Trees and Order Conditions

Transcription

1 Trees and Order Condtons Constructon of Runge-Kutta order condtons usng Butcher trees and seres. Paul Tranqull 1 1 Computatonal Scence Laboratory CSL) Department of Computer Scence Vrgna Tech. Trees and Order Condtons. [1/14] Dfferental Equatons group.

2 The general process for constructng order condtons s to: Construct a truncated Taylor expanson of the exact soluton comng from y = fy). Construct a truncated Taylor expanson of the numercal soluton comng from the general form of the Runge-Kutta ntegrator. Match terms n each expanson up to the specfed order p.. Order Condtons. [2/14] Dfferental Equatons group.

3 Start wth and y = fy) yt n + h) = yt n ) + hy t n ) + h2 2! y t n ) +... Note that each term s comprsed of a dervatve of y evaluated at t = t n,.e. when h = 0), a power of h, and a constant coeffcent. Snce the power of h and the constant coeffcent are easy to generalze, we wll loo at how to construct the terms whch come from the dervatves of y.. Order Condtons. Exact Soluton [3/14] Dfferental Equatons group.

4 For the scalar case we have that y = y y = fy) y = f y) y = f f y = f y) y y + f y) f y) y however, n our case y s a vector. So we use the notaton that captal superscrpts represent the ndex of the vector. Moreover, the appearance of a subscrpt represents a dervatve and the letter s the ndex of that dervatve. In ths way f K s the element, K) of the acoban matrx of f, and f KL s element, K, L) of the Hessan. We wll also mae use of ensten notaton, so that repeated ndces are summed over the range of that ndex. So that c x = 3 c x = c 1 x 1 + c 2 x 2 + c 3 x 3 =1. Order Condtons. Exact Soluton [4/14] Dfferental Equatons group.

5 Now we can construct each of the terms appearng n the taylor expanson of the exact soluton yt n + h) as ) y 0) h=0 ) y 1) h=0 y ) 1) h=0 ) y 2) h=0 y ) 2) h=0 y ) 2) h=0 ) y 3) h=0 y ) 3) h=0 ) y 3) h=0 = y = f y) ) 0) h=0 = f = f y) ) 1) h=0 = fk y) y K) ) 1) h=0 = f K f K = f y) ) 2) h=0 = f ) KL y K 1) ) y L 1) ) ) + f K y K 2) h=0 = f KL f K f L + f K f K L f L. Order Condtons. Exact Soluton [5/14] Dfferental Equatons group.

6 Collectng only the fnal results we have that up to thrd order, we have that ) y 0) h=0 ) y 1) h=0 ) y 2) h=0 ) y 3) h=0 = y = f = fk f K = fkl f K f L + fk f L Kf L are the terms, along wth the correspondng power of h and Taylor coeffcent, comprsng the Taylor seres expanson of the exact soluton.. Order Condtons. Exact Soluton [6/14] Dfferental Equatons group.

7 We can now represent the terms n the taylor seres usng labeled trees. l l f b = 1 fk f K b a = 1/2 fkl f K f L b a a l = 1/3 fk f L Kf L b a a l = 1/6 Where the rghtmost column comes from matchng the exact and numercal whch we wll derve later) expansons.. Order Condtons. Exact Soluton [7/14] Dfferental Equatons group.

8 For the elementary dfferentals superscrpts represent component ndces, and subscrpts represent ndces of varables wth respect to whch partal dervatves are taen. For example, f s the -th component of f, and f K f K = K f / y K f K. A node represents a dervatve of f. The order of the f dervatve equals the number of chldren the node has.. Order Condtons. Exact Soluton [8/14] Dfferental Equatons group.

9 Numercal Expanson Recall that an s-stage Runge-Kutta method has the general form = hfy n + a ) y n+1 = y n + b We wll construct the ndvdual terms n the expanson of the numercal soluton by expandng each stage value,, and then addng them together n the end. We wll mae heavy use of the formula hφh)) q) h=0 = q φh)) q 1) h=0. Numercal Soluton. [9/14] Dfferental Equatons group.

10 ) 0) ) 1) ) 1) ) 2) ) 2) ) 2) ) 2) = 0 = hf y n + 1) a )) = f = hf y n + 2) a )) = 2f y n + 1) a )) ) 1) = 2fK a = 2 a f K f K h=0 h=0 h=0 h=0. Numercal Soluton. [10/14] Dfferental Equatons group.

11 ) 3) ) 3) ) 3) = 3f y n + 2) a )) h=0 ) 1) = 3fKL a K a ) L 1) ) 2) +3fK a K h=0 = 3 a a fkl f K f L + 6a a fk f L Kf L. Numercal Soluton. [11/14] Dfferental Equatons group.

12 Note that each dervatve of recurses onto the prevous dervatves. We wll mae use of ths fact to generalze the constructon of the coeffcents from the tree structure. Collectng only the fnal results we have that ) 0) ) 1) ) 2) ) 3) = 0 = f = 2 a fk f K = 3 a a fkl f K f L + 6a a fk f L Kf L. Numercal Soluton. [12/14] Dfferental Equatons group.

13 We have now constructed the general set of condtons up to order three. l l f b = 1 fk f K b a = 1/2 fkl f K f L b a a l = 1/3 fk f L Kf L b a a l = 1/6 Generalzng to a hgher order requres some extra wor, however.. Numercal Soluton. [13/14] Dfferental Equatons group.

14 We can now phrase the order condtons n the followng way Theorem A Runge-Kutta method s of order p ff s =1 b φ τ) = 1 γτ) wth 1 f τ = [ ] φ τ) = a,m φ 1 τ 1 )... φ m τ m ) f τ = [τ 1,... τ m ] 1,..., m where [τ 1,..., τ m ] s a tree wth ts root havng chldren as the root of the subtrees τ 1,..., τ m, and γτ) s defned fully n the text.. Numercal Soluton. [14/14] Dfferental Equatons group.