{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

27b on each subinterval and then solving for the

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: 7 5 (9.2.10b) J Eliminating k in (9.2.10a) using (9.2.9b) yields i Y =; Y 1+g i i i; i = 1 2 ::: J i (9.2.11a) where ; = I + h D W 1V ; i i 19 i i i (9.2.11b) g = h DW 1q : (9.2.11c) ; i i i i The resulting linear algebraic system is 2L 6 ;;1 I 6 6 ;; 2 6 6 6 6 4 3 23 7 2 Y0 3 6 l 7 7 6 7 6 g1 7 7 6 Y1 7 6 . 7 7 6 . 7 = 6 .. 7 : 7 4 .. 5 6 7 7 4g 5 5 ;; I 7 Y r ... (9.2.11d) J R J J Thus, once again, the algebraic system is block bidiagonal and may be solved by the methods of Section 8.5. 9.3 Convergence and Stability Let us consider the linear rst-order BVP Ly = y ; A(x)y = b(x) a<x<b 0 Ly(a) = l (9.3.1a) Ry(b) = r: (9.3.1b) Using (9.2.4 - 9.2.6) we have X J y( ) = y(x 1) + h ij i; i k X J y(x ) = y(x 1 ) + h i i; =1 i k where E= j and E= Z j 0 =1 0 jk ik i b y( )+h E 0 k ik i j (9.3.2a) (9.3.2b) R( )d (9.3.2c) R( )d : (9.3.2d) 0 Z1 a y( )+hE 20 The function R(x) is the interpolation error. In general, if we interpolate a function f (x) at J distinct points f 1 < 2 < < g then J R(x) = f (x) ; P (x) = f...
View Full Document

{[ snackBarMessage ]}