Recall that simpsons rule is z the mass matrix is 0 h

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: s and A is an N N tridiagonal matrix having the form 2 a1 6 b2 6 A=6 6 6 4 c1 a2 c2 ... ... bN ... 1 ; aN 1 cN 1 bN aN ; ; 3 7 7 7: 7 7 5 (1.3.24b) Assume that pivoting is not necessary and factor A as A = LU (1.3.25a) where L and U are lower and upper bidiagonal matrices having the form 2 3 1 6 l2 1 7 6 7 7 L = 6 l3 1 (1.3.25b) 6 7 6 ... ... 7 4 5 lN 1 2 3 u1 v1 6 u2 v2 7 6 7 6 7: ... ... U=6 (1.3.25c) 7 6 7 4 uN 1 vN 1 5 uN Once the coe cients lj , j = 2 3 : : : N , uj , j = 1 2 : : : N , and vj , j = 1 2 : : : N ; ; ; 1, have been determined, the system (1.3.24a) may easily be solved by forward and backward substitution. Thus, using (1.3.25a) in (1.3.24a) gives LUX = F: (1.3.26a) UX = Y (1.3.26b) LY = F: (1.3.26c) Let then, 2.1. Using (1.3.24) and (1.3.25), show lj = bj =uj 1 ; u1 = a1 uj = aj ; lj cj 1 j = 2 3 ::: N vj = cj j = 2 3 : : : N: ; 1.3. A Simple Finite Element Problem 21 2.2. Show that Y and X are computed as Y1 = F1 Yj = Fj ; lj Yj 1 j = 2 3 ::: N XN = yN =uN Xj = (Yj ; vj Xj+1)=uj j = N ; 1 N ; 2 : : : 1: ; 2.3. Develop a procedure to implement this scheme for solving tridi...
View Full Document

Ask a homework question - tutors are online