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

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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...
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