differenial equaions of his form canno be found b analical means. Therefore i is imporan o be able o approach he problem in oher was.

93 Direcion Fields For he firs order iniial value problem f (, ), ( ), we can skech a direcion field and visualize he behavior of soluions. This has he advanage of being a relaivel simple process, even for complicaed equaions. However, direcion fields do no lend hemselves o quaniaive compuaions or comparisons.

94 Numerical Mehods For our firs order iniial value problem f, ), ( ), ( an alernaive is o compue approimae values of he soluion = f() a a seleced se of -values. Ideall, he approimae soluion values will be accompanied b error bounds ha ensure he level of accurac. There are man numerical mehods ha produce numerical approimaions o soluions of differenial equaions, some of which are discussed in Chaper 8. In his secion, we eamine he angen line mehod, which is also called Euler s Mehod.

95 Euler s Mehod: Tangen Line Approimaion For he iniial value problem f (, ), ( ), we begin b approimaing soluion = f() a iniial poin. The soluion passes hrough iniial poin (, ) wih slope f (, ). The line angen o he soluion a his iniial poin is The angen line is a good approimaion o soluion curve on an inerval shor enough. Thus if is close enough o, we can approimae = f( ) b f, f,

96 Euler s Formula For a poin close o, we approimae using he line passing hrough (, ) wih slope f (, ): Thus we creae a sequence n of approimaions : where f n = f ( n, n ). For a uniform sep size n+ = n + h, Euler s formula becomes n n n n n f f f, f,,,, n h f n n n = f( ) = f( n )

97 Euler Approimaion To graph an Euler approimaion, we plo he poins (, ), (, ),, ( n, n ), and hen connec hese poins wih line segmens., where f f, n n fn n n n n n

98 Eample : Euler s Mehod ( of 3) For he iniial value problem d 3.5, () d we can use Euler s mehod wih h =. o approimae he soluion a =.,.4,.6,.8, and. as shown below f f f f f 3 4 h h h h h (3.5)(.).5(.) (.) (.) (.) (.)

99 Eample : Eac Soluion ( of 3) We can find he eac soluion o our IVP, as in Chaper.: 3.5, ().5 3 e.5.5e.5 3e.5 e e 4e 4e 4 4 ke.5.5 k..5 () k e

100 Eample : Error Analsis (3 of 3) From able below, we see ha he errors sar small, bu ge larger. This is mos likel due o he fac ha he eac soluion is no linear on [, ]. Noe: Percen Relaive Error Eac Appro Error % Rel Error eac eac appro Eac in red Approimae in blue

101 Eample : Euler s Mehod ( of 3) For he iniial value problem d d 3.5, () we can use Euler s mehod wih various sep sizes o approimae he soluion a =.,., 3., 4., and 5. and compare our resuls o he eac soluion e a hose values of.

102 Eample : Euler s Mehod ( of 3) Comparison of eac soluion wih Euler s Mehod for h =.,.5,.5,. h =. h =.5 h =.5 h =. EXACT

103 Percenage Error Eample : Euler s Mehod (3 of 3) 5 Percenage Error Decreases as Sep Size Decreases h=. h=.5 h=.5 h=. - -5

104 Eample 3: Euler s Mehod ( of 3) For he iniial value problem d 4, () d we can use Euler s mehod wih h =. o approimae he soluion a =,, 3, and 4, as shown below f f f f 3 h h h h Eac soluion (see Chaper.): ()() (.).6 4. ()(.6) ()(3.5) (.) 4. ()(.3) (.) 3.5 e 4 (.).3 4.5

105 Eample 3: Error Analsis ( of 3) The firs en Euler approimaionss are given in able below on lef. A able of approimaions for =,,, 3 is given on righ for h =.. See e for numerical resuls wih h =.5,.5,.. The errors are small iniiall, bu quickl reach an unaccepable level. This suggess a nonlinear soluion. Eac Appro Error % Rel Error Eac Appro Error % Rel Error Eac Soluion : 7 4 e 4

106 Eample 3: Error Analsis & Graphs (3 of 3) Given below are graphs showing he eac soluion (red) ploed ogeher wih he Euler approimaion (blue). Eac Appro Error % Rel Error Eac Soluion : 7 4 e 4

107 General Error Analsis Discussion ( of ) Recall ha if f and f / are coninuous, hen our firs order iniial value problem d f (, ), ( ) d has a soluion = f() in some inerval abou. In fac, he equaion has infiniel man soluions, each one indeed b a consan c deermined b he iniial condiion. Thus f() is he member of an infinie famil of soluions ha saisfies f( ) =.

108 General Error Analsis Discussion ( of ) The firs sep of Euler s mehod uses he angen line o f a he poin (, ) in order o esimae f( ) wih. The poin (, ) is picall no on he graph of f, because is an approimaion of f( ). Thus he ne ieraion of Euler s mehod does no use a angen line approimaion o f, bu raher o a nearb soluion ha passes hrough he poin (, ). f Thus Euler s mehod uses a succession of angen lines o a sequence of differen soluions f(), f (), f (),... of he differenial equaion.

109 Error Bounds and Numerical Mehods In using a numerical procedure, keep in mind he quesion of wheher he resuls are accurae enough o be useful. In our eamples, we compared approimaions wih eac soluions. However, numerical procedures are usuall used when an eac soluion is no available. Wha is needed are bounds for (or esimaes of) errors, which do no require knowledge of eac soluion. More discussion on hese issues and oher numerical mehods is given in Chaper 8. Since numerical approimaions ideall reflec behavior of soluion, a member of a diverging famil of soluions is harder o approimae han a member of a converging famil. Also, direcion fields are ofen a relaivel eas firs sep in undersanding behavior of soluions.

110 Boce/DiPrima/Meade h ed, Ch.8: The Eisence and Uniqueness Theorem Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. The purpose of his secion is o prove Theorem.4., he fundamenal eisence and uniqueness heorem for firs order iniial value problems. This heorem saes ha under cerain condiions on f(, ), he iniial value problem ' f (, ), ( ) has a unique soluion in some inerval conaining. Firs, we noe ha i is sufficien o consider he problem in which he poin (, ) is he origin. If some oher iniial poin is given, we can alwas make a preliminar change of variables, corresponding o a ranslaion of he coordinae aes, ha will ake he given poin ino he origin.

111 Theorem.8. If f and f / are coninuous in a recangle R: a, b, hen here is some inerval h a in which here eiss a unique soluion () of he iniial value problem ' f (, ), () We will begin he proof b ransforming he differenial equaion ino an inegral equaion. If we suppose ha here is a differeniable funcion () ha saisfies he iniial value problem, hen f [, ( )] is a coninuous funcion of onl. Hence we can inegrae ' f (, ) f (, ( )) from he iniial value = o an arbirar value, obaining ( ) f [ s, ( s)] ds

112 Proving he Theorem for he Inegral Equaion I is more convenien o show ha here is a unique soluion o he inegral equaion in a cerain inerval h han o show ha here is a unique soluion o he corresponding differenial equaion. The inegral equaion also saisfies he iniial condiion. f() = ò f (s,f(s))ds Þ f() = s is a dumm variable The same conclusion will hen hold for he iniial value problem ' f (, ), () as holds for he inegral equaion. ( )

113 The Mehod of Successive Approimaions One mehod of showing ha he inegral equaion has a unique soluion is known as he mehod of successive approimaions or Picard s ieraion mehod. We begin b choosing an iniial funcion ha in some wa approimaes he soluion. The simples choice uilizes he iniial condiion The ne approimaion is obained b subsiuing for f(s) ino he righ side of he inegral equaion. Thus Similarl, f () = And in general, f () = ò f (s,f (s))ds = ò f (s,)ds f () = ò f n+ () = f (s,f (s))ds ò f (s,f n (s))ds f (s)

114 n ( ) Eamining he Sequence f [ s, ( s)] ds n As described on he previous slide, we can generae he sequence wih {f n } = f,f,f,...,f n,... f () = and f n+ () = f (s,f n (s))ds Each member of he sequence saisfied he iniial condiion, bu in general none saisfies he differenial equaion. However, if for some n = k, we find k ( ) ( ), hen k k () is a soluion of he inegral equaion and hence of he iniial value problem, and he sequence is erminaed. In general, he sequence does no erminae, so we mus consider he enire infinie sequence. Then o prove he heorem, we answer four principal quesions. ò

115 Four Principal Quesions abou he Sequence. Do all members of he sequence { n } eis, or ma he process break down a some sage?. Does he sequence converge? n ( ) 3. Wha are he properies of he limi funcion? In paricular, does i saisf he inegral equaion and hence he corresponding iniial value problem? 4. Is his he onl soluion or ma here be ohers? f [ s, ( s)] ds n To gain insigh ino how hese quesions can be answered, we will begin b considering a relaivel simple eample.

116 Eample : An Iniial Value Problem ( of 6) We will use successive approimaions o solve he iniial value problem ' ( ), () Noe firs ha he corresponding inegral equaion becomes f() = ò s(+f(s)])ds ( ) The iniial approimaion generaes he following: ò f () = s(+ )ds = s ds = ò ò f () = s(+ s )ds = (s + s 3 )ds = + 4 ò f 3 () = ò s(+ s + s 4 / )ds = ò (s+ s 3 + s 5 )ds =

117 Eample : An Inducive Proof ( of 6) The evolving sequence suggess ha ( ) s[ ( s)] ds f n () = + 4! + 6 n ! n! This can be proved rue for all n b mahemaical inducion. I was alread esablished for n = and if we assume i is rue for n = k, we can prove i rue for n = k+: Thus, he inducive proof is complee.

118 Eample : The Limi of he Sequence (3 of 6) A plo of he firs five ieraes suggess evenual convergence o a limi funcion: Taking he limi as n and recognizing he Talor series and he funcion o which i converges, we have: ) ( ) ( ) ( 5! / /4! /3! /! ) ( n n n!! lim ) ( lim k k n k k n n n e k k

119 Eample : The Soluion (4 of 6) Now ha we have an epression for ( ) le us eamine k inerval of convergence: lim ( ) n ( ) ( ) n lim n n k k k! lim ( ) e n k for increasing values of k in order o ge a sense of he n k e k!..5 k= k=. k= The inerval of convergence increases as k increases, so he erms of he sequence provide a good approimaion o he soluion abou an inerval conaining =.

120 ( ) s[ ( s)] ds Eample : The Soluion Is Unique (5 of 6) To deal wih he quesion of uniqueness, suppose ha he IVP has wo soluions ( ) and ( ). Boh funcions mus saisf he inegral equaion. We will show ha heir difference is zero: ( ) ( ) A s[ ( s)] ds For he las inequali, we resric o A/, where A is arbirar, hen A. s[ ( s) ( s)] ds ( s) ( s) ds s[ ( s)] ds s( s) ( s) ds

121 ( ) ( ) A ( s) ( s) Eample : The Soluion Is Unique (6 of 6) ds I is now convenien o define a funcion U such ha U( ) ( s) ( s) ds Noice ha U() = and U() for and U() is differeniable wih U' ( ) ( ) ( ). This gives: U '() - AU() and mulipling b e A (e A U())' Þ e A U() ÞU() for ³ The onl wa for he funcion U() o be boh greaer han and less han zero is for i o be idenicall zero. A similar argumen applies in he case where. Thus we can conclude ha our soluion is unique.

122 Theorem.8.: The Firs Sep in he Proof Reurning o he general problem, do all members of he sequence eis? In he general case, he coninui of f and is parial wih respec o were assumed onl in he recangle R: a, b. Furhermore, he members of he sequence canno usuall be eplicil deermined. A heorem from calculus saes ha a funcion coninuous in a closed region is bounded here, so here is some posiive number M such ha f(,) M for (, ) in R. ( ) and '( ) f (, ( )) n ( ) f [ s, n( s)] ds Since n n n, he maimum slope for an funcion in he sequence is M. The graphs on page 88 of he e indicae how his ma impac he inerval over which he soluion is defined. ' f (, ), M ()

123 Theorem.8.: The Second Sep in he Proof The erms in he sequence can be wrien in he form The convergence of his sequence depends on being able o bound he value of. This can be esablished based on he fac ha is coninuous over a closed region and hence bounded here. Problems 5 hrough 8 in he e lead ou hrough his validaion. { n } n 3 )] ( ) ( [ ) ( ) ( lim and )] ( ) ( [ )] ( ) ( [ )] ( ) ( [ ) ( ) ( k k k n n n n ) ( ) ( k k f / n n ds s s f f )] (, [ ) ( () ),, ( '

124 Theorem.8.: The Third Sep in he Proof There are deails in his proof ha are beond he scope of he e. If we assume uniform convergence of our sequence over some inerval h a and he coninui of f and is firs parial derivaive wih respec o for h a, he following seps can be jusified: f() = limf n+ ()= lim n n ò f (s,f n (s))ds = ò lim f (s,f n (s))ds = ò f (s,limf n (s))ds ò n = f (s,f(s))ds n ' f (, ), () n ( ) f [ s, n( s)] ds

125 Theorem.8.: The Fourh Sep in he Proof () The seps oulined esablish he fac ha he funcion is a soluion o he inegral equaion and hence o he iniial value problem. To esablish is uniqueness, we would follow he seps oulined in Eample. We conjecure ha he IVP has wo soluions: ( ) and ( ). Boh funcions have o saisf he inegral equaion and we show ha heir difference is zero using he inequali: ò f() - () A f(s) - (s) ds ( ) () If he assumpions of his heorem are no saisfied, ou canno be guaraneed a unique soluion o he IVP. There ma be no soluion or here ma be more han one soluion. ' f (, ), f [ s, ( s)] ds

126 Boce/DiPrima/Meade h ed, Ch.9: Firs Order Difference Equaions Elemenar Differenial Equaions and Boundar Value Problems, h ediion, b William E. Boce, Richard C. DiPrima, and Doug Meade 7 b John Wile & Sons, Inc. Alhough a coninuous model leading o a differenial equaion is reasonable and aracive for man problems, here are some cases in which a discree model ma be more appropriae. Eamples of his include accouns where ineres is paid or charged monhl raher han coninuousl, applicaions involving drug dosages, and cerain populaion growh problems where he populaion one ear depends on he populaion in he previous ear. For eample, n f ( n, n), n,,,... Noice here ha he independen variable n is discree. Such equaions are classified according o order, as linear or nonlinear, as homogeneous or nonhomogeneous. There is frequenl an iniial condiion describing he firs erm.

127 Difference Equaion and Equilibrium Soluion Assume for now ha he sae a ear n + depends onl on he sae a ear n, and no on he value of n iself n+ = f (n, n ), n =,,,... Then = f ( ), = f ( ) = f ( f ( )), 3 = f ( ) = f 3 ( ),..., n = f n ( ) This procedure is referred o as ieraing he difference and i is ofen of ineres o deermine he behavior of as n. An equilibrium soluion eiss when f ( n ) n and his is ofen of special ineres, jus as i is in differenial equaions. n

128 Linear Homogeneous Difference Equaions Suppose ha he populaion of a cerain species in a region in ear n + is a posiive muliple of he populaion in ear n:, n,,,... Noice ha he reproducion rae ma differ from ear o ear.,, If he reproducion rae has he same value ρ for all n: n If he iniial value is zero, hen he equilibrium soluion = Oherwise lim n = n n n n, n ì, if r < ; Þ asmpoicall sable ï í, if r = ; ï îdoes no eis, oherwise. Þ asmpoicall unsable n n