lecture 35: Linear Multistep Mehods: Truncation Error

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1 88 lecture 5: Linear Multistep Meods: Truncation Error 5.5 Linear ultistep etods One-step etods construct an approxiate solution x k+ x(t k+ ) using only one previous approxiation, x k. Tis approac enoys te virtue tat te step size can be canged at every iteration, if desired, tus providing a ecanis for error control. Tis flexibility coes at a price: For eac order of accuracy in te truncation error, eac step ust perfor at least one new evaluation of te derivative function f (t, x). Tis igt not sound particularly onerous, but in any practical probles, f evaluations are terribly tie-consuing. A classic exaple is te N-body proble, wic arises in odels ranging fro olecular dynaics to galactic evolution. Suc odels give rise to N coupled second order nonlinear differential equations, were te function f easures te forces between te N different particles. An evaluation of f requires O(N 2 ) aritetic operations to copute, costly indeed wen N is in te illions. Every f evaluation counts. A landark iproveent to tis N 2 One could potentially iprove tis situation by re-using f evaluations fro previous steps of te ODE integrator, rater tan always requiring f to be evaluated at ultiple new (t, x) values at eac step (as is te case, for exaple, wit iger order Runge Kutta etods). Consider te etod x k+ = x k + 2 f (t k, x k ) 2 f (t k, x k ), approac, te fast ultipole etod, was developed by Leslie Greengard and Vladiir Roklin in te late 980s. were is te step size, = t k+ t k = t k t k. Here x k+ is deterined fro te two previous values, x k and x k. Unlike Runge Kutta etods, f is not evaluated at points between t k and t k+. Rater, eac step requires only one new f evaluation, since f (t k, x k ) would ave been coputed already at te previous step. Hence tis etod as rougly te sae coputational requireents as Euler s etod, toug soon we will see tat its truncation error is O( 2 ). Te Heun and idpoint rules attained tis sae truncation error, but required two new f evaluations at eac step. Several drawbacks to tis new scee are evident: te step size is fixed trougout te iteration, and values for bot x 0 and x are needed before starting te etod. Te forer concern can be addressed in practice troug interpolation tecniques. To andle te latter concern, initial data can be generated using a one-step etod wit sall step size. In soe applications, including soe probles in celestial ecanics, an asyptotic series expansion of te solution, accurate near t t 0, can provide suitable initial data. Step-size adustent is possible, e.g., interpolating (t k, x k ) and (t k, x k ) to get interediate values, but step-size adustent still requires ore care tan wit one-step etods.

2 General linear ultistep etods Tis section considers a general class of integrators known as linear ultistep etods. Definition 5.2. general -step linear ultistep etod as te for a x k+ = b f (t k+, x k+ ), Tese notes on linear ultistep etods draw eavily fro te excellent presentation in Süli and Mayers, Nuerical Analysis: An Introduction, Cabridge University Press, 200. wit a 6= 0. If b 6= 0, ten te forula for x k+ involves x k+ on te rigt and side, so te etod is iplicit; oterwise, te etod is explicit. A final convention requires a 0 + b 0 6= 0, for if a 0 = b 0 = 0, ten we actually ave an step etod asquerading as a -step etod. As f is only evaluated at (t, x ), we adopt te abbreviation f = f (t, x ). By tis definition ost Runge Kutta etods, toug one-step etods, are not ultistep etods. Euler s etod is an exaple of a one-step etod tat also fits tis ultistep teplate. Here are a few exaples of linear ultistep etods: Euler s etod: x k+ x k = f k a 0 =, a = ; b 0 =, b = 0. Trapezoid rule: x k+ x k = 2 ( f k + f k+ ) a 0 =, a = ; b 0 = 2, b = 2. Adas Basfort: x k+2 x k+ = 2 ( f k+ f k ) a 0 = 0, a =, a 2 = ; b 0 = 2, b = 2, b 2 = 0. Te Adas Basfort etod presented above is te 2-step exaple of a broader class of Adas Basfort forulas. Te -step Adas Basfort etod takes te for for wic x k+ = x k f k+ 59 f k f k+ 9 f k a 0 = 0, a = 0, a 2 = 0, a =, a = ; b 0 = 9, b = 7, b 2 = 59, b = 55, b = 0. Te Adas Moulton etods are a parallel class of iplicit forulas. Te -step version of tis etod is giving x k+ = x k f k+ + 9 f k+2 5 f k+ + f k, a 0 = 0, a = 0, a 2 =, a = ; b 0 =, b = 5, b 2 = 9, b = 9.

3 Truncation error for linear ultistep etods Recall tat te truncation error of one-step etods of te for x k+ = x k + F(t k, x k ; ) was given by T k = x(t k+) x(t k ) F(t k, x k ; ). Wit general linear ultistep etods is associated an analogous forula, based on substituting te exact solution x(t k ) for te approxiation x k, and rearranging ters. Definition 5.. Te truncation error for te linear ultistep etod is given by te forula T k = a x k+ = b f (t k+, x k+ ) i a x(t k+ ) b f (t k+, x(t k+ )) b. Te b ter in te denoinator is a noralization ter; were it absent, ultiplying te entire ultistep forula by a constant would alter te truncation error, but te not te iterates x k. To get a siple for of te truncation error, we turn to Taylor series: x(t k+ )= x(t k + ) = x(t k ) + x 0 (t k ) + 2 x00 (t k ) + x000 (t k ) +! x() (t k ) + x(t k+2 )= x(t k + 2) = x(t k ) + 2x 0 (t k ) x 00 (t k ) + 2 x 000 (t k ) +! x () (t k ) + x(t k+ )= x(t k + ) = x(t k ) + x 0 (t k ) x 00 (t k ) + x 000 (t k ) +! x () (t k ) +. x(t k+ )= x(t k + ) = x(t k ) + x 0 (t k ) x 00 (t k ) + x 000 (t k ) +! x () (t k ) + and also f (t k+, x(t k+ )) = x 0 (t k + ) = x 0 (t k ) + x 00 (t k ) + 2 x000 (t k ) + x() (t k ) + f (t k+2, x(t k+2 )) = x 0 (t k + 2) = x 0 (t k ) + 2x 00 (t k ) x 000 (t k ) + 2 x () (t k ) + f (t k+, x(t k+ )) = x 0 (t k + ) = x 0 (t k ) + x 00 (t k ) x 000 (t k ) + x () (t k ) +.. f (t k+, x(t k+ )) = x 0 (t k + ) = x 0 (t k ) + x 00 (t k ) x 000 (t k ) + x () (t k ) +. Substituting tese expansions into te expression for T k (eventually)

4 9 yields a convenient forula: i b T k = a x(t k+ ) b f (t k+, x(t k+ )) = = a ix(t k `=0 i a x(t k a b ix 0 (t k ) a a + a b ix 00 (t k ) ` a `+ ` i b (` + )! x (`+) (t `! k ) 2 i 2 b x 000 (t k ) i 6 b x () (t k. In particular, te coefficient of te ` ter is siply `+ (` + )! a ` `! b for all nonnegative integers `. Definition 5.. A linear ultistep etod is consistent if T k! 0 as! 0. Te forula for T k. gives an easy condition to ceck te consistency of a etod. Teore 5.2. An -step linear ultistep etod of te for is consistent if and only if a x k+ = b f k+ a = 0 and a = b. If one of te conditions in Teore 5.2 are violated, ten te forula for te truncation error contains eiter a ter tat grows like / or reains constant as! 0. Te Taylor analysis of te truncation error yields even ore inforation, toug: inspecting te coefficients ultiplying, 2, etc. reveals easy conditions for deterining te overall truncation error of a linear ultistep etod.

5 92 Definition 5.5. A linear ultistep etod is order-p accurate if T k = O( p ) as! 0. Teore 5.. An -step linear ultistep etod is order-p accurate if and only if it is consistent and for all ` =,..., p. `+ (` + )! a = ` `! b Te next few exaples sow tis teore in action, applying it to soe of te linear ultistep etods discussed earlier. Exaple 5.7 (Forward Euler etod). a 0 =, a = ; b 0 =, b = 0. Clearly a 0 + a = + = 0 and (0a 0 + a ) (b 0 + b )=0. Tus te etod is consistent. Wen analyzed as a one-step etod, te forward Euler etod ad truncation error T k = O(). Te sae result sould old wen we analyze te algorit as a linear ultistep etod. Indeed, Tus, T k = O() a a 0b 0 + b Exaple 5.8 (Trapezoid etod). = 2 6= 0. a 0 =, a = ; b 0 = 2, b = 2. Again, consistency is easy to verify: a 0 + a = + = 0 and (0a 0 + a ) (b 0 + b )= = 0. Furterore, a a (0b 0 + b )= 2 2 = 0, so T k = O( 2 ), but 6 0 a a b b )= 6 6= 0, so te trapezoid rule is not tird order accurate: T k = O( 2 ). Exaple 5.9 (2-step Adas Basfort). a 0 = 0, a =, a 2 = ; b 0 = 2, b = 2, b 2 = 0.

6 9 Does tis explicit 2-step etod deliver O( 2 ) accuracy, like te (iplicit) Trapezoid etod? Consistency follows easily: a 0 + a + a 2 = 0 + = 0 and (0a 0 + a + 2a 2 ) (b 0 + b )= = 0. Te second order condition is also satisfied, a a a 2 0b 0 + b = 2 2 = 0, but not te tird order, 6 0 a a a b b = 7 6 6= 0. Tus T k = O( 2 ): te etod is second order. Exaple 5.0 (-step Adas Basfort). a 0 = 0, a = 0, a 2 = 0, a =, a = ; b 0 = 9, b = 7, b 2 = 59, b = 55, b = 0. Consistency olds, since a = + = 0 and a b = ( () = = 0. Te coefficients of, 2, and in te expansion for T k all vanis: 2 2 a b = ( 2 () 9 0( ( 7 2( 59 ( 55 ) = = 0; 6 a 2 2 b = a 6 b = 6 ( 6 () 2 ( () 2 ( ( 59 6 ( ( ( 55 ) = ( 55 ) = 75 8 = 0; 050 = 0. However, te O( ) ter is not eliinated: 20 5 a b = ( 20 () ( 7 ( 59 ( 55 ) = = 0. Tus T k = O( ): te etod is fourt order. A siilar coputation establises fourt-order accuracy for te -step (iplicit) Adas Moulton forula x k+ = x k f k+ + 9 f k+2 5 f k+ + f k.

7 9 In te last lecture, we proved tat consistency iplies convergence for one-step etods. Essentially, provided te differential equation is sufficiently well-beaved (in te sense of Picard s Teore), ten te nuerical solution produced by a consistent one-step etod on te fixed interval [t 0, t final ] will converge to te true solution as! 0. Of course, tis is a key property tat we ope is sared by ultistep etods. Weter tis is true for general linear ultistep etods is te subect of te next lecture. For now, we erely present soe coputational evidence tat, for certain etods, te global error at t final beaves in te sae anner as te truncation error. Consider te odel proble x 0 (t) = x(t) for t 2 [0, ] wit x(0) =x 0 =, wic as te exact solution is x(t) =e t. We sall approxiate tis solution using Euler s etod, te second-order Adas Basfort forula, and te fourt-order Adas Basfort forula. Te latter two etods require data not only at t = 0, but also at several additional values, t =, t = 2, and t =. For tis siple experient, we can use te value of te exact solution, x = x(t ), x 2 = x(t 2 ), and x = x(t ). We close by offering evidence tat tere is ore to te analysis of linear ultistep etods tan truncation error. Here are two explicit etods tat are bot second order: x k+2 2 x k+ + 2 x k = ( 5 f k+ x k+2 x k+ + 2x k = ( 2 f k+ f k) 2 f k). We apply tese etods to te odel proble x 0 (t) =x(t) wit x(0) =, wit exact initial data x 0 = and x = e. Te results of tese two etods are sown below. Te first etod tracks te exact solution x(t) =e t very nicely. Te second etod, owever, sows a disturbing property: wile it atces up quite well for te initial steps, it soon starts to fall far fro te solution. Wy does tis second-order etod do so poorly for suc a siple proble? Does tis reveal a general proble wit linear ultistep etods? If not, ow do we identify suc ill-annered etods?