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# 27b on each subinterval and then solving for the

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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...
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