Unformatted text preview: s said to converge to order p. The computations are performed on a sequence of meshes having nerand ner spacing, but always ending at the same time . Although the de nition has been stated with
sequences of uniform meshes, it could easily be revised to include more general mesh
sequences.
With these preliminaries behind us, we are in a position to state and prove our main
result. Once again, a Lipschitz condition on ( ) will come to our aid.
T fty 7 Theorem 2.1.1. Suppose ( ) exists and ( ) satis es a Lipschitz condition on the
strip f( ) j 0
;1
1g. Then Euler's method (2.1.2) converges to the
y ty t 00 t T fty <y< solution of the IVP (2.1.1). Proof. From (2.1.4b), the exact solution of the IVP (2.1.1) satis es ( ) = ( n 1) + y tn yt ; ( ( hf tn;1 y tn;1 )) + dn : Subtracting this from (2.1.2) and using (2.1.3)
= en en;1 + ( ( h f tn;1 y tn;1 )) ; ( n
ft yn;1 1 ; )] + (2.1.7) dn : Since ( ) satis es a Lipschitz condition,
fty j (n
ft ; 1 ( y tn;1 )) ; ( n
ft yn;1 1 ; )j j( Ln y tn;1 ); yn;1 j = nj n 1j
L e ; : Taking the absolute value of (2.1.7) and using the triangular inequality and the above
Lipschitz condition yields j nj (1 +
e hLn )j en;1 Let
= 1max
nN L and write (2.1.8) as Ln d j + j nj
d = 1max j nj
nN
d j nj (1 + )j n 1j +
e Since the inequality holds for all hL (2.1.8) : e ; d: n j nj (1 + ) (1 + )j n 2 j + ] + = (1 + )2 j n 2j + 1 + (1 + )]
e hL hL e d ; d hL e ; d hL Continuing the iteration j nj (1 + )n j 0j + 1 + (1 + ) +
e hL e d hL ::: + (1 + Using the formula for the sum of a geometric series X(1 +
n;1
k=0 )k = (1 + hL 8 hL )n ; 1 hL hL )n 1 ]
; : : yields j nj (1 + )nj 0j +
e hL (1 + d e hL )n ; 1] hL : (2.1.9) Equation (2.1.9) can be written in a more revealing form by using the following
Lemma, which is common throughout numerical analysis. Lemma 2.1.1. For all real z 1+ (2.1.10) z z e: Proof. Using a Taylor's series expansion
2 1+ + 2
z z = e 2 (0 ) z e z: Neglecting the positive third term on the left leads to (2.1.10).
Now, using (2.1.10) with =
z , hL (1 +
since nh = tn T for n hL )n e hLn LT e . Using the above expression in (2.1.9) N j nj LT e e j 0j + d e hL ( LT e ; 1) (2.1.11) : Since ( ) exists for 2 0 ], Euler's method is consistent and, using (2.1.4c), we
can bound on as
y 00 t t T d 2 d = 0max j nj = 2 0max j ( )j
nN
tT
h d y 00 t (2.1.12) : Substituting (2.1.12) into (2.1.11) j nj LT e e Roundo error is neglected, so e0 j 0 j + 2 ( LT ; 1) 0max j ( )j
tT
h e L = (0) ;
y e y0 y h L e y Thus, limh 0 j nj ! 0 and Euler's method converges.
! e 9 t : = 0, and we have j nj 2 ( LT ; 1) 0max j ( )j
tT
e 00 00 t : (2.1.13) Remark 1. The assumption that y (t) exists is not necessary. A proof without this
assumption appears in Hairer et al. 2], Section 1.7. They also show that the Lipschitz
condition on f (t y) need only be satis ed on a compact domain rather than an in nite
strip. Gear 1], Section 1.3, presents a proof with the additional assumption that f (t y)
satisfy a Lipschitz condition on t instead of requiring y (t) to be bounded.
Example 2.1.3. Consider the simple problem...
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 Spring '14
 JosephE.Flaherty
 Numerical Analysis, yn, Euler

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