1 ode.mcd. Find solution to ODE dy/dx=f(x,y). Instructor: Nam Sun Wang

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1 Fin solution to ODE /=f(). Instructor: Nam Sun Wang oe.mc Backgroun. Wen a sstem canges wit time or wit location, a set of ifferential equations tat contains erivative terms "/" escribe suc a namic sstem. In contrast, algebraic equations escribe a static sstem. Static: algebraic equation: f( ) Dnamic: ifferential equation: g or f( )... stanar form were is te inepenent variable, an is te epenent variable tat varies wit. Wen tere is onl one inepenent variable te equation is calle an orinar ifferential equation (ODE); te function f() tat appears on te RHS in te equation above escribes ow canges (i.e.,) wit canges in (i.e., ). Wen tere are more tan one inepenent variables (sa t for time an z for spatial coorinates), te equation is calle a partial ifferential equation, wic is not aresse ere. Partial ifferential equation: g t,, t... not aresse ere e number of equations can be one or man. In te former case, is a scalar function of an f similarl is a scalar function of bot an. In te latter case, is a vector function of an f too is a vector function of bot an. us, if we allow to be eiter a scalar function or a vector function, te same matematical equation escribes bot a one equation case an a multiple couple equation case. A special case of an ODE is wen f is a function of onl ; in tat case, te solution is a efinite integral. f( ) ( ) f( ξ ) ξ Fin () b evaluating te efinite integral Anoter special case of an ODE is wen f is a function of onl ; in tat case, an implicit solution eists in te form of an integral algebraic equation. f( ) ψ f( ψ) A more general case is wen f is a function of bot an. Fin () b solving te algebraic equation. f( ) Initial Conition (I.C.)... initial value problem f(, ) Bounar Conition (B.C.),... bounar value problem

2 oe.mc In contrast to a set of algebraic equations f()= were a single point is a solution, in ODE, a function () efine in te interval of interest =[a b] constitutes a solution. Wen all te conitions are given at, one single value of te ODE is classifie as an initial value problem (even wen is not te initial point, sa te en point or at a mi-poinn te interval of interest [a b]). Wen te conitions are given at more tan one point of sa at bot an, te ODE is classifie as a bounar value problem (even wen is not locate at te two bounaries a & b). In general, tere are infinite number of functions () tat satisf te ODE /=f(). However, wen an initial conition is impose, tere is onl one unique solution, accoring to te eistence an uniqueness teorem. is is analogous to linear algebraic equations aving one unique solution. Wen bounar conitions are impose instea, anting is possible: no solution, one solution, or multiple solutions. is is analogous to nonlinear algebraic equations aving te possibilities of no solution, one solution, or multiple solutions. In aition, te total number of conitions tat can be impose equals te number of first-orer ifferential equations. is is analogous to aving to matc te number of unknowns to te number of inepenent equations in solving algebraic equations. f( ) I.C. analogous A. b f(, ) B.C. f( ( a), '( a), ( b), '( b ) ) analogous f( ) Stanar notation: /=f(). o evelop a sstematic approac to solving orinar ifferential equations, we use te "stanar" first-orer /=f() notation. o re-write a given ifferential equation into tis stanar form, is a matter of moving all te first erivative terms to te LHS of te equation an leaving everting else on te RHS. All n-orer or iger orer erivatives are first transforme to te "stanar" first-orer /=f()= form. Eample: One 3r-orer oe. 3 α(. z. β( z γ(. 3 z δ( Let z z oe z z z oe z z 3 z 3 Substituting back into te z/ z/, an 3 z/ 3 terms in te original ODE, α(. z β(. z γ(. z δ( Furter substituting back into te z/ term above, oe3 α(. z β(. z γ(. z δ( z ) z. α( z ) ( β(. z γ(. z δ( ) In summar we re-write te original 3r-orer ODE as tree st-orer ODEs.

3 z z z z oe 3 oe.mc oe z. α( z ) ( β(. z γ(. z δ( ) oe3 Or, in a vector notation. z z z z z z f ( z, z, f ( z, z, f 3 ( z, z, z z. α( z ) ( β(. z γ(. z δ( ) Or, te above epresse wit te "prime" notation for erivative. z z' z'' z z' z'' f ( z, z', z'' ) f ( z, z', z'' ) f 3 ( z, z', z'' ) z' z''. α( z ) ( β(. z'' γ(. z' δ( ) z z z f( ) f,, 3 f,, 3 f 3,, 3 3. β 3 γ δ α is last "=" efines te vector function f in one step, bus absolutel not recommene! oo confusing an almosmpossible to ebug! We also transform initial/bounar conitions accoringl. Note tat, te transformation operation is not unique; tere are oter was to transform to te first-orer stanar form. r for eample, z=z/+z (rater tan z=z/). Eample: wo couple n-orer oes. α(. u β(. u γ( δ(. v φ(. v χ(

4 4 oe.mc Let z u oe z u z v oe z v Substituting back into te u/ u/, v/ an v/ terms in te original set of ODEs, α(. z β(. z γ( v ) z. α( v ) ( β(. z γ( ) oe3 δ(. z φ(. z χ( v ) z. δ( v ) ( φ(. z χ( ) oe4 In summar we re-write te original two r-orer ODE as four couple st-orer ODEs. u z v z oe oe z. α( v ) ( β(. z γ( ) oe3 z. δ( v ) ( φ(. z χ( ) oe4 Or, in a vector notation. u v z z u v z z u v z z f( ) f ( v, z, f ( v, z, f 3 ( v, z, f 4 ( v, z, f,, 3, 4 f,, 3, 4 f 3,, 3, 4 f 3,, 3, 4 z z. α( v ) ( β(. z γ( ).( φ(. z χ( ) δ( 3 4. β, 3 γ, α,. φ, 4 χ, δ, is last "=" efines te vector function f in one step, bus absolutel not recommene! oo confusing an almosmpossible to ebug!

5 5 oe.mc Numerical Algoritm #. Euler's Meto. Numericall we approimate te solution curve () wit a series of paire points ( i, i ), were te ine i represents te it point; i=..n (i.e., n steps intervals). We start te solution curve () wit te given value of = at =. We ten approimate te value of at te net point = +, were is te step size Δ=. An estimate of cange in (Δ) over tis interval Δ is given b f(, ), i.e., te function evaluate at te starting point (, ). In oter wors, te slope witin tis interval is estimate b te slope at te starting point. Δ Δ f, Δ f, Δ f,. f, Basicall we estimate =( ). Once we know (te value of at ), we can regar as te starting point an fin =( ) (te value of at ). f, We repeat tis process to avance te curve () step after step until we cover te entire range of interest. i i f i, i... Euler's iteration formula i E f.5 Euler's Meto slope = k =f(, ) In Euler's meto, te slope witin an interval is estimate b te one slope evaluate at te starting point (, ) of te interval. e given function f() epens on bot an ; tus, its value is constantl canging witin te interval an f(, ) oes not best represent of te average slope in tis interval. Man moifications eist. Basicall ifferent versions mainl iffer in te location were f() is evaluate (i.e., were slopes are estimate), te number of locations (i.e., te number of slopes), an ow te evaluate values of f() are weigte in fining te average value (i.e., ow to average slopes). k

6 oe.mc n-orer Meto. e first slope is evaluate at te starting point, as in te Euler's Meto. A secon slope is evaluate at te mi-point of te interval, wit te value at te mi-point estimate base on te st slope. An te average slope is a combination of te first slope an te secon slope. Sometimes, te two slopes are weigte equall. Sometimes, te secon slope evaluate at te mi-poins weigte more eavil sa X as muc as te first slope evaluate at te starting point, because te mi-point value is likel to be more representative of te slope in te interval. st slope, evaluate at (, ): k f, n slope, evaluate at mi-point of interval, i.e., at ( +/, +k /): k f, Average slope: k ave k k Estimate at :. k ave or k ave k. k 3 or some oter weigting sceme k. i n-orer Meto slope = k =f( +/, +k /).5 slope = k =f(, ) slope = k ave =(k +k )/3 k 3.5 3r-Orer Meto. We continue te n-orer meto above an evaluate te tir slope at te mi-point, wit te value at te mi-point estimate base on te n slope. st slope, evaluate at (, ): n slope, evaluate at mi-point of interval, i.e., at ( +/, +k /): k f, k f, k. 3r slope, evaluate at mi-point of interval, i.e., at ( +/, +k /): k 3 f, k k k k k. k. k 3 Average slope: k ave or k ave or some oter weigting sceme 3 5 Estimate at :. k ave i

7 7 oe.mc 3r-Orer Meto slope = k =f( +/, +k /).5 slope = k =f(, ) slope = k ave =(k + k + k 3 )/5 slope = k 3 =f( +/, +k /) slope = k.5 Numerical Algoritm #. Runge-Kutta Meto. 4t-Orer Meto. We continue te 3r-orer meto above an evaluate te fourt slope at, te en of te step, wit te value at te en of te step estimate base on te 3r slope. k f, st slope, evaluate at (, ): k f, k n slope, evaluate at mi-point of interval, i.e., at ( +/, +k /): 3r slope, evaluate at mi-point of interval, i.e., at ( +/, +k /): k 3 f, k. 3r slope, evaluate at en of interval, i.e., at ( +, +k 3 ): k 4 f, k 3 k k k 4 Average slope: k ave 4 Estimate at : k ave k. k. k 3 k 4 or k ave or some oter weigting sceme k 4t-Orer Meto slope = k 4 =f( +, +k 3 ) slope = k =f( +/, +k /).5 slope = k =f(, ) slope = k ave =(k + k + k 3 )/5 slope = k 3 =f( +/, +k /) slope = k.5

8 8 oe.mc Runge-Kutta's 4t-Orer Meto. We can go on an evaluate more slopes at more points witin te interval =[ ]. However, past stuies ave sown tat slopes more tan four cease to improve results. us, is common to stop at evaluating four slopes. In Runge-Kutta's 4t orer meto, te slopes at te mi-point (k an k 3 ) are generall weigte twice as eavil as tose evaluate at te starting point (k ) an at te en point (k 4 ). As in te Euler's meto, te en point of te previous step becomes te starting point of te net step, an we repeat te process until we cover te entire interval of interest. Summar. k f i, st slope, evaluate at (i, i i ): k f i, k n slope, evaluate at mi-point of interval, i.e., at ( i i +/, i +k /): 3r slope, evaluate at mi-point of interval, i.e., at ( i +/, i +k /): k 3 f i, i k. 3r slope, evaluate at en of interval, i.e., at ( i +, i +k 3 ): k 4 f i, i k 3 Average slope: k ave k. k. k 3 k 4 Start at te given I.C. (, ), iterate: i i. k ave Avantages & Disavantages. e avantage of te Euler's meto is tat requires onl one slope evaluation an is simple to appl especiall for iscretel sample (eperimental) ata points; te isavantage is tat errors accumulate uring successive iterations an te results are not ver accurate. e avantage of te Runge-Kutta's metos is tas accurate, but requires four slope evaluations at four ifferent points of (); tese slope evaluations are not possible for iscretel sample ata points, because all we ave is was given to us an we o not get to coose at will were to evaluate slopes. is is analogous to estimating a efinite integral wit te trapezoial rule -- if iscrete points are given to us, we o not get to ivie into finer steps at will to improve accurac sa we o not get to evaluate at a point alfwa between an wen we are given te table below. Given: : : n n Multiple couple set of ifferential equations. Appl te same Euler's formula or Runge-Kutta's formula as above, ecept te smbols f, k, k, k 3, k 4, k ave, are (column) vectors.

9 9 oe.mc Eample: Numerical implementation of Euler's Meto I.C. ( ) ( ) step size Number of steps n i.. n start at i i i i i, i i.. n 8 i 4 5 i Given /=f(), efine a function tat returns te net step solution base on Euler's meto witout programming. euler( f, ) f( ). Given /=f(), efine a function tat returns te net step solution base on Runge-Kutta's meto witout programming. -- efine in multiple program lines. Note tat programming is not necessar. k ( f, ) f( ) k ( f, ) f, k. ( f, ) k 3 ( f, ) f, k. ( f, ) k 4 ( f, ) f, k. 3 ( f, ) k ave ( f, ). k ( f, ). k ( f, ). k 3 ( f, ) k 4 ( f, ) rk( f, ) k. ave ( f, ) Given /=f(), efine a function tat returns te net step solution base on Runge-Kutta's meto witout programming. -- combine te above steps all in one program line (but toug to eciper). rk( f, ) f( ). f, f( ).. f, f,. f( ). f, f,

10 oe.mc Eample. call te "euler" & "rk" functions, one oe. ( ) step size Number of steps n i.. n start at i i i euler i, i, euler i rk i, i, rk i.. n 8 One ODE euleri rki 4 5 Euler Runge-Kutta i

11 Eample. call te "euler" & "rk" functions, two couple oes. oe.mc a b c rt( r, f ) a. r b. r. f... rabbit (pre) I.C. t r ft( r, f ) c. f. r. f... fo (preator) f vector notation r f t( t, ) rt, ft, I.C.... reset previous junk r f step size. Number of steps n i.. n & t are column vectors. If te Euler formula an te Runge-Kutta formula remain uncange, ten all & te slopes etc (k, k etc) are also column vectors. us, we store ifferent time points in ifferent columns. avance troug t & wit te "euler" function i euler t,, i, ecompose back m own variables (in column vectors) r euler avance troug t & wit te "rk" function i rk t,, i, f euler ecompose back m own variables (in column vectors) r rk f rk i.. n Couple ODEs (Preator-Pre Dnamics) r euleri 4 f euleri r rki f rki 4 8 rabbit - Euler fo - Euler rabbit - Runge-Kutta fo - Runge-Kutta

12 oe.mc Effect of number of function evaluations (i.e., number of slopes). n-orer meto k ave ( f, ). 3 k ( ). k ( f, ) rk( f, ) k. ave ( f, ) < i > rk t,, i, r rk f rk 3r-orer meto k ave3 ( f, ). 5 k ( ). k ( f, ). k 3 ( f, ) rk3( f, ) k. ave3 ( f, ) < i > rk3 t,, i, r rk3 f rk3 Couple ODEs (Preator-Pre Dnamics) r euleri 5 f euleri r rki 4 f rki r rk3i 3 f rk3i r rki f rki 4 8 rabbit - Euler fo - Euler rabbit - n-orer fo - n-orer rabbit - 3r-Orer fo - 3r-Orer rabbit - Runge-Kutta fo - Runge-Kutta

13 3 oe.mc "Initial Conition" given at te en of interval of interest, =b. We appl te same formula to integrate backwar from =b to =a (a<b) witout cange, ecept tat te step size is now a negative number. Eample, I.C., at t=, r=, f=; we seek solution in t=[ ]. reset previous junk r euler f euler r rk f rk negative step size. Number of steps n i.. n I.C. ("initial conition", altoug given not at te initial point, but at te en of te interval t=[ ] r t f We iterate eactl te same wa as before for bot te Euler meto an te Runge-Kutta meto. avance backwar troug t & wit te "euler" function i euler t,, i, ecompose back m own variables (in column vectors) r euler avance backwar troug t & wit te "rk" function i rk t,, i, f euler ecompose back m own variables (in column vectors) r rk f rk i.. n 5 Integration Backwar r euleri 4 f euleri r rki 3 f rki r "Initial" conition at =b f "Initial" conition at =b 4 8 t, t, t, t, t, i i i rabbit - Euler fo - Euler rabbit - Runge-Kutta fo - Runge-Kutta rabbit - "Initial" (reall means "Final") fo - "Initial" (reall means "Final")

14 4 oe.mc "Initial Conition" given at one point witin te interval of interest, sa at =c, a<c<b. We appl te same formula to integrate backwar from =c towar =a witout cange, ecept tat te step size is now a negative number. We also integrate forwar from =c towar =b. Eample, I.C., at t=5, r=, f=; we seek solution in =[ ]. reset previous junk r euler f euler r rk f rk I.C. ("initial conition", altoug given not at te initial point, but at a point witin te interval t=[ ] t 5 r f t 5 t 5 r f Euler meto: we iterate eactl te same wa as before. negative step size. Number of steps backwar. m 5 i m, m.. avance backwar troug t & wit te "euler" function i euler t,, i, positive step size. Number of steps forwar n 5 i m.. m n avance forwar troug t & wit te "euler" function i euler t,, i, ecompose back m own variables (in column vectors) r euler Runge-Kutta meto: we iterate eactl te same wa as before. f euler negative step size. Number of steps backwar. m 5 i m, m.. avance backwar troug t & wit te "euler" function i rk t,, i, positive step size. Number of steps forwar n 5 i m.. m n avance forwar troug t & wit te "euler" function i rk t,, i, ecompose back m own variables (in column vectors) r rk f rk i.. m n

15 5 oe.mc 4 Integration Backwar & Forwar r euleri 3 f euleri r rki f rki "Initial" conition at a< <b witin te interval of interest t=[a,b] r f "Initial" conition at a< <b witin te interval of interest t=[a,b] 4 8 t, t, t, t, t, i i i i t rabbit - Euler fo - Euler rabbit - Runge-Kutta fo - Runge-Kutta rabbit - "Initial" (reall means "Mile" fo - "Initial" (reall means "Mile"

16 oe.mc Convergence Ceck # -- Decrease Step Size (Increase # Steps). How o we gain confience te solution is a correct one? We tpicall ecrease te step size or, equivalentl increase te number of points to cover te same interval of interest. We ten eamine if te solution as converge. Below, we eamine te ifference from oubling te number of steps. reset previous junk r euler f euler r rk f rk Euler's Meto. n Cange te number of steps (te curves look inistinguisable at n=5) Base calculation i.. n step size n t r f i euler t,, i, r euler Double te number of points i... n step size. n i euler t,, i, r euler f euler i.. n r t t i... n f f euler Convergence Ceck r euleri f euleri 4 r euleri f euleri 4 8 rabbit - Euler fo - Euler rabbit - Euler (X number of steps) fo - Euler (X number of steps) t, t, t, i i i

17 Runge-Kutta's Meto. 7 oe.mc n Cange te number of steps (te curves look inistinguisable at n=5) Base calculation i.. n step size i rk t,, i, r rk n t... reset previous junk f rk r f i.. n Double te number of points i... n step size. n i rk t,, i, r rk t t... reset previous junk f rk r f i... n 3 Convergence Ceck r rki f rki r rki f rki 4 8 t, t, t, i i i rabbit - Runge-Kutta fo - Runge-Kutta rabbit - Runge-Kutta (X number of steps) fo - Runge-Kutta (X number of steps) In canging te number of steps above, we see tat Euler's meto nees ~5 steps to converge; wereas, Runge-Kutta nees onl ~5 steps (wic is equivalent to 54= function evaluations). us, even factoring in te 4 function evaluations per step in Runge-Kutta's meto, we see tat Runge-Kutta's meto gives better results in significantl less number of steps. As we mentione before, more number of function evaluations (i.e., slopes) for eac step beon te four in Runge-Kutta's meto ceases to improve results.

18 8 oe.mc Convergence Ceck # -- Integrate forwar ten reverse. Anoter wa to ceck for convergence is to integrate forwar first, ten traverse te steps backwar. Eample, I.C., at t=, r=, f=; we seek solution in t=[ ]. reset previous junk r euler f euler r rk f rk Euler Meto. Number of steps n Cange te number of steps to see convergence integrate forwar wit positive step size i.. n n I.C. t t i i euler t,, i, integrate backwar wit negative step size i n... n n i euler t,, i, r euler f euler Runge-Kutta Meto. integrate forwar wit positive step size I.C. t n i rk t,, i, integrate backwar wit negative step size n i rk t,, i, r rk Integration Backwar i.. n i n... n f rk i... n r euleri 4 f euleri r rki f rki 4 8 rabbit - Euler fo - Euler rabbit - Runge-Kutta fo - Runge-Kutta

19 9 oe.mc Eample: Manuall calculate for one iteration to avance from = to = in one step. I.C. ( ) In te stanar form, te function f() is f( ) Algoritm. Euler's Meto. step size Δ start at value of at f,. Algoritm. Runge-Kutta's Meto. st slope, evaluate at: k f, f(, ) n slope, evaluate at.5 k f, k. 3r slope, evaluate at.5 k 3 f, k. k...5 f(.5,.5) k f(.5,.444 ) t slope, evaluate at. k Average of 4 slopes: k 4 f, k 3 f(,.93 ) k. k. k 3 k 4 k ave value of at. k ave Eample: couple preator-pre namics. Manuall avance from t= to t=. in one step. rt( r, f ) a. r b. r. f... rabbit (pre) I.C. r( ) r ft( r, f ) c. f. r. f... fo (preator) f( ) f were a b c In te stanar form, te function f() is r a. r b. r. f f r ( t, r, f ). r. r. f t f t c. f. r. f f f ( t, r, f ). f. r. f

20 Algoritm. Euler's Meto. step size Δt. start at t r f oe.mc value of at t. r r. f r t, r, f r.. r.. r f (... )...8 Algoritm. Runge-Kutta's Meto. f f. f f t, r, f f.. f.. r f (... ).. st slope, evaluate at: t t r r f f n slope, evaluate at t t. k r f r t, r, f f r (,, )... k f f f t, r, f f f (,, )... k r f r t,, r. k r f k f f f t,, r. k r f 3r slope, evaluate at t t. k r3 f r t,, r. k r f k f3 f f t,, r. k r f.5 r r. k r.. f f. k f.. k. f k. f.5 r r. k r.8.. f f. k f... k. f k. f.9 f r (.5,.9, ) f f (.5,.9, ) t slope, evaluate at t t.. r r. k r f r (.5,.9,.99 ) f f (.5,.9,.99 ) f f. k f k r4 f r t, r k. r3, f. k f3 f r (.,.8,.98 ) k f4 f f t, r k. r3, f. k f3 f f (.,.8,.98 ) Average of 4 slopes: k r.ave k r. k r. k r3 k r4 k f.ave k f. k f. k f3 k f4 value of at t. r r. k r.ave f f. k f.ave