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# 13 approaches zero as the mesh is re ned de nition 214

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Unformatted text preview: s said to converge to order p. The computations are performed on a sequence of meshes having ner-and- 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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