Thus the rst block of 3310c is solved for y1 knowing

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: bleau c1 a11 0 c2 a21 a22 b1 b2 and it could be of third order. According to Theorem 3.2.1, the conditions for secondorder accuracy are (3.2.5c) with l = 1 2 when (3.2.5b) is satis ed, i.e., b1 + b2 = 1 c1 = a11 c2 = a21 + a22 b1 c1 + b2c2 = 1=2: (As noted earlier, satisfaction of (3.2.5b) is not necessary, but it simpli es the algebraic manipulations.) We might guess that the remaining conditions necessary for third order accuracy are (3.2.5c) with l = 3 and (3.2.7) with k = 3, i.e., b1 c2 + b2 c2 = 1=3 1 2 and bT A2l = bAc = b1 a11 c1 + b2 (a21 c1 + a22 c2 ) = 1=6 27 where (3.2.5b) was used to simplify the last expression. After some e ort, this system of six equations in seven unknowns can be solved to yield 2 ; 3c c2 = 3 ; 6c1 2 b1 = cc ; 1c=2 ; 1 2 ; b2 = 1c=2; cc1 1 2 2 1 a22 = b =(6 ;;b1cc1) c a11 = c1 2 2 1 a21 = c2 ; a22 : 1 As written, the solution is parameterized by c1 . Choosing c1 = 1=3 gives 1/3 1/3 0 1 1/2 1/2 3/4 1/4 Using (3.2.3), the method is Y1 = yn 1 + h f (tn 1 + h Y...
View Full Document

Ask a homework question - tutors are online