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
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,
ju(tn) ; y(tn)j = jEpj jC jhp+1 maxt ] j @@ p xu(tn u( )) u ( ) ; f ( u)]j:
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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- Spring '14
- Numerical Analysis, yn, Tn, Numerical ordinary differential equations