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# 41b and to begin well show that one step methods are

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Unformatted text preview: le which has the form of (3.4.1a) with (t y h) = f (t + h=2 y + hf (t y)=2): Then, j (t y h) ; (t z h)j = jf (t + h=2 y + hf (t y)=2) ; f (t + h=2 z + hf (t z)=2)j Using the Lipschitz condition on f j (t y h) ; (t z h)j Ljy + hf (t y)=2 ; z ; hf (t z)=2j or or j (t y h) ; (t z h)j L jy ; zj + (h=2)jf (t y) ; f (t z)j] j (t y h) ; (t z h)j L(1 + hL=2)jy ; zj: ^ Thus, we can take the Lipschitz constant for to be L(1 + hL=2) for h 2 (0 ^ ]. h In addition to a Lipschitz condition, convergence of the one-step method (3.4.1a) requires consistency. Recall (De nition 2.1.3), that consistency implies that the local discretization error limh 0 n = 0. Consistency is particularly simple for a one-step method. ! Lemma 3.4.1. The one-step method (3.4.1a) is consistent with the ODE y = f (t y) if 0 (t y 0) = f (t y): (3.4.3) Proof. The local discretization error of (3.4.1a) satis es n = y(tn) ;hy(tn 1) ; (tn ; ; 1 y(tn 1) h): ; Letting h tend to zero lim n = y (tn 1) ; (tn 0 h 0 ; ! 1 ; Using the ODE to replace y yields the result. 0 39 y(tn 1) 0): ; Theorem 3.4.2. Let (t y h) be a continuous function of t, y, and h on 0 t T , ^ ;1 < y < 1, and 0 h h, respectively, and satisfy a Lipschitz condition on y. Then the one-step method (3.4.1a) converges to the solution of (3.4...
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