The solution utn 1 h may be used as the initial

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Unformatted text preview: here Lj ( ) = s Y ; ci i=1 i=j cj ; ci (3.3.13b) 6 = t ; tn 1 : (3.3.13c) h The polynomials Lj ( ), j = 1 2 : : : s, are a product of s ; 1 linear factors and are, hence, of degree s ; 1. They satisfy ; Lj (ci) = ji j i = 1 2 ::: s 29 (3.3.13d) where ji is the Kronecker delta. Using (3.3.13a), we see that u (t) satis es the interpolation conditions 0 u (tn 1 + cih) = ki i = 1 2 : : : s: 0 ; Transforming variables in (3.3.13a) using (3.3.13c) Z u(tn 1 + h) = yn 1 + h u (tn 1 + h)d : 0 ; ; ; 0 (3.3.13e) (3.3.14) By construction, (3.3.14) satis es (3.3.12b). Substituting (3.3.13e) and (3.3.14) into (3.3.12c), we have s X Zc ki = f (tn 1 + cih yn 1 + h kj Lj ( )d ): i ; ; j =1 0 This formula is identical to the typical Runge-Kutta formula (3.2.2b) provided that Zc aij = Lj ( )d : (3.3.15a) i 0 Similarly, using (3.3.13a) in (3.3.14) and evaluating the result at = 1 yields s X Z1 u(tn 1 + h) = yn = yn 1 + h kj Lj ( )d : ; ; j =1 This formula is identical to (3.2.2a) provided that Z1 bj = Lj ( )d : 0 0 (3.3.15b) This view of a Runge-Kutta method a...
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This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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