Consider the two two stage runge kutta methods based

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Unformatted text preview: r 2 readily extend to a general class of (explicit or implicit) one-step methods having the form yn = yn 1 + h (tn 1 yn 1 h): ; ; (3.4.1a) ; Again, consider the scalar IVP y = f (t y ) y(0) = y0 0 (3.4.1b) and, to begin, we'll show that one-step methods are stable when satis es a Lipschitz condition on y. Theorem 3.4.1. If (t y h) satis es a Lipschitz condition on y then the one-step method (3.4.1a) is stable. Proof. The analysis follows the lines of Theorem 2.1.1. Let yn and zn satisfy method (3.4.1) and zn = zn 1 + h (tn 1 zn 1 h) ; ; z0 = y0 + ; (3.4.2) 0 respectively. Subtracting (3.4.2) from (3.4.1) yn ; zn = yn 1 ; zn 1 + h (tn 1 yn 1 h) ; (tn 1 zn 1 h)]: ; ; ; ; ; Using the Lipschitz condition jyn ; zn j (1 + hL)jyn 1 ; zn 1j: ; ; Iterating the above inequality leads to jyn ; zn j (1 + hL)n jy0 ; z0 j: Using (2.1.10) since nh T and j 0j jyn ; zn j enhLj 0j eLT . 38 k ; Example 3.4.1. The function satis es a Lipschitz condition whenever f does. Consider, for example, the explicit midpoint ru...
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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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