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 onestep 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 onestep
method.
! Lemma 3.4.1. The onestep 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 onestep method (3.4.1a) converges to the solution of (3.4...
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 Spring '14
 JosephE.Flaherty
 Numerical Analysis, yn, Tn, Numerical ordinary differential equations

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