# 18 of the exact solution yields 325c with l 1 325c

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Unformatted text preview: s section with a brief discussion of the absolute stability of explicit methods. We will present a more detailed analysis in Section 3.4 however, the present material will serve to motivate the need for implicit methods. Thus, consider an s-stage explicit Runge-Kutta method applied to the test equation y = y: (3.2.8) 0 Using (3.2.8) in (3.2.3) with the simpli cation that aij = 0, j yields yn = yn 1 + z ; s X i=1 bi Yi = yn 1 + zbT Y ; 14 i, for explicit methods (3.2.9a) where Yi = yn 1 + z ; i1 X ; j =1 aij Yj i = 1 2 ::: s (3.2.9b) and z=h : (3.2.9c) Y = yn 1l + zAY: (3.2.9d) The vector form of (3.2.9) is ; Using this to eliminate Y in (3.2.9a), we have yn = yn 1 1 + zbT (I ; zA) 1l]: ; ; Expanding the inverse yn = yn 1 1 + zbT (I + zA + : : : + zk Ak + : : : )l] ; Using (3.2.7) yn = R(z)yn (3.2.10a) 1 ; where 2 p X jT j1 R(z) = 1 + z + z2 + : : : + z ! + p j=p+1 z b A l: The matrix A is strictly lower triangular for an s-stage explicit Runge-Kutta method thus, Aj 1 = 0, j > s. Therefore, 1 ; ; s 2 p X jT j1 R(z) = 1 + z + z2 + : : : + z ! + z b A l: p ; j =p+1 In particular, for explicit s-stage methods with p = s...
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## This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

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