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 RungeKutta 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 RungeKutta method a...
View
Full Document
 Spring '14
 JosephE.Flaherty
 Numerical Analysis, yn, Tn, Numerical ordinary differential equations

Click to edit the document details