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# Then z t z z y0 y0 z0 y0 3317a 3317b 3317c xt

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Unformatted text preview: 1 + cih) + Ep (3.3.18a) 2 (tn 1 tn) (3.3.18b) ; where Ep = Chp+1F (p)( n) n ; F 2 C p(tn 1 tn), and C is a constant. Then the collocation method (3.3.12) has order p. ; Proof. Consider the identity u = f (t u) + u ; f (t u)] 0 and use Theorem 3.3.5 on tn ; 1 u(tn) ; y(tn) = 0 tn] with z(t) = u(t) and g(t u) = u ; f (t u) to obtain 0 Z tn tn;1 xu (tn u( )) u ( ) ; f ( u( ))]d : 0 Replace this integral by the quadrature rule (3.3.18) to obtain s X u(tn) ; y(tn) = h bi xu(tn tn 1 + cih u(tn 1 + cih)) u (tn 1 + cih); 0 ; i=1 ; ; f (tn 1 + ci h u(tn 1 + ci h))] + Ep: ; ; All terms in the summation vanish upon use of the collocation equations (3.3.12) thus, p ju(tn) ; y(tn)j = jEpj jC jhp+1 maxt ] j @@ p xu(tn u( )) u ( ) ; f ( u)]j: t ;1 It remains to show that the derivatives in the above expression are bounded as h ! 0. We'll omit this detail which is proven in Hairer et al. 16], Section II.7. Thus, 0 2 n n ^ jy(tn) ; u(tn)j Chp+1 and the collocation method (3.3.12) is of order p. 32 (3.3.19) At last, our task is clear. We should select the collocation points ci , i = 1 2 : : : s, to maximize the order p of the quadrature rule (3.3.18...
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