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

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: 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...
View Full Document

This document was uploaded on 03/16/2014 for the course CSCI 6820 at Rensselaer Polytechnic Institute.

Ask a homework question - tutors are online